# Quantitative Aptitude DI - Dec 08 2020

Here we are providing new series of Quantitative Aptitude Questions for upcoming exams, so the aspirants can practice it on a daily basis.

Study the following data carefully and answer the questions:

In a cloth shop, there are 3 sections X, Y and Z with difference sizes small, medium and large Jeans. The total number of small Jeans in all the sections together is 90.

In section X, the probability of choosing a small Jeans is 1/3, medium Jeans is 1/5 and the number of large Jeans is 20 more than the number of small Jeans.

In section Y, the total number of Jeans is 120 and the sum of small and large Jeans is 2 times the number of medium Jeans. Also, the ratio of probability of choosing small and medium Jeans is 3:4.

In section Z, total jeans are half the total number of Jeans in section Y and the probability of getting large jeans is 1/6 more than probability of getting medium Jeans.

1) What is total number of Jeans in all three sections together?

a) 300

b) 320

c) 350

d) 330

e) None of these

2) What is the difference between the number of medium Jeans from section X and Y together to the number of Large Jeans from section Y and Z together?

a) 15

b) 20

c) 10

d) 5

e) None of these

3)‘K’ number of small Jeans are shifted from section X to section Z and the probability of getting medium Jeans from section Z decreased by 1/12. How many small Jeans are shifted from section X to section Z?

a) 20

b) 10

c) 30

d) 15

e) 25

4) If a customer takes 3/5th of the small Jeans from section X, all the medium Jeans from section Y and 1/3rd of the large Jeans from section Z and placed in his shopping basket. What is the probability of choosing an exactly 3 small size Jeans from shopping basket?

a) 203/4108

b) 223/4180

c) 203/4107

d) 222/4107

e) None of these

5) One Jeans each is selected from Section X, Y and Z. Find the probability that all are small size Jeans.

a) 1/78

b) 1/36

c) 1/72

d) 1/70

e) None of these

Section X:

Let number of small Jeans = S, medium Jeans = M and Large Jeans = L

P(S) = 1/3

P(M) = 1/5

So, P(L) = 1 – 1/3 – 1/5 = 7/15

Given L = 20 + S

Let total number of Jeans in section X is T.

Now, P(S) = S/T = 1/3

And, P(L) = L/T = 7/15

=> (20 + S)/T = 7/15

=> 20/T + S/T = 7/15

=> 20/T = 7/15 – 1/3 = 2/15

=> T = 20 * 15/2 = 150

Thus, S = 150 * 1/3 = 50, M = 150 * 1/5 = 30 and, L = 150 * 7/15 = 70

Section Y:

T = 120

Given, 2M = S + L

So, T = S + M + L = 2M + M = 3M

=> 120 = 3M

=> M = 40

P(M) = M/T = M/3M = 1/3

Also, P(S) : P(M) = 3 : 4

=> P(S) = 3/4 * 1/3 = ¼

Thus, S = 120 * 1/4 = 30

Thus, P(L) = 1 – 1/3 – 1/4 = (12 – 4 – 3)/12 = 5/12

=> L = 120 * 5/12 = 50

Section Z:

Since, the total number of small Jeans in all the sections together are 90.

So, number of small Jeans in section Z = 90 – 50 – 30 = 10

Total Jeans in Section Z = ½ * total number of Jeans in Section Y

=> T = ½ * 120 = 60

P(S) = 10/60 = 1/6

P(L) + P(M) = 1 – 1/6 = 5/6      ----(1)

Given, P(L) = 1/6 + P(M)      ----(2)

Adding equation (1) and (2), we get

2P(L) = 1

=> P(L) = ½

=> L/60 = 1/2

=> L = 30

So, M = T – S – L = 60 – 10 – 30 = 20

So, total number of Jeans = 150 + 120 + 60 = 330

The number of medium Jeans from section X and Y together = 30 + 40 = 70

The number of large Jeans from section Y and Z together = 50 + 30 = 80

So, required difference = 80 – 70 = 10

Probability of getting medium Jeans from section Z was 1/3

Now it reduces by 1/12

So new probability P(M) = 1/3 - 1/12 = 1/4

Now, 20/(60 + K) = 1/4

=> 60 + K = 80

=> K = 20

Small size Jeans = 3/5 * 50 = 30

Medium size Jeans = 40

Large size Jeans = 1/3 * 30 = 10

Total Jeans = 30 + 40 + 10 = 80

Required probability = (30C3)/(80C3) = 203/4108