CAT LRDI MCQ - 1


32 Questions MCQ Test MBA Exams Mock Test Series and Past Year Question Papers | CAT LRDI MCQ - 1


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This mock test of CAT LRDI MCQ - 1 for CAT helps you for every CAT entrance exam. This contains 32 Multiple Choice Questions for CAT CAT LRDI MCQ - 1 (mcq) to study with solutions a complete question bank. The solved questions answers in this CAT LRDI MCQ - 1 quiz give you a good mix of easy questions and tough questions. CAT students definitely take this CAT LRDI MCQ - 1 exercise for a better result in the exam. You can find other CAT LRDI MCQ - 1 extra questions, long questions & short questions for CAT on EduRev as well by searching above.
QUESTION: 1

Group Question 

Answer the following question based on the information given below.

Five girls and four boys of a class were to be sent for a sports camp. Their teacher was asked to measure the weight of these students. But the weighing machine available in the school was defective and only measured weights greater than 100 kg. Hence, the teacher decided to measure the weight of three students at a time, but all three students selected would be of the same gender. He took all the possible weight combinations of girls as well as of boys. For girls, the measurements were 122 kg, 127 kg, 130 kg, 130 kg, 133 kg, 134 kg, 137 kg, 138 kg, 142 kg and 145 kg. For boys, the measurements were 174 kg, 170 kg, 165 kg and 163 kg. No two boys or no two girls had the same weight.

Q. What is the average weight of the nine students?  

Solution:

Let the weight of the girls, in ascending order (in kgs), be p, q, r, s, t and the weight of the boys, also in ascending order (in kgs), be a, b, c, d.
Thus, p < q < r < s < t and a < b < c < d.
Number of weight combinations for three girls at a time = 5C3 = 10

Hence, the weight combinations for three girls at a time are: p + q + r, p + q + s, p + q + t; p + r+ s, p + r + t, p + s + t, q + r + s , q + r +t , q + s + t and r + s + t.

Addition of all these combinations gives: 6(p + q + r + s + t) Total of all these combinations = 122 + 127 + 130 + 130 + 133 + 134 + 137 + 138 + 142 + 145= 1338 kgs.

6 (p + q + r + s + t) = 1338

Total weight o f the girls = p + g + r + s + t = 1338/ 6 = 223 kgs.

Similarly, for the boys: 3(a + b + c + d ) = 163 + 165 + 170 + 174 = 672

Total weight of the boys = a + b + c + d = 672/3 = 224 kgs. Average weight of the nine students = (223 + 224)/(5 + 4) = 447/9 = 49.67 kgs
Hence, option 4.

QUESTION: 2

Five girls and four boys of a class were to be sent for a sports camp. Their teacher was asked to measure the weight of these students. But the weighing machine available in the school was defective and only measured weights greater than 100 kg. Hence, the teacher decided to measure the weight of three students at a time, but all three students selected would be of the same gender. He took all the possible weight combinations of girls as well as of boys. For girls, the measurements were 122 kg, 127 kg, 130 kg, 130 kg, 133 kg, 134 kg, 137 kg, 138 kg, 142 kg and 145 kg. For boys, the measurements were 174 kg, 170 kg, 165 kg and 163 kg. No two boys or no two girls had the same weight.

Q. What is the difference in the weights of the second heaviest boy and the  second lightest girl?

Solution:

Consider the solution to the first question. p < q < r < s < t and a < b <c <d Consider the boys. a+ b+ c< a + b + d < a + c+ d < b+ c+ d

a + b + c = 163 ; a + b + d = 165 ;a + c + d = 170 and b + c + d = 1 7 4 From the earlier solution, a + b + c + d = 224 ... (i) Hence, subtracting each of the above equations from (i), a = 50, b = 54, c = 59 and d = 61 

Similarly, consider the combinations for the girls.
The three lightest girls will have the least total and the three heaviest girls will have the greatest total. p + q + r = 122 and r + s + t = 145 Adding the two equations above, (p + q + r  + s + t ) + r = 122 + 145

Considering the total weight of the girls as obtained in the earlier question, 223 + r = 267 r = 44.  p + q = 78 and s + t = 101. 

Considering the eight remaining combinations, it can be said that the second lowest sum is for p + q + s. p + q + s = 127. So  s = 127 - 78 = 49, t = 101 - 4 9 = 52 Similarly, the second highest sum is from q + s + t. q + s + t = 142. q= 142- 101 = 41 ... p = 78 - 41 = 37. Second heaviest boy ~ Second lightest girl = c - q = 59 - 41 = 18 kg 

Hence, option 3.

QUESTION: 3

Five girls and four boys of a class were to be sent for a sports camp. Their teacher was asked to measure the weight of these students. But the weighing machine available in the school was defective and only measured weights greater than 100 kg. Hence, the teacher decided to measure the weight of three students at a time, but all three students selected would be of the same gender. He took all the possible weight combinations of girls as well as of boys. For girls, the measurements were 122 kg, 127 kg, 130 kg, 130 kg, 133 kg, 134 kg, 137 kg, 138 kg, 142 kg and 145 kg. For boys, the measurements were 174 kg, 170 kg, 165 kg and 163 kg. No two boys or no two girls had the same weight.

Q. What is the weight of the heaviest boy?

Solution:

Consider the solution to the earlier questions. Weight of the heaviest boy = d = 61 kg. Hence, option 4. 

QUESTION: 4

Five girls and four boys of a class were to be sent for a sports camp. Their teacher was asked to measure the weight of these students. But the weighing machine available in the school was defective and only measured weights greater than 100 kg. Hence, the teacher decided to measure the weight of three students at a time, but all three students selected would be of the same gender. He took all the possible weight combinations of girls as well as of boys. For girls, the measurements were 122 kg, 127 kg, 130 kg, 130 kg, 133 kg, 134 kg, 137 kg, 138 kg, 142 kg and 145 kg. For boys, the measurements were 174 kg, 170 kg, 165 kg and 163 kg. No two boys or no two girls had the same weight.

