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Since 10 is not a prime number, we have to first find the maximum power of 2 and 5 in 125!.
Maximum power of 2 = Multiples of 2 in 125! + Multiples of 4 in 125! + Multiples of 8 in 125! + Multiples of 16 in 125! + Multiples of 32 in 125! + Multiples of 64 in 125!
Where [x] = Maximum integer less than or equal to x. = 62 + 31 + 15 + 7 + 3 + 1 = 119
Similarly the maximum power of 5
= 25 + 5 + 1 = 31
To get the number 10, we need one 2 and one 5.
As the number of 5's are lesser, the highest power of 10 in 125! will be 31.
Hence, option 5.
S is the set of all possible permutations of the letters of the word APPLE. In how many of these permutations does the letter A come before the letter P?
A can take 4 places  first to fourth  to have a P after it.
Case 1: A is first i.e. A
There are 4 letters, out of which P is present twice.
∴ No. of permutations possible =
Case 2: A is second i . e . __A _______ Within this, there are two possibilities: 1) Both P’s are after A.
Here, one of L and E is also after A.
So, there are three letters after A i.e. P, P and E or L
∴ No. of permutations possible =
2) Exactly one P is after A.
Here, the letters after A are P, E and L, and there is no repetition.
∴ No. of permutations possible = 3! = 6
∴ Total no. of permutations when A is second = 6 + 6=12
Case 3: A is third i . e . ____ A _____ Again, within this, there are two possibilities: 1) Both P’s are after A.
Hence, L and E are before A, without any repetition.
These L and A can be arranged in 2! ways No. of permutations possible = 2! = 2 2) Exactly one P is after A.
So, one of L and E is before A and one is after A.
The two letters before A can be arranged in 2! ways.
Similarly, the two letters after A can also be arranged in 2! ways. Finally, L or E can be after A in another 2 ways.
∴No. of permutations possible = 2! *2! x 2 = 8
∴Total no. of permutations when A is third = 2 + 8 = 10 Case 4: A is fourth i . e . _______ A __
Now, exactly one P is after A.
So, the letters before A are P, L and E, without any repetition.
∴No. of permutations possible = 3! = 6
∴Total number of permutations = 12 + 12 + 10 + 6 = 40 Hence, option 5.
Answer the following question based on the information given below.
The first table shows the birth rate and natural growth rate per thousand in a country.
The second table shows the infant mortality rate.
Natural growth rate = Birth rate  Death rate and Infant Mortality rate = number of infant deaths per thousand births
Q.
If the population of Maharashtra on 1^{st} January, 2009 was 100 crores, what was the population of Maharashtra on 31^{st} December 2010 (in crores)?
The natural growth rate gives the growth in population for every 1000 people. Natural growth rate for Maharashtra in 2009 = 17.4
∴ Population at the end of 2009 =
Natural growth rate for Maharashtra in 2010 = 17.3
∴ Population at the end of 2010 =
Hence, option 4.
If an infant death is defined as the death of a child on the very day that it is bom, then approximately what percent of the total deaths in Maharashtra are infant deaths in the year 2009?
In 2009, the growth rate in Maharashtra =17.4 per 1000
Birth rate = 26 per 1000
∴ Overall death rate = 26  17.4 = 8.6 per 1000
Infant mortality in Maharashtra in 2009 = 70 ∴ Infant mortality for 26 births on 1000 = (70 x 26)/1000 = 1.82
∴ Required percentage = (1.82/8.6) x 100 = 21.16%
Hence, option 3.
Which of the following states recorded the highest percentage decrease in death rate over the entire period?
The death rate for each state for each year is as shown below:
The death rate for Karnataka is the same in 2007 and 2010.
Hence, the percentage decrease for Karnataka is 0.
Observe that the death rate in West Bengal is greater than 10% while in the remaining states, it is less than 10%.
Hence, the greatest death rate is in West Bengal over the given period.
Hence, option 4.
Which state never saw a decrease in birth rates over the given period?
The birth rate in Tamil Nadu increased from 2007 to 2008 and then remained constant throughout. Hence, it never decreased.
The birth rate in each of the other states decreased atleast once.
Hence, option 2.
Group Question Answer the following question based on the information given below.
The chart below shows the income and expenditure (in Rs. crores) of a firm for four successive quarters in an year.
Q.
The profit earned by the firm in Q1 is invested at the start of Q2 in Tbills at simple interest at 8% per annum for three months. What is the total profit made by the firm (in Rs. crores) at the end of Q2?
