Comprehension:
You are presented with a paragraph with options (1), (2), (3), (4) and (5). Read the paragraph carefully and identify which of the given options correctly explains the paragraph or correctly summarizes it, if none of the options correctly explains it mark (5) as your answer.
Question:
Fossils found deep in South Africa's Rising Star cave complex in 2013 have been dated by several expert teams with their findings suggesting the hominids, called Homo naledi, may have lived alongside Homo sapiens. It had earlier been thought that the hominids were millions of years old. A team of 20 scientists from laboratories and institutions around the world, including in South Africa and Australia, established the age of the fossils which suggests that Homo naledi may well have lived at the same time as humans. The focus of the team's research has been South Africa's barely accessible Rising Star Cave system, part of the Cradle of Humankind World Heritage Site, about 50 kilometers northwest of Johannesburg. The area has been an incredibly rich source of artifacts.
Comprehension:
You are presented with a paragraph with options (1), (2), (3), (4) and (5). Read the paragraph carefully and identify which of the given options correctly explains the paragraph or correctly summarizes it, if none of the options correctly explains it mark (5) as your answer.
Question:
Under normal market conditions, for customers debt funds have nearly the same liquidity as term deposits with banks. However, given illiquid bond markets, they can have serious liquidity problems. The banking system has access to a liquidity window run by the regulator that allows it to tide over any short-term liquidity challenge, including, at its extreme, a bank run. Currently, no such strong liquidity support exists for credit funds. There is the collateralized lending and borrowing operation (CBLO) run by Clearing Corporation of India but with rapidly growing credit funds, this window is grossly inadequate. So we should consider providing liquidity adjustment facility (LAF) access to credit funds too. This would ensure that in the event of a "fund run" there is enough access to liquidity for funds.
Comprehension:
You are presented with a paragraph with options (1), (2), (3), (4) and (5). Read the paragraph carefully and identify which of the given options correctly explains the paragraph or correctly summarizes it, if none of the options correctly explains it mark (5) as your answer.
Question:
Rational choice theory assumes that all people try to actively maximize their advantage in any situation and therefore consistently try to minimize their losses. The theory is based on the idea that all humans base their decisions on rational calculations, act with rationality when choosing, and aim to increase either pleasure or profit. Rational choice theory also stipulates that all complex social phenomena are driven by individual human actions. Therefore, if an economist wants to explain social change or the actions of social institutions, he needs to look at the rational decisions of the individuals that make up the whole.
Comprehension:
Directions: Twelve children A, B, C, D, E, F, G, H, I, J, K and L were seated around a square table facing the centre. They were aged 1-12 not necessarily in the same order. No two children were of the same age. There was one child at each of the corner and 2 children were seated at each of the sides.
• D was at one of the sides and was 3rd to the left of the one who was 10 years old
• K was to the immediate right of E and he was aged twice as that of E's age
• K and J were seated at a gap of 3 persons
• J was aged 4 times as that of A
• A was seated opposite to D
• D was 6 years elder to G who was to the immediate left of J
• J was opposite to the person who was 11 years old
• B was 2nd to the left of the one who was 3 years old
• The one who was 7 years old was seated opposite to C who was not the youngest
• K and the one who was 1 year old was seated at a gap of 1 and neither of them was at the corner
• K and H were born in consecutive years
• L was seated 3rd to the left of I who was opposite to the one who was 6 years old
• B was 9 years old and G was not seated beside the child who was aged 10.
• B was not neighbouring the child who was 8 years old.
Question:
Who among the following was 2 years old?
• K was to the immediate right of E and he was aged twice as that of E's age.
• K and J were seated at a gap of 3 persons.
• G was to the immediate left of J.
• J as opposite to the person who was 11 years old.
• K and the one who was 1 year old was seated at a gap of 1 and neither of them was at the corner. Thus, the child between them was at one of the corners.
• K and H were born in consecutive years.
• K was twice as that of E's age. K can be 2, 4, 6, 8, 10 or 12 years old. From the figures, E cannot be 1 year old, thus, E cannot be 2 years old.
• D was at one of the sides and was 3rd to the left of the one who was 10 years old.
• D was 6 years elder to G who was to the immediate left of J. D can be 7, 8, 9, 10, 11 or 12 years old. G can be 1, 2, 3, 4, 5 or 6 years old.
• G was not seated beside the child who was aged 10.
Case 1:
Case 2:
3rd situation is invalid as J cannot be 10 years old as his age is 4 times as that of A.
Case 3:
1st situation is invalid as J cannot be 10 years old as his age is 4 times as that of A.
Case 4:
In cases where K is 10 years old, H will be 11 years old as they were born in consecutive years. H cannot be 9 years old as B is 9 years old.
• J was aged 4 times as that of A.
• A was seated opposite to D.
• The one who was 7 years old was seated opposite to C who was not the youngest.
• L was seated 3rd to the left of I who was opposite to the one who was 6 years old.
Case 1:
Scenarios 1 and 2 are invalid as K and H cannot be born in consecutive years and scenario 4th is invalid as we cannot place the C.
Case 2:
1st scenario is invalid as B cannot be 2nd to the left of the one who was 3 years old and scenario 2 is invalid as if G is 3 years old then D should have been 9 years old which is not possible.
Case 3:
• K was aged twice as that of E's age.
• J was aged 4 times as that of A.
• D was 6 years elder to G.
• B was 2nd to the left of the one who was 3 years old.
• K and H were born in consecutive years.
• B was 9 years old.
If G is 3 years old then D should have been 9 years old which is not possible, thus, 1st scenario is invalid. In scenario 3, K and H cannot be born in consecutive years, thus it is invalid.
• B was not neighbouring the child who was 8 years old.
Thus, the final arrangement is as follows:
L was 2 years old.
Comprehension:
Directions: Twelve children A, B, C, D, E, F, G, H, I, J, K and L were seated around a square table facing the centre. They were aged 1-12 not necessarily in the same order. No two children were of the same age. There was one child at each of the corner and 2 children were seated at each of the sides.
• D was at one of the sides and was 3rd to the left of the one who was 10 years old
• K was to the immediate right of E and he was aged twice as that of E's age
• K and J were seated at a gap of 3 persons
• J was aged 4 times as that of A
• A was seated opposite to D
• D was 6 years elder to G who was to the immediate left of J
• J was opposite to the person who was 11 years old
• B was 2nd to the left of the one who was 3 years old
• The one who was 7 years old was seated opposite to C who was not the youngest
• K and the one who was 1 year old was seated at a gap of 1 and neither of them was at the corner
• K and H were born in consecutive years
• L was seated 3rd to the left of I who was opposite to the one who was 6 years old
• B was 9 years old and G was not seated beside the child who was aged 10.
• B was not neighbouring the child who was 8 years old.
Question:
Who among the following was seated opposite to the one who was 8 years old?
• K was to the immediate right of E and he was aged twice as that of E's age.
• K and J were seated at a gap of 3 persons.
• G was to the immediate left of J.
• J as opposite to the person who was 11 years old.
• K and the one who was 1 year old was seated at a gap of 1 and neither of them was at the corner. Thus, the child between them was at one of the corners.
• K and H were born in consecutive years.
• K was twice as that of E's age. K can be 2, 4, 6, 8, 10 or 12 years old. From the figures, E cannot be 1 year old, thus, E cannot be 2 years old.
• D was at one of the sides and was 3rd to the left of the one who was 10 years old.
• D was 6 years elder to G who was to the immediate left of J. D can be 7, 8, 9, 10, 11 or 12 years old. G can be 1, 2, 3, 4, 5 or 6 years old.
• G was not seated beside the child who was aged 10.
Case 1:
Case 2:
3rd situation is invalid as J cannot be 10 years old as his age is 4 times as that of A.
Case 3:
1st situation is invalid as J cannot be 10 years old as his age is 4 times as that of A.
Case 4:
In cases where K is 10 years old, H will be 11 years old as they were born in consecutive years. H cannot be 9 years old as B is 9 years old.
• J was aged 4 times as that of A.
• A was seated opposite to D.
• The one who was 7 years old was seated opposite to C who was not the youngest.
• L was seated 3rd to the left of I who was opposite to the one who was 6 years old.
Case 1:
Scenarios 1 and 2 are invalid as K and H cannot be born in consecutive years and scenario 4th is invalid as we cannot place the C.
Case 2:
1st scenario is invalid as B cannot be 2nd to the left of the one who was 3 years old and scenario 2 is invalid as if G is 3 years old then D should have been 9 years old which is not possible.
Case 3:
• K was aged twice as that of E's age.
• J was aged 4 times as that of A.
• D was 6 years elder to G.
• B was 2nd to the left of the one who was 3 years old.
• K and H were born in consecutive years.
• B was 9 years old.
If G is 3 years old then D should have been 9 years old which is not possible, thus, 1st scenario is invalid. In scenario 3, K and H cannot be born in consecutive years, thus it is invalid.
• B was not neighbouring the child who was 8 years old.
Thus, the final arrangement is as follows:
F was seated opposite to C who was 8 years old.
Comprehension:
Directions: Twelve children A, B, C, D, E, F, G, H, I, J, K and L were seated around a square table facing the centre. They were aged 1-12 not necessarily in the same order. No two children were of the same age. There was one child at each of the corner and 2 children were seated at each of the sides.
• D was at one of the sides and was 3rd to the left of the one who was 10 years old
• K was to the immediate right of E and he was aged twice as that of E's age
• K and J were seated at a gap of 3 persons
• J was aged 4 times as that of A
• A was seated opposite to D
• D was 6 years elder to G who was to the immediate left of J
• J was opposite to the person who was 11 years old
• B was 2nd to the left of the one who was 3 years old
• The one who was 7 years old was seated opposite to C who was not the youngest
• K and the one who was 1 year old was seated at a gap of 1 and neither of them was at the corner
• K and H were born in consecutive years
• L was seated 3rd to the left of I who was opposite to the one who was 6 years old
• B was 9 years old and G was not seated beside the child who was aged 10.
• B was not neighbouring the child who was 8 years old.
Question:
Who among the following was the youngest?
• K was to the immediate right of E and he was aged twice as that of E's age.
• K and J were seated at a gap of 3 persons.
• G was to the immediate left of J.
• J as opposite to the person who was 11 years old.
• K and the one who was 1 year old was seated at a gap of 1 and neither of them was at the corner. Thus, the child between them was at one of the corners.
• K and H were born in consecutive years.
• K was twice as that of E's age. K can be 2, 4, 6, 8, 10 or 12 years old. From the figures, E cannot be 1 year old, thus, E cannot be 2 years old.
• D was at one of the sides and was 3rd to the left of the one who was 10 years old.
• D was 6 years elder to G who was to the immediate left of J. D can be 7, 8, 9, 10, 11 or 12 years old. G can be 1, 2, 3, 4, 5 or 6 years old.
• G was not seated beside the child who was aged 10.
Case 1:
Case 2:
3rd situation is invalid as J cannot be 10 years old as his age is 4 times as that of A.
Case 3:
1st situation is invalid as J cannot be 10 years old as his age is 4 times as that of A.
Case 4:
In cases where K is 10 years old, H will be 11 years old as they were born in consecutive years. H cannot be 9 years old as B is 9 years old.
• J was aged 4 times as that of A.
• A was seated opposite to D.
• The one who was 7 years old was seated opposite to C who was not the youngest.
• L was seated 3rd to the left of I who was opposite to the one who was 6 years old.
Case 1:
Scenarios 1 and 2 are invalid as K and H cannot be born in consecutive years and scenario 4th is invalid as we cannot place the C.
Case 2:
1st scenario is invalid as B cannot be 2nd to the left of the one who was 3 years old and scenario 2 is invalid as if G is 3 years old then D should have been 9 years old which is not possible.
Case 3:
• K was aged twice as that of E's age.
• J was aged 4 times as that of A.
• D was 6 years elder to G.
• B was 2nd to the left of the one who was 3 years old.
• K and H were born in consecutive years.
• B was 9 years old.
If G is 3 years old then D should have been 9 years old which is not possible, thus, 1st scenario is invalid. In scenario 3, K and H cannot be born in consecutive years, thus it is invalid.
• B was not neighbouring the child who was 8 years old.
Thus, the final arrangement is as follows:
A was the youngest child.
Comprehension:
Directions: Twelve children A, B, C, D, E, F, G, H, I, J, K and L were seated around a square table facing the centre. They were aged 1-12 not necessarily in the same order. No two children were of the same age. There was one child at each of the corner and 2 children were seated at each of the sides.
• D was at one of the sides and was 3rd to the left of the one who was 10 years old
• K was to the immediate right of E and he was aged twice as that of E's age
• K and J were seated at a gap of 3 persons
• J was aged 4 times as that of A
• A was seated opposite to D
• D was 6 years elder to G who was to the immediate left of J
• J was opposite to the person who was 11 years old
• B was 2nd to the left of the one who was 3 years old
• The one who was 7 years old was seated opposite to C who was not the youngest
• K and the one who was 1 year old was seated at a gap of 1 and neither of them was at the corner
• K and H were born in consecutive years
• L was seated 3rd to the left of I who was opposite to the one who was 6 years old
• B was 9 years old and G was not seated beside the child who was aged 10.
• B was not neighbouring the child who was 8 years old.
Question:
In which of the following groups was the 3rd person neighbouring the 1st and the 2nd person?
• K was to the immediate right of E and he was aged twice as that of E's age.
• K and J were seated at a gap of 3 persons.
• G was to the immediate left of J.
• J as opposite to the person who was 11 years old.
• K and the one who was 1 year old was seated at a gap of 1 and neither of them was at the corner. Thus, the child between them was at one of the corners.
• K and H were born in consecutive years.
• K was twice as that of E's age. K can be 2, 4, 6, 8, 10 or 12 years old. From the figures, E cannot be 1 year old, thus, E cannot be 2 years old.
• D was at one of the sides and was 3rd to the left of the one who was 10 years old.
• D was 6 years elder to G who was to the immediate left of J. D can be 7, 8, 9, 10, 11 or 12 years old. G can be 1, 2, 3, 4, 5 or 6 years old.
• G was not seated beside the child who was aged 10.
Case 1:
Case 2:
3rd situation is invalid as J cannot be 10 years old as his age is 4 times as that of A.
Case 3:
1st situation is invalid as J cannot be 10 years old as his age is 4 times as that of A.
Case 4:
In cases where K is 10 years old, H will be 11 years old as they were born in consecutive years. H cannot be 9 years old as B is 9 years old.
• J was aged 4 times as that of A.
• A was seated opposite to D.
• The one who was 7 years old was seated opposite to C who was not the youngest.
• L was seated 3rd to the left of I who was opposite to the one who was 6 years old.
Case 1:
Scenarios 1 and 2 are invalid as K and H cannot be born in consecutive years and scenario 4th is invalid as we cannot place the C.
Case 2:
1st scenario is invalid as B cannot be 2nd to the left of the one who was 3 years old and scenario 2 is invalid as if G is 3 years old then D should have been 9 years old which is not possible.
Case 3:
• K was aged twice as that of E's age.
• J was aged 4 times as that of A.
• D was 6 years elder to G.
• B was 2nd to the left of the one who was 3 years old.
• K and H were born in consecutive years.
• B was 9 years old.
If G is 3 years old then D should have been 9 years old which is not possible, thus, 1st scenario is invalid. In scenario 3, K and H cannot be born in consecutive years, thus it is invalid.
• B was not neighbouring the child who was 8 years old.
Thus, the final arrangement is as follows:
G was neighbouring J and A.
Comprehension:
Directions: Twelve children A, B, C, D, E, F, G, H, I, J, K and L were seated around a square table facing the centre. They were aged 1-12 not necessarily in the same order. No two children were of the same age. There was one child at each of the corner and 2 children were seated at each of the sides.
• D was at one of the sides and was 3rd to the left of the one who was 10 years old
• K was to the immediate right of E and he was aged twice as that of E's age
• K and J were seated at a gap of 3 persons
• J was aged 4 times as that of A
• A was seated opposite to D
• D was 6 years elder to G who was to the immediate left of J
• J was opposite to the person who was 11 years old
• B was 2nd to the left of the one who was 3 years old
• The one who was 7 years old was seated opposite to C who was not the youngest
• K and the one who was 1 year old was seated at a gap of 1 and neither of them was at the corner
• K and H were born in consecutive years
• L was seated 3rd to the left of I who was opposite to the one who was 6 years old
• B was 9 years old and G was not seated beside the child who was aged 10.
• B was not neighbouring the child who was 8 years old.
Question:
Four of the following bear a similar relationship and hence form a group, who among the following is not a part of that group?
• K was to the immediate right of E and he was aged twice as that of E's age.
• K and J were seated at a gap of 3 persons.
• G was to the immediate left of J.
• J as opposite to the person who was 11 years old.
• K and the one who was 1 year old was seated at a gap of 1 and neither of them was at the corner. Thus, the child between them was at one of the corners.
• K and H were born in consecutive years.
• K was twice as that of E's age. K can be 2, 4, 6, 8, 10 or 12 years old. From the figures, E cannot be 1 year old, thus, E cannot be 2 years old.
• D was at one of the sides and was 3rd to the left of the one who was 10 years old.
• D was 6 years elder to G who was to the immediate left of J. D can be 7, 8, 9, 10, 11 or 12 years old. G can be 1, 2, 3, 4, 5 or 6 years old.
• G was not seated beside the child who was aged 10.
Case 1:
Case 2:
3rd situation is invalid as J cannot be 10 years old as his age is 4 times as that of A.