Q. How many girls weigh less than all the boys?

Solution:

Consider the solution to the earlier questions. Lightest boy = a = 50 kg Among the girls, (p, q, r, s)< a Hence, four girls weigh less than all the boys.
Hence, option 2.

QUESTION: 5

All Saints High School is organizing a Parents-Teacher meet for students of class V. For this purpose, every parent has been given a fixed 10 minute time slot to meet five different teachers, each of which teaches a different subject among Maths, Science, Social Studies, English, and Hindi. Mr. Balu, Mr. Lal, Mr. Johri, Mr. Patnaik, and Mr. Ahmed are fathers of five children who studies in class V of All Saints High School. Each of these five fathers meet all the teachers between 2 p.m. and 2:50 p.m. Each parent can meet only one teacher in a slot and each teacher can meet only one parent in a slot. Each parent can meet the teachers only in the assigned slot.
Further, the following information is also known-

(i) Except for one teacher, Mr. Johri met all the five teachers immediately after Mr. Ahmed met them.
(ii) Mr. Lal was the second parent to meet Social Studies teacher and Mr. Johri was the first parent to meet the Hindi Teacher.
(iii) Mr. Balu met the English teacher immediately before he met the Social Studies teacher and Mr. Ahmed was the third parent to meet the Social Studies teacher.
(iv) Mr. Johri was not the last parent to meet the Science teacher and Mr. Lal met the English teacher before he met the Social Studies teacher.

Who met the maths teacher just before Mr. Lal?

Solution:

Let us name the slots for our convenience-
2-2:10 -> 1, 2:10-2:20 -> 2, 2:20-2:30 -> 3, 2:30-2:40 -> 4, 2:40-2:50 -> 5

So, according to (ii), Mr. Lal met Social Studies teacher in slot 2 and Mr. Johri met the Hindi Teacher in slot 1. Also by (iii), Mr. Ahmed met the Social Studies teacher in slot 3.

We know that except for one teacher, Mr. Johri met all the five teachers after Mr. Ahmed met them. This is only possible when Mr. Johri met all the five teachers immediately after Mr. Ahmed met them. This means that the teacher that Mr. Johri visits in slot 1 must be visited by Mr. Ahmed in slot 5. This means Mr. Ahmed met the Hindi teacher in slot 5 and Mr. Johri must have met the Social studies teacher in slot 4.

Since by (iii) we know that Mr. Balu met the English teacher immediately before he met the Social Studies teacher, we can conclude that Mr. Balu must have met Social teacher in slot 5 and consequently English teacher in slot 4. This means Mr. Patnaik met the Social teacher in slot 1. Now, Mr. Johri doesn’t meet the Science teacher in slot 5 [Given in (iv)]. Also, he cannot meet English teacher in slot 5 as it would mean Ahmed met English teacher in slot 4 which is not possible. So, Mr. Johri met Maths teacher in last slot which means Mr. Ahmed met Maths teacher in Slot 4.

Thus, Johri can meet English teacher only in Slot 3 which means Ahmed met English teacher in slot 2. This means Johri met Science teacher in slot 2 and Ahmed met Science teacher in slot 1. Also, Lal can now only meet English teacher in slot 1 which leaves Patnaik to meet English teacher in slot 5. This means that Balu met Maths teacher in Slot 1.

Now, Patnaik met Maths teacher in 2 which means Lal met him in 3. Similarly, all other slots can be assigned uniquely as follows-

QUESTION: 6

Read the following information carefully and answer the questions that follow.

All Saints High School is organizing a Parents-Teacher meet for students of class V. For this purpose, every parent has been given a fixed 10 minute time slot to meet five different teachers, each of which teaches a different subject among Maths, Science, Social Studies, English, and Hindi. Mr. Balu, Mr. Lal, Mr. Johri, Mr. Patnaik, and Mr. Ahmed are fathers of five children who studies in class V of All Saints High School. Each of these five fathers meet all the teachers between 2 p.m. and 2:50 p.m. Each parent can meet only one teacher in a slot and each teacher can meet only one parent in a slot. Each parent can meet the teachers only in the assigned slot.
Further, the following information is also known-

(i) Except for one teacher, Mr. Johri met all the five teachers immediately after Mr. Ahmed met them.
(ii) Mr. Lal was the second parent to meet Social Studies teacher and Mr. Johri was the first parent to meet the Hindi Teacher.
(iii) Mr. Balu met the English teacher immediately before he met the Social Studies teacher and Mr. Ahmed was the third parent to meet the Social Studies teacher.
(iv) Mr. Johri was not the last parent to meet the Science teacher and Mr. Lal met the English teacher before he met the Social Studies teacher.
 

The Hindi teacher was meeting which parent at 2:37 p.m.?

Solution:

Let us name the slots for our convenience-
2-2:10 -> 1, 2:10-2:20 -> 2, 2:20-2:30 -> 3, 2:30-2:40 -> 4, 2:40-2:50 -> 5

So, according to (ii), Mr. Lal met Social Studies teacher in slot 2 and Mr. Johri met the Hindi Teacher in slot 1. Also by (iii), Mr. Ahmed met the Social Studies teacher in slot 3.

We know that except for one teacher, Mr. Johri met all the five teachers after Mr. Ahmed met them. This is only possible when Mr. Johri met all the five teachers immediately after Mr. Ahmed met them. This means that the teacher that Mr. Johri visits in slot 1 must be visited by Mr. Ahmed in slot 5. This means Mr. Ahmed met the Hindi teacher in slot 5 and Mr. Johri must have met the Social studies teacher in slot 4.

Since by (iii) we know that Mr. Balu met the English teacher immediately before he met the Social Studies teacher, we can conclude that Mr. Balu must have met Social teacher in slot 5 and consequently English teacher in slot 4. This means Mr. Patnaik met the Social teacher in slot 1. Now, Mr. Johri doesn’t meet the Science teacher in slot 5 [Given in (iv)]. Also, he cannot meet English teacher in slot 5 as it would mean Ahmed met English teacher in slot 4 which is not possible. So, Mr. Johri met Maths teacher in last slot which means Mr. Ahmed met Maths teacher in Slot 4.

