IIFT 2013 - SOLVED PAPERS - How to Prepare for Quantitative Aptitude for CAT

How to Prepare for Quantitative Aptitude for CAT (2014)

SOLVED PAPERS

IIFT 2013

1.

Suppose there are 4 bags. Bag 1 contains 1 black and a2 – 6a + 9 red balls, bag 2 contains 3 black and a2 – 6a + 7 red balls, bag 3 contains 5 black and a2 – 6a + 5 red balls and bag 4 contains 7 black and a2 – 6a + 3 red balls. A ball is drawn at random from a randomly chosen bag. The maximum value of probability that the selected ball is black, is

(a) 16/a2 – 6a + 10

(b) 20/ a2 – 6a + 10

(c) 1/16

(d) None of the above

2.

If the product of the integers a, b, c and d is 3094 and if 1 < a < b < c < d, what is the product of b and c?

(a) 26

(b) 91

(c) 133

(d) 221

3.

Mrs. Sonia buys ` 249.00 worth of candies for the children of a school. For each girl she gets a strawberry flavoured candy priced at ` 3.30 per candy; each boy receives a chocolate flavoured candy priced at ` 2.90 per candy. How many candies of each type did she buy?

(a) 21, 57

(b) 57, 21

(c) 37, 51

(d) 27, 51

4.

There is a triangular building (ABC) located in the heart of Jaipur, the Pink City. The length of the wall in east (BC) direction is 397 feet. If the length of south wall (AB) is perfect cube, the length of southwest wall (AC) is a power of three, and the length of wall in southwest (AC) is thrice the length of side AB, determine the perimeter of this triangular building.

(a) 3209 feet

(b) 3213 feet

(c) 3773 feet

(d) 3313 feet

5.

Out of 8 consonants and 5 vowels how many words can be made, each containing 4 consonants and 3 vowels?

(a) 700

(b) 504000

(c) 3528000

(d) 7056000

6.

If x2 + 3x – 10 is a factor of 3x4 + 2x3ax2 + bx a + b – 4, then the closest approximate values of a and b are

(a) 25, 43

(b) 52, 43

(c) 52, 67

(d) None of the above

7.

If the product of n positive integers is n n, then their sum is

(a) A negative integer

(b) Equal to n

(c) Equal to n + 1/n

(d) Never less than n 2

8.

A tennis ball is initially dropped from a height of 180 m. After striking the ground, it rebounds (3/5)th of the height from which it has fallen. The total distance that the ball travels before it comes to rest is

(a) 540 m

(c) 600 m

(c) 720 m

(d) 900 m

9.

In a sports meet for senior citizens organised by the Rotary Club in Kolkata, 9 married couples participated in table tennis mixed double event. The number of ways in which the mixed double team can be made, so that no husband and wife play in the same set, is

(a) 1512

(b) 1240

(c) 960

(d) 640

10.

Two trains P and Q are scheduled to reach New Delhi railway station at 10.00 AM. The probability that train P and train Q will be late is 7/9 and 11/27 respectively. The probability that train Q will be late, given that train Pis late, is 8/9. Then the probability that neither train will be late on a particular day is

(a) 40/81

(b) 41/81

(c) 77/81

(d) 77/243

11.

A survey was conducted to test relative aptitudes in quantitative and logical reasoning of MBA applicants. It is perceived (prior to the survey) that 80 percent of MBA applicants are extremely good in logical reasoning, while only 20 percent are extremely good in quantitative aptitude. Further, it is believed that those with strong quantitative knowledge are also sound in data interpretation, with conditional probability as high as 0.87. However, some MBA applicants who are extremely good in logical reasoning can be also good in data interpretation, with conditional probability 0.15. An applicant surveyed is found to be strong in data interpretation. The probability that the applicant is also strong in quantitative aptitude is

(a) 0.4

(b) 0.6

(c) 0.8

(d) 0.9

12.

Your friend’s cap is in the shape of a right circular cone of base radius 14 cm and height 26.5 cm. The approximate area of the sheet required to make 7 such caps is

(a) 6750 sq cm

(b) 7280 sq cm

(c) 8860 sq cm

(d) 9240 sq cm

13.