Case 3:
1st situation is invalid as J cannot be 10 years old as his age is 4 times as that of A.
Case 4:
In cases where K is 10 years old, H will be 11 years old as they were born in consecutive years. H cannot be 9 years old as B is 9 years old.
• J was aged 4 times as that of A.
• A was seated opposite to D.
• The one who was 7 years old was seated opposite to C who was not the youngest.
• L was seated 3rd to the left of I who was opposite to the one who was 6 years old.
Case 1:
Scenarios 1 and 2 are invalid as K and H cannot be born in consecutive years and scenario 4th is invalid as we cannot place the C.
Case 2:
1st scenario is invalid as B cannot be 2nd to the left of the one who was 3 years old and scenario 2 is invalid as if G is 3 years old then D should have been 9 years old which is not possible.
Case 3:
• K was aged twice as that of E's age.
• J was aged 4 times as that of A.
• D was 6 years elder to G.
• B was 2nd to the left of the one who was 3 years old.
• K and H were born in consecutive years.
• B was 9 years old.
If G is 3 years old then D should have been 9 years old which is not possible, thus, 1st scenario is invalid. In scenario 3, K and H cannot be born in consecutive years, thus it is invalid.
• B was not neighbouring the child who was 8 years old.
Thus, the final arrangement is as follows:
Apart from I others were seated at the corners.
Comprehension:
Directions: Twelve children A, B, C, D, E, F, G, H, I, J, K and L were seated around a square table facing the centre. They were aged 1-12 not necessarily in the same order. No two children were of the same age. There was one child at each of the corner and 2 children were seated at each of the sides.
• D was at one of the sides and was 3rd to the left of the one who was 10 years old
• K was to the immediate right of E and he was aged twice as that of E's age
• K and J were seated at a gap of 3 persons
• J was aged 4 times as that of A
• A was seated opposite to D
• D was 6 years elder to G who was to the immediate left of J
• J was opposite to the person who was 11 years old
• B was 2nd to the left of the one who was 3 years old
• The one who was 7 years old was seated opposite to C who was not the youngest
• K and the one who was 1 year old was seated at a gap of 1 and neither of them was at the corner
• K and H were born in consecutive years
• L was seated 3rd to the left of I who was opposite to the one who was 6 years old
• B was 9 years old and G was not seated beside the child who was aged 10.
• B was not neighbouring the child who was 8 years old.
What was the difference between the age of D and C?
• K was to the immediate right of E and he was aged twice as that of E's age.
• K and J were seated at a gap of 3 persons.
• G was to the immediate left of J.
• J as opposite to the person who was 11 years old.
• K and the one who was 1 year old was seated at a gap of 1 and neither of them was at the corner. Thus, the child between them was at one of the corners.
• K and H were born in consecutive years.
• K was twice as that of E's age. K can be 2, 4, 6, 8, 10 or 12 years old. From the figures, E cannot be 1 year old, thus, E cannot be 2 years old.
• D was at one of the sides and was 3rd to the left of the one who was 10 years old.
• D was 6 years elder to G who was to the immediate left of J. D can be 7, 8, 9, 10, 11 or 12 years old. G can be 1, 2, 3, 4, 5 or 6 years old.
• G was not seated beside the child who was aged 10.
Case 1:
Case 2:
3rd situation is invalid as J cannot be 10 years old as his age is 4 times as that of A.
Case 3:
1st situation is invalid as J cannot be 10 years old as his age is 4 times as that of A.
Case 4:
In cases where K is 10 years old, H will be 11 years old as they were born in consecutive years. H cannot be 9 years old as B is 9 years old.
• J was aged 4 times as that of A.
• A was seated opposite to D.
• The one who was 7 years old was seated opposite to C who was not the youngest.
• L was seated 3rd to the left of I who was opposite to the one who was 6 years old.
Case 1:
Scenarios 1 and 2 are invalid as K and H cannot be born in consecutive years and scenario 4th is invalid as we cannot place the C.
Case 2:
1st scenario is invalid as B cannot be 2nd to the left of the one who was 3 years old and scenario 2 is invalid as if G is 3 years old then D should have been 9 years old which is not possible.
Case 3:
• K was aged twice as that of E's age.
• J was aged 4 times as that of A.
• D was 6 years elder to G.
• B was 2nd to the left of the one who was 3 years old.
• K and H were born in consecutive years.
• B was 9 years old.
If G is 3 years old then D should have been 9 years old which is not possible, thus, 1st scenario is invalid. In scenario 3, K and H cannot be born in consecutive years, thus it is invalid.
• B was not neighbouring the child who was 8 years old.
Thus, the final arrangement is as follows:
D was 12 years old and C was 8 years old.
Comprehension:
Directions: Twelve children A, B, C, D, E, F, G, H, I, J, K and L were seated around a square table facing the centre. They were aged 1-12 not necessarily in the same order. No two children were of the same age. There was one child at each of the corner and 2 children were seated at each of the sides.
• D was at one of the sides and was 3rd to the left of the one who was 10 years old
• K was to the immediate right of E and he was aged twice as that of E's age
• K and J were seated at a gap of 3 persons
• J was aged 4 times as that of A
• A was seated opposite to D
• D was 6 years elder to G who was to the immediate left of J
• J was opposite to the person who was 11 years old
• B was 2nd to the left of the one who was 3 years old
• The one who was 7 years old was seated opposite to C who was not the youngest
• K and the one who was 1 year old was seated at a gap of 1 and neither of them was at the corner
• K and H were born in consecutive years
• L was seated 3rd to the left of I who was opposite to the one who was 6 years old
• B was 9 years old and G was not seated beside the child who was aged 10.
• B was not neighbouring the child who was 8 years old.
Question:
What was the position of L with respect to the one who was 4 years old?
• K was to the immediate right of E and he was aged twice as that of E's age.
• K and J were seated at a gap of 3 persons.
• G was to the immediate left of J.
• J as opposite to the person who was 11 years old.
• K and the one who was 1 year old was seated at a gap of 1 and neither of them was at the corner. Thus, the child between them was at one of the corners.
• K and H were born in consecutive years.
• K was twice as that of E's age. K can be 2, 4, 6, 8, 10 or 12 years old. From the figures, E cannot be 1 year old, thus, E cannot be 2 years old.
• D was at one of the sides and was 3rd to the left of the one who was 10 years old.
• D was 6 years elder to G who was to the immediate left of J. D can be 7, 8, 9, 10, 11 or 12 years old. G can be 1, 2, 3, 4, 5 or 6 years old.
• G was not seated beside the child who was aged 10.
Case 1:
Case 2:
3rd situation is invalid as J cannot be 10 years old as his age is 4 times as that of A.
Case 3:
1st situation is invalid as J cannot be 10 years old as his age is 4 times as that of A.
Case 4:
In cases where K is 10 years old, H will be 11 years old as they were born in consecutive years. H cannot be 9 years old as B is 9 years old.
• J was aged 4 times as that of A.
• A was seated opposite to D.
• The one who was 7 years old was seated opposite to C who was not the youngest.
• L was seated 3rd to the left of I who was opposite to the one who was 6 years old.
Case 1:
Scenarios 1 and 2 are invalid as K and H cannot be born in consecutive years and scenario 4th is invalid as we cannot place the C.
Case 2:
1st scenario is invalid as B cannot be 2nd to the left of the one who was 3 years old and scenario 2 is invalid as if G is 3 years old then D should have been 9 years old which is not possible.
Case 3:
• K was aged twice as that of E's age.
• J was aged 4 times as that of A.
• D was 6 years elder to G.
• B was 2nd to the left of the one who was 3 years old.
• K and H were born in consecutive years.
• B was 9 years old.
If G is 3 years old then D should have been 9 years old which is not possible, thus, 1st scenario is invalid. In scenario 3, K and H cannot be born in consecutive years, thus it is invalid.
• B was not neighbouring the child who was 8 years old.
Thus, the final arrangement is as follows:
L was to the immediate right of J who was 4 years old.
Comprehension:
Directions: Twelve children A, B, C, D, E, F, G, H, I, J, K and L were seated around a square table facing the centre. They were aged 1-12 not necessarily in the same order. No two children were of the same age. There was one child at each of the corner and 2 children were seated at each of the sides.
• D was at one of the sides and was 3rd to the left of the one who was 10 years old
• K was to the immediate right of E and he was aged twice as that of E's age
• K and J were seated at a gap of 3 persons
• J was aged 4 times as that of A
• A was seated opposite to D
• D was 6 years elder to G who was to the immediate left of J
• J was opposite to the person who was 11 years old
• B was 2nd to the left of the one who was 3 years old
• The one who was 7 years old was seated opposite to C who was not the youngest
• K and the one who was 1 year old was seated at a gap of 1 and neither of them was at the corner
• K and H were born in consecutive years
• L was seated 3rd to the left of I who was opposite to the one who was 6 years old
• B was 9 years old and G was not seated beside the child who was aged 10.
• B was not neighbouring the child who was 8 years old.
Question:
How many persons were younger than the person who was seated opposite to H?
• K was to the immediate right of E and he was aged twice as that of E's age.
• K and J were seated at a gap of 3 persons.
• G was to the immediate left of J.
• J as opposite to the person who was 11 years old.
• K and the one who was 1 year old was seated at a gap of 1 and neither of them was at the corner. Thus, the child between them was at one of the corners.
• K and H were born in consecutive years.
• K was twice as that of E's age. K can be 2, 4, 6, 8, 10 or 12 years old. From the figures, E cannot be 1 year old, thus, E cannot be 2 years old.
• D was at one of the sides and was 3rd to the left of the one who was 10 years old.
• D was 6 years elder to G who was to the immediate left of J. D can be 7, 8, 9, 10, 11 or 12 years old. G can be 1, 2, 3, 4, 5 or 6 years old.
• G was not seated beside the child who was aged 10.
Case 1:
Case 2:
3rd situation is invalid as J cannot be 10 years old as his age is 4 times as that of A.
Case 3:
1st situation is invalid as J cannot be 10 years old as his age is 4 times as that of A.
Case 4:
In cases where K is 10 years old, H will be 11 years old as they were born in consecutive years. H cannot be 9 years old as B is 9 years old.
• J was aged 4 times as that of A.
• A was seated opposite to D.
• The one who was 7 years old was seated opposite to C who was not the youngest.
• L was seated 3rd to the left of I who was opposite to the one who was 6 years old.
Case 1:
Scenarios 1 and 2 are invalid as K and H cannot be born in consecutive years and scenario 4th is invalid as we cannot place the C.
Case 2:
1st scenario is invalid as B cannot be 2nd to the left of the one who was 3 years old and scenario 2 is invalid as if G is 3 years old then D should have been 9 years old which is not possible.
Case 3:
• K was aged twice as that of E's age.
• J was aged 4 times as that of A.
• D was 6 years elder to G.
• B was 2nd to the left of the one who was 3 years old.
• K and H were born in consecutive years.
• B was 9 years old.
If G is 3 years old then D should have been 9 years old which is not possible, thus, 1st scenario is invalid. In scenario 3, K and H cannot be born in consecutive years, thus it is invalid.
• B was not neighbouring the child who was 8 years old.
Thus, the final arrangement is as follows:
3 persons were younger than J who was seated opposite to H.
Comprehension:
Directions: 12 friends A, B, C, D, E, F, G, H, I, J, K and L went to see a movie. They were seated in two different rows each having six seats. They were seated facing the north. The seats were numbered 1-6 from left to right and were of different colour red, green, yellow, white, grey, black, purple, pink, orange, golden, silver and brown not necessarily in the same order.
• L was seated 3rd to the left of the one on the silver seat, L was not at the extreme end
• D's seat number was twice as that of L's seat number and they were in different rows
• D and H were seated at a gap of 2 seats
• H's neighbour was seated on the yellow seat
• The red and pink seats had the same seat number
• F was seated neighbouring the one who was on the red seat and the one who was seated exactly in front of I.
• K was seated 2nd to the left of C
• The seat number of the green seat was 3 more than the seat number of the grey seat and they were not in the same row
• G's seat number was twice as that of J's seat number. J was on the white seat
• A was seated exactly in front of the golden seat
• B was 2nd to the right of the one seated on the black seat
• E was on the orange seat and was at either of the extreme ends
• K was seated either on the pink or the brown seat
• F was not seated neighbouring L or H
Question:
Four of the following bear a similar relationship and hence form a group, who among the following is not a part of that group?
• L was seated 3rd to the left of the one on the silver seat, L was not at the extreme end
• D's seat number was twice as that of L's seat number and they were in different rows
• D and H were seated at a gap of 2 seats
• H's neighbour was seated on the yellow seat
• The red and pink seats had the same seat number.
• F was seated neighbouring the one who was on the red seat and the one who was seated exactly in front of I.
• F was not seated neighbouring L.
Case 1:
Case 2:
Invalid as I cannot be placed in the arrangement.
Case 3:
Case 4:
• K was seated 2nd to the left of C
• G's seat number was twice as that of J's seat number. J was on the white seat
• E was on the orange seat and was at either of the extreme ends
• K was seated either on the pink or the brown seat
Case 1:
1st scenario is invalid as G and J cannot be placed in those arrangements. 2nd scenario is invalid as E cannot be placed in that arrangement.
Case 2:
Scenario 2nd is invalid as J should be on white seat
Case 3:
• The seat number of the green seat was 3 more than the seat number of the grey seat and they were not in the same row
• A was seated exactly in front of the golden seat
• B was 2nd to the right of the one seated on the black seat
Case 1:
The cases are invalid as A cannot be placed in these arrangements.
Case 2:
2nd scenario is invalid as green and grey seats cannot be placed as per the conditions.
Case 3:
Invalid as E cannot be placed in the arrangements.
Thus, the final arrangement is as follows:
Apart from K others are seated at the corner seats.
Comprehension:
Directions: 12 friends A, B, C, D, E, F, G, H, I, J, K and L went to see a movie. They were seated in two different rows each having six seats. They were seated facing the north. The seats were numbered 1-6 from left to right and were of different colour red, green, yellow, white, grey, black, purple, pink, orange, golden, silver and brown not necessarily in the same order.
• L was seated 3rd to the left of the one on the silver seat, L was not at the extreme end
• D's seat number was twice as that of L's seat number and they were in different rows
• D and H were seated at a gap of 2 seats
• H's neighbour was seated on the yellow seat
• The red and pink seats had the same seat number
• F was seated neighbouring the one who was on the red seat and the one who was seated exactly in front of I.
• K was seated 2nd to the left of C
• The seat number of the green seat was 3 more than the seat number of the grey seat and they were not in the same row
• G's seat number was twice as that of J's seat number. J was on the white seat
• A was seated exactly in front of the golden seat
• B was 2nd to the right of the one seated on the black seat
• E was on the orange seat and was at either of the extreme ends
• K was seated either on the pink or the brown seat
• F was not seated neighbouring L or H
Question:
On which of the following seats was C seated?
• L was seated 3rd to the left of the one on the silver seat, L was not at the extreme end
• D's seat number was twice as that of L's seat number and they were in different rows
• D and H were seated at a gap of 2 seats
• H's neighbour was seated on the yellow seat
• The red and pink seats had the same seat number.
• F was seated neighbouring the one who was on the red seat and the one who was seated exactly in front of I.
• F was not seated neighbouring L.
Case 1:
Case 2:
Invalid as I cannot be placed in the arrangement.
Case 3:
Case 4:
• K was seated 2nd to the left of C
• G's seat number was twice as that of J's seat number. J was on the white seat
• E was on the orange seat and was at either of the extreme ends
• K was seated either on the pink or the brown seat
Case 1:
1st scenario is invalid as G and J cannot be placed in those arrangements. 2nd scenario is invalid as E cannot be placed in that arrangement.
Case 2:
Scenario 2nd is invalid as J should be on white seat
Case 3:
• The seat number of the green seat was 3 more than the seat number of the grey seat and they were not in the same row
• A was seated exactly in front of the golden seat
• B was 2nd to the right of the one seated on the black seat
Case 1:
The cases are invalid as A cannot be placed in these arrangements.
Case 2:
2nd scenario is invalid as green and grey seats cannot be placed as per the conditions.
Case 3:
Invalid as E cannot be placed in the arrangements.
Thus, the final arrangement is as follows:
C was seated on the silver coloured seat.
Comprehension:
Directions: 12 friends A, B, C, D, E, F, G, H, I, J, K and L went to see a movie. They were seated in two different rows each having six seats. They were seated facing the north. The seats were numbered 1-6 from left to right and were of different colour red, green, yellow, white, grey, black, purple, pink, orange, golden, silver and brown not necessarily in the same order.
• L was seated 3rd to the left of the one on the silver seat, L was not at the extreme end
• D's seat number was twice as that of L's seat number and they were in different rows
• D and H were seated at a gap of 2 seats
• H's neighbour was seated on the yellow seat
• The red and pink seats had the same seat number
• F was seated neighbouring the one who was on the red seat and the one who was seated exactly in front of I.
• K was seated 2nd to the left of C
• The seat number of the green seat was 3 more than the seat number of the grey seat and they were not in the same row
• G's seat number was twice as that of J's seat number. J was on the white seat
• A was seated exactly in front of the golden seat
• B was 2nd to the right of the one seated on the black seat
• E was on the orange seat and was at either of the extreme ends
• K was seated either on the pink or the brown seat
• F was not seated neighbouring L or H
Question:
Who among the following was seated on the purple seat?
• L was seated 3rd to the left of the one on the silver seat, L was not at the extreme end
• D's seat number was twice as that of L's seat number and they were in different rows
• D and H were seated at a gap of 2 seats
• H's neighbour was seated on the yellow seat
• The red and pink seats had the same seat number.