Total profit made by firm = profit made in Q1 + profit made in Q2 + interest income
Profit made in Q1 = 4.3  3.1 = Rs. 1.2 crores
Profit made in Q2 = 5.2  4.4 = Rs. 0.8 crores
Rs. 1.2 crores is invested at 8% per annum for 3 months
∴ Interestin come =
∴ Total profit = 1.2 + 0.8 + 0.024 = Rs. 2.024 crores
Hence, option 1
Group Question Answer the following question based on the information given below.
The chart below shows the income and expenditure (in Rs. crores) of a firm for four successive quarters in an year.
Q.
For Q1 of the next year, the income is the average quarterly income of the previous year and the expenditure is the average quarterly expenditure of the previous year. What is the profit generated (in Rs. crores) in Q1 of the next year?
Average quarterly income of previous year =
Average quarterly expenditure of previous year =
∴ Profit in Q1 of next year = 4.375  3.525 = Rs. 0.85 crores
Hence, option 2.
Group Question Answer the following question based on the information given below.
The chart below shows the income and expenditure (in Rs. crores) of a firm for four successive quarters in an year.
Q.
What is the maximum percentage profit earned by the company in any of the given quarters?
This question can be solved faster by observation.
Note that the income is less than expenditure in Q3. Hence, there is a percentage loss in
Q3. Hence, Q3 can be ignored in calculations.
Now, compare Q1 and Q4.
The income in Q4 is more than the income in Ql, while the expenditure in Q4 is less than the expenditure in Ql.
Hence, the percentage profit in Q4 is more than that in Ql and so, Ql can be ignored in calculations.
For Q2: percentage profit =
For Q4: percentage profit =
Thus, the maximum percentage profit is approximately 61%.
Hence, option 4.
Group Question Answer the following question based on the information given below.
The chart below shows the income and expenditure (in Rs. crores) of a firm for four successive quarters in an year.
Q.
What is the income (in Rs. crores) in Ql of the next year, if the expenditure in that quarter is the same as the expenditure in Q4 this year, while the percentage profit has gone up by 5 percentage points from Q4 this year?
Expenditure in Ql of next year = expenditure in Q4 of this year = Rs. 2.8 crores From the solution to the previous question, percentage profit in Q4 this year = 60.71%
∴ Percentage profit in Ql next year = (60.71 + 5)% i.e. 65.71%.
Hence, option 2.
Group Question Answer the following question based on the information given below.
The chart below shows the income and expenditure (in Rs. crores) of a firm for four successive quarters in an year.
Q.
A circle of maximum area is cut out from a square cardboard of side a cm. The cardboard left is thrown away. Now, a right angled triangle is cut out from the circle such that all three vertices of the triangle lie on the circumference of the circle and touch the sides of the square. Again the cardboard left is thrown away. What is the ratio of the area of the triangular cardboard to the area of the cardboard that is thrown away?
The given data can be represented as shown below:
Let the area of the square, circle and triangle be S, C and T respectively.
Area o f cardboard thrown away when the circle is cut = A  C
And, area o f cardboard thrown away when the triangle is cut = C  T
∴ Total area o f cardboard thrown away = A  C + C  T = A  T
Area of square =A = a^{2 }sq.cm
Since all three vertices of the right triangle lie on the circle, the hypotenuse of the triangle is the same as the diameter of the circle (which in turn is the same as the side of the square i.e. a cm).
If the centre of the circle is joined to the third vertex of the triangle, it becomes the height of the triangle (as shown in the figure). The hypotenuse becomes the base of the triangle.
∴ Area of triangle =
∴ Required ratio =
Hence, option 4.
Group Question Answer the following question based on the information given below.
The chart below shows the income and expenditure (in Rs. crores) of a firm for four successive quarters in an year.
Q.
The weight of a circular disc is proportional to the square of the radius of the disc when the thickness of the disc is kept constant, whereas the weight of the disc is proportional to its thickness when the radius of the disc is kept constant.
What is the ratio of the radii of two circular discs, if their thicknesses are in the ratio 9 : 8 and their weights are in the ratio 2 : 1 ?
Let w, r and t be the weight, radius and thickness of the circular disc respectively.
∴ w ∝ r_{2} x t
∴ w = k x r_{2 }x t
∴ W_{1} = k x r1_{2} x t_{1 }
_{Hence, ratio of radius of two discs is 4 : 3. Hence, option 2.}
What is the value of x of log_{x } log_{2 }
Instead of trying to solve this equation, substitute the value of x from the options and check which value satisfies this equation.