In an engineering college there is a rectangular garden of dimensions 34 m by 21 m. Two mutually perpendicular walking corridors of 4 m width have been made in the central part and flowers have been grown in the rest of the garden. The area under the flowers is

(a) 320 sq m

(b) 400 sq m

(c) 510 sq m

(d) 630 sq m

14.

If decreasing 70 by X percent yields the same result as increasing 60 by X percent, then X percent of 50 is

(a) 3.84

(b) 4.82

(c) 7.10

(d) The data is insufficient to answer the question

15.

A rod is cut into 3 equal parts. The resulting portions are then cut into 12, 18 and 32 equal parts, respectively. If each of the resulting portions have integer length, the minimum length of the rod is

(a) 6912 units

(b) 864 units

(c) 288 units

(d) 240 units

16.

If log10x – log10 = 6 logx10 then the value of x is

(a) 10

(b) 30

(c) 100

(d) 1000

17.

A mother along with her two sons is entrusted with the task of cooking biryani for a family get-together. It takes 30 minutes for all three of them cooking together to complete 50 percent of the task. The cooking can also be completed if the two sons start cooking together and the elder son leaves after 1 hour and the younger son cooks for further 3 hours. If the mother needs 1 hour less than the elder son to complete the cooking, how much cooking does the mother complete in an hour?

(a) 33.33%

(b) 50%

(c) 66.67%

(d) None of the above

18.

It was a rainy morning in Delhi when Rohit drove his mother to a dentist in his Maruti Alto. They started at 8.30 AM from home and Rohit maintained the speed of the vehicle at 30 km/hr. However, while returning from the doctor’s chamber, rain intensified and the vehicle could not move due to severe water logging. With no other alternative, Rohit kept the vehicle outside the doctor’s chamber and returned home along with his mother in a rickshaw at a speed of 12 km/hr. They reached home at 1.30 PM. If they stayed at the doctor’s chamber for the dental check-up for 48 minutes, the distance of the doctor’s chamber from Rohit’s house is

(a) 15 km

(b) 30 km

(c) 36 km

(d) 45 km

19.

Two alloys of aluminium have different percentages of aluminium in them. The first one weighs 8 kg and the second one weighs 16 kg. One piece each of equal weight was cut off from both the alloys and first piece was alloyed with the second alloy and the second piece alloyed with the first one. As a result, the percentage of aluminium became the same in the resulting two new alloys. What was the weight of each cut-off piece?

(a) 3.33 kg

(b) 4.67 kg

(c) 5.33 kg

(d) None of the above

20.

Three years ago, your close friend had won a lottery of ` 1 crore. He purchased a flat for ` 40 lakhs, a car for ` 20 lakhs and shares worth ` 10 lakhs. He put the remaining money in a bank deposit that pays compound interest @ 12 percent per annum. If today, he sells off the flat, the car and the shares at certain percentage of their original value and withdraws his entire money from the bank, the total gain in his assets is 5%. The closest approximate percentage of the original value at which he sold off the three items is

(a) 60 percent

(b) 75 percent

(c) 90 percent

(d) 105 percent

21.

If log13log21 = 0, then the value of x is

(a) 21

(b) 13

(c) 81

(d) None of the above

22.

If x is real, the smallest value of the expression

3x2 – 4x + 7 is :

(a) 2/3

(b) 3/4

(c) 7/9

(d) None of the above

23.

The average of 7 consecutive numbers is P. If the next three numbers are also added, the average shall

(a) Remain unchanged

(b) Increase by 1

(c) Increase by 1.5

(d) Increase by 2

24.

The duration of the journey from your home to the College in the local train varies directly as the distance and inversely as the velocity. The velocity varies directly as the square root of the diesel used per km., and inversely as the number of carriages in the train. If, in a journey of 70 km. in 45 minutes with 15 carriages, 10 litres of diesel is required, then the diesel that will be consumed in a journey of 50 km in half an hour with 18 carriages is

(a) 2.9 litres

(b) 11.8 litres

(c) 15.7 litres

(d) None of the above

25.