Thus, Johri can meet English teacher only in Slot 3 which means Ahmed met English teacher in slot 2. This means Johri met Science teacher in slot 2 and Ahmed met Science teacher in slot 1. Also, Lal can now only meet English teacher in slot 1 which leaves Patnaik to meet English teacher in slot 5. This means that Balu met Maths teacher in Slot 1.

Now, Patnaik met Maths teacher in 2 which means Lal met him in 3. Similarly, all other slots can be assigned uniquely as follows-

QUESTION: 7

Read the following information carefully and answer the questions that follow.

All Saints High School is organizing a Parents-Teacher meet for students of class V. For this purpose, every parent has been given a fixed 10 minute time slot to meet five different teachers, each of which teaches a different subject among Maths, Science, Social Studies, English, and Hindi. Mr. Balu, Mr. Lal, Mr. Johri, Mr. Patnaik, and Mr. Ahmed are fathers of five children who studies in class V of All Saints High School. Each of these five fathers meet all the teachers between 2 p.m. and 2:50 p.m. Each parent can meet only one teacher in a slot and each teacher can meet only one parent in a slot. Each parent can meet the teachers only in the assigned slot.
Further, the following information is also known-

(i) Except for one teacher, Mr. Johri met all the five teachers immediately after Mr. Ahmed met them.
(ii) Mr. Lal was the second parent to meet Social Studies teacher and Mr. Johri was the first parent to meet the Hindi Teacher.
(iii) Mr. Balu met the English teacher immediately before he met the Social Studies teacher and Mr. Ahmed was the third parent to meet the Social Studies teacher.
(iv) Mr. Johri was not the last parent to meet the Science teacher and Mr. Lal met the English teacher before he met the Social Studies teacher.

Mr. Johri met the Maths teacher in the same time slot as the time slot in which

Solution:

Let us name the slots for our convenience-
2-2:10 -> 1, 2:10-2:20 -> 2, 2:20-2:30 -> 3, 2:30-2:40 -> 4, 2:40-2:50 -> 5

So, according to (ii), Mr. Lal met Social Studies teacher in slot 2 and Mr. Johri met the Hindi Teacher in slot 1. Also by (iii), Mr. Ahmed met the Social Studies teacher in slot 3.

We know that except for one teacher, Mr. Johri met all the five teachers after Mr. Ahmed met them. This is only possible when Mr. Johri met all the five teachers immediately after Mr. Ahmed met them. This means that the teacher that Mr. Johri visits in slot 1 must be visited by Mr. Ahmed in slot 5. This means Mr. Ahmed met the Hindi teacher in slot 5 and Mr. Johri must have met the Social studies teacher in slot 4.

Since by (iii) we know that Mr. Balu met the English teacher immediately before he met the Social Studies teacher, we can conclude that Mr. Balu must have met Social teacher in slot 5 and consequently English teacher in slot 4. This means Mr. Patnaik met the Social teacher in slot 1. Now, Mr. Johri doesn’t meet the Science teacher in slot 5 [Given in (iv)]. Also, he cannot meet English teacher in slot 5 as it would mean Ahmed met English teacher in slot 4 which is not possible. So, Mr. Johri met Maths teacher in last slot which means Mr. Ahmed met Maths teacher in Slot 4.

Thus, Johri can meet English teacher only in Slot 3 which means Ahmed met English teacher in slot 2. This means Johri met Science teacher in slot 2 and Ahmed met Science teacher in slot 1. Also, Lal can now only meet English teacher in slot 1 which leaves Patnaik to meet English teacher in slot 5. This means that Balu met Maths teacher in Slot 1.

Now, Patnaik met Maths teacher in 2 which means Lal met him in 3. Similarly, all other slots can be assigned uniquely as follows-

QUESTION: 8

Read the following information carefully and answer the questions that follow.

All Saints High School is organizing a Parents-Teacher meet for students of class V. For this purpose, every parent has been given a fixed 10 minute time slot to meet five different teachers, each of which teaches a different subject among Maths, Science, Social Studies, English, and Hindi. Mr. Balu, Mr. Lal, Mr. Johri, Mr. Patnaik, and Mr. Ahmed are fathers of five children who studies in class V of All Saints High School. Each of these five fathers meet all the teachers between 2 p.m. and 2:50 p.m. Each parent can meet only one teacher in a slot and each teacher can meet only one parent in a slot. Each parent can meet the teachers only in the assigned slot.
Further, the following information is also known-

(i) Except for one teacher, Mr. Johri met all the five teachers immediately after Mr. Ahmed met them.
(ii) Mr. Lal was the second parent to meet Social Studies teacher and Mr. Johri was the first parent to meet the Hindi Teacher.
(iii) Mr. Balu met the English teacher immediately before he met the Social Studies teacher and Mr. Ahmed was the third parent to meet the Social Studies teacher.
(iv) Mr. Johri was not the last parent to meet the Science teacher and Mr. Lal met the English teacher before he met the Social Studies teacher.
 

Which of the following is definitely true?

Solution:

Let us name the slots for our convenience-
2-2:10 -> 1, 2:10-2:20 -> 2, 2:20-2:30 -> 3, 2:30-2:40 -> 4, 2:40-2:50 -> 5

So, according to (ii), Mr. Lal met Social Studies teacher in slot 2 and Mr. Johri met the Hindi Teacher in slot 1. Also by (iii), Mr. Ahmed met the Social Studies teacher in slot 3.

We know that except for one teacher, Mr. Johri met all the five teachers after Mr. Ahmed met them. This is only possible when Mr. Johri met all the five teachers immediately after Mr. Ahmed met them. This means that the teacher that Mr. Johri visits in slot 1 must be visited by Mr. Ahmed in slot 5. This means Mr. Ahmed met the Hindi teacher in slot 5 and Mr. Johri must have met the Social studies teacher in slot 4.