• F was seated neighbouring the one who was on the red seat and the one who was seated exactly in front of I.
• F was not seated neighbouring L.
Case 1:
Case 2:
Invalid as I cannot be placed in the arrangement.
Case 3:
Case 4:
• K was seated 2nd to the left of C
• G's seat number was twice as that of J's seat number. J was on the white seat
• E was on the orange seat and was at either of the extreme ends
• K was seated either on the pink or the brown seat
Case 1:
1st scenario is invalid as G and J cannot be placed in those arrangements. 2nd scenario is invalid as E cannot be placed in that arrangement.
Case 2:
Scenario 2nd is invalid as J should be on white seat
Case 3:
• The seat number of the green seat was 3 more than the seat number of the grey seat and they were not in the same row
• A was seated exactly in front of the golden seat
• B was 2nd to the right of the one seated on the black seat
Case 1:
The cases are invalid as A cannot be placed in these arrangements.
Case 2:
2nd scenario is invalid as green and grey seats cannot be placed as per the conditions.
Case 3:
Invalid as E cannot be placed in the arrangements.
Thus, the final arrangement is as follows:
F was seated on the purple seat.
Comprehension:
Directions: 12 friends A, B, C, D, E, F, G, H, I, J, K and L went to see a movie. They were seated in two different rows each having six seats. They were seated facing the north. The seats were numbered 1-6 from left to right and were of different colour red, green, yellow, white, grey, black, purple, pink, orange, golden, silver and brown not necessarily in the same order.
• L was seated 3rd to the left of the one on the silver seat, L was not at the extreme end
• D's seat number was twice as that of L's seat number and they were in different rows
• D and H were seated at a gap of 2 seats
• H's neighbour was seated on the yellow seat
• The red and pink seats had the same seat number
• F was seated neighbouring the one who was on the red seat and the one who was seated exactly in front of I.
• K was seated 2nd to the left of C
• The seat number of the green seat was 3 more than the seat number of the grey seat and they were not in the same row
• G's seat number was twice as that of J's seat number. J was on the white seat
• A was seated exactly in front of the golden seat
• B was 2nd to the right of the one seated on the black seat
• E was on the orange seat and was at either of the extreme ends
• K was seated either on the pink or the brown seat
• F was not seated neighbouring L or H
Question:
Who among the following was/were seated on seat number 4?
I. B
II. I
III. K
• L was seated 3rd to the left of the one on the silver seat, L was not at the extreme end
• D's seat number was twice as that of L's seat number and they were in different rows
• D and H were seated at a gap of 2 seats
• H's neighbour was seated on the yellow seat
• The red and pink seats had the same seat number.
• F was seated neighbouring the one who was on the red seat and the one who was seated exactly in front of I.
• F was not seated neighbouring L.
Case 1:
Case 2:
Invalid as I cannot be placed in the arrangement.
Case 3:
Case 4:
• K was seated 2nd to the left of C
• G's seat number was twice as that of J's seat number. J was on the white seat
• E was on the orange seat and was at either of the extreme ends
• K was seated either on the pink or the brown seat
Case 1:
1st scenario is invalid as G and J cannot be placed in those arrangements. 2nd scenario is invalid as E cannot be placed in that arrangement.
Case 2:
Scenario 2nd is invalid as J should be on white seat
Case 3:
• The seat number of the green seat was 3 more than the seat number of the grey seat and they were not in the same row
• A was seated exactly in front of the golden seat
• B was 2nd to the right of the one seated on the black seat
Case 1:
The cases are invalid as A cannot be placed in these arrangements.
Case 2:
2nd scenario is invalid as green and grey seats cannot be placed as per the conditions.
Case 3:
Invalid as E cannot be placed in the arrangements.
Thus, the final arrangement is as follows:
Among the given persons only I was seated on seat number 4.
Comprehension:
Directions: 12 friends A, B, C, D, E, F, G, H, I, J, K and L went to see a movie. They were seated in two different rows each having six seats. They were seated facing the north. The seats were numbered 1-6 from left to right and were of different colour red, green, yellow, white, grey, black, purple, pink, orange, golden, silver and brown not necessarily in the same order.
• L was seated 3rd to the left of the one on the silver seat, L was not at the extreme end
• D's seat number was twice as that of L's seat number and they were in different rows
• D and H were seated at a gap of 2 seats
• H's neighbour was seated on the yellow seat
• The red and pink seats had the same seat number
• F was seated neighbouring the one who was on the red seat and the one who was seated exactly in front of I.
• K was seated 2nd to the left of C
• The seat number of the green seat was 3 more than the seat number of the grey seat and they were not in the same row
• G's seat number was twice as that of J's seat number. J was on the white seat
• A was seated exactly in front of the golden seat
• B was 2nd to the right of the one seated on the black seat
• E was on the orange seat and was at either of the extreme ends
• K was seated either on the pink or the brown seat
• F was not seated neighbouring L or H
Question:
How many persons were seated to the left (same row) of the golden seat?
• L was seated 3rd to the left of the one on the silver seat, L was not at the extreme end
• D's seat number was twice as that of L's seat number and they were in different rows
• D and H were seated at a gap of 2 seats
• H's neighbour was seated on the yellow seat
• The red and pink seats had the same seat number.
• F was seated neighbouring the one who was on the red seat and the one who was seated exactly in front of I.
• F was not seated neighbouring L.
Case 1:
Case 2:
Invalid as I cannot be placed in the arrangement.
Case 3:
Case 4:
• K was seated 2nd to the left of C
• G's seat number was twice as that of J's seat number. J was on the white seat
• E was on the orange seat and was at either of the extreme ends
• K was seated either on the pink or the brown seat
Case 1:
1st scenario is invalid as G and J cannot be placed in those arrangements. 2nd scenario is invalid as E cannot be placed in that arrangement.
Case 2:
Scenario 2nd is invalid as J should be on white seat
Case 3:
• The seat number of the green seat was 3 more than the seat number of the grey seat and they were not in the same row
• A was seated exactly in front of the golden seat
• B was 2nd to the right of the one seated on the black seat
Case 1:
The cases are invalid as A cannot be placed in these arrangements.
Case 2:
2nd scenario is invalid as green and grey seats cannot be placed as per the conditions.
Case 3:
Invalid as E cannot be placed in the arrangements.
Thus, the final arrangement is as follows:
Only one person was to the left of L who was on the golden seat.
Comprehension:
Directions: 12 friends A, B, C, D, E, F, G, H, I, J, K and L went to see a movie. They were seated in two different rows each having six seats. They were seated facing the north. The seats were numbered 1-6 from left to right and were of different colour red, green, yellow, white, grey, black, purple, pink, orange, golden, silver and brown not necessarily in the same order.
• L was seated 3rd to the left of the one on the silver seat, L was not at the extreme end
• D's seat number was twice as that of L's seat number and they were in different rows
• D and H were seated at a gap of 2 seats
• H's neighbour was seated on the yellow seat
• The red and pink seats had the same seat number
• F was seated neighbouring the one who was on the red seat and the one who was seated exactly in front of I.
• K was seated 2nd to the left of C
• The seat number of the green seat was 3 more than the seat number of the grey seat and they were not in the same row
• G's seat number was twice as that of J's seat number. J was on the white seat
• A was seated exactly in front of the golden seat
• B was 2nd to the right of the one seated on the black seat
• E was on the orange seat and was at either of the extreme ends
• K was seated either on the pink or the brown seat
• F was not seated neighbouring L or H
Question:
Who among the following was seated exactly behind J?
• L was seated 3rd to the left of the one on the silver seat, L was not at the extreme end
• D's seat number was twice as that of L's seat number and they were in different rows
• D and H were seated at a gap of 2 seats
• H's neighbour was seated on the yellow seat
• The red and pink seats had the same seat number.
• F was seated neighbouring the one who was on the red seat and the one who was seated exactly in front of I.
• F was not seated neighbouring L.
Case 1:
Case 2:
Invalid as I cannot be placed in the arrangement.
Case 3:
Case 4:
• K was seated 2nd to the left of C
• G's seat number was twice as that of J's seat number. J was on the white seat
• E was on the orange seat and was at either of the extreme ends
• K was seated either on the pink or the brown seat
Case 1:
1st scenario is invalid as G and J cannot be placed in those arrangements. 2nd scenario is invalid as E cannot be placed in that arrangement.
Case 2:
Scenario 2nd is invalid as J should be on white seat
Case 3:
• The seat number of the green seat was 3 more than the seat number of the grey seat and they were not in the same row
• A was seated exactly in front of the golden seat
• B was 2nd to the right of the one seated on the black seat
Case 1:
The cases are invalid as A cannot be placed in these arrangements.
Case 2:
2nd scenario is invalid as green and grey seats cannot be placed as per the conditions.
Case 3:
Invalid as E cannot be placed in the arrangements.
Thus, the final arrangement is as follows:
K was seated exactly behind J.
Comprehension:
Directions: 12 friends A, B, C, D, E, F, G, H, I, J, K and L went to see a movie. They were seated in two different rows each having six seats. They were seated facing the north. The seats were numbered 1-6 from left to right and were of different colour red, green, yellow, white, grey, black, purple, pink, orange, golden, silver and brown not necessarily in the same order.
• L was seated 3rd to the left of the one on the silver seat, L was not at the extreme end
• D's seat number was twice as that of L's seat number and they were in different rows
• D and H were seated at a gap of 2 seats
• H's neighbour was seated on the yellow seat
• The red and pink seats had the same seat number
• F was seated neighbouring the one who was on the red seat and the one who was seated exactly in front of I.
• K was seated 2nd to the left of C
• The seat number of the green seat was 3 more than the seat number of the grey seat and they were not in the same row
• G's seat number was twice as that of J's seat number. J was on the white seat
• A was seated exactly in front of the golden seat
• B was 2nd to the right of the one seated on the black seat
• E was on the orange seat and was at either of the extreme ends
• K was seated either on the pink or the brown seat
• F was not seated neighbouring L or H
Question:
On which of the following seats was D seated?
• L was seated 3rd to the left of the one on the silver seat, L was not at the extreme end
• D's seat number was twice as that of L's seat number and they were in different rows
• D and H were seated at a gap of 2 seats
• H's neighbour was seated on the yellow seat
• The red and pink seats had the same seat number.
• F was seated neighbouring the one who was on the red seat and the one who was seated exactly in front of I.
• F was not seated neighbouring L.
Case 1:
Case 2:
Invalid as I cannot be placed in the arrangement.
Case 3:
Case 4:
• K was seated 2nd to the left of C
• G's seat number was twice as that of J's seat number. J was on the white seat
• E was on the orange seat and was at either of the extreme ends
• K was seated either on the pink or the brown seat
Case 1:
1st scenario is invalid as G and J cannot be placed in those arrangements. 2nd scenario is invalid as E cannot be placed in that arrangement.
Case 2:
Scenario 2nd is invalid as J should be on white seat
Case 3:
• The seat number of the green seat was 3 more than the seat number of the grey seat and they were not in the same row
• A was seated exactly in front of the golden seat
• B was 2nd to the right of the one seated on the black seat
Case 1:
The cases are invalid as A cannot be placed in these arrangements.
Case 2:
2nd scenario is invalid as green and grey seats cannot be placed as per the conditions.
Case 3:
Invalid as E cannot be placed in the arrangements.
Thus, the final arrangement is as follows:
D was seated on the black seat.
Comprehension:
Directions: 12 friends A, B, C, D, E, F, G, H, I, J, K and L went to see a movie. They were seated in two different rows each having six seats. They were seated facing the north. The seats were numbered 1-6 from left to right and were of different colour red, green, yellow, white, grey, black, purple, pink, orange, golden, silver and brown not necessarily in the same order.
• L was seated 3rd to the left of the one on the silver seat, L was not at the extreme end
• D's seat number was twice as that of L's seat number and they were in different rows
• D and H were seated at a gap of 2 seats
• H's neighbour was seated on the yellow seat
• The red and pink seats had the same seat number
• F was seated neighbouring the one who was on the red seat and the one who was seated exactly in front of I.
• K was seated 2nd to the left of C
• The seat number of the green seat was 3 more than the seat number of the grey seat and they were not in the same row
• G's seat number was twice as that of J's seat number. J was on the white seat
• A was seated exactly in front of the golden seat
• B was 2nd to the right of the one seated on the black seat
• E was on the orange seat and was at either of the extreme ends
• K was seated either on the pink or the brown seat
• F was not seated neighbouring L or H
Question:
Who were the immediate neighbours of the one who was on the purple seat?
• L was seated 3rd to the left of the one on the silver seat, L was not at the extreme end
• D's seat number was twice as that of L's seat number and they were in different rows
• D and H were seated at a gap of 2 seats
• H's neighbour was seated on the yellow seat
• The red and pink seats had the same seat number.
• F was seated neighbouring the one who was on the red seat and the one who was seated exactly in front of I.
• F was not seated neighbouring L.
Case 1:
Case 2:
Invalid as I cannot be placed in the arrangement.
Case 3:
Case 4:
• K was seated 2nd to the left of C
• G's seat number was twice as that of J's seat number. J was on the white seat
• E was on the orange seat and was at either of the extreme ends
• K was seated either on the pink or the brown seat
Case 1:
1st scenario is invalid as G and J cannot be placed in those arrangements. 2nd scenario is invalid as E cannot be placed in that arrangement.
Case 2:
Scenario 2nd is invalid as J should be on white seat
Case 3:
• The seat number of the green seat was 3 more than the seat number of the grey seat and they were not in the same row
• A was seated exactly in front of the golden seat
• B was 2nd to the right of the one seated on the black seat
Case 1:
The cases are invalid as A cannot be placed in these arrangements.
Case 2:
2nd scenario is invalid as green and grey seats cannot be placed as per the conditions.
Case 3:
Invalid as E cannot be placed in the arrangements.
Thus, the final arrangement is as follows:
B and D were neighbouring F who was on the purple seat.
Comprehension:
Directions: Kiran, Sushma, Jaya, Pratik, Sonal and Mayank stay in a six floored apartment (from first floor to sixth floor). Each of them lives at different floor. Due to earthquake, the residents of the apartment panic and decide that for next two days two persons would stay at one floor. None of them stay at a floor in which they were already staying.
i) None of them live at top floor after the earthquake. Sonal lives below Kiran before shifting. Mayank lives exactly between Pratik and Sushma before shifting.
ii) Mayank shifts at a gap of one floor below. Pratik shifts at a gap of three floors below.
iii) Kiran and Sonal stays together after shifting. Jaya shifts at a gap of three floors above.
iv) Pratik lives below Sonal after shifting.
Question:
At which floor does Sonal shifts after earthquake?
Pratik shifts four floors below. There are two cases possible:
Case 1:
Case 2:
Mayank lives exactly between Pratik and Sushma before shifting. Mayank shifts two floors below.
Case 1a:
Sushma lives below Sonal before shifting. But fifth floor and sixth floor is already occupied. So case 1a will be rejected.
Case 1b:
Kiran and Sonal stays together after shifting. That means Sushma lives with jaya after shifting
Case 2:
Kiran and Sonal stays together after shifting. Jaya, Mayank and Pratik live at different floor. There can be only three floors occupied after earthquake. So case 2 will be rejected.
Case 1b can be modified as follows:
Sonal lives below Kiran before shifting.
Pratik lives below Sonal after shifting. None of them stay at a floor in which they were already staying. So Sonal and Kiran can stay at 4th floor.
Sonal shifts to fourth floor after earthquake.
Comprehension:
Directions: Kiran, Sushma, Jaya, Pratik, Sonal and Mayank stay in a six floored apartment (from first floor to sixth floor). Each of them lives at different floor. Due to earthquake, the residents of the apartment panic and decide that for next two days two persons would stay at one floor. None of them stay at a floor in which they were already staying.
i) None of them live at top floor after the earthquake. Sonal lives below Kiran before shifting. Mayank lives exactly between Pratik and Sushma before shifting.
ii) Mayank shifts at a gap of one floor below. Pratik shifts at a gap of three floors below.
iii) Kiran and Sonal stays together after shifting. Jaya shifts at a gap of three floors above.
iv) Pratik lives below Sonal after shifting.
Question:
At which floor does Kiran lived before shifting?
Jaya shifts four floors above. No one lives at top floor after the earthquake. That means Jaya lives at first floor before earthquake.
Pratik shifts four floors below. There are two cases possible:
Case 1:
Case 2:
Mayank lives exactly between Pratik and Sushma before shifting. Mayank shifts two floors below.
Case 1a:
Sushma lives below Sonal before shifting. But fifth floor and sixth floor is already occupied. So case 1a will be rejected.
Case 1b:
Kiran and Sonal stays together after shifting. That means Sushma lives with jaya after shifting
Case 2:
Kiran and Sonal stays together after shifting. Jaya, Mayank and Pratik live at different floor. There can be only three floors occupied after earthquake. So case 2 will be rejected.
Case 1b can be modified as follows:
Sonal lives below Kiran before shifting.
Pratik lives below Sonal after shifting. None of them stay at a floor in which they were already staying. So Sonal and Kiran can stay at 4th floor.
Kiran lived at fifth floor before shifting.
Comprehension:
Directions: Kiran, Sushma, Jaya, Pratik, Sonal and Mayank stay in a six floored apartment (from first floor to sixth floor). Each of them lives at different floor. Due to earthquake, the residents of the apartment panic and decide that for next two days two persons would stay at one floor. None of them stay at a floor in which they were already staying.
i) None of them live at top floor after the earthquake. Sonal lives below Kiran before shifting. Mayank lives exactly between Pratik and Sushma before shifting.
ii) Mayank shifts at a gap of one floor below. Pratik shifts at a gap of three floors below.
iii) Kiran and Sonal stays together after shifting. Jaya shifts at a gap of three floors above.
iv) Pratik lives below Sonal after shifting.