The capacity of tap Y is 60% more than that of X. If both the taps are opened simultaneously, they take 40 hours to fill the rank. The time taken by Y alone to fill the tank is

(a) 60 hours

(b) 65 hours

(c) 70 hours

(d) 75 hours

ANSWER KEY

1. (d)

2. (b)

3. (b)

4. (d)

5. (c)

6. (c)

7. (d)

8. (c)

9. (a)

10. (b)

11. (b)

12. (d)

13. (c)

14. (d)

15. (c)

16. (d)

17. (b)

18. (c)

19. (c)

20. (c)

21. (d)

22. (d)

23. (c)

24. (b)

25. (b)

Solutions

1.The number of balls every bag is a2 – 6a + 10. The probability of picking up a black ball would be defined as:

Probability of picking up bag 1 × Probability of picking up a black ball + Probability of picking up bag 2 × Probability of picking up a black ball + Probability of picking up bag 3 × Probability of picking up a black ball + Probability of picking up bag 4 × Probability of picking up a black ball

= ¼ × [1/(a2 – 6a + 10)] + ¼ × [3/( a2 – 6a +10)] + ¼ × [5/( a2 – 6a + 10)] + ¼ × [7/( a2 – 6a + 10)] = ¼ [16/( a2 – 6a + 10)] = [4/( a2 – 6a + 10)].

Hence, Option (d) is correct.

2.The key in this question is to look for factor pairs of 3094. If we do a prime factor search of 3094, we get 3094 = 2 × 7 × 13 × 17. These are clearly the values of a, b, c and d in that order. Thus, the value of the product of band c is 13 × 7 =91. Hence, Option (b) is correct.

3.Solve this one through options. If you use Option (a) you get:

21 × 3.3 + 57 × 2.9 = 234.6 π 249.

Option (b) gives us: 57 × 3.3 + 21 × 2.9 = 249.

Hence, Option (b) is correct.

4.The perimeter P = AB + BC +AC.

Of these BC = 397, AC = 3n and AB = a3.

Thus, P = 397 + 3n + a3

We also know that AC = 3 × AB

3n = 3 × a3 Æ

3n–1 = a3.

So, P = 397 + 3n + 3n–1

Æ (P – 397) = 3n–1 × (3 + 1) = 3n–1 × 4

This means that (P – 397) should be a multiple of 3 and 4 simultaneously.

Checking the options, only Option (d) meets the criteria because if we put P = 3313 we get P – 397 = 2916 which is a multiple of 3 and 4. The other options do not satisfy this requirement and hence Option (d) is the correct answer.

5.We would need to select 4 consonants and 3 vowels and arrange them. This can be done in 8C4 × 5C3 × 7! = 70 × 10 × 5040 = 3528000 words.

6.The equation x2 + 3x – 10 has two factors 2 and –5. Putting x = 2 in the expression 3x4 + 2x3ax2 + bx a + b – 4, we get: 48 + 16 – 4a + 2ba + b – 4 = 0 Æ 5a – 3b = 0...(i)

We also have: x = –5.

Putting x = 2 in the expression 3x4 + 2x3ax2 + bx a + b – 4, we get: 26a + 4b = 1621...(ii)

Solving (i) and (ii) simultaneously we get a = 52 and b = 67. Option (c) is correct.

7.Since we are talking about the calculation of the sum of n positive integers, it cannot be negative or a fraction. Thus, Options (a) and (c) are ruled out. Further, Option (b) can be rejected by taking a sample value of n as 4 – it gives us nn as 256. A possible value of the 4 integers could be: 1, 2, 16 and 8. Obviously their sum is not equal to the value of n and hence we can also reject Option (b). This leaves us only with Option (d) as a possible answer. Hence, Option (d) is correct.

8.The total distance traveled would be the infinite sum of the ‘falling down’ distance plus the infinite sum of the ‘bouncing up’ distance traveled by the ball. The answer would be got by:

180 ÷ (1 – 3/5) + 108 ÷ (1 – 3/5) = 450 + 270 = 720. Hence, Option (c) is correct.