Since by (iii) we know that Mr. Balu met the English teacher immediately before he met the Social Studies teacher, we can conclude that Mr. Balu must have met Social teacher in slot 5 and consequently English teacher in slot 4. This means Mr. Patnaik met the Social teacher in slot 1. Now, Mr. Johri doesn’t meet the Science teacher in slot 5 [Given in (iv)]. Also, he cannot meet English teacher in slot 5 as it would mean Ahmed met English teacher in slot 4 which is not possible. So, Mr. Johri met Maths teacher in last slot which means Mr. Ahmed met Maths teacher in Slot 4.

Thus, Johri can meet English teacher only in Slot 3 which means Ahmed met English teacher in slot 2. This means Johri met Science teacher in slot 2 and Ahmed met Science teacher in slot 1. Also, Lal can now only meet English teacher in slot 1 which leaves Patnaik to meet English teacher in slot 5. This means that Balu met Maths teacher in Slot 1.

Now, Patnaik met Maths teacher in 2 which means Lal met him in 3. Similarly, all other slots can be assigned uniquely as follows-

QUESTION: 9

Group Question

Answer the following question based on the information given below.


Shantanu plays an online fantasy game during a cricket tournament. In any game, one has to pick players and the player earns Shantanu as many points as the runs scored by him in the actual match. For a particular game, Shantanu is allowed to choose six players.
For Bangladesh vs Sri Lanka, the players that he chose were Chandimal, Dilshan, Matthews, Rahman, Shakib, and Tamim. After the match, an interesting feature that he noticed was that the scores of the players were such that the difference in the squares of the scores of three pairs of players was 75 each. No player featured in more than one more pair. Dilshan scored one more run than Rahman, while Tamim scored one more than Chandimal. Neither was Tamim the top scorer nor was Matthews the lowest scorer.

 

 

Q. What was the average score of the six players?  

Solution:

Let the score of the six players be a, b, c, d, e and f.

Since difference between square of score of two players is 75 for three different pairs, let 

a2 - b2 = c2 - d2 = e2 - f2 = 75
where, a > b; c > d and e > f
∴ (a + b)(a - b) = 75 x 1
Since a > b, a + b = 75 and a - b = 1
On solving, a = 38 and b = 37
Similarly, let (c + d)(c - d) = 25 x 3
a2 – b2 = c2 – d2 = e2 – f2 = 75
where, a > b; c > d and e > f
∴ (a + b)(a – b) = (c + d)(c – d) = (e + f)(e – f) = 75
Also, 75 = 75 x 1 = 25 x 3 = 15 x 5
Hence, each pair corresponds to one of these products.
Let (a + b)(a – b) = 75 x 1
Since a > b, a + b = 75 and a – b = 1
On solving, a = 38 and b = 37
Similarly, let (c + d)(c – d) = 25 x 3
Since c > d, c + d = 25 and c - d = 3 On solving, c = 38 and d = 37 Similarly, using the last pair, e = 10 and d = 5 Thus, the six scores, in descending order are 38, 37, 14, 11, 10, 5.
Dilshan = Rahman + 1 and Tamim = Chandimal + 1 Hence, Dilshan and Tamim have scored 38 and 11, in no specific order. Since Tamim is not the top scorer, Dilshan = 38, Rahman = 37, Tamim = 11 and Chandimal = 10
Hence, Mathews and Shakib have score 14 and 5, in no specific order.
Since Mathews is not the lowest scorer, Mathews = 14 and Shakib = 5 Average score = (38 + 37 + 14 + 11 + 10 + 5)/6 = 115/6 = 19.17 Hence, option 4.

QUESTION: 10

Shantanu plays an online fantasy game during a cricket tournament. In any game, one has to pick players and the player earns Shantanu as many points as the runs scored by him in the actual match. For a particular game, Shantanu is allowed to choose six players.
For Bangladesh vs Sri Lanka, the players that he chose were Chandimal, Dilshan, Matthews, Rahman, Shakib and Tamim. After the match, an interesting feature that he noticed was that the scores of the players were such that the difference in the squares of the scores of three pairs of players was 75 each. No player featured in more than one more pair. Dilshan scored one more run than Rahman, while Tamim scored one more than Chandimal. Neither was Tamim the top scorer nor was Matthews the lowest scorer.

 

 

Q. What was the least number of runs scored by any of these batsmen?

Solution:

The least runs by any of the players were by Shakib (5 runs).
Hence, option 3.

QUESTION: 11

Shantanu plays an online fantasy game during a cricket tournament. In any game, one has to pick players and the player earns Shantanu as many points as the runs scored by him in the actual match. For a particular game, Shantanu is allowed to choose six players.
For Bangladesh vs Sri Lanka, the players that he chose were Chandimal, Dilshan, Matthews, Rahman, Shakib, and Tamim. After the match, an interesting feature that he noticed was that the scores of the players were such that the difference in the squares of the scores of three pairs of players was 75 each. No player featured in more than one more pair. Dilshan scored one more run than Rahman, while Tamim scored one more than Chandimal. Neither was Tamim the top scorer nor was Matthews the lowest scorer.

 

 

Q. As per the rules of the game, one has to choose a captain and a vice-captain in his team. The captain gets twice the points as runs scored and the vice-captain gets 1.5 times the points as runs scored. If Shantanu chose Tamim as the captain and Chandimal as the vice-captain, how many points did he get?

Solution:

Correct Answer :- a

Explanation : Points scored by Shantanu = (11 x 2) + (10 x 1.5) + (38 + 37 + 14 + 5) 

= 22 + 15 + 94

= 131

QUESTION: 12

Shantanu plays an online fantasy game during a cricket tournament. In any game, one has to pick players and the player earns Shantanu as many points as the runs scored by him in the actual match. For a particular game, Shantanu is allowed to choose six players.
For Bangladesh vs Sri Lanka, the players that he chose were Chandimal, Dilshan, Matthews, Rahman, Shakib, and Tamim. After the match, an interesting feature that he noticed was that the scores of the players were such that the difference in the squares of the scores of three pairs of players was 75 each. No player featured in more than one more pair. Dilshan scored one more run than Rahman, while Tamim scored one more than Chandimal. Neither was Tamim the top scorer nor was Matthews the lowest scorer.