Question:
Who lives at 4th floor after earthquake?
Jaya shifts four floors above. No one lives at top floor after the earthquake. That means Jaya lives at first floor before earthquake.
Pratik shifts four floors below. There are two cases possible:
Case 1:
Case 2:
Mayank lives exactly between Pratik and Sushma before shifting. Mayank shifts two floors below.
Case 1a:
Sushma lives below Sonal before shifting. But fifth floor and sixth floor is already occupied. So case 1a will be rejected.
Case 1b:
Kiran and Sonal stays together after shifting. That means Sushma lives with jaya after shifting
Case 2:
Kiran and Sonal stays together after shifting. Jaya, Mayank and Pratik live at different floor. There can be only three floors occupied after earthquake. So case 2 will be rejected.
Case 1b can be modified as follows:
Sonal lives below Kiran before shifting.
Pratik lives below Sonal after shifting. None of them stay at a floor in which they were already staying. So Sonal and Kiran can stay at 4th floor.
Kiran lived at 4th floor after earth quake.
Comprehension:
Directions: Kiran, Sushma, Jaya, Pratik, Sonal and Mayank stay in a six floored apartment (from first floor to sixth floor). Each of them lives at different floor. Due to earthquake, the residents of the apartment panic and decide that for next two days two persons would stay at one floor. None of them stay at a floor in which they were already staying.
i) None of them live at top floor after the earthquake. Sonal lives below Kiran before shifting. Mayank lives exactly between Pratik and Sushma before shifting.
ii) Mayank shifts at a gap of one floor below. Pratik shifts at a gap of three floors below.
iii) Kiran and Sonal stays together after shifting. Jaya shifts at a gap of three floors above.
iv) Pratik lives below Sonal after shifting.
Question:
How many floors does Kiran shift?
Jaya shifts four floors above. No one lives at top floor after the earthquake. That means Jaya lives at first floor before earthquake.
Pratik shifts four floors below. There are two cases possible:
Case 1:
Case 2:
Mayank lives exactly between Pratik and Sushma before shifting. Mayank shifts two floors below.
Case 1a:
Sushma lives below Sonal before shifting. But fifth floor and sixth floor is already occupied. So case 1a will be rejected.
Case 1b:
Kiran and Sonal stays together after shifting. That means Sushma lives with jaya after shifting
Case 2:
Kiran and Sonal stays together after shifting. Jaya, Mayank and Pratik live at different floor. There can be only three floors occupied after earthquake. So case 2 will be rejected.
Case 1b can be modified as follows:
Sonal lives below Kiran before shifting.
Pratik lives below Sonal after shifting. None of them stay at a floor in which they were already staying. So Sonal and Kiran can stay at 4th floor.
Kiran shifts one floor.
Comprehension:
Directions: Kiran, Sushma, Jaya, Pratik, Sonal and Mayank stay in a six floored apartment (from first floor to sixth floor). Each of them lives at different floor. Due to earthquake, the residents of the apartment panic and decide that for next two days two persons would stay at one floor. None of them stay at a floor in which they were already staying.
i) None of them live at top floor after the earthquake. Sonal lives below Kiran before shifting. Mayank lives exactly between Pratik and Sushma before shifting.
ii) Mayank shifts at a gap of one floor below. Pratik shifts at a gap of three floors below.
iii) Kiran and Sonal stays together after shifting. Jaya shifts at a gap of three floors above.
iv) Pratik lives below Sonal after shifting.
Question:
Who lives at 3rd floor(before shifting)?
Jaya shifts four floors above. No one lives at top floor after the earthquake. That means Jaya lives at first floor before earthquake.
Pratik shifts four floors below. There are two cases possible:
Case 1:
Case 2:
Mayank lives exactly between Pratik and Sushma before shifting. Mayank shifts two floors below.
Case 1a:
Sushma lives below Sonal before shifting. But fifth floor and sixth floor is already occupied. So case 1a will be rejected.
Case 1b:
Kiran and Sonal stays together after shifting. That means Sushma lives with jaya after shifting
Case 2:
Kiran and Sonal stays together after shifting. Jaya, Mayank and Pratik live at different floor. There can be only three floors occupied after earthquake. So case 2 will be rejected.
Case 1b can be modified as follows:
Sonal lives below Kiran before shifting.
Pratik lives below Sonal after shifting. None of them stay at a floor in which they were already staying. So Sonal and Kiran can stay at 4th floor.
Sonal lived at third floor before shifting.
Comprehension:
Directions: In a family of 8 persons, every child has either none or both parents alive. Family doctor visits every month twice i.e. on 8th and 27th of the month. On every visit, doctor does routine check-up of only 1 person. Check-up of every person is repeated after every 4 months.
• Check-up of C is done immediately after his wife D and immediately before his son F.
• Check-up of both, F and his sister is done in April.
• G is daughter-in-law of C and she is checked in May.
• Every couple in the family has only one son and only one daughter.
• Sister-in-law of D is checked on 27th June.
• A is aunt of B, who is aunt of H.
• Check-up of E is done in in October. E is not the male member of the family.
Question:
Check-up of __ is done on 27th November?
Since there are only 8 persons in the family, the person checked on 8th of January will again be checked on 8th of May and thereafter again on 8th of September and cycle will keep repeating.
Now,
Check-up of C is done immediately after his wife D and immediately before his son F.
Check-up of F and his sister is done in April.
G is daughter-in-law of C and she is checked in May.
And,
Every couple in the family has only one son and only one daughter.
Sister-in-law of D is checked on 27th June. The sister-in-law of D has to be the sister of C otherwise the number of persons in the family will become more than 8.
And,
A is aunt of B, who is aunt of H.
So, A is the sister of C and B is the sister of F.
Check-up of E is done in October. E is female.
So, E is the daughter of F and H is the son of F.
And,
C, who is brother of A is checked on every 27th of March/July/November.
Comprehension:
Directions: In a family of 8 persons, every child has either none or both parents alive. Family doctor visits every month twice i.e. on 8th and 27th of the month. On every visit, doctor does routine check-up of only 1 person. Check-up of every person is repeated after every 4 months.
• Check-up of C is done immediately after his wife D and immediately before his son F.
• Check-up of both, F and his sister is done in April.
• G is daughter-in-law of C and she is checked in May.
• Every couple in the family has only one son and only one daughter.
• Sister-in-law of D is checked on 27th June.
• A is aunt of B, who is aunt of H.
• Check-up of E is done in in October. E is not the male member of the family.
Question:
Find the odd one out.
Since there are only 8 persons in the family, the person checked on 8th of January will again be checked on 8th of May and thereafter again on 8th of September and cycle will keep repeating.
Now,
Check-up of C is done immediately after his wife D and immediately before his son F.
Check-up of F and his sister is done in April.
G is daughter-in-law of C and she is checked in May.
And,
Every couple in the family has only one son and only one daughter.
Sister-in-law of D is checked on 27th June. The sister-in-law of D has to be the sister of C otherwise the number of persons in the family will become more than 8.
And,
A is aunt of B, who is aunt of H.
So, A is the sister of C and B is the sister of F.
Check-up of E is done in October. E is female.
So, E is the daughter of F and H is the son of F.
And,
F is the male member of the family. Rest are female members of the family.
Comprehension:
Directions: In a family of 8 persons, every child has either none or both parents alive. Family doctor visits every month twice i.e. on 8th and 27th of the month. On every visit, doctor does routine check-up of only 1 person. Check-up of every person is repeated after every 4 months.
• Check-up of C is done immediately after his wife D and immediately before his son F.
• Check-up of both, F and his sister is done in April.
• G is daughter-in-law of C and she is checked in May.
• Every couple in the family has only one son and only one daughter.
• Sister-in-law of D is checked on 27th June.
• A is aunt of B, who is aunt of H.
• Check-up of E is done in in October. E is not the male member of the family.
Question:
How H is related to B?
Since there are only 8 persons in the family, the person checked on 8th of January will again be checked on 8th of May and thereafter again on 8th of September and cycle will keep repeating.
Now,
Check-up of C is done immediately after his wife D and immediately before his son F.
Check-up of F and his sister is done in April.
G is daughter-in-law of C and she is checked in May.
And,
Every couple in the family has only one son and only one daughter.
Sister-in-law of D is checked on 27th June. The sister-in-law of D has to be the sister of C otherwise the number of persons in the family will become more than 8.
And,
A is aunt of B, who is aunt of H.
So, A is the sister of C and B is the sister of F.
Check-up of E is done in October. E is female.
So, E is the daughter of F and H is the son of F.
And,
H is the nephew of B.
Comprehension:
Directions: In a family of 8 persons, every child has either none or both parents alive. Family doctor visits every month twice i.e. on 8th and 27th of the month. On every visit, doctor does routine check-up of only 1 person. Check-up of every person is repeated after every 4 months.
• Check-up of C is done immediately after his wife D and immediately before his son F.
• Check-up of both, F and his sister is done in April.
• G is daughter-in-law of C and she is checked in May.
• Every couple in the family has only one son and only one daughter.
• Sister-in-law of D is checked on 27th June.
• A is aunt of B, who is aunt of H.
• Check-up of E is done in in October. E is not the male member of the family.
Question:
How many persons are checked after third check-up of D and before 31st December of the same year?
Since there are only 8 persons in the family, the person checked on 8th of January will again be checked on 8th of May and thereafter again on 8th of September and cycle will keep repeating.
Now,
Check-up of C is done immediately after his wife D and immediately before his son F.
Check-up of F and his sister is done in April.
G is daughter-in-law of C and she is checked in May.
And,
Every couple in the family has only one son and only one daughter.
Sister-in-law of D is checked on 27th June. The sister-in-law of D has to be the sister of C otherwise the number of persons in the family will become more than 8.
And,
A is aunt of B, who is aunt of H.
So, A is the sister of C and B is the sister of F.
Check-up of E is done in October. E is female.
So, E is the daughter of F and H is the son of F.
And,
B, C and F are checked after D and before 31st December of the same year.
Comprehension:
Directions: In a family of 8 persons, every child has either none or both parents alive. Family doctor visits every month twice i.e. on 8th and 27th of the month. On every visit, doctor does routine check-up of only 1 person. Check-up of every person is repeated after every 4 months.
• Check-up of C is done immediately after his wife D and immediately before his son F.
• Check-up of both, F and his sister is done in April.
• G is daughter-in-law of C and she is checked in May.
• Every couple in the family has only one son and only one daughter.
• Sister-in-law of D is checked on 27th June.
• A is aunt of B, who is aunt of H.
• Check-up of E is done in in October. E is not the male member of the family.
Question:
B is married to __.
Since there are only 8 persons in the family, the person checked on 8th of January will again be checked on 8th of May and thereafter again on 8th of September and cycle will keep repeating.
Now,
Check-up of C is done immediately after his wife D and immediately before his son F.
Check-up of F and his sister is done in April.
G is daughter-in-law of C and she is checked in May.
And,
Every couple in the family has only one son and only one daughter.
Sister-in-law of D is checked on 27th June. The sister-in-law of D has to be the sister of C otherwise the number of persons in the family will become more than 8.
And,
A is aunt of B, who is aunt of H.
So, A is the sister of C and B is the sister of F.
Check-up of E is done in October. E is female.
So, E is the daughter of F and H is the son of F.
And,
B is unmarried.
Comprehension:
Directions: In a family of 8 persons, every child has either none or both parents alive. Family doctor visits every month twice i.e. on 8th and 27th of the month. On every visit, doctor does routine check-up of only 1 person. Check-up of every person is repeated after every 4 months.
• Check-up of C is done immediately after his wife D and immediately before his son F.
• Check-up of both, F and his sister is done in April.
• G is daughter-in-law of C and she is checked in May.
• Every couple in the family has only one son and only one daughter.
• Sister-in-law of D is checked on 27th June.
• A is aunt of B, who is aunt of H.
• Check-up of E is done in in October. E is not the male member of the family.
Question:
How C is related to H?
Since there are only 8 persons in the family, the person checked on 8th of January will again be checked on 8th of May and thereafter again on 8th of September and cycle will keep repeating.
Now,
Check-up of C is done immediately after his wife D and immediately before his son F.
Check-up of F and his sister is done in April.
G is daughter-in-law of C and she is checked in May.
And,
Every couple in the family has only one son and only one daughter.
Sister-in-law of D is checked on 27th June. The sister-in-law of D has to be the sister of C otherwise the number of persons in the family will become more than 8.
And,
A is aunt of B, who is aunt of H.
So, A is the sister of C and B is the sister of F.
Check-up of E is done in October. E is female.
So, E is the daughter of F and H is the son of F.
And,
C is the grandfather of H.
Comprehension:
Directions: 7 poles A, B, C, D, E, F & G are located in a city. Pole A is 12 km west of pole E. pole C is north-east of pole A and north-west of pole E. Distance between pole E & C is equal to distance between pole A & C. Distance between pole A & C is 3√13 km. Pole G is 10 km south of pole F. Pole G is south-east of pole A. Pole D is north-east of pole A but north-west of pole C. pole B is 12 km east of pole D. Distance between pole E & F is 5 km. distance between pole E & G is √45 km.
Question:
What is the direction of B with respect to F?
• Pole A is 12 km west of pole E.
• Pole C is north-east of pole A but north-west of pole E.
• Distance between pole E & C is equal to distance between pole A & C.
• Distance between pole A & C is 3√13 km.
on solving, we get, y = 6 m
Putting y=6 min X =(3√13)2-(12-y)2
we get, X = 9 m
• Pole D is north-east of pole A but north-west of pole C.
• Pole B is 12 km east of pole D.
• Pole G is 10 km south of pole F which is north-east of pole B.
• Pole G is south-east of pole A.
• Distance between pole E & F is 5 km.
• Distance between pole E & G is √45 km.
y = 4 km
using y = 4 km
we get, x = 3 km
So we have,
B will be north-east or north-west or north of F.
B will be north-east or north-west or north of F.
Comprehension:
Directions: 7 poles A, B, C, D, E, F & G are located in a city. Pole A is 12 km west of pole E. pole C is north-east of pole A and north-west of pole E. Distance between pole E & C is equal to distance between pole A & C. Distance between pole A & C is 3√13 km. Pole G is 10 km south of pole F. Pole G is south-east of pole A. Pole D is north-east of pole A but north-west of pole C. pole B is 12 km east of pole D. Distance between pole E & F is 5 km. distance between pole E & G is √45 km.
Question:
What is the direction of F with respect to C?
• Pole A is 12 km west of pole E.
• Pole C is north-east of pole A but north-west of pole E.
• Distance between pole E & C is equal to distance between pole A & C.
• Distance between pole A & C is 3√13 km.
on solving, we get, y = 6 m
Putting y=6 min X =(3√13)2-(12-y)2
we get, X = 9 m
• Pole D is north-east of pole A but north-west of pole C.
• Pole B is 12 km east of pole D.
• Pole G is 10 km south of pole F which is north-east of pole B.
• Pole G is south-east of pole A.
• Distance between pole E & F is 5 km.
• Distance between pole E & G is √45 km.
y = 4 km
using y = 4 km
we get, x = 3 km
So we have,
B will be north-east or north-west or north of F.
F is towards south-east of C.
Comprehension:
Directions: 7 poles A, B, C, D, E, F & G are located in a city. Pole A is 12 km west of pole E. pole C is north-east of pole A and north-west of pole E. Distance between pole E & C is equal to distance between pole A & C. Distance between pole A & C is 3√13 km. Pole G is 10 km south of pole F. Pole G is south-east of pole A. Pole D is north-east of pole A but north-west of pole C. pole B is 12 km east of pole D. Distance between pole E & F is 5 km. distance between pole E & G is √45 km.
Question:
What is the shortest distance between F & A?
• Pole A is 12 km west of pole E.
• Pole C is north-east of pole A but north-west of pole E.
• Distance between pole E & C is equal to distance between pole A & C.
• Distance between pole A & C is 3√13 km.
on solving, we get, y = 6 m
Putting y=6 min X =(3√13)2-(12-y)2
we get, X = 9 m
• Pole D is north-east of pole A but north-west of pole C.
• Pole B is 12 km east of pole D.
• Pole G is 10 km south of pole F which is north-east of pole B.
• Pole G is south-east of pole A.
• Distance between pole E & F is 5 km.
• Distance between pole E & G is √45 km.
y = 4 km
using y = 4 km
we get, x = 3 km
So we have,
B will be north-east or north-west or north of F.
AF2 = 42 + 152
AF2 = 16 + 225
AF2 = 241
AF = √241 km
Comprehension:
Directions: In the family of 8 persons viz A, B, C, D, E, F, G, and H, there are more male members than female members. Height of each member in the family is different.
• Every female member except C is shorter than all male members.
• D is female but is not shortest among all.
• Number of family members taller than C are more than members shorter than C.
• Number of male members taller than A are same as the number of male members shorter than A.
• Neither F nor H is the female member in the family.
• B is taller than G, who is not the shortest member of the family.
• F is shorter than G, but not the shortest male member of the family.
Question:
Find the odd one out.
Number of family members taller than C are more than members shorter than C.
So, C is 4th shortest person in the family.
Equal number of male members are taller and shorter than A.
So, A is 3rd tallest person.
Neither F nor H is the female member in the family.