9.The number of ways of selecting 2 men and 2 women (such that there would be no husband wife pair in the same set) would be 9C2 × 7C2. Further, there are two ways in which we can form the two teams for the same set of players. Thus, the correct answer would be: 9C2 × 7C2 × 2 = 36 × 21 × 2 = 1512. Hence, Option (a) is correct.

10.Let the probability of P coming late = P(P); the probability of Q coming late = P(Q). Then:

P(P » Q) = P(P) + P(Q) – P(P « Q)

Where P(P » Q) means the situation where at least one of the two trains P and Q is late and P(P « Q) means the situation where both the trains P and Q are late.

The value of P(P « Q) = 7/9 × 8/9 = 56/81.

This gives us:

P (P » Q) = 7/9 + 11/27 – 56/81 = (63 + 33 – 56)/81 = 40/81.

Thus, the probability that none of the two trains is late = 1 – Probability that at least 1 train is late.

The required answer = 1– 40/81 = 41/81.

Hence, Option (b) is correct.

11.Out of every 100 applicants, 80 are extremely good at Logical Reasoning while 20 are extremely good at Quantitative Aptitude. Further, since the probability of someone who is extremely good at Quantitative Aptitude also being strong at Data Interpretation is 0.87, we can expect 20 × 0.87 = 17.4 people out of 20 to be strong at DI.

Also, since the related probability for someone being strong at DI if he/she is strong at LR is 0.15, we get 80 × 0.15 =12 people who would be expected to be strong at DI (from the 80 who are strong at LR).

Thus, we get a total of 29.4 people in every 100 who would be strong at DI. Naturally, if we find someone who is strong at DI, the probability he/she would also be strong at QA = 17.4/29.4 ª 0.6. Hence, Option (b) is correct.

12.The surface area of the sheet required to make the cap would be equal to the lateral surface area of the cap which is given by the formula p × r × l. Since the base radius is 14 cm and height is 26.5 cm, the value of ‘l’ can be calculated using r2 + h2 = l 2 Æ l = 29.97 ª 30.

Thus, 7 caps would require 7 × p × 14 × 30 = 9240 cm2.

Hence, Option (d) is correct.

13.The area of the garden is 34 × 21 = 714 sq. m. Out of this the area covered by the paths = 4 × 34 + 4 × 21 – 4 × 4 = 204 sq. m. The remaining area being covered by flowers would be equal to: 714 – 204 = 510 sq.m.

14.Solve this one through options. One reading of the question sentence should tell you clearly that there would be a value of ‘X’ that would satisfy this – and hence Option (d) can be rejected. This leaves us to check the first three options. Rather than trying to solve the question by creating complex mathematical processes, it is better to solve this using the options. The following process would help you identify whether an option is correct or not.

For Option (c): X % of 50 = 7.10 implies that the value of X must be 14.2%.

If we now visualise the equation 70 – 14.2% of 70 = 60 + 14.2% of 60, we can clearly realise that the LHS and the RHS of the above equation do not match. Hence, this option can be rejected.

For Option (b): X % of 50 = 4.82 implies that the value of X must be 9.64 %.

If we now visualise the equation 70 – 9.64 % of 70 = 60 + 9.64 % of 60, we can clearly realise that the LHS and the RHS of the above equation do not match. Hence, this option can be rejected.

15.Solve this question using options. The correct option would be the smallest number amongst the given options, whose 1/3rd value would be divisible by 12, 18 as well as by 32. If we try the lowest option first our thought process would go as follows:

Option (d): If rod length = 240, then 1/3rd of the rod length = 80. But 80 is not divisible by 12. Hence, option (d) is wrong.

Option (c): If rod length = 288, then 1/3rd of the rod length = 96. But 96 is not divisible by 18. Hence, option (c) is wrong.

For Option (b), rod length = 864, we get 1/3rd of the rod length = 288 which is a multiple of 12, 18 and 32 simultaneously. Hence, Option (c) is correct.

16.log10x – log10 = 6 logx10 Æ log10 x – ⅓ × log10 x

= 6/(log10 x);

Put log10x = m, gives us:

m – ⅓ × m = 6/m Æ m2 – ⅓ × m2 = 6 Æ m2 = 9 Æ m = ± 3.