 

Q. Considering the scoring system as given in the previous question, what is the maximum number of points that Shantanu could have scored?

Solution:

Shantanu’s score could have been maximised had he selected the top two scorers as the captain and vice captain respectively.
Hence, Dilshan (38) and Rahman (37) were his captain and vice captain respectively.
Maximum points = (38 x 2) + (37 x 1.5) + (14 + 11 + 10 + 5) = 171.5 Hence, option 2.

*Answer can only contain numeric values
QUESTION: 13

Group Question

Answer the following question based on the information given below.


In a hockey tournament, India, Australia, New Zealand and Pakistan have been the top four teams for the past seven years (2009-15). Each year, after the semi-final, two matches were played. The winners of the two semi finals faced each other in the finals for the top two spots. The losers of the two semi finals faced each other for the third spot play-off. It is also known that:

1. India did not play against Australia from 2009-12 and played against Australia from 2013-15.

2. No team became the champion for two times in three consecutive years.

3. No team ended at the same position in two consecutive years.

4. Pakistan was the champion only in 2014 and New Zealand was always in the top three.

5. The runner up of 2012 slipped by one position in the next year.

6. The champion of 2009 stood fourth in 2015 and the winner of 2010 was fourth in 2009.

7. In 2009, Pakistan beat Australia in an eventful semi-final.

8. India was the runner up in 2011 and 2013.

 

Q. How many times did India have a better finish compared to Australia?


Solution:

Consider the ranks in any particular year as: Winner = 1, Runner up = 2, Third Place winner = 3, last place = 4 Denote teams by their initials.
India played Australia in 2013-15.
Hence, in all three years from 2013 to 2105, the matches were India vs Australia and New Zealand vs Pakistan.
Since New Zealand always finished in the top three, NZ  4.

In 2013, India was the runner up.
Hence, the rankings for 2013 are : Aus = 1, Ind = 2, NZ = 3, Pak = 4 Pakistan was the winner in 2014.
Hence, in 2014: Pak = 1, NZ = 2 India was the runner up in 2011.
Hence, in 2011, Ind = 2 Since Pakistan beat Australia in the semi final in 2009, Pakistan would have played the final and been the winner or runner up. Also, Australia would have been at third or fourth place.  Also, since Australia and India did not play each other from 2009-2011, Australia would have played New Zealand for the third and fourth place.
Hence, India and Pakistan would have been the winner and runner up (in no specific order).
But Pakistan was the winner only in 2014. Hence, India was the winner in 2009.
Also, since New Zealand was always in the top three, Australia was fourth placed in 2009.
Hence, in 2009: Ind = 1, Pak = 2, NZ = 3, Aus = 4 The information so far can be tabulated as shown below: 


Now, the champion of 2009 finished fourth in 2015. Hence, India was last in 2015.
In 2015, the matches were Ind vs Aus and NZ vs Pak. Hence, Australia was third.
Since no two teams ended at the same position in consecutive years, New Zealand and Pakistan were the winners and runner up in 2015. Hence, in 2015: NZ = 1, Pak = 2, Aus = 3, Ind = 4 Also, the winner of 2010 was last in 2009 Hence, in 2010, Aus = 1 Since a team could not have finished in the same position in two consecutive years, NZ  3 or 4 in 2010 (as it was always in the top three).
Hence, in 2010: NZ = 2 The runner up of 2012, slipped by one position in the next year.
Thus, the team ranked 3 in 2013 would have been ranked 2 in 2012.
Hence, in 2012: NZ = 2 The information so far is as tabulated below: 

Consider 2011.
Since no team had the same position in consecutive years, Aus  1 in 2011. Since Pakistan were champions only in 2014, Pak  1 in 2011.
Hence, in 2011, NZ = 1 Using the same logic as the above step, in 2012, Ind = 1 

Now, using the rule that a team cannot have the same rank in two consecutive years, the rest of the table can be filled as shown below.

Thus, India had a better finish than Australia in four years (2009, 2011, 2012 and 2014).
Answer: 4

*Answer can only contain numeric values
QUESTION: 14

In a hockey tournament, India, Australia, New Zealand and Pakistan have been the top four teams for the past seven years (2009-15). Each year, after the semi-final, two matches were played. The winners of the two semi finals faced each other in the finals for the top two spots. The losers of the two semi finals faced each other for the third spot play-off. It is also known that:

1. India did not play against Australia from 2009-12 and played against Australia from 2013-15.

2. No team became the champion for two times in three consecutive years.

3. No team ended at the same position in two consecutive years.

4. Pakistan was the champion only in 2014 and New Zealand was always in the top three.

5. The runner up of 2012 slipped by one position in the next year.

6. The champion of 2009 stood fourth in 2015 and the winner of 2010 was fourth in 2009.

7. In 2009, Pakistan beat Australia in an eventful semi-final.

8. India was the runner up in 2011 and 2013.

 

Q. What was the maximum number of ‘final’ appearances for any team?


Solution:

‘Final’ appearances imply that a team would have been ranked 1 or 2. The number of ‘final’ appreances for each team is: Ind = 4, Pak = 3, Aus = 2 and NZ = 5 Thus, the maximum number of ‘final’ appreances for any team were 5.

*Answer can only contain numeric values
QUESTION: 15

In a hockey tournament, India, Australia, New Zealand and Pakistan have been the top four teams for the past seven years (2009-15). Each year, after the semi-final, two matches were played. The winners of the two semi finals faced each other in the finals for the top two spots. The losers of the two semi finals faced each other for the third spot play-off. It is also known that:

1. India did not play against Australia from 2009-12 and played against Australia from 2013-15.

2. No team became the champion for two times in three consecutive years.

3. No team ended at the same position in two consecutive years.

4. Pakistan was the champion only in 2014 and New Zealand was always in the top three.

5. The runner up of 2012 slipped by one position in the next year.

6. The champion of 2009 stood fourth in 2015 and the winner of 2010 was fourth in 2009.

7. In 2009, Pakistan beat Australia in an eventful semi-final.

8. India was the runner up in 2011 and 2013.

 

Q. In how many instances was the champion of a certain year, the runner up of the previous year?


Solution:

The condition that the champion of a certain year was the runner up of the previous year was satisfied thrice (in 2011, 2012 and 2015).