Only x = 81 satisfies this equation.
Hence, option 3.
Nine members of a family are standing in a row. Their average age is 70. The average age of the first four members in the row is 64 and the average age of the last seven members in the row is 80. What is the average age of the third and fourth member in the row?
Let the age of the members in the row be A, B, C, D, E, F, G, H and I.
∴ A + B + C + D + E + F + G + H + I = 1 0 * 9 = 630 ...(i)
Since the average age of the first four members is 64 years,
A + B + C + D = 6 4 x 4 = 256 ...(ii)
§jnce the average age of the last seven members is 80 years,
C + D + E + F + G + H + I = 80 x 7 = 560 ...(iii)
Subtracting (iii) from (i),
A + B = 70 ... (iv)
Subtracting (iv) from (ii),
C + D= 186
∴ Averageage of third and four the member =
Hence, option 3.
Group Question
Answer the following question based on the information given below.
The charts below show the performance of cricket teams from various cities. The matches won, drawn and lost are given in percentage terms in the bar graph. The total number of matches played by various teams is not necessarily the same.
Q.
What is the total number of matches played by Indore, Chennai, Ludhiana, Bangalore and Kolkata?
Using the given data, the number of matches played, won, drawn and lost can be found as shown below.
∴ Total number of matches played by Indore, Chennai, Ludhiana, Banglore and Kolkata = 80 + 80 + 40 + 20 + 60 = 280
Hence, option 1.
Group Question
Answer the following question based on the information given below.
The charts below show the performance of cricket teams from various cities. The matches won, drawn and lost are given in percentage terms in the bar graph. The total number of matches played by various teams is not necessarily the same.
Q.
By what percent is the number of matches drawn by Banglore more/less than the number of matches lost by Chandigarh?
Correct Answer : b
Explanation : Number of matches drawn by Bangalore = 4
Number of matches drawn by Chandigarh = 18
% = (184)/2 * 100
= 77.77%
Group Question
Answer the following question based on the information given below.
The charts below show the performance of cricket teams from various cities. The matches won, drawn and lost are given in percentage terms in the bar graph. The total number of matches played by various teams is not necessarily the same.
Q.
What is the difference between the number of matches lost by Pune and Ludhiana? 1)12
2 ) 14
3 ) 10
4 ) 9
5 ) 13
Consider the solution to the first question.
Number of matches lost by Pune =14 Number of matches lost by Ludhiana = 2
∴Required difference = 1 4  2 = 12
Hence, option 1.
Group Question
Answer the following question based on the information given below.
The charts below show the performance of cricket teams from various cities. The matches won, drawn and lost are given in percentage terms in the bar graph. The total number of matches played by various teams is not necessarily the same.
Q.
What is the difference between the total number of matches won by all the teams and the total number of matches drawn by all the teams?
Consider the solution to the first question.
Total number of matches won by all the teams = 16 + 30 + 28 + 2+ 12 + 26 + 56 + 8 = 178
Total number of matches drawn by all the teams = 10 + 9 + 12 + 4 + 10 + 12 + 20 + 3 = 80
∴ Difference = 178  80 = 98
Hence, option 5
Group Question
Answer the following question based on the information given below.
The charts below show the performance of cricket teams from various cities. The matches won, drawn and lost are given in percentage terms in the bar graph. The total number of matches played by various teams is not necessarily the same.
Q.
There is a match box of dimensions 10 cm x 6 cm x 2 cm. What is the maximum length of the matchstick that can be kept inside it?
The maximum length of the match stick will be equal to the biggest possible diagonal of the cube (match box).
Dimensions of the box are given as 10 x 6 x 2
∴ The biggest possible diagonal =
Hence, option 4.
Group Question
Answer the following question based on the information given below.
The charts below show the performance of cricket teams from various cities. The matches won, drawn and lost are given in percentage terms in the bar graph. The total number of matches played by various teams is not necessarily the same.
Q.
A and B work on a project on alternate days. If A starts the project, they complete 7/10^{th }of the work in 9 days. If B starts the project, they complete 5/6^{th} of the work in 10 days. If B were to work alone, how many days would he take to complete the project?
Let A alone do a units of work per day and B alone do b units of work per day.
When A starts, A works for 5 of the 9 days and B works for the remaining 4.
∴ Work done in 9 days when A starts = 5a + 4b
Let the total work be the LCM of 10 and 6 i.e. 30 units.