B is taller than G, who is not the shortest member of the family.
So, E must be the female member of the family.
F is shorter than G, but not the shortest male member of the family. So, we get: -
E is female member and rest are the male members of the family.
Comprehension:
Directions: In the family of 8 persons viz A, B, C, D, E, F, G, and H, there are more male members than female members. Height of each member in the family is different.
• Every female member except C is shorter than all male members.
• D is female but is not shortest among all.
• Number of family members taller than C are more than members shorter than C.
• Number of male members taller than A are same as the number of male members shorter than A.
• Neither F nor H is the female member in the family.
• B is taller than G, who is not the shortest member of the family.
• F is shorter than G, but not the shortest male member of the family.
Question:
Who is the 2nd shortest male member of the family?
Number of family members taller than C are more than members shorter than C.
So, C is 4th shortest person in the family.
Equal number of male members are taller and shorter than A.
So, A is 3rd tallest person.
Neither F nor H is the female member in the family.
B is taller than G, who is not the shortest member of the family.
So, E must be the female member of the family.
F is shorter than G, but not the shortest male member of the family. So, we get: -
F is the 2nd shortest male member of the family.
Comprehension:
Directions: In the family of 8 persons viz A, B, C, D, E, F, G, and H, there are more male members than female members. Height of each member in the family is different.
• Every female member except C is shorter than all male members.
• D is female but is not shortest among all.
• Number of family members taller than C are more than members shorter than C.
• Number of male members taller than A are same as the number of male members shorter than A.
• Neither F nor H is the female member in the family.
• B is taller than G, who is not the shortest member of the family.
• F is shorter than G, but not the shortest male member of the family.
Question:
How many persons are taller than H but shorter than G?
Number of family members taller than C are more than members shorter than C.
So, C is 4th shortest person in the family.
Equal number of male members are taller and shorter than A.
So, A is 3rd tallest person.
Neither F nor H is the female member in the family.
B is taller than G, who is not the shortest member of the family.
So, E must be the female member of the family.
F is shorter than G, but not the shortest male member of the family. So, we get: -
3 persons are taller than H but shorter than G.
Comprehension:
Directions: In the family of 8 persons viz A, B, C, D, E, F, G, and H, there are more male members than female members. Height of each member in the family is different.
• Every female member except C is shorter than all male members.
• D is female but is not shortest among all.
• Number of family members taller than C are more than members shorter than C.
• Number of male members taller than A are same as the number of male members shorter than A.
• Neither F nor H is the female member in the family.
• B is taller than G, who is not the shortest member of the family.
• F is shorter than G, but not the shortest male member of the family.
Question:
Who among the following is taller than A?
Number of family members taller than C are more than members shorter than C.
So, C is 4th shortest person in the family.
Equal number of male members are taller and shorter than A.
So, A is 3rd tallest person.
Neither F nor H is the female member in the family.
B is taller than G, who is not the shortest member of the family.
So, E must be the female member of the family.
F is shorter than G, but not the shortest male member of the family. So, we get: -
B is taller than A. Other persons given in the option are shorter than A.
Comprehension:
Directions: In the family of 8 persons viz A, B, C, D, E, F, G, and H, there are more male members than female members. Height of each member in the family is different.
• Every female member except C is shorter than all male members.
• D is female but is not shortest among all.
• Number of family members taller than C are more than members shorter than C.
• Number of male members taller than A are same as the number of male members shorter than A.
• Neither F nor H is the female member in the family.
• B is taller than G, who is not the shortest member of the family.
• F is shorter than G, but not the shortest male member of the family.
Question:
Who is the tallest person of the family?
Number of family members taller than C are more than members shorter than C.
So, C is 4th shortest person in the family.
Equal number of male members are taller and shorter than A.
So, A is 3rd tallest person.
Neither F nor H is the female member in the family.
B is taller than G, who is not the shortest member of the family.
So, E must be the female member of the family.
F is shorter than G, but not the shortest male member of the family. So, we get: -
B is the tallest person of the family.
Comprehension:
Directions: In a 3-generation family of 11 members, there is at least one couple in each generation. Z's mother is married to L's brother and, B's mother is married to P's brother. P is female. N is K's daughter-in-law and has two children, who are married. T, who is A's wife is D's sister-in-law. C is P's unmarried brother. D is B's daughter-in-law. G is L's mother.
Question:
How is G related to B?
→ Z's mother is married to L's brother and, B's mother is married to P's brother.
→ P is female.
→ D is B's daughter-in-law.
→ G is L's mother.
We get,
Also,
→ N is K's daughter-in-law and has two children, who are married.
We get,
Now, we have eleven members and, there is at least one couple in each generation.
Only possibility is that, B is married in second generation.
B must be brother of L and G is their mother.
Also, B must be married to N and K must be the father of B.
→ C is P's unmarried brother.
All the above three diagrams are combined and,
We get,
Now,
→ T, who is A's wife is D's sister-in-law.
Thus, we get, D is Z's wife.
Final arrangement is as follows:
G is B's mother.
Comprehension:
Directions: In a 3-generation family of 11 members, there is at least one couple in each generation. Z's mother is married to L's brother and, B's mother is married to P's brother. P is female. N is K's daughter-in-law and has two children, who are married. T, who is A's wife is D's sister-in-law. C is P's unmarried brother. D is B's daughter-in-law. G is L's mother.
Question:
Z is K's ___________.
→ Z's mother is married to L's brother and, B's mother is married to P's brother.
→ P is female.
→ D is B's daughter-in-law.
→ G is L's mother.
We get,
Also,
→ N is K's daughter-in-law and has two children, who are married.
We get,
Now, we have eleven members and, there is at least one couple in each generation.
Only possibility is that, B is married in second generation.
B must be brother of L and G is their mother.
Also, B must be married to N and K must be the father of B.
→ C is P's unmarried brother.
All the above three diagrams are combined and,
We get,
Now,
→ T, who is A's wife is D's sister-in-law.
Thus, we get, D is Z's wife.
Final arrangement is as follows:
Z is K's grandson.
Comprehension:
Directions: Read the following passages/statements and answer the questions that follow them.
Question:
Statement: Working from the comforts of their own homes should, in theory, improve employee well-being. However, new research suggests that working from home may have more disadvantages than perks.
Which of the following can be inferred from the given statement?
1. There are other research which suggest that working from home have more perks than advantages.
2. There is an option of working from home provided by companies/organisations.
3. Working from the comforts of their own homes doesn't improve employee's well-being in practicality.
Comprehension:
Directions: Read the following passages/statements and answer the questions that follow them.
Question:
Statement: Working from the comforts of their own homes should, in theory, improve employee well-being. However, new research suggests that working from home may have more disadvantages than perks.
Which of the following can be a possible reason for conducting such a research regarding pros and cons of working from home?
1. Companies are required to frame new policies regarding work from home for their respective employees.
2. Bringing work home can be more distressing compared with simply working from home.
3. To compare the stress levels between those who worked at home and those who worked at offices.
Comprehension:
Directions: When line of input is given to the machine, it operates the input in several steps and gives the output as below;
Input: Strand incline Grind Plan Eclipse Operation
Step-1: Eclipse incline Grind Plan Operation Strand
Step-2: Eclipser incline Grind Plan Operation Stranded
Step-3: Grind Eclipser incline Operation Stranded Plan
Step-4: Grinded Eclipser incline Operation Stranded Planed
Step-5: incline Grinded Eclipser Stranded Planed Operation
Step-6: incliner Grinded Eclipser Stranded Planed Operationr
Step-6 is the final step.
Apply the same logic to the following line of input and answer the questions.
Input: Ellipse Incline Clinch Defer Out Glimpse
Question:
What is the rightmost element in the last step?
In next step, alphabetically 2nd and 2nd last words are operated.
Applying the same logic to the given line of input, we get: -
Input: Ellipse Incline Clinch Defer Out Glimpse
Step-1: Clinch Ellipse Incline Defer Glimpse Out
Step-2: Clinched Ellipse Incline Defer Glimpse Outr
Step-3: Defer Clinched Ellipse Glimpse Outr Incline
Step-4: Defered Clinched Ellipse Glimpse Outr Incliner
Step-5: Ellipse Defered Clinched Outr Incliner Glimpse
Step-6: Ellipser Defered Clinched Outr Incliner Glimpseed
This is the final step.
Step-6: Ellipser Defered Clinched Outr Incliner Glimpseed.
Comprehension:
Directions: When line of input is given to the machine, it operates the input in several steps and gives the output as below;
Input: Strand incline Grind Plan Eclipse Operation
Step-1: Eclipse incline Grind Plan Operation Strand
Step-2: Eclipser incline Grind Plan Operation Stranded
Step-3: Grind Eclipser incline Operation Stranded Plan
Step-4: Grinded Eclipser incline Operation Stranded Planed
Step-5: incline Grinded Eclipser Stranded Planed Operation
Step-6: incliner Grinded Eclipser Stranded Planed Operationr
Step-6 is the final step.
Apply the same logic to the following line of input and answer the questions.
Input: Ellipse Incline Clinch Defer Out Glimpse
Question:
What is the 3rd element from left end in step-5?
In next step, alphabetically 2nd and 2nd last words are operated.
Applying the same logic to the given line of input, we get: -
Input: Ellipse Incline Clinch Defer Out Glimpse
Step-1: Clinch Ellipse Incline Defer Glimpse Out
Step-2: Clinched Ellipse Incline Defer Glimpse Outr
Step-3: Defer Clinched Ellipse Glimpse Outr Incline
Step-4: Defered Clinched Ellipse Glimpse Outr Incliner
Step-5: Ellipse Defered Clinched Outr Incliner Glimpse
Step-6: Ellipser Defered Clinched Outr Incliner Glimpseed
This is the final step.
Step-5: Ellipse Defered Clinched Outr Incliner Glimpse.
Comprehension:
Directions: When line of input is given to the machine, it operates the input in several steps and gives the output as below;
Input: Strand incline Grind Plan Eclipse Operation
Step-1: Eclipse incline Grind Plan Operation Strand
Step-2: Eclipser incline Grind Plan Operation Stranded
Step-3: Grind Eclipser incline Operation Stranded Plan
Step-4: Grinded Eclipser incline Operation Stranded Planed
Step-5: incline Grinded Eclipser Stranded Planed Operation
Step-6: incliner Grinded Eclipser Stranded Planed Operationr
Step-6 is the final step.
Apply the same logic to the following line of input and answer the questions.
Input: Ellipse Incline Clinch Defer Out Glimpse
Question:
What is the 3rd element from left end in step-3?
In next step, alphabetically 2nd and 2nd last words are operated.
Applying the same logic to the given line of input, we get: -
Input: Ellipse Incline Clinch Defer Out Glimpse
Step-1: Clinch Ellipse Incline Defer Glimpse Out
Step-2: Clinched Ellipse Incline Defer Glimpse Outr
Step-3: Defer Clinched Ellipse Glimpse Outr Incline
Step-4: Defered Clinched Ellipse Glimpse Outr Incliner
Step-5: Ellipse Defered Clinched Outr Incliner Glimpse
Step-6: Ellipser Defered Clinched Outr Incliner Glimpseed
This is the final step.
Step-3: Defer Clinched Ellipse Glimpse Outr Incline.
Comprehension:
Directions: When line of input is given to the machine, it operates the input in several steps and gives the output as below;
Input: Strand incline Grind Plan Eclipse Operation
Step-1: Eclipse incline Grind Plan Operation Strand
Step-2: Eclipser incline Grind Plan Operation Stranded
Step-3: Grind Eclipser incline Operation Stranded Plan
Step-4: Grinded Eclipser incline Operation Stranded Planed
Step-5: incline Grinded Eclipser Stranded Planed Operation
Step-6: incliner Grinded Eclipser Stranded Planed Operationr
Step-6 is the final step.
Apply the same logic to the following line of input and answer the questions.
Input: Ellipse Incline Clinch Defer Out Glimpse
Question:
What is the 3rd element from right end in step-6?
In next step, alphabetically 2nd and 2nd last words are operated.
Applying the same logic to the given line of input, we get: -
Input: Ellipse Incline Clinch Defer Out Glimpse
Step-1: Clinch Ellipse Incline Defer Glimpse Out
Step-2: Clinched Ellipse Incline Defer Glimpse Outr
Step-3: Defer Clinched Ellipse Glimpse Outr Incline
Step-4: Defered Clinched Ellipse Glimpse Outr Incliner
Step-5: Ellipse Defered Clinched Outr Incliner Glimpse
Step-6: Ellipser Defered Clinched Outr Incliner Glimpseed
This is the final step.
Step-6: Ellipser Defered Clinched Outr Incliner Glimpseed.
Comprehension:
Directions: When line of input is given to the machine, it operates the input in several steps and gives the output as below;
Input: Strand incline Grind Plan Eclipse Operation
Step-1: Eclipse incline Grind Plan Operation Strand
Step-2: Eclipser incline Grind Plan Operation Stranded
Step-3: Grind Eclipser incline Operation Stranded Plan
Step-4: Grinded Eclipser incline Operation Stranded Planed
Step-5: incline Grinded Eclipser Stranded Planed Operation
Step-6: incliner Grinded Eclipser Stranded Planed Operationr
Step-6 is the final step.
Apply the same logic to the following line of input and answer the questions.
Input: Ellipse Incline Clinch Defer Out Glimpse
Question:
What is the 2nd element from right end in step-3?
In next step, alphabetically 2nd and 2nd last words are operated.
Applying the same logic to the given line of input, we get: -
Input: Ellipse Incline Clinch Defer Out Glimpse
Step-1: Clinch Ellipse Incline Defer Glimpse Out
Step-2: Clinched Ellipse Incline Defer Glimpse Outr
Step-3: Defer Clinched Ellipse Glimpse Outr Incline
Step-4: Defered Clinched Ellipse Glimpse Outr Incliner
Step-5: Ellipse Defered Clinched Outr Incliner Glimpse
Step-6: Ellipser Defered Clinched Outr Incliner Glimpseed
This is the final step.
Step-3: Defer Clinched Ellipse Glimpse Outr Incline.
Comprehension:
Directions: When line of input is given to the machine, it operates the input in several steps and gives the output as below;
Input: Strand incline Grind Plan Eclipse Operation
Step-1: Eclipse incline Grind Plan Operation Strand
Step-2: Eclipser incline Grind Plan Operation Stranded
Step-3: Grind Eclipser incline Operation Stranded Plan
Step-4: Grinded Eclipser incline Operation Stranded Planed
Step-5: incline Grinded Eclipser Stranded Planed Operation
Step-6: incliner Grinded Eclipser Stranded Planed Operationr
Step-6 is the final step.
Apply the same logic to the following line of input and answer the questions.
Input: Ellipse Incline Clinch Defer Out Glimpse
Question:
What is the 2nd element from left end in step-6?
In next step, alphabetically 2nd and 2nd last words are operated.
Applying the same logic to the given line of input, we get: -
Input: Ellipse Incline Clinch Defer Out Glimpse
Step-1: Clinch Ellipse Incline Defer Glimpse Out
Step-2: Clinched Ellipse Incline Defer Glimpse Outr
Step-3: Defer Clinched Ellipse Glimpse Outr Incline
Step-4: Defered Clinched Ellipse Glimpse Outr Incliner
Step-5: Ellipse Defered Clinched Outr Incliner Glimpse
Step-6: Ellipser Defered Clinched Outr Incliner Glimpseed
This is the final step.
Step-6: Ellipser Defered Clinched Outr Incliner Glimpseed.
By means of a military intervention, the United States has helped Afghanistan organize democratic elections and a centralized government, which on their own they likely would have been unable to do.
Which of the following statements is the author most likely to agree with?
Theologian Harvey Cox, echoing Simmel, has pointed out that in an urban environment the attempt to "involve" oneself fully with everyone can lead only to self-destruction and emotional emptiness. Urban man, he writes, "must have more or less impersonal relationships with most of the people with whom he comes in contact precisely in order to choose certain friendships to nourish and cultivate.
Which of the following statements best sums up the main idea of the passage?
Praveen started moving towards his south and stops after walking certain kilometers. He then moves to the south-west direction and travels the same distance and stops, at another point he moved to his north. Finally, he moves to his north-west and stops at a point. If the distance between each point is same (say x), then, in which direction is the point where he took his final turn with respect to his starting point?
Clearly, that point is to the south-west of the initial point.
D said, 'A's father is the only brother of my sister's son.' How is A's father related to D?
Option 2
Comprehension:
In each of the questions below are given some statements followed by some conclusions. You have to take the given statements to be true even if they seem to be at variance from commonly known facts. Read all the conclusions and then decide which of the given conclusions logically follows from the given statements disregarding commonly known facts.
Question:
Statements:
Some men are women.
Some women are kids.
All kids are educated.
No educated is poor.
Conclusions:
I. Some educated are kids.
II. Some poor may be both educated and women.
III. Some men are educated.
IV. Some women are educated.
Thus conclusions I and IV follow.
Comprehension:
In each of the questions below are given some statements followed by some conclusions. You have to take the given statements to be true even if they seem to be at variance from commonly known facts. Read all the conclusions and then decide which of the given conclusions logically follows from the given statements disregarding commonly known facts.
Question:
Statements:
Some reds are blues.
Some blues are whites.
All blacks are reds.
All greens are whites.
Conclusions:
I. Some reds are whites.
II. Some blues are greens.
III. No white is red.
IV. Some reds are blacks.
Case I: Some reds are whites
Case II: No white is a red
In either case, some reds are black.
