Daily Practice Questions Questions for CAT with Answers PDF

Treat the 5 Engineering books as 1 unit
Treat the 4 Mathematics books as 1 unit
Treat the 2 Physics books as 1 unit
Then there are 3 objects, which can be arranged in 3! ways, but amongst each subject, the books can be arranged in 5!4!2! ways, so
Required Probability

Total odd numbers between 1 to 25 = 13
Hence, probability = 13/25

Total number of possible cases = 36
Favourable events = (4, 6) (5, 5) (6, 4) (5, 6) (6, 5) (6, 6) = 6
Required probability = 6/36 = 1/6

Suppose, X denotes the number of coins showing heads up.
Then, X is a binomial variate with n = 100 and probability of success p.
We have, P (X = 51) = P (X = 50)
¹