When B starts, B works for 5 of the 10 days and A works for the remaining 5.
∴ Work done in 10 days when A starts = 5a + 5b
(ii)  (i) gives b = 4
Hence, option 2.
Group Question
Answer the following question based on the information given below.
The charts below show the performance of cricket teams from various cities. The matches won, drawn and lost are given in percentage terms in the bar graph. The total number of matches played by various teams is not necessarily the same.
Q.
What will be the value of the following expression if x = 2^{1/2 }?
The numerator is equal to:
Similarly, the denominator is equal to:
Hence, option 1.
Each question is followed by two statements, A and B. Answer each question using the following instructions:
Mark option (1) if the question can be answered by using statement A alone but not by using statement B alone.
Mark option (2) if the question can be answered by using statement B alone but not by using statement A alone.
Mark option (3) if the question can be answered by using either statement alone.
Mark option (4) if the question can be answered by using both the statements together but not by either of the statements alone.
Mark option (5) if the question cannot be answered on the basis of the two statements.
A boy plucked 15 mangoes in 4 days. On the first day, he plucked 4 mangoes. Find the number of mangoes that he plucked on each day (ignoring the order in which they were plucked), if he plucked at least one mango every day.
A. He plucked an odd number of mangoes on each of the remaining days.
B. He plucked an equal number of mangoes on no two days.
The boy picked 11 mangoes in the remaining 3 days.
Using statement A alone: The number of mangoes plucked could be (1, 1, 9), (1, 3, 7), (1, 5, 5) and (3, 3, 5).
Thus, the question cannot be answered using statement A alone.
Using statement B alone: The number of mangoes plucked could be (1, 3, 7), (1,2, 8) and so on.
Thus, the question cannot be answered using statement B alone.
Using both the statements together: When conditions are combined, (1, 3, 7) is the only combination satisfying both conditions.
Thus, the boy picked 1, 3 and 7 mangoes (in no particular order) on the remaining 3 days.
Thus, the question can be answered using both the statements together but not by using either statement alone.
Hence, option 4.
Each question is followed by two statements, A and B. Answer each question using the following instructions:
Mark option (1) if the question can be answered by using statement A alone but not by using statement B alone.
Mark option (2) if the question can be answered by using statement B alone but not by using statement A alone.
Mark option (3) if the question can be answered by using either statement alone.
Mark option (4) if the question can be answered by using both the statements together but not by either of the statements alone.
Mark option (5) if the question cannot be answered on the basis of the two statements. The ages of Jai and Veeru are in the ratio of 2 : 1 .What is the age of Veeru?
A. The ages of Veeru and Gabbar are in the ratio of 2 : 1
B. After 4 years the ratio of Jai and Veeru’s ages will be 3 : 2
Let Jai and Veeru’s current age be 2x and jc respectively.
Using Statement A alone: Though the ratio of Veeru and Gabbar’s age is known, there is no additional information on Jai. So, none of the ages can be found.
Thus, the question cannot be answered using statement A alone.
Using Statement B alone:
This equation can be solved for x. Once x is found, Veeru’s age is also found.
Thus, the question can be answered using statement B alone.
Thus, the question can be answered using statement B alone but not by using statement A alone.
Hence, option 2.
Each question is followed by two statements, A and B. Answer each question using the 3 following instructions:
Mark option (1) if the question can be answered by using statement A alone but not by using statement B alone.
Mark option (2) if the question can be answered by using statement B alone but not by
using statement A alone.
Mark option (3) if the question can be answered by using either statement alone.
Mark option (4) if the question can be answered by using both the statements together but not by either of the statements alone.
Mark option (5) if the question cannot be answered on the basis of the two statements. What is the profit percent obtained by selling each pen?
A. By selling x pens at Rs. x per pen, shopkeeper earns a profit of Rs. 150.
B. Pens are brought at Rs. x per gross and are sold at Rs. x/8 per pen. (1 gross = 12 pens)
Using statement A alone: Let the cost price of each pen be Rs. y.
∴ Profit = (x x x)  (x x y) = x2  xy = Rs. 150
Since we don’t know the value of x, we cannot find the profit percent.
Thus, statement A alone is not sufficient to answer the question.
Using statement B alone: Cost price of 12 pens = Rs.
∴Cost price of one pen = Rs. x/12
Selling price of one pen = Rs. x/8
Thus, statement B alone is sufficient to answer the question.
Hence, option 2.
Each question is followed by two statements, A and B. Answer each question using the following instructions:
Mark option (1) if the question can be answered by using statement A alone but not by using statement B alone.