Comprehension:
In each of the questions below are given some statements followed by some conclusions. You have to take the given statements to be true even if they seem to be at variance from commonly known facts. Read all the conclusions and then decide which of the given conclusions logically follows from the given statements disregarding commonly known facts.
Question:
Statements:
All trees are woods.
Some woods are panes.
All huts are panes.
Some panes are saw.
Conclusions:
I. No woods are saws.
II. No huts are saws.
III. No trees are panes.
IV. All panes are trees.
Thus none of the conclusions follow.
Comprehension:
In each of the questions below are given some statements followed by some conclusions. You have to take the given statements to be true even if they seem to be at variance from commonly known facts. Read all the conclusions and then decide which of the given conclusions logically follows from the given statements disregarding commonly known facts.
Question:
Statements:
Some American are Chinese.
Some Chinese are Japanese.
All Japanese are Indian.
No Indian is Pakistani.
Conclusions:
I. Some Indian are Japanese.
II. Some Pakistani may be both Indian and Chinese.
III. Some American are Indian.
IV. Some Chinese are Indian.
Hence conclusions I and IV follow.
Comprehension:
Directions [Set of 5 questions]: In a certain language:
'The craft play role' is written as '12 65 78 99'
'All the very best' is written as '21 56 78 84'
'Role of fair play' is written as '65 77 12 41'
'Is the craft good too' is written as '78 99 66 33 31'
'Too much good role' is written as '33 27 31 65'
Question:
How can 'Best Fair Play Award' be possibly written as?
'All the very best' = '21 56 78 84' -(2)
'Role of fair play' = '65 77 12 41' -(3)
'Is the craft good too' = '78 99 66 33 31' -(4)
'Too much good role' = '33 27 31 65' -(5)
From 1 and 2, the common word is 'the' and common code is 78. Hence the=78
From 1 and 3, the common words are play/role and the common codes are 12/65
Now in 5, the word role is used and the code present is 65. Hence role=65 and play=12.
The remaining word in 1, i.e. craft will have the code 99.
From 4 and 5, the common words are good/too and the common codes are 33/31
In 4, the=78, craft=99, good/too=33/31, hence is=66
In 5, good/too=33/31, role=65, hence much = 27
The final table is:
Best can be written as 21, 56 or 84, fair = 77 or 41, play = 12. Award is a new word and hence shall have a new code. 96 can be such a code.
Comprehension:
Directions [Set of 5 questions]: In a certain language:
'The craft play role' is written as '12 65 78 99'
'All the very best' is written as '21 56 78 84'
'Role of fair play' is written as '65 77 12 41'
'Is the craft good too' is written as '78 99 66 33 31'
'Too much good role' is written as '33 27 31 65'
Question:
What does '77' stand for?
'All the very best' = '21 56 78 84' -(2)
'Role of fair play' = '65 77 12 41' -(3)
'Is the craft good too' = '78 99 66 33 31' -(4)
'Too much good role' = '33 27 31 65' -(5)
From 1 and 2, the common word is 'the' and common code is 78. Hence the=78
From 1 and 3, the common words are play/role and the common codes are 12/65
Now in 5, the word role is used and the code present is 65. Hence role=65 and play=12.
The remaining word in 1, i.e. craft will have the code 99.
From 4 and 5, the common words are good/too and the common codes are 33/31
In 4, the=78, craft=99, good/too=33/31, hence is=66
In 5, good/too=33/31, role=65, hence much = 27
The final table is:
77 stands for of or fair.
Comprehension:
Directions [Set of 5 questions]: In a certain language:
'The craft play role' is written as '12 65 78 99'
'All the very best' is written as '21 56 78 84'
'Role of fair play' is written as '65 77 12 41'
'Is the craft good too' is written as '78 99 66 33 31'
'Too much good role' is written as '33 27 31 65'
Question:
How is 'Play' written in that code?
'All the very best' = '21 56 78 84' -(2)
'Role of fair play' = '65 77 12 41' -(3)
'Is the craft good too' = '78 99 66 33 31' -(4)
'Too much good role' = '33 27 31 65' -(5)
From 1 and 2, the common word is 'the' and common code is 78. Hence the=78
From 1 and 3, the common words are play/role and the common codes are 12/65
Now in 5, the word role is used and the code present is 65. Hence role=65 and play=12.
The remaining word in 1, i.e. craft will have the code 99.
From 4 and 5, the common words are good/too and the common codes are 33/31
In 4, the=78, craft=99, good/too=33/31, hence is=66
In 5, good/too=33/31, role=65, hence much = 27
The final table is:
Play is written as 12
Comprehension:
Directions [Set of 5 questions]: In a certain language:
'The craft play role' is written as '12 65 78 99'
'All the very best' is written as '21 56 78 84'
'Role of fair play' is written as '65 77 12 41'
'Is the craft good too' is written as '78 99 66 33 31'
'Too much good role' is written as '33 27 31 65'
Question:
How is 'Much' written in that code?
'All the very best' = '21 56 78 84' -(2)
'Role of fair play' = '65 77 12 41' -(3)
'Is the craft good too' = '78 99 66 33 31' -(4)
'Too much good role' = '33 27 31 65' -(5)
From 1 and 2, the common word is 'the' and common code is 78. Hence the=78
From 1 and 3, the common words are play/role and the common codes are 12/65
Now in 5, the word role is used and the code present is 65. Hence role=65 and play=12.
The remaining word in 1, i.e. craft will have the code 99.
From 4 and 5, the common words are good/too and the common codes are 33/31
In 4, the=78, craft=99, good/too=33/31, hence is=66
In 5, good/too=33/31, role=65, hence much = 27
The final table is:
Mush is written as 27
Comprehension:
Directions [Set of 5 questions]: In a certain language:
'The craft play role' is written as '12 65 78 99'
'All the very best' is written as '21 56 78 84'
'Role of fair play' is written as '65 77 12 41'
'Is the craft good too' is written as '78 99 66 33 31'
'Too much good role' is written as '33 27 31 65'
Question:
What may be the representation for '27 84 66' in the above language?
'All the very best' = '21 56 78 84' -(2)
'Role of fair play' = '65 77 12 41' -(3)
'Is the craft good too' = '78 99 66 33 31' -(4)
'Too much good role' = '33 27 31 65' -(5)
From 1 and 2, the common word is 'the' and common code is 78. Hence the=78
From 1 and 3, the common words are play/role and the common codes are 12/65
Now in 5, the word role is used and the code present is 65. Hence role=65 and play=12.
The remaining word in 1, i.e. craft will have the code 99.
From 4 and 5, the common words are good/too and the common codes are 33/31
In 4, the=78, craft=99, good/too=33/31, hence is=66
In 5, good/too=33/31, role=65, hence much = 27
The final table is:
27=much, 84=all/very/best, 66=is
South Korean President Lee Myung-bak himself has argued "A textbook of modern history should be written in a way that does not hurt our national pride," when criticising a South Korean textbook's interpretation of the dividing of Korea.
Which of the following statements is the President most likely to agree with?
A spurt of surveys over recent years has shown that youngsters world over are choosing to switch from tequila, rum and vodka to Scotch whiskey.
What can be inferred from the passage given above?
Google has said there was a dramatic spike in searches for Irish passport applications as news of the UK's decision to leave the EU broke.
What can be inferred from the passage?
On Thursday, Uber taxi announced that it is expanding its upfront fares globally, meaning you'll know the exact price before you request your ride.
What is the essential assumption made by the author?
Mobile app revenue is growing and is expected to reach $100 billion dollars by 2020. These numbers are amazing; we are in an unprecedented era of growth when it comes to revenue from mobile apps.
Which of the following can weaken the conclusion of the passage above?
The majority of innovation in mobiles over the past 10 years in the start-up space has been around software but it's time for entrepreneurs to think differently for innovation to continue at this pace.
Which of the following can best explain the author's conclusion?
Chinese-backed electric car maker Faraday Future has started working on a self-driving vehicle which it will sell to its existing customers. Hence, it will create a new line of profit.
Which of the following statements strengthens the conclusion by Faraday Future?
Choose the most appropriate option after reading the following statements.
1. Whether due to haste or design, the new laws are marked by vagueness, leaving officials all down the organization's bureaucratic chain great latitude in enforcing them.
2. The opacity of the language leaves the law open to manipulation on political grounds.
Choose the most appropriate option after reading the following statements.
1. If there is any endeavor whose fruits should be freely available, that endeavor is surely publicly financed science.
2. There is a widespread feeling that the journal publishers who have mediated the scientific exchange for the past century or more are becoming an impediment to free distribution of knowledge.
3. Internet revolution is happening, making knowledge transfer cheaper. Technology permits it; researchers and politicians want it, more public money can be spent on it.
Choose the most appropriate option after reading the following statements.
1. Business schools are ideally positioned to point out when an action that provides a benefit for an individual comes at a cost to society, but in reality they rarely bother.
2. It is part of the malaise that has befallen the political debate on capitalism, which has been taken over by special interest and people who have no faith in a real market-based system.
3. When governments favours the private sector it is all too often by being ''probusiness' rather than ''pro-market', meaning that favourable conditions are provided to particular institutions rather than to institutions broadly.
Comprehension:
Directions: In the given questions, assuming the given statements to be true. Find which of the given four conclusions numbered I, II, III and IV is/are definitely true and give your answer accordingly.
Question:
Statement:
T > N ≥ E > I ≤ S ≤ U < X
Conclusions:
I. T > I
II. X > I
III. S < X
IV. N > U
I. T > I is true.
II. I < X or X > I is true.
III. S < X is true.
IV. N and U cannot be compared. So, IV is not true.
Therefore, only I, II and III are true. Hence, option a.
Comprehension:
Directions: In the given questions, assuming the given statements to be true. Find which of the given four conclusions numbered I, II, III and IV is/are definitely true and give your answer accordingly.
Question:
Statements:
N > V > M ≤ O; M ≥ Z > B
Conclusions:
I. Z< O
II. N > O
III.O = Z
IV. N > B
N > V > M ≤ O ... (i)
M ≥ Z > B ... (ii)
Combining (i) and (ii), we get
N > V ≥ Z > B and B < Z ≤ M ≤ O
Thus, Z ≤ O is true.
Hence Z < O may be true.
And Z = O may be true.
Thus, either I or III is true.
Again, we can’t compare N and O. So, II is not true.
But N > B is true. So, IV is true.
Therefore, only IV and either I or III are true. Hence, option b.
Comprehension:
Directions: In the given questions, assuming the given statements to be true. Find which of the given four conclusions numbered I, II, III and IV is/are definitely true and give your answer accordingly.
Question:
Statements:
K ≥ B > E = F; M < B < N
Conclusions:
I. N < K
II. K > M
III. E > M
IV. F < N
M < B < N ... (ii)
Combining (i) and (ii), we get
N > B > E = F ... (iii)
K≥ B < N ... (iv)
K≥B > M ... (v)
M < B > E ... (vi)
Thus, from (iv), we can’t compare J and M. Hence I is
not true.
From (v), K > M is true. Hence II is true.
Again, from (vi), we can’t compare D and L. Hence III
is not true.
Now, from (iii), N > F or F < N is true. Hence IV is true.
Thus, only II and IV are true.
Hence, option d.
Comprehension:
Directions: In the given questions, assuming the given statements to be true. Find which of the given four conclusions numbered I, II, III and IV is/are definitely true and give your answer accordingly.
Question:
Statements:
Z > G ≤ P ≤ Q; G ≥ V < U
Conclusions:
I. Z > Q
II. U < G
III. P > U
IV. Q < V
G ≥ V < U ... (ii)
U > V ≤ G ≤ P ≤ Q
Thus, from (i) we can’t compare Z and Q. Therefore, I is not true.
Again, from (ii) we can’t compare G and U. So, II is not true.
Thus, we can’t compare P and U. Therefore, III is not true.
Again, V≤Q is true. So, IV (Q < V) is not true.
Therefore, none is true. Hence, option d.
Select a figure from amongst the Answer Figures which will continue the same series as established by the four Problem Figures.
In each of the following questions series begins with unnumbered figure on the extreme left. One and only one of the five numbered figures in the series do not fit into the series. The two unnumbered figures one each on the extreme left and the extreme right fit into the series. You have to take as many aspects into account as possible of the figures in the series and find out the one and only one of the five numbered figures which does not fit into the series. The number of that figure is the answer.
In this question the number of lines in the figures goes on increasing one from left to right. If we go by this aspect of number of lines' only then there is no wrong figure in the series. But if we also consider the 'manner' in which the 'number of lines' in the figures goes on increasing from left to right we come to know that the figure No. 4 does not fit into the series. Therefore, 4 is the answer.
Solve the following question accordingly:
In each of the following questions series begins with unnumbered figure on the extreme left. One and only one of the five numbered figures in the series do not fit into the series. The two unnumbered figures one each on the extreme left and the extreme right fit into the series. You have to take as many aspects into account as possible of the figures in the series and find out the one and only one of the five numbered figures which does not fit into the series. The number of that figure is the answer.
In this question the number of lines in the figures goes on increasing one from left to right. If we go by this aspect of number of lines' only then there is no wrong figure in the series. But if we also consider the 'manner' in which the 'number of lines' in the figures goes on increasing from left to right we come to know that the figure No. 4 does not fit into the series. Therefore, 4 is the answer.
Solve the following question accordingly:
In each of the following questions, a related pair of figures (unnumbered) is followed by five numbered pairs of figures. Out of these five, four have relationship similar to that in the unnumbered pair. Only one pair of figures does not have a similar relationship. Select that pair of figures which does not have a similar relationship to that in the unnumbered pair. Number of that pair is your answer.
Element at the top is rotated by 900 in clockwise direction and the shaded part is mirrored.
Remaining elements are rotated by 90o in anti-clockwise direction.
Only figure 3 satisfies this.
In each of the following questions, a related pair of figures (unnumbered) is followed by five numbered pairs of figures. Out of these five, four have relationship similar to that in the unnumbered pair. Only one pair of figures does not have a similar relationship. Select that pair of figures which does not have a similar relationship to that in the unnumbered pair. Number of that pair is your answer.
Figure 3 does not satisfy this condition.
Select a figure from amongst the Answer Figures which will continue the same series as established by the four Problem Figures.
Choose a figure which would most closely resemble first pair?
Select a figure from amongst the Answer Figures which will continue the same series as established by the four Problem Figures.
In each of the questions given below which one of the five answer figures on the right should come after the problem figures on the left, if the sequence were continued?
Remaining elements are shifted by two positions in clockwise and anti-clockwise direction alternately.
New element is added in the empty row or column.
Figure A and B are related in a particular manner. Establish the same relationship between Figures C and D by choosing a figure from the five alternates which would replace the question mark.
Select a figure from amongst the Answer Figures which will continue the same series as established by the four Problem Figures.
In each of the questions given below which one of the five answer figures on the right should come after the problem figures on the left, if the sequence were continued?
Top left and lower middle elements are interchanged. Top middle and lower right elements are interchanged. Remaining elements shift down along the diagonal.
From figure II to figure III:
Top right and lower-middle elements are interchanged. Top middle and lower left elements are interchanged. Remaining elements shift down along the diagonal.
These two steps are repeated alternately.
Which of the following will be the mirror image of the given question figure, if a mirror is placed along the line MN.
Choose a figure which would most closely resemble first pair?
Select a figure from amongst the Answer Figures which will continue the same series as established by the four Problem Figures.
Each of the following questions consists of five figures marked A, B, C, D and E called the Problem Figures followed by five other figures marked 1, 2, 3, 4 and 5 called the Answer Figures. Select a figure from amongst the Answer Figures which will continue the same series as established by the five Problem Figures.
In each of the questions given below which one of the five answer figures on the right should come after the problem figures on the left, if the sequence were continued?
Triangle shifts right by one position and then will be in same position in next step. Then it will be shifted left by one position. This sequence of shifting is repeated.
Hence, triangle should be in the middle in next step.
Among the options 2, 3 and 4, only option 4 has traingle at the middle psoition.
In each of the questions there are two sets of figures. The figures on the left are problem Figures (four figures and one question-marked space) and those on the right are Answer Figures indicated by numbers 1, 2, 3, 4 and 5. A series is established if one of the five Answer Figures is placed at the"Question-marked space"
Alternately, upper and lower elements are replaced with a new element after shifting.
In each of the questions there are two sets of figures. The figures on the left are problem Figures (four figures and one question-marked space) and those on the right are Answer Figures indicated by numbers 1, 2, 3, 4 and 5. A series is established if one of the five Answer Figures is placed at the"Question-marked space"
In each of the questions there are two sets of figures. The figures on the left are problem Figures (four figures and one question-marked space) and those on the right are Answer Figures indicated by numbers 1, 2, 3, 4 and 5. A series is established if one of the five Answer Figures is placed at the"Question-marked space"
Lower element and right element are interchanged. And upper element and left element are interchanged.
After interchanging, left element is replaced with a new element.
From figure II to figure III:
Lower and left elements are interchanged. And upper and right elements are interchanged.
In each of the questions there are two sets of figures. The figures on the left are problem Figures (four figures and one question-marked space) and those on the right are Answer Figures indicated by numbers 1, 2, 3, 4 and 5. A series is established if one of the five Answer Figures is placed at the"Question-marked space"
Or
Arcs shift by one position in anti-clockwise direction. And similar arc is added to the existing arc sequence in clockwise direction.
In each of the questions there are two sets of figures. The figures on the left are problem Figures (four figures and one question-marked space) and those on the right are Answer Figures indicated by numbers 1, 2, 3, 4 and 5. A series is established if one of the five Answer Figures is placed at the"Question-marked space"
Remaining elements shift in anti-clockwise direction.