We can reject the value of m = –3 because that would involve x to be a decimal value between 0 and 1 (which is ruled out from the options).

Thus, m = 3, which means that log10 x = 3 Æ x =103

= 1000.

Hence, Option (d) is correct.

17.The total work for them is 100% per hour. If we try to go through the options and try Option (b) which is the easiest to check – we get: The mother’s work per hour = 50%. This would mean that the mother requires 2 hours to complete the task. Since it is given that the elder son takes one hour more than his mother to complete the task, it would imply that the elder son’s time taken for the task = 3 hours Æ elder son’s work per hour = 33.33%. Then the younger son’s work per hour would be 100–50–33.33= 16.66% per hour.

With these working rates, we need to confirm whether the last condition mentioned in the problem matches with these working rates for the two sons. The condition states that – “The cooking can also be completed if the two sons start cooking together and the elder son leaves after 1 hour and the younger son cooks for further 3 hours.” If we go by this – we will get that the elder son would do 33.33% of the work in the first hour, while the younger son would cook for 4 hours and do 66.66% of the work – thus completing 100% of the work together (since 33.333% + 66.666% = 100%).

Thus, Option (b) is the correct option.

18.The equation that one needs to solve for this question is:

d/30 + d/12 = 4.2. (4.2 here, means that the total time taken for the travel is 4 hours 12 minutes – since he comes home in exactly 5 hours, which includes the 48 minutes at the doctor’s chamber.)

The value of ‘d’ can then be checked from the options – the best option which fits the value of d = 36. Hence, Option (c) is correct.

19.The main focus in this problem should be on understanding that the cut off piece’s weight should be such that the alloyed pieces that are created should have the same ratio weight wise for each of the two alloys.

Thus, if the weight of the cut off pieces is ‘w’ each, then:

(8–w)/w = w/(16–w).

Again to solve this, it is better to try to use the options given in the question to see which one fits the above equation.

Using Option (c), we can see that if we take w = 5.33, we get:

2.66/5.33 = 5.33/10.66 = ½.

Hence, Option (c) is correct.

20.30 lakh invested @ 12% per annum compounded, would approximately become 42 lakh in three years. Since, he gets back 1.05 crore – his return on his remaining 70 lakh would be equal to 1.05 crore – 42 lakh ª 63 lakh – which is around 90% of the value of his investment of 70 lakh. Thus, Option (c) is the correct answer.

21.log13log21 = 0 implies:

log21 = 130 = 1 implies:

= 211

The value of x can be visualised to be 100 because at x = 100 , LHS = RHS.

Hence, Option (d) is correct.

22.The given expression would have its minima at the value of x got by differentiating the quadratic expression with respect to x and equating the resultant linear expression to 0. We get 6x–4 = 0, which means that the minima would occur at x = 2/3.

The minimum value would be got by putting x = 2/3 in the given expression.

We get, minimum value = 12/9 – 8/3 + 7 = 17/3.

Hence, Option (d) is correct.

23.It can be experimentally verified by taking the values of the 7 integers as 1, 2, 3, 4, 5, 6 and 7 to get an average of 4, while the average of the group if the next three numbers are included would become 5.5 (the group of numbers would be 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10).

Thus, the average increases by 1.5.

Hence, Option (c) is correct.

24.Let Distance of the journey = D, Diesel used per km = A and the number of carriages = N. Then we have Time, T =

From the first set of values given in the question, we have T = 45, D = 70, N = 15 and A = 1/7. Putting these values in the equation we get: K =

The equation then becomes, T = .

In the second case, D = 50 km, N = 18 and T = 30 minutes.

Solving for A, we get the value of A = 11.8.

Hence, Option (b) is correct.

25.Let the work done by Tap X in an hour be w. Then the work done by Tap Y in an hour would be 1.6 w. Together, they would do 2.6 w work in an hour. Thus, the total work in 40 hours = 40 × 2.6 w.

This amount of work would be done by Y alone in hours = 65 hours.

Hence, Option (b) is correct.