Mark option (2) if the question can be answered by using statement B alone but not by using statement A alone.
Mark option (3) if the question can be answered by using either statement alone.
Mark option (4) if the question can be answered by using both the statements together but not by either of the statements alone.
Mark option (5) if the question cannot be answered on the basis of the two statements. What is the speed of the cruise ship when sailing in still water?
A. It passes a lighthouse tower (of negligible length) in 10 seconds.
B. It passes another cruise ship of length 600 metres in 30 seconds
Let the speed of the cruise ship in still water = s m/s and length of the cruise ship = ℓ m Using Statement A alone:
Since the length of the cruise ship is not known, the speed cannot be determined Thus, the question cannot be answered using statement A alone.
Using Statement B alone:
Since the length of the cruise ship is not known, the speed cannot be determined.
Thus, the question cannot be answered using statement B alone.
Using Statements A and B together: Solving equations i and ii, 5 = 30 m/s.
Thus, the question can be answered using both the statements together but not by using either statement alone.
Hence, option 4.
The question is followed by two statements, A and B. Answer each question using the following instructions: Mark option (1) if the question can be answered by using the statement A alone but not by using the statement II alone.
Mark option (2) if the question can be answered by using the statement B alone but not by using the statement I alone.
Mark option (3) if the question can be answered by using either of the statements alone. Mark option (4) if the question can be answered by using both the statements together but not by either of the statements alone.
Mark option (5) if the question cannot be answered on the basis of the two statements.
If p and m  8 are positive integers, is p/m less than one?
A. p  m > 0
B.p < 8
(P/m)< 1 is possible when p < m .
Using statement A alone:
∴ p  m > 0
∴ p > m
∴ p/m > 1
Thus, p/m is not less than one.
Thus, the question can be answered using statement A alone. Using statement B alone:
p < 8 and m  8 > 0
∴ m > 8
∴ p < m
∴ p/m < 1
Thus, the question can be answered using statement B alone.
Thus, the question can be answered using either statement alone.
Hence, option 3.
Each question is followed by two statements, A and B. Answer each question using the following instructions: Mark option (1) if the question can be answered by using statement A alone but not by using statement B alone.
Mark option (2) if the question can be answered by using statement B alone but not by using statement A alone.
Mark option (3) if the question can be answered by using either statement alone.
Mark option (4) if the question can be answered by using both the statements together but not by either of the statements alone.
Mark option (5) if the question cannot be answered on the basis of the two statements. What is the value of the positive integer N.
A. Sum of all the factors of N, excluding itself, is 56.
B. The total number of factors A has is 4.
Statements A and B alone cannot give a unique solution.
Using both the statements, the number is in the form of (a^{1} x b^{1}) or a^{3}, where a and b are prime.
If the number is of the form a3, its factors will be 1, a, a^{2} and a^{3}.
Therefore the number cannot be in the form of a^{3} since a^{2} + a + 1 = 56 does not have any integer solutions.
∴ The number is of the form a^{1}x b^{1}
Its factors are 1, a, b and ab.
Sum of the factors excluding N is given to be 56.
a + b = 56  1 = 55
∴ a and b should be two primes such that,
a + b = 55
∴ a = 2, b = 53
∴ N= 53 x 2 = 106
∴ Statements A and B together are together sufficient to answer the question.
Hence, option 4.
What is the remainder when 1! + 2! + 3! + 4! + ... + 55! is divided by 12?
Remainder when (a + b + c) is divided by n = rem (a/n) + rem (bln) + rem (dn) x! = X x (x  1) X (x  2) x ... X 3 X 2 x 1 Thus, every term from 12! to 55! will have a ‘12’ in it. Thus, every term from 12! to 55! is divisible by 12.
Since 12 = 4 x 3 , any x! that has a term (4 x 3) in it will also be divisible by 12. 4! =4x 3x 2xl
Since every subsequent term from 4! to 11! will have a 4! in it, each term from 4! to 11! is divisible by 12.
∴ Required remainder = rem(l !/12) + rem (2!/12) + rem (3!/12) = rem(l/12) + rem (2/12) + rem (6/12) = 1 + 2 + 6
= 9
Hence, option 5.
If m and n are two natural numbers, ratio of m to n is x and ratio of n to m is y, then (x + y) will be
∴ x +y > 1
Even then x + y > 1
∴ x + y is always greater than 1.
Hence, option 2.