Choose a figure which would most closely resemble first pair?
In each of the questions given below which one of the five answer figures on the right should come after the problem figures on the left, if the sequence were continued?
Elements in the middle row interchange their positions. Remaining elements shift in clockwise direction. And the element at the top is replaced with a new element after shifting.
From figure II to III:
Bottom row elements interchange their positions. Middle row elements are interchanged. Then right middle element is removed and top element is shifted to this position. New element is added at the top.
These two steps are repated alternately.
In each question, five figures are given. Four are similar in a certain way and so form a group. Find out the figure which does not belong to that group?
In an election among three contestants A,B and C, A gets 120% more votes than B. A beats C by 3,50,000 votes. B beats C by 5% of the total votes. What is the total number of votes polled (in lakh)?
A gets 120% more votes than B.
We know that,
Total number of votes = Votes got by (A+B+C)
Now,
Let the number of votes B gets be x,
Number of votes A gets =2.2x
Number of votes C gets =2.2x−3,50,000
Now, according to the question:
x−2.2x+3,50,000=(5/100)×(x+2.2x+2.2x−3,50,000)
⇒−1.2x+3,50,000=(1/20)×(5.4x−3,50,000)
⇒−24x+70,00,000=(5.4x−3,50,000)
⇒24x+5.4x=70,00,000+3,50,000
⇒29.4x=73,50,000
⇒x=2,50,000
Number of votes polled =5.4x−3,50,000=10,00,000=10 lakh
∴ The required vote is 10 lakh.
Hence, the correct option is (B).
Direction: Read the given table carefully and answer the following question.
Given table shows the data of number of seats in 5 different buses and the percentage of the seats filled by passengers. Some values are missing here.
Note : Total number of seats in Buses D and E together is 65 .
(Total number of vacant seats in all the buses is 40)
The number of vacant seat of bus C is what percent more or less than the number of vacant seat of bus A?
The required percentage = × 100 = 40 %
∴ The number of vacant seat of bus C is 40% less than the number of vacant seat of bus A.
Hence, the correct option is (A).
Direction: Read the given table carefully and answer the following question.
Given table shows the data of number of seats in 5 different buses and the percentage of the seats filled by passengers. Some values are missing here.
Note : Total number of seats in Buses D and E together is 65.
(Total number of vacant seats in all the buses is 40)
If filled seat of bus E is 3 more than the filled seats of bus B. Find the ratio of vacant seat of buses D and E.
The ratio = 14 : 6 = 7 : 3
Hence, the correct option is (E).
Direction: Read the given table carefully and answer the following question.
Given table shows the data of number of seats in 5 different buses and the percentage of the seats filled by passengers. Some values are missing here.
Note : Total number of seats in Buses D and E together is 65 .
(Total number of vacant seats in all the buses is 40)
If in another bus F, number of filled seats is twice the filled seat of bus A, which is 60% of total seats in bus F. Then find the difference in vacant seats of buses F and A.
Required diffrence = 40 - 10
=30
Hence, the correct option is (D).
Direction: Read the given table carefully and answer the following question.
Given table shows the data of number of seats in 5 different buses and the percentage of the seats filled by passengers. Some values are missing here.
Note : Total number of seats in Buses D and E together is 65.
(Total number of vacant seats in all the buses is 40)
If the ratio of number of seats of D and E is 7 : 6. Find the average of number of filled seats in all buses.
The average =
∴ The average number of filled seats in all buses is 28.
Hence, the correct option is (C).
In the figure, G is the center of the circle and ∠AGB=150∘. Find the value of ∠ACB.
∠AGB=150∘
We know that,
The angle made by the chord at center is double the angle made by it in a major segment of the circle.
∠AGB=2×∠ACB
⇒150∘=2×∠ACB
⇒∠ACB=75∘
∴ The value of ∠ACB is 75∘.
Hence, the correct option is (B).
Direction: Below given tabular data shows details of enrollment of students from various schools during 2015 and 2016. While preparing tabular data few data are missing. Read the following data carefully and answer the question that follows based on the table.
If the ratio of pass male to female students from JNV and DPS during 2016 is 8 : 7 and 5 : 4 respectively then, the total number of failed male students from JNV in 2016 is approximately what percent of the total number of failed male students from DPS during 2016?
From the given data,
Passed students from JNV during 2016 = 80% of total students
So, according to the question,
Passed male students from JNV
Therefore, fail male students from JNV = (4080 - 3584) = 496
Passed students from DPS = 84% of total students
So, passed male students from DPS
Fail male students from DPS = (4050 - 3640) = 410
Therefore, required percent 496/410 x 100 = 120.97% ≈ 121%
Hence, the correct option is (A).
Directions [Set of 6 questions]: Below given tabular data shows details of enrollment of students from various schools during 2015 and 2016. While preparing tabular data few data are missing. Read the following data carefully and answer the following questions based on that.
If the average of total students enrolled from DPS, St. Marry and Holy Cross during 2016 is 7020, then what is the percent change in number of passed students from Holy Cross during 2015 to 2016?
Students from DPS during 2016 = (4050 + 3750) = 7800
Students from St. Marry during 2016 = (2250 + 3750) = 6000
Students from Holy Cross during 2016 = 21060 - (7800 + 6000) = 7260
Passed students from Holy cross during 2016 = (7260 x 75/100) = 5445
Passed students from Holy cross during 2015 = (4050 + 3600) x 84/100 = 6426
Required percent = (6426 - 5445)/6426 x 100 = 15.2%
Directions [Set of 6 questions]: Below given tabular data shows details of enrollment of students from various schools during 2015 and 2016. While preparing tabular data few data are missing. Read the following data carefully and answer the following questions based on that.
What is the ratio of total passed male students of JNV to DAV during 2016, if 45% and 60% of passed students from JNV and DAV respectively are males?
Male pass students from JNV during 2016 =
Passed students from DAV during 2016 =
Passed male students from DAV during 2016 =
Required ratio =
Hence, the correct option is (B).
Directions [Set of 6 questions]: Below given tabular data shows details of enrollment of students from various schools during 2015 and 2016. While preparing tabular data few data are missing. Read the following data carefully and answer the following questions based on that.
If the ratio of passed male to female students from DAV during 2015 is 5:7, then total female students enrolled from CHS and DAV are approximately what percent of that from DPS and St. Marry during 2015 taken together, if number of male passed students from DAV during 2015 is 2640?
Let female students from DAV during 2015 be Y
Then,
(3600 + Y) x 80/100 = 6336
Y = 4320
Female students enrolled from CHS and DAV = (3240 + 4320) = 7560
Female students enrolled from DPS and St. Marry = (3600 + 2850) = 6450
Required percent = 7560/6450 x 100 = 117.2%
Directions [Set of 6 questions]: Below given tabular data shows details of enrollment of students from various schools during 2015 and 2016. While preparing tabular data few data are missing. Read the following data carefully and answer the following questions based on that.
What is the difference between average of total male students enrolled from JNV, CHS and DAV together and average of total female students enrolled from DPS, St. Marry and Holy Cross together during 2016?
Female students from DPS, St. Marry and Holy Cross = (3750 + 3750 + 3060) = 10560
Required average difference = (12900 - 10560)/3 = 780
1000, 100, 20, 6, 2.4, ?
1000 * (1/10) = 100
100 * (2/10) = 20
20 * (3/10) = 6
6 * (4/10) = 2.4
Missing term of the series would be: 2.4 * (5/10) = 1.2
1, 3, 7, 15, 31, 63, ?
(1 × 2) + 1 = 3
(3 ×2) + 1 = 7
(7 × 2) + 1 = 15
(15 × 2) + 1 = 31
(31 ×2) + 1 = 63
(63× 2) + 1 = 127
What number will come in place of the question mark?
15, 22, 48, 111, ?
15 + 13 - 1 = 15
15 + 23 - 1 = 22
22 + 33 - 1 = 48
48 + 43 - 1 = 111
111 + 53 - 1 = 235
Hence, the correct option is (D).
Direction: In the following question, a number series is given with one of the terms missing. Choose the correct alternative that will continue the same pattern and replace the question mark (?) in the given series.
44, 41, 36, 29, 20, ?
44 - 3 = 41
41 - 5 = 36
36 - 7 = 29
29 - 9 = 20
20 - 11 = 9
Hence, the correct option is (C).
45, 39, 49, 35, 53, ?
45 + 4 = 49
49 + 4 = 53
And,
39 - 4 = 35
35 - 4 = 31
Direction: In the following question, various terms of a number series are given with one term missing as shown by (?). Choose the missing term out of the given alternatives.
10, 3, 17, 8, ?,18, 31
One is: 10, 17, ?, 31
Here to get a new number add 7 to the previous number.
10 + 7 = 17
17 + 7 = 24
24 + 7 = 31
Other one is: 3, 8, 18
Here to get the next number to multiply the previous number by 2 and add 2.
3 × 2 + 2 = 8
8 × 2 + 2 = 18
So, the missing term is 24.
Hence, the correct option is (A).
A solid piece of iron in the form of a cuboid of dimensions 24.5 cm×16.5 cm×12 cm, is melted to form a solid sphere.
What is the radius (in cm ) of the sphere?
Dimensions of cuboid =24.5 cm×16.5 cm×12 cm
According to question:
Volume of cuboid = Volume of sphere
∴ The radius is 10.5 cm.
Hence, the correct option is (A).
There is a triangle with three circles drawn such that the three corners of the triangle are the centers of the three circles. If the radius of the circle is given as 14 units and the side of the triangle is equal to the diameter of the circles. Then, find the area of the triangle formed?
Radius of circles = 14 cm
Side of the triangle = Diameter of circle = 28 cm
We know that.
Area of the equilateral triangle =
Here, AB = BC = CA = 28 cm
Hence, the correct option is (A).
In trapezium ABCD, AD‖BC, O is intersection point of diagonals BD and AC. Points F and E are on AB and CD, respectively such that FE passes through point O. If BC=5 cm and AD=20 cm, then, the length of EF is:
AD //BC
BC = 5 cm and AD = 20 cm
We know that.
Therefore,
⇒ 200/25
⇒ 8 cm
Hence, the correct option is (A).
ABCD is a trapezium and ABCE is a parallelogram. ∠ADC=75∘. Find the value of ∠BAE.
Given:
∠ADC=75∘
ABCD is a trapezium.
ABCE is a parallelogram.
Now,
∠ADC+∠ABC=180∘ (The opposite angles in a cyclic quadrilateral are supplementary)
⇒75+∠ABC=180∘
⇒∠ABC=105∘
∠ABC=∠AEC=105∘ (Opposite angles of parallelogram are equal)
∠ABC+∠BCE+∠CEA+∠EAB=360∘ (Sum of all the angles in a quadrilateral is 360∘)
⇒105∘+∠BCE+105∘+∠EAB=360∘
∠BCE=∠EAB (Opposite angles of parallelogram are equal)
⇒105∘+∠BAE+105∘+∠BAE=360∘
⇒2∠BAE=360∘−210∘
⇒∠BAE=75∘
Hence, the correct option is (A).
A sum was borrowed at 20% compound interest. It was repaid in 3 annual installments with each installment being paid at the end of the year. The first, second and third installments were Rs 1200, Rs 1440 and Rs 1728 respectively. Find the total amount borrowed by him?
The present value of 1st installment which is paid at the end of 1 year = I1/(1+ r/100)1 = 1200/(1+0.2)1 = Rs 1000
The present value of 2nd installment which is paid at the end of 2nd year = I2/(1+r/100)2 = 1440/ 1.22 = 1000
The present value of 3rd installment which is paid at the end of 3rd year = I3/(1+ r/100)3 = 1728/1.23 = Rs 1000
Therefore, total sum borrowed = 1000 + 1000 + 1000 = Rs 3000
Directions [Set of 5 questions]: Read the data carefully and answer the following questions.
Mr. Sharma and Mr. Singh spend his monthly salary on various items says House rent, Grocery, Education, Interest payment and Bills in such a way that total monthly expenditure of Mr. Singh is ₹40500. Expenditure of Mr. Sharma on Bills is ₹1020 more than expenditure of Mr. Singh on Bills and together both spend ₹18240 on Interest payment. Expenditure of Mr. Sharma on Grocery is 75% more than expenditure on education and expenditure of Mr. Singh on House rent is 25% more than total expenditure of Bills taken both (Mr. Sharma and Mr. Singh) together and expenditure of Mr. Sharma on Education is 60% more than expenditure of Mr. Singh on Bills. Ratio of expenditure of Mr. Singh on Education to Interest payment is 2:3 and expenditure of Mr. Singh on Grocery is 25% more than twice expenditure of Mr. Sharma on Bills. Expenditure of Mr. Singh on Bills is ₹4050 and expenditure of Mr. Sharma on House rent is 20% less than thrice the expenditure of Mr. Singh on Education.
What is the ratio of monthly expenditure of Mr. Sharma on education to monthly expenditure of Mr. Singh on interest payment?
Expenditure on different items can be calculated as:
Total expenditure of Mr. Singh = 40500 .......... (i)
Bills (Mr. Singh) = 4050 ......... (ii)
Let Bills (Mr. Sharma) be Y
Then,
Y = Bills (Mr. Singh) + 1020
Y = 4050 + 1020 .......... (from (ii))
Bills (Mr. Sharma) = 5070
Bills (total) = (5070 + 4050) = 9120 ....... (iii)
House Rent (Mr. Singh) = Bills (Total) x 125/100
House Rent (Mr. Singh) = 9120 x 125/100 = 11400 ........ (from (iii))
Interest Payment (total) = 18240 ........ (iv)
Education (Mr. Sharma) = Bills (Mr. Singh) x 160/100
Education (Mr. Sharma) = 4050 x 160/100 = 6480
Grocery (Mr. Sharma) = Education (Mr. Sharma) x 175/100
Grocery (Mr. Sharma) = 6480 x 175/100 = 11340
Grocery (Mr. Singh) = 2 x Bills (Mr. Sharma) x 125/100
Grocery (Mr. Singh) = 2 x 5070 x 125/100 = 12675
Expenditure of Mr. Singh on Education to Interest payment = 2:3 ........ (v)
Let Education (Mr. Singh) be 2Y
Interest Payment (Mr. Singh) = 3Y
Then,
(2Y + 3Y) = 40500 - (11400 + 12675 + 4050) ........ (from (i))
5Y = 12375
Y = 2475
Thus, Education (Mr. Singh) = (2 x 2475) = 4950
Interest Payment (Mr. Singh) = (3 x 2475) = 7425
Interest Payment (Mr. Sharma) = (18240 - 7425) = 10815
House Rent (Mr. Sharma) = 3 x Education (Mr. Singh) x 80/100
House Rent (Mr. Sharma) = 3 x 4950 x 80/100 = 11880
Based on the above data we get:
Monthly expenditure of Mr. Sharma on Education = 6480
Monthly expenditure of Mr. Singh on Interest payment = 7425
Required ratio = 6480/7425 = 48:55
Directions [Set of 5 questions]: Read the data carefully and answer the following questions.
Mr. Sharma and Mr. Singh spend his monthly salary on various items says House rent, Grocery, Education, Interest payment and Bills in such a way that total monthly expenditure of Mr. Singh is ₹40500. Expenditure of Mr. Sharma on Bills is ₹1020 more than expenditure of Mr. Singh on Bills and together both spend ₹18240 on Interest payment. Expenditure of Mr. Sharma on Grocery is 75% more than expenditure on education and expenditure of Mr. Singh on House rent is 25% more than total expenditure of Bills taken both (Mr. Sharma and Mr. Singh) together and expenditure of Mr. Sharma on Education is 60% more than expenditure of Mr. Singh on Bills. Ratio of expenditure of Mr. Singh on Education to Interest payment is 2:3 and expenditure of Mr. Singh on Grocery is 25% more than twice expenditure of Mr. Sharma on Bills. Expenditure of Mr. Singh on Bills is ₹4050 and expenditure of Mr. Sharma on House rent is 20% less than thrice the expenditure of Mr. Singh on Education.
What is the average monthly expenditure of Mr. Sharma for all the items together?
Expenditure on different items can be calculated as:
Total expenditure of Mr. Singh = 40500 .......... (i)
Bills (Mr. Singh) = 4050 ......... (ii)
Let Bills (Mr. Sharma) be Y
Then,
Y = Bills (Mr. Singh) + 1020
Y = 4050 + 1020 .......... (from (ii))
Bills (Mr. Sharma) = 5070
Bills (total) = (5070 + 4050) = 9120 ....... (iii)
House Rent (Mr. Singh) = Bills (Total) x 125/100
House Rent (Mr. Singh) = 9120 x 125/100 = 11400 ........ (from (iii))
Interest Payment (total) = 18240 ........ (iv)
Education (Mr. Sharma) = Bills (Mr. Singh) x 160/100
Education (Mr. Sharma) = 4050 x 160/100 = 6480
Grocery (Mr. Sharma) = Education (Mr. Sharma) x 175/100
Grocery (Mr. Sharma) = 6480 x 175/100 = 11340
Grocery (Mr. Singh) = 2 x Bills (Mr. Sharma) x 125/100
Grocery (Mr. Singh) = 2 x 5070 x 125/100 = 12675
Expenditure of Mr. Singh on Education to Interest payment = 2:3 ........ (v)
Let Education (Mr. Singh) be 2Y
Interest Payment (Mr. Singh) = 3Y
Then,
(2Y + 3Y) = 40500 - (11400 + 12675 + 4050) ........ (from (i))
5Y = 12375
Y = 2475
Thus, Education (Mr. Singh) = (2 x 2475) = 4950
Interest Payment (Mr. Singh) = (3 x 2475) = 7425
Interest Payment (Mr. Sharma) = (18240 - 7425) = 10815
House Rent (Mr. Sharma) = 3 x Education (Mr. Singh) x 80/100
House Rent (Mr. Sharma) = 3 x 4950 x 80/100 = 11880
Based on the above data we get:
Total expenditure = (11880 + 11340 + 6480 + 10815 + 5070) = 45585
Required average = 45585/5 = 9117
Directions [Set of 5 questions]: Read the data carefully and answer the following questions.