*Answer can only contain numeric values
QUESTION: 16

In a hockey tournament, India, Australia, New Zealand and Pakistan have been the top four teams for the past seven years (2009-15). Each year, after the semi-final, two matches were played. The winners of the two semi finals faced each other in the finals for the top two spots. The losers of the two semi finals faced each other for the third spot play-off. It is also known that:

1. India did not play against Australia from 2009-12 and played against Australia from 2013-15.

2. No team became the champion for two times in three consecutive years.

3. No team ended at the same position in two consecutive years.

4. Pakistan was the champion only in 2014 and New Zealand was always in the top three.

5. The runner up of 2012 slipped by one position in the next year.

6. The champion of 2009 stood fourth in 2015 and the winner of 2010 was fourth in 2009.

7. In 2009, Pakistan beat Australia in an eventful semi-final.

8. India was the runner up in 2011 and 2013.

 

 

Q. How many times did India beat Pakistan in the last two matches?


Solution:

India beat Pakistan exactly once (i.e. in 2009)

QUESTION: 17

Group Question

Answer the following question based on the information given below.

An entrance test had only two types of questions: 1-mark questions and 2-mark questions. The test had six sections and the sectional distribution of the 1-mark and 2- mark questions was as shown in the table below: 


The table below shows the percentage distribution (on a sectional basis) of the marks scored by one particular student, Karan, in this test.


Also it is known that

1. No negative marks were given for any wrong attempt or for any unattempted question.

2. There were 500 questions in the test, of which 80 were from QA.

3. Karan scored 300 marks in the test.

 

 

Q. What are the maximum marks that a student could have got in the test?  

Solution:

The test had 500 questions in all, which were either 1-mark and 2-mark questions. Hence, if the total number of 1-mark questions was a, the number of 2-mark questions was (500 - a) There were 80 questions in QA.
Also, 15% of the total 1-mark questions and 20% of the total 2-mark questions were in QA.

0 .15 a + 0 .2 (500 - a ) = 80 .

-. 0.15a + 100 - 0 . 2a = 80 .

-. a = 20/0.05 = 400 Hence, there are 400 1-mark questions and 100 2-mark questions in all. Thus, the number of 1-mark and 2-mark questions per section and the maximum marks for each section can be calculated as shown below: 

Maximum marks in the test = 400(1) + 100(2) = 600 Hence, option 2.

QUESTION: 18

An entrance test had only two types of questions: 1-mark questions and 2-mark questions. The test had six sections and the sectional distribution of the 1-mark and 2- mark questions was as shown in the table below: 


The table below shows the percentage distribution (on a sectional basis) of the marks scored by one particular student, Karan, in this test.


Also it is known that

1. No negative marks were given for any wrong attempt or for any unattempted question.

2. There were 500 questions in the test, of which 80 were from QA.

3. Karan scored 300 marks in the test.

Q. What is the minimum number of 1-mark questions that Karan could have attempted?

Solution:

Solution: Total marks of Karan = 300 Karan’s marks in RC = 25% of 300 = 75 Similarly, Karan’s marks in each section are LR = 75; QA = 60; Dl = 30; GK = 15 and VA = 45 Now, consider the solution to the first question. A student does not lose marks for incorrect and unattempted questions.
Consider RC. It has 80 1-mark questions and 10 2-mark questions.
To minimize the number of 1-mark attempts in this section, Karan should attempt as many 2-mark questions as possible.
If Karan correctly attempts all 10 2-mark questions in RC, maximum marks obtained through these = 10(2) = 20 Minimum marks obtained through 1-mark questions = minimum number of 1- mark questions attempted = 75 - 20 = 55 Similarly, the minimum number of 1-mark questions attempted in each section is as tabulated below. 

Observe that in GK and VA, maximum possible marks through 2-mark questions > Karan’s total score in that section Let the maximum number of 2-mark questions attempted by Karan in GK be p. 

∴ 2p ≤ 15 i.e. p ≤ 7.5
Thus, maximum 2-mark questions attempted in Karan in GK is 7.  Minimum 1-mark questions attempted by Karan in GK = 15 - 2(7) = 1 The same logic also applies to the VA section. Minimum number of 1-mark questions that he could have attempted = 55 + 35 + 20 + 20 + 1 + 1 = 132
Hence, option 4.

QUESTION: 19

An entrance test had only two types of questions: 1-mark questions and 2-mark questions. The test had six sections and the sectional distribution of the 1-mark and 2- mark questions was as shown in the table below: 


The table below shows the percentage distribution (on a sectional basis) of the marks scored by one particular student, Karan, in this test.


Also it is known that

1. No negative marks were given for any wrong attempt or for any unattempted question.

2. There were 500 questions in the test, of which 80 were from QA.

3. Karan scored 300 marks in the test.

 

Q. The minimum number of questions that Karan could have attempted in Dl is 

Solution:

Let Karan attempt a and b 1-mark and 2-mark questions respectively in Dl. a + 2b = 30 . Number of questions attempted by Karan in Dl = a + b = 30 - b For a + b to be minimum, (30 - b) should also be minimum.
This happens when b takes its maximum value.
Now, b can take a value from 0 (where Karan does not attempt any 2-mark questions) to 5.
Minimum value of (a + b) = 30 - 5 = 25 Hence, option 3.

QUESTION: 20

An entrance test had only two types of questions: 1-mark questions and 2-mark questions. The test had six sections and the sectional distribution of the 1-mark and 2- mark questions was as shown in the table below: 


The table below shows the percentage distribution (on a sectional basis) of the marks scored by one particular student, Karan, in this test.