Esha is 28 years old. She is twice as old as Ankita was when Esha was as old as Ankita is now. How old is Ankita now?
Let the present age of Esha and Ankita be E and A years respectively.
Number of years between Esha’s current age and the age when she was as old as Ankita’s current age = E  A Ankita’s age when her age was reduced by this number = A  ( E  A ) = 2 A  E
∴ E = 2(2AE)
i.e. 28 = 4A  56
∴ 4A = 84
∴ A = 21
Hence, option 3.
Between 4 pm and 5 pm, Raju sees that the angle between the minute hand and hour hand is 82°. Sometime later, but between 4 pm and 5 pm itself, Shyam also sees the same angle between the minute hand and hour hand. How many minutes after Raju did Shyam see the clock?
When the hour hand is ahead of the minute hand, the angle is positive and when the hour hand is behind the minute hand, the angle is negative.
∴ Difference between Raju and Shyam’s viewing = 36.72  6.91 30 minutes
Hence, option 2.
There are 35 students in a hostel. If the number of students increases by 7, the expenses of the mess increase by Rs.42 per day while the average expenditure per head per week diminishes by Rs.7. What is the original weekly expenditure of the mess (in Rs.)?
Let the average daily expenditure per student be Rs. a.
∴ Total weekly expenditure for 35 students = 35 x a x 7 = Rs. 245a Also, average expenditure per head per week = Rs. la
When the number of students increases by 7 to 42, average expenditure per head per week reduces by Rs. 7
∴ Total weekly expenditure for 42 students = 42 x (la  7) = Rs. 294(a  1)
However, it is also known that in the same case of 42 students , total expenditure increases compared to the previous case by Rs. 42 per day.
∴ Total weekly expenditure for 42 students as per this data = 245a + 42(7) = Rs. 49(5a + 6)
Equating the two, 294(a  1) = 49(5a + 6)
∴ 5a + 6 = 6(a  1)
∴ a = 12
∴ Original weekly expenditure of the mess = 245 x 12 = Rs. 2,940
Hence, option 4.
ABC is a right angled isosceles triangle such that AQ = QP = PC. If AC = 12 cm, find the value of:
where x is the radius of the circle.
Since AQ = QP = PC and AC = 12, AQ = QP = PC = 4 cm
Let the radius of the circle be jc cm as shown above.
Since OSBR forms a square, BR = BS = x cm
Also, AR^{2} = AQ.AP
∴ AR^{2} = 4(8) = 32
Since ΔAEC is a right isosceles triangle, AE = EC = x + 4√2
Hence, option 3.
A triangle ABC is such that a line from A on side BC meets BC at point P and divides it in the ratio 2 : 5 . Another line is drawn parallel to AP at a distance o f 2 units from it. This line touches BC at Q and divides it in the ratio 1:2. If BPQ as well as PQC and Q is farther from C compared to B, what is the length of side BC?
The data given in the question can be represented as shown below.
The order of B, P, Q and C is because BPQ and PQC.
Based on the figure,
BP = 2x, PC = 5x, BQ =y and QC = 2y
BC = BP + PC = BQ + QC
∴ 2x + 5 x = y + 2y
∴3y7x = 0 ...(i)
Since the two lines are drawn parallel to each other, PQ = 2 units ... QB  PB = 2
∴ y  2x = 2
∴ 3 y  6 x = 6 ...(ii)
Solving (i) and (ii), x = 6 and y = 14
∴ BC = lx = 42 units
Hence, option 1.
Aman bought 3 apples, 5 oranges and 7 guavas for Rs. 317 from a fruit shop. Meena bought 2 apples, 3 oranges and 4 guavas for Rs. 191 from the same shop. How much it will cost if Aman buys 7 apples, 11 oranges and 15 guavas?
Let the price of 1 apple, 1 orange and 1 guava be x, y and z respectively.
∴ 3x + 5y + 7z = 317 ...(i)
2x + 3y + 4z=191 (ii)
We need to find the value of 7x + 11y + 15z
Multiplying equation (ii) by 2 and then adding to equation (i),
∴ 7x + Uy + 15z = 317 + 191 x 2 = 699
Hence, option 1.
Answer the following question based on the information given below.
The pie chart below shows the market share (by value) of different cosmetic companies in the US market in 2010.
The pie chart below shows the market share (by volume) of different cosmetic companies in the US market in 2010.
Number of units sold in 2010 in US = 1500 and value of these units sold = USD 1,20,000.
The average unit selling price sale price of a Lakme product in the US 2010 was approximately
Average units elling price of a Lakme product =
Hence, option 3.