Mr. Sharma and Mr. Singh spend his monthly salary on various items says House rent, Grocery, Education, Interest payment and Bills in such a way that total monthly expenditure of Mr. Singh is ₹40500. Expenditure of Mr. Sharma on Bills is ₹1020 more than expenditure of Mr. Singh on Bills and together both spend ₹18240 on Interest payment. Expenditure of Mr. Sharma on Grocery is 75% more than expenditure on education and expenditure of Mr. Singh on House rent is 25% more than total expenditure of Bills taken both (Mr. Sharma and Mr. Singh) together and expenditure of Mr. Sharma on Education is 60% more than expenditure of Mr. Singh on Bills. Ratio of expenditure of Mr. Singh on Education to Interest payment is 2:3 and expenditure of Mr. Singh on Grocery is 25% more than twice expenditure of Mr. Sharma on Bills. Expenditure of Mr. Singh on Bills is ₹4050 and expenditure of Mr. Sharma on House rent is 20% less than thrice the expenditure of Mr. Singh on Education.
If Mr. Singh spends 7.5% of his monthly salary on Bills payment then the expenditure on House rent and Grocery together is approximately what percentage of his salary?
Expenditure on different items can be calculated as:
Total expenditure of Mr. Singh = 40500 .......... (i)
Bills (Mr. Singh) = 4050 ......... (ii)
Let Bills (Mr. Sharma) be Y
Then,
Y = Bills (Mr. Singh) + 1020
Y = 4050 + 1020 .......... (from (ii))
Bills (Mr. Sharma) = 5070
Bills (total) = (5070 + 4050) = 9120 ....... (iii)
House Rent (Mr. Singh) = Bills (Total) x 125/100
House Rent (Mr. Singh) = 9120 x 125/100 = 11400 ........ (from (iii))
Interest Payment (total) = 18240 ........ (iv)
Education (Mr. Sharma) = Bills (Mr. Singh) x 160/100
Education (Mr. Sharma) = 4050 x 160/100 = 6480
Grocery (Mr. Sharma) = Education (Mr. Sharma) x 175/100
Grocery (Mr. Sharma) = 6480 x 175/100 = 11340
Grocery (Mr. Singh) = 2 x Bills (Mr. Sharma) x 125/100
Grocery (Mr. Singh) = 2 x 5070 x 125/100 = 12675
Expenditure of Mr. Singh on Education to Interest payment = 2:3 ........ (v)
Let Education (Mr. Singh) be 2Y
Interest Payment (Mr. Singh) = 3Y
Then,
(2Y + 3Y) = 40500 - (11400 + 12675 + 4050) ........ (from (i))
5Y = 12375
Y = 2475
Thus, Education (Mr. Singh) = (2 x 2475) = 4950
Interest Payment (Mr. Singh) = (3 x 2475) = 7425
Interest Payment (Mr. Sharma) = (18240 - 7425) = 10815
House Rent (Mr. Sharma) = 3 x Education (Mr. Singh) x 80/100
House Rent (Mr. Sharma) = 3 x 4950 x 80/100 = 11880
Based on the above data we get:
Let the monthly salary of Mr. Singh be Y
Then,
Y x 7.5/100 = 4050
Y = 4050 x 100/7.5 = 54000
Expenditure on house rent and grocery = (11400 + 12675) = 24075
Required percentage = 24075/54000 x 100 ≈ 44.5%
Directions [Set of 5 questions]: Read the data carefully and answer the following questions.
Mr. Sharma and Mr. Singh spend his monthly salary on various items says House rent, Grocery, Education, Interest payment and Bills in such a way that total monthly expenditure of Mr. Singh is ₹40500. Expenditure of Mr. Sharma on Bills is ₹1020 more than expenditure of Mr. Singh on Bills and together both spend ₹18240 on Interest payment. Expenditure of Mr. Sharma on Grocery is 75% more than expenditure on education and expenditure of Mr. Singh on House rent is 25% more than total expenditure of Bills taken both (Mr. Sharma and Mr. Singh) together and expenditure of Mr. Sharma on Education is 60% more than expenditure of Mr. Singh on Bills. Ratio of expenditure of Mr. Singh on Education to Interest payment is 2:3 and expenditure of Mr. Singh on Grocery is 25% more than twice expenditure of Mr. Sharma on Bills. Expenditure of Mr. Singh on Bills is ₹4050 and expenditure of Mr. Sharma on House rent is 20% less than thrice the expenditure of Mr. Singh on Education.
If monthly expenditure of Mr. Sharma on House rent and Grocery increased by 15% and 25% respectively while expenditure on Education and Bills decreased by 25% and 20% respectively during next month and expenditure on interest payment remains the same then what is the difference of expenditure of Mr. Sharma from the previous month?
Expenditure on different items can be calculated as:
Total expenditure of Mr. Singh = 40500 .......... (i)
Bills (Mr. Singh) = 4050 ......... (ii)
Let Bills (Mr. Sharma) be Y
Then,
Y = Bills (Mr. Singh) + 1020
Y = 4050 + 1020 .......... (from (ii))
Bills (Mr. Sharma) = 5070
Bills (total) = (5070 + 4050) = 9120 ....... (iii)
House Rent (Mr. Singh) = Bills (Total) x 125/100
House Rent (Mr. Singh) = 9120 x 125/100 = 11400 ........ (from (iii))
Interest Payment (total) = 18240 ........ (iv)
Education (Mr. Sharma) = Bills (Mr. Singh) x 160/100
Education (Mr. Sharma) = 4050 x 160/100 = 6480
Grocery (Mr. Sharma) = Education (Mr. Sharma) x 175/100
Grocery (Mr. Sharma) = 6480 x 175/100 = 11340
Grocery (Mr. Singh) = 2 x Bills (Mr. Sharma) x 125/100
Grocery (Mr. Singh) = 2 x 5070 x 125/100 = 12675
Expenditure of Mr. Singh on Education to Interest payment = 2:3 ........ (v)
Let Education (Mr. Singh) be 2Y
Interest Payment (Mr. Singh) = 3Y
Then,
(2Y + 3Y) = 40500 - (11400 + 12675 + 4050) ........ (from (i))
5Y = 12375
Y = 2475
Thus, Education (Mr. Singh) = (2 x 2475) = 4950
Interest Payment (Mr. Singh) = (3 x 2475) = 7425
Interest Payment (Mr. Sharma) = (18240 - 7425) = 10815
House Rent (Mr. Sharma) = 3 x Education (Mr. Singh) x 80/100
House Rent (Mr. Sharma) = 3 x 4950 x 80/100 = 11880
Based on the above data we get:
Total monthly expenditure of Mr. Sharma = 11880+11340+6480+10815+5070 = 45585
Expenditure of Mr. Sharma on House rent during the next month = 11880 x 115/100 = 13662
Expenditure of Mr. Sharma on Grocery during the next month = 11340 x 125/100 = 14175
Expenditure of Mr. Sharma on Education during the next month = 6480 x 75/100 = 4860
Expenditure of Mr. Sharma on Bills during next month = 5070 x 80/100 = 4056
Expenditure of Mr. Sharma on Interest payment during next month = 10815
Total expenditure = (13662 + 14175 + 4860 + 4056 + 10815) = 47568
Required difference = (47568 - 45585) = 1983
Directions [Set of 5 questions]: Read the data carefully and answer the following questions.
Mr. Sharma and Mr. Singh spend his monthly salary on various items says House rent, Grocery, Education, Interest payment and Bills in such a way that total monthly expenditure of Mr. Singh is ₹40500. Expenditure of Mr. Sharma on Bills is ₹1020 more than expenditure of Mr. Singh on Bills and together both spend ₹18240 on Interest payment. Expenditure of Mr. Sharma on Grocery is 75% more than expenditure on education and expenditure of Mr. Singh on House rent is 25% more than total expenditure of Bills taken both (Mr. Sharma and Mr. Singh) together and expenditure of Mr. Sharma on Education is 60% more than expenditure of Mr. Singh on Bills. Ratio of expenditure of Mr. Singh on Education to Interest payment is 2:3 and expenditure of Mr. Singh on Grocery is 25% more than twice expenditure of Mr. Sharma on Bills. Expenditure of Mr. Singh on Bills is ₹4050 and expenditure of Mr. Sharma on House rent is 20% less than thrice the expenditure of Mr. Singh on Education.
Monthly expenditure on Grocery is approximately what percentage more than the expenditure on interest payment for both the persons together?
Expenditure on different items can be calculated as:
Total expenditure of Mr. Singh = 40500 .......... (i)
Bills (Mr. Singh) = 4050 ......... (ii)
Let Bills (Mr. Sharma) be Y
Then,
Y = Bills (Mr. Singh) + 1020
Y = 4050 + 1020 .......... (from (ii))
Bills (Mr. Sharma) = 5070
Bills (total) = (5070 + 4050) = 9120 ....... (iii)
House Rent (Mr. Singh) = Bills (Total) x 125/100
House Rent (Mr. Singh) = 9120 x 125/100 = 11400 ........ (from (iii))
Interest Payment (total) = 18240 ........ (iv)
Education (Mr. Sharma) = Bills (Mr. Singh) x 160/100
Education (Mr. Sharma) = 4050 x 160/100 = 6480
Grocery (Mr. Sharma) = Education (Mr. Sharma) x 175/100
Grocery (Mr. Sharma) = 6480 x 175/100 = 11340
Grocery (Mr. Singh) = 2 x Bills (Mr. Sharma) x 125/100
Grocery (Mr. Singh) = 2 x 5070 x 125/100 = 12675
Expenditure of Mr. Singh on Education to Interest payment = 2:3 ........ (v)
Let Education (Mr. Singh) be 2Y
Interest Payment (Mr. Singh) = 3Y
Then,
(2Y + 3Y) = 40500 - (11400 + 12675 + 4050) ........ (from (i))
5Y = 12375
Y = 2475
Thus, Education (Mr. Singh) = (2 x 2475) = 4950
Interest Payment (Mr. Singh) = (3 x 2475) = 7425
Interest Payment (Mr. Sharma) = (18240 - 7425) = 10815
House Rent (Mr. Sharma) = 3 x Education (Mr. Singh) x 80/100
House Rent (Mr. Sharma) = 3 x 4950 x 80/100 = 11880
Based on the above data we get:
Monthly expenditure on grocery for both = (11340 + 12675) = 24015
Monthly expenditure on Interest payment for both = 7425+10815 = 18240
Required percentage = (24015 - 18240)/18240 x 100 ≈ 31.6%
The following question, two statements numbered I and II are given. Solve for the Quantity I and II, then choose correct answer from the given options.
Quantity I: In U.P. there is 960 females per thousand males and in Bihar there are 945 females per thousand males. If the ratio of population of U.P. and Bihar is 11: 6 then what is the average number of females per thousand males in both the states taken together?
Quantity II: Infant mortality is 36 per thousand in Karnataka and corresponding figure is 64 per thousand in Andhra Pradesh. If the ratio of population of Karnataka and Andhra Pradesh is 2 : 5 then what is the average infant mortality rate in both the states taken together?
Average number of females per thousand in both the states = 960 X 11)+ (945 X 6/(11 + 6)
= (10560 +5670)/17 =954.7
Quantity - II
Average infant mortality rate in Andhra and Karnataka taken together = 36 X 2)+ (64 X 5/(2 + 5) = (72+ 320)/7 =392/7 =56
In the following questions, two statements numbered I and II are given. On solving them, we get quantities I and II respectively. Solve for both the quantities and choose the correct option.
Quantity I: Raj and Ravi starts a business with initial capital of ₹4800 and ₹5400 respectively at the end of 6 month Raj increases his investment by 25% while Ravi withdraw 20% of his initial investment then what is profit share of Ravi out of total profit of ₹71250?
Quantity II: What is the initial sum for which CI is ₹486 more than SI at the rate of 12% per annum in two years?
Quantity I:
Capital investment of Raj = (4800 x 6) + (4800 125/100 × 6) = 64800
Capital investment of Ravi = (5400 x 6) + (5400 × 80/100 × 6) = 58320
Ratio of investment of Raj to Ravi = 64800:58320
Ratio of investment of Raj to Ravi = 10:9
Total profit = ₹71250
Profit share of Ravi = 71250 × 9/19 = ₹33750
Quantity II:
Difference of CI to SI for two years
CI - SI = P × (R/100)2
P × (12/100)2 = 486
P = (4860000/144) = 33750
Thus, initial sum = ₹33750
Hence, Quantity I = Quantity II
If the total paid amount by a person on his zomato order is Rs 550, when he receives a discount of 20% on base price of food, 10% incurred as delivery charge, 5% incurred as packaging charge and Rs 75 is taken by Delivery Boy as tip, what is the base price of the food?
Given,
100x(1 - 20/100) + 100x*(10/100) + 100x*(5/100) + 75 = 550
=> 80x + 10x + 5x = 475
=> 95x = 475
=> x = 5
Base price of the food = Rs 500
In the following question, two statements, numbered as I and II are given. On solving them, we get quantities I and II respectively. Solve for both the quantities and choose the correct option.
Quantity I: Ratio of age of Ravi after 4 years and that of Jay after 8 years is 13:20. Riya is 12 years younger to Jay and 12 years elder to Ravi then what is the present age of Ravi(in years)?
Quantity II: Train A crosses a platform of length 240 meter and pole in 25 sec and 15 sec respectively, another train B crosses train A running in same direction in 12.5 sec. what is speed (m/sec) of train B?
Let present age of Ravi = (13Y - 4)
Thus, present age of Riya = (13Y - 4 + 12) = (13Y + 8) ........ (i)
Let present age of Jay = (20Y - 8)
Thus, present age of Riya = (20Y - 8 - 12) = (20Y - 20) .......... (ii)
By comparing (i) and (ii) we get:
(13Y + 8) = (20Y - 20)
7Y = 28
Y = 4
Thus, present age of Ravi = (13 x 4 - 4) = 48 years
Quantity II:
We have:
Let length of train A be Y meter
Length of platform crossed by train A = 240 meter
Thus, speed of train crossing platform and pole must be equal.
(Y + 240)/25 = Y/15
15Y + 3600 = 25Y
10Y = 3600
Y = 360 meter
Speed of train A = 360/15 = 24 m/sec
Let length of train B be K meter
Speed of train B be S m/sec
Relative speed of train A to B = (24 + S)
Then,
(24 + S) = (K + 360)/12.5
Since, length of train B is not known.
Thus, speed of train B can't be calculated.
Hence, relationship can't be determined.
Directions: Following graph shows distance in kilometres covered by 5 different bikes at different ranges of average speeds.
(40 to 50) km/h means above 40 km/h and up to 50 km/h. Similarly (50 to 60) km/h means above 50 km/h and up to 60 km/h and (60 to 70) km/h means above 60 km/h and up to 70 km/h.
What will be the sum of distances covered by all 5 bikes in a litre each if average speed of each is 60 km/h?
Distance covered by all 5 bikes= (80 +60 +85 +90 +92) = 407 km.
Directions: Following graph shows distance in kilometres covered by 5 different bikes at different ranges of average speeds.
(40 to 50) km/h means above 40 km/h and up to 50 km/h. Similarly (50 to 60) km/h means above 50 km/h and up to 60 km/h and (60 to 70) km/h means above 60 km/h and up to 70 km/h.
If in 4 hours Mahindra Centuro covered 260 km and in 5 hours Bajaj CT100 covered 320 km, then how many litres of petrol consumed by these two bikes together to cover the mentioned distance?
In 1 litre Mahindra Centuro at 65 km/h covers 40 km.
Petrol required for Mahindra Centuro = 260/40 = 6.5 litres.
Speed of Mahindra Bajaj CT100 = 320/5 =64 km/h
In 1 litre Bajaj CT100 at 64 km/h covers 80 km.
Petrol required for Bajaj CT100 = 320/80 = 4 litres.
Total petrol required = 6.5 +4 = 10.5 litres.
Directions: Following graph shows distance in kilometres covered by 5 different bikes at different ranges of average speeds.
(40 to 50) km/h means above 40 km/h and up to 50 km/h. Similarly (50 to 60) km/h means above 50 km/h and up to 60 km/h and (60 to 70) km/h means above 60 km/h and up to 70 km/h.
What is the average of distances covered in 1 litre of petrol at 45 km/h by all 5 given bikes together?
Explanation Required average= (70 +50 +75 +65 +80)/5 = 340/5 =68 km.
Directions: Following graph shows distance in kilometres covered by 5 different bikes at different ranges of average speeds.
(40 to 50) km/h means above 40 km/h and up to 50 km/h. Similarly (50 to 60) km/h means above 50 km/h and up to 60 km/h and (60 to 70) km/h means above 60 km/h and up to 70 km/h.
Ram is travelling with Mahindra Centuro consuming 20.4 litres petrol. If ratio of journey times of Ram at 50 km/h, at 54 km/h and at 65 km/h is 1:3:4, then what is the total distance covered by Ram?