Also it is known that

1. No negative marks were given for any wrong attempt or for any unattempted question.

2. There were 500 questions in the test, of which 80 were from QA.

3. Karan scored 300 marks in the test.

Q. In which section is Karan’s score, as a percentage of the maximum possible score for that section, the worst?

Solution:

Consider the solution to the earlier questions.Maximum possible score in each section is: RC = 100; LR = 100; QA = 100; Dl = 50; GK= 110; VA= 140 Karan’s score in each section is: RC = 75; LR = 75; QA = 60; Dl = 30; GK = 15 and VA = 45.By observation, since Karan’s actual score in GK is the least and the maximum possible marks in the GK are the highest, his percentage performance in GK is the worst.Hence, option 3.

*Answer can only contain numeric values
QUESTION: 21

Group Question

Answer the following question based on the information given below.


The table below shows the journey schedule of the Agra-Kolkata Express that leaves Agra at 5 p.m. every day and reaches Kolkata the next day at 5:20 p.m. The distance in the table below is cumulative in nature.

Q. What is the difference in the speed of the train excluding and including stoppages? Give your answer up to one decimal.   


Solution:

Total distance = 1405 km With stoppages, the train leaves at 5:00 p.m. today and reaches at 5:20 p.m.the next day.
Total time (including stoppages) = 24(60) + 20 = 1460 minutes Speed including stoppages (in kmph) = 1405/(1460/60) = 57.74 kmph Total time (excluding stoppages) = Total time (including stoppages) - Total stoppage time = 1460 - (10 + 5 + 10 + 10 + 20 + 5 + 10 + 15 + 2)
= 1373 minutes Speed excluding stoppages (in kmph) = 1405/(1373/60) = 61.39 kmph Required difference = 61.39 - 57.74 = 3.65 kmph

*Answer can only contain numeric values
QUESTION: 22

The table below shows the journey schedule of the Agra-Kolkata Express that leaves Agra at 5 p.m. everyday and reaches Kolkata the next day at 5:20 p.m. The distance in the table below is cumulative in nature.


Q. What is the slowest running speed of the train between any two consecutive stations? Assume that the train runs at a constant speed between stations and give your answer up to one decimal.


Solution:

Solution: The running speed (in kmph) between two consecutive stations can be calculated as shown in the table below: 

Thus, the lowest speed is in the Farrukhabad-Kanpur section (47.7 kmph).
This can even be done by observation once the set of distances and time taken are found.

*Answer can only contain numeric values
QUESTION: 23

The table below shows the journey schedule of the Agra-Kolkata Express that leaves Agra at 5 p.m. everyday and reaches Kolkata the next day at 5:20 p.m. The distance in the table below is cumulative in nature.


Q. If the train runs the entire distance at the maximum speed with which it runs  between any two stations, approximately how much time will it save? Assume that the stoppage time remains the same. Give an integral answer in minutes.


Solution:

Consider the solution to the earlier questions.
Maximum running speed of the train = 84 kmph (between Asansol and Durgapur). /. Time in this case = 1405/(84/60) » 1004 minutes Original running time taken = 1373 minutes Time saved = 1373 - 1004 = 369 minutes

*Answer can only contain numeric values
QUESTION: 24

The table below shows the journey schedule of the Agra-Kolkata Express that leaves Agra at 5 p.m. everyday and reaches Kolkata the next day at 5:20 p.m. The distance in the table below is cumulative in nature.


Q. On a certain day, the train is delayed due to fog. It leaves Kanpur 2 hours late than the usual time. What   should be its average running speed (in km/min) for the remaining distance so that it arrives at Kolkata at the scheduled time? The stoppage time at the stations remain the same. Give your answer to two decimal points.


Solution:

Consider the solution to the earlier questions.
Distance between Kanpur and Kolkata = 1405 - 406 = 999 km Time taken by the train to travel from Kanpur to Kolkata = (17:20 - 1 : 1 5 ) - (20 + 5 + 10+15 + 2) [stoppage time in between] = 16 hour 5 min - 52 min = 15 hour 13 min = 913 minutes Since the train is delayed by 2 hours, it should take cover the journey in 913 — 120 = 793 minutes Average speed for the remaining distance = 999/793 = 1.26 km/min Answer: 1.26 

QUESTION: 25

Group Question

Answer the following question based on the information given below.


A placement agency assists students in getting placed in MNCs. The table below shows the month-on-month placement record of the agency in the last year.  


Conversion ratio is the ratio of the number of students who clear the placement process of a company to the number of the test takers. However, even after clearing the placement process, some students do not accept the offer in pursuit of higher studies or better offers. Acceptance ratio is the ratio of students who finally join the company they were placed in, to the students who cleared the placement process.
Q. What was the conversion percentage for the entire year?   

Solution:

Solution: This question requires number of conversions for each month. Hence, it is helpful to find number of converted students, number of students accepting offers and number of students rejecting offers right away. 

Total test takers = 3410 and conversions = 1914. Conversion percentage = (1914/3410) x 100 = 56.13% Hence, option 3.

QUESTION: 26

A placement agency assists students in getting placed in MNCs. The table below shows the month-on-month placement record of the agency in the last year.  

Conversion ratio is the ratio of the number of students who clear the placement process of a company to the number of the test takers. However, even after clearing the placement process, some students do not accept the offer in pursuit of higher studies or better offers. Acceptance ratio is the ratio of students who finally join the company they were placed in, to the students who cleared the placement process.

Q. What percent of students getting an offer from the agency do not accept it?

Solution:

Consider the solution to the first question of the set.
Number of students getting an offer = number of converts = 1914 Number of students not accepting the offer = number of rejections = 178 Required percentage = (178/1914) x 100 = 9.3% Hence, option 1.

QUESTION: 27

A placement agency assists students in getting placed in MNCs. The table below shows the month-on-month placement record of the agency in the last year.  


Conversion ratio is the ratio of the number of students who clear the placement process of a company to the number of the test takers. However, even after clearing the placement process, some students do not accept the offer in pursuit of higher studies or better offers. Acceptance ratio is the ratio of students who finally join the company they were placed in, to the students who cleared the placement process.
Q. Starting from January, a set of three consecutive months is called a quarter e.g. Jan to Mar is the first quarter. If the success rate of the agency is measured in terms of what percentage of its test takers accepted a job offer, which quarter was the most successful for the agency?