Answer the following question based on the information given below.
The pie chart below shows the market share (by value) of different cosmetic companies in the US market in 2010.
The pie chart below shows the market share (by volume) of different cosmetic companies in the US market in 2010.
Number of units sold in 2010 in US = 1500 and value of these units sold = USD 1,20,000.
Q.
Which cosmetic company has realized the lowest average unit sales price for the year 2010 for its products?
Note that the number of companies in the Others category is not known. Also, the breakup of sales volume and sales value within this category is not known.
Hence, the company that has realized the lowest average unit sales price in 2010 cannot be found.
Hence, option 5.
Answer the following question based on the information given below.
The pie chart below shows the market share (by value) of different cosmetic companies in the US market in 2010.
The pie chart below shows the market share (by volume) of different cosmetic companies in the US market in 2010.
Number of units sold in 2010 in US = 1500 and value of these units sold = USD 1,20,000.
Q.
From 2010 to 11, if the sales of Lakme went up by 50% (by value) and the sales of L’Oreal went down by 16% (by value), by what percentage did the US market grow (by value)? Assume that the sales (by value) for all other companies remained constant over the period.
Sales of all companies, except Lakme and L’Oreal, in 2010 (by value)
= (6 + 6 + 8 + 30 + 8)% of 120000 = 58% of 120000 = USD 69,600
Sales of Lakme in 2011 (by value) = 1.5 x 0.28 x 120000 = USD 50,400
Sales of L’Oreal in 2011 (by value) = 0.84 x 0.14 x 120000 = USD 14,112
∴Total size of US market (by value) in 2011 = 69600 + 50400 + 14112 = USD 134112
∴ Percentage growth in market size (b y value) =
Hence, option 3.
Answer the following question based on the information given below.
The pie chart below shows the market share (by value) of different cosmetic companies in the US market in 2010.
The pie chart below shows the market share (by volume) of different cosmetic companies in the US market in 2010.
Number of units sold in 2010 in US = 1500 and value of these units sold = USD 1,20,000.
Q.
What is the ratio of the average un it selling price of an Elle 18 product to the a verage unit selling price of a Nivea product?
AveragB e unit selling price o f an Elle IS Kp roduct =
Average unit selling price o f a Nivea product =
∴ Required ratio = 44 : 54 = 22 : 27
Hence, option 5.
Raika the owner of a boutique, marked each saree 60% above its cost price. What can be the maximum number of successive discounts each being 10%, Mr. Raika can give on each saree such that he doesn’t make any loss? (Assume the cost price of each saree to be the same)
Let the cost price of each saree be Rs. 100. ∴ Marked price = Rs. 160
We can say that,
160(0.9)^{n }≥ 100, where n is the number of successive discounts.
∴ (0.9)^{n }≥ 0.625
0.9^{2 }= 0.81
0.9^{3} = 0.729
0.9^{4} = 0.6561
0. 9^{5} = 0.59 < 0.625
∴ n = 4 Hence, option 3.
A Chinese chessboard is o f the size 6 x 6 . The probability that a square selected at random is of the dimension 2 x 2 or 4 x 4 is:
In a chessboard of dimensions 6x6, Number of squares of dimension 1 x 1 = 6^{2} Number of squares of dimension 2 x 2 = 5^{2} Number of squares of dimension 3 x 3 = 4^{2 }Number of squares of dimension 4 x 4 = 3^{2 }
^{and so on... ∴Total number o f squares =}
^{∴ The required probability is:}
i. e. 0.37
Hence, option 5.
4 white shirts, 2 red shirts and 2 black shirts are to be arranged in a wardrobe in a straight line from left to right. No two shirts of the same colour can be adjacent to each other. Also the shirts at the two extremes cannot be of the same color. In how many ways can the shirts be arranged?
There are 8 shirts out of which 4 are white, 2 are red and 2 are black. No two shirts of the same colour are adjacent.
So, the white shirts should be alternately placed at either the 1^{st}, 3^{rd}, 5^{th} and 7^{th }positions or the 2^{nd}, 4^{th}, 6^{th} and 8^{th} positions from the left.
Thus, the white shirts can be arranged in 2 ways.
Now, the remaining 4 shirts can be placed in the remaining 4 positions in 4! ways. However, there are 2 red and 2 black shirts.
Therefore number of ways of arranging remaining 4 shirts =
∴ Total number o f ways o f arranging the shirts = 2 x 6 = 1 2
Hence, option 1.
What is the remainder when 256^{874!} is divided by 231?
231 = 7 x 11 x3
Similarly, it can be shown that the remainder when 256^{874!} is divided by 11 = 1 and the remainder when 256^{874! }is divided by 3 = 1 ∴ The remainder is of the form:
11/7+1= 3q + 1 = 7r + 1 where p, q, r are integers The smallest value which satisfies all three is 1.
Hence, option 2.
Group Question
Answer the following question based on the information given below.
The table below shows the top four coffee producers in India, in terms of production and sales. All other smaller coffee producers are clubbed under ‘Others’.
Q.
What is the total production capacity of the companies falling under ‘Others’ (in ‘000 3 tonnes)?
Total capacity of coffee industry in India =
Total capacity of Brooke Bond =
Total capacity of Nestle =
Total capacity of Lipton =
Total capacity of MAC =
All the figures above are in ‘000 tonnes.
∴ Total capacity of companies under Others (in '000 tonnes) = 18.92  (3.88 + 3.48 + 3
2.53 + 2.59) = 6.44
Hence, option 4.
Group Question
Answer the following question based on the information given below.
The table below shows the top four coffee producers in India, in terms of production and sales. All other smaller coffee producers are clubbed under ‘Others’.
Q.
Which of these has the minimum unutilised production capacity (in ’000 tonnes)?
Unutilised capacity (in ’000 tonnes) = Total capacity  Actual production
Total capacity of coffee industry in India =
Total capacity of Brooke Bon d =
Total capacity of Nestle =
Total capacity of Lipton =
Total capacity of MAC =
All the figures above are in ‘000 tonnes.
∴ Total capacity of companies under Others (in '000 tonnes) = 18.92  (3.88 + 3.48 + 3
2.53 + 2.59) = 6.44
Actual production of Others (in ’000 tonnes) = 11.6 (2.97 + 2.48 + 1.64 + 1.54) = 2.97
Hence, the unutilised capacity for each company is:
Lipton: 2.53  1.64 = 0.89
Nestle: 3.48  2.48 = 1
Brooke Bond: 3.88  2.97 = 0.91
MAC: 2.59 1.54= 1.05
Others: 6.44  2.97 = 3.47
Thus, the minimum unutilised production capacity is for Lipton.
Hence, option 1.
Group Question
Answer the following question based on the information given below.
The table below shows the top four coffee producers in India, in terms of production and sales. All other smaller coffee producers are clubbed under ‘Others’.
Q.
Which is the most expensive coffee in India (in Rs. per tonne)?
The cost of coffee (in Rs. per tonne) can be found for the top four producers.
However, within Others, there may exist some producer that sells coffee at a higher rate.
Hence, the most expensive coffee cannot be identified.
Hence, option 5.
Group Question
Answer the following question based on the information given below.
The table below shows the top four coffee producers in India, in terms of production and sales. All other smaller coffee producers are clubbed under ‘Others’.
Q.
A producer “CoffeeBean” contributes to 25% of the sales volume and 20% of the sales 3 value of Others? At what price (in Rs. per tonne), does CoffeeBean sell coffee?
Sales volume of Others (in ‘000 tonnes) = 10.67  (2.55 + 2.03 + 1.26 + 1.47) = 3.36
Sales value of Others (in Rs. Cr.) = 132.8  (31.15+ 26.75 + 15.25 + 17.45) = 42.2
CoffeeBean contributes to 25% of the sales volume and 20% of the sales value.
∴ Cost of CoffeeBean (in Rs. pertonne ) =
Hence, option 2.
In the figure below, ΔPQR is an equilateral triangle of side 7 cm and radius of the circle is 3.5 cm. What is the ratio of area of the circle to that of the shaded region?
Length of the diameter of the circle = 2 x 3.5 = 7 cm
Since the chord of maximum length is the diameter, QR passes through the centre of the circle.
Let O be the centre of the circle and the triangle intersects the circle at the points S and T.
Draw OS and OT.
In ΔQOS,
OQ = OS and m∠OQS = 60°
∴ ΔQOS is an equilateral triangle and OQ = 3.5 cm
Similarly, we can prove that ΔSOT and ΔTOR are equilateral triangles and since OQ = OS = OT = OR = 3.5 cm, ΔQOS, ASOT and ATOR are congruent.
A(n QRTS) = 3 x A(ΔOQS)
Area of shaded region = Area of semicircle  A(ΔQRTS)
Area of the circle
Hence, option 1
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