Solution:

First Quarter:
Test takers = 168 + 203 + 216 = 587 and
Acceptances = 78 + 105 + 120 = 303
Success rate = (303/587) x 100 = 51.62%
Second Quarter:
Test takers = 240 + 320 + 441 = 1001 and
Acceptances = 112 + 180 + 272 = 564
Success rate = (564/1001) x 100 = 56.34%
Third Quarter:
Test takers = 456 + 368 + 295 = 1119 and
Acceptances = 208 + 154 + 160 = 522 
Success rate = (522/1119) x 100 = 46.65%
Fourth Quarter:
Test takers = 276 + 235 + 192 = 703 and
Acceptances = 138 + 110 + 99 = 347. 
Success rate = (347/703) x 100 = 49.36% Thus, the second quarter was the most successful.
Hence, option 2.

QUESTION: 28

A placement agency assists students in getting placed in MNCs. The table below shows the month-on-month placement record of the agency in the last year.  


Conversion ratio is the ratio of the number of students who clear the placement process of a company to the number of the test takers. However, even after clearing the placement process, some students do not accept the offer in pursuit of higher studies or better offers. Acceptance ratio is the ratio of students who finally join the company they were placed in, to the students who cleared the placement process.
Q. If all the months were arranged in ascending order of accepted offers, how many month(s) would be at the same position as in the chronological order of months in a year?

Solution:

Consider the solution to the first question. The sequence of months in ascending order of accepted offers is: Jan - Dec - Feb - Nov - Apr - Mar - Oct - Aug - Sep - May - Jul - Jun The sequence in normal chronological order is: Jan - Feb - Mar - Apr - May - Jun - Jul - Aug - Sep - Oct - Nov - Dec Thus, three months (Jan, Aug and Sep) would be at the same position in both lists.
Hence, option 2.

QUESTION: 29

Group Question

Answer the following question based on the information given below.


The Committee for North-East Indian Sports Development organizes an annual sports meet for the seven northeastern states of India. For every sporting event, the top three winners are awarded medals viz. gold, silver and bronze. However, more or less than three medals are never awarded. The graph below shows the number of medals won by these seven states in the last three years. The total gold medals given out in this period were 139. Assam’s performance in 2013 was better than that in 2014 but not as good as that in 2015.

 

Q. Which state got the second highest number of medals over the given period?

Solution:

For each event, one gold, one silver and one bronze medal is given.
Total medals across all three years = total gold medals x 3 = 139 x 3 = 417

This is also equal to the sum of all the values given in the chart (as they represent medals won by each state). (3x + 18 + 20 + 25 + 21) + (20 + 14 + 25 + 24 + 17 + 22 + 19) + (2y + 23 + 18 + 21 + 24 + 18) = 3x + 2y + 329. 3x + 2y+ 329 = 417. 

 3x + 2y = 88.

 2y = 88 - 3x ... (i)

Observe that x is the number of medals won by Assam (as well as Mizorma and Tripura) in 2013.
Assam’s performance in 2013 was better than that in 2014 but not as good as in 2015. 14 < x < 18 i.e. x = 15, 16 or 17 In (i), if y has to be integral, (88 - 3x) has to be even.
Hence, 3x has to be even.
Hence, from the given range, x can only be 16. x = 16 and y = 20 Thus, number of medals won by each state in each year is: 

Thus, Meghalaya (65) got the second highest number of medals over the given period.
Hence, option 4.

QUESTION: 30

The Committee for North-East Indian Sports Development organizes an annual sports meet for the seven north eastern states of India. For every sporting event, the top three winners are awarded medals viz. gold, silver and bronze. However, more or less than three medals are never awarded. The graph below shows the number of medals won by these seven states in the last three years. The total gold medals given out in this period were 139. Assam’ performance in 2013 was better than that in 2014 but not as good as that in 2015.

 

 

 

Q. What percentage of the total medals given out over these three years were won by Manipur?

Solution:

Solution: Consider the solution to the first question.
Total medals = 417 and medals won by Manipur = 70 Required % = (70/417) x 100 = 16.79% Hence, option 3.

QUESTION: 31

The Committee for North-East Indian Sports Development organizes an annual sports meet for the seven north eastern states of India. For every sporting event, the top three winners are awarded medals viz. gold, silver and bronze. However, more or less than three medals are never awarded. The graph below shows the number of medals won by these seven states in the last three years. The total gold medals given out in this period were 139. Assam’s performance in 2013 was better than that in 2014 but not as good as that in 2015.

 

 

 

Q. The Arunachal Pradesh government awarded cash prizes to the medal winners. Each gold, silver and bronze winner was given Rs.5 lakh, Rs.3 lakh and Rs.2 lakh respectively. If Arunachal Pradesh won 5 golds and 6 silvers in each of the three years, what amount (in Rs. lakhs) was spent by the government over the given period to reward the winners?

Solution:

Solution: Consider the solution to the first question.
Total medals won by Arunachal Pradesh = 61 Arunachal Pradesh won 5 golds and 6 silvers in each of the three years of the given period.
Total bronze medals won = 61 - 3(5) - 3(6) = 28. Total reward amount = 5(15) + 3(18) + 2(28) = 75 + 54 + 56 = Rs.185 lakhs Hence, option 2.

QUESTION: 32

The Committee for North-East Indian Sports Development organizes an annual sports meet for the seven north eastern states of India. For every sporting event, the top three winners are awarded medals viz. gold, silver and bronze. However, more or less than three medals are never awarded. The graph below shows the number of medals won by these seven states in the last three years. The total gold medals given out in this period were 139. Assam’s performance in 2013 was better than that in 2014 but not as good as that in 2015.

 

 

 

Q. What is the difference in the medal tallies of Arunachal Pradesh and Mizoram in 2013, 2014 and 2015 respectively?

Solution:

Solution:
Consider the solution to the first question.
Difference in the medal tallies for each year is: 2013: 18 - 16 = 2
2014:20- 17 = 3
2015: 24 - 23 = 1
Hence, option 4.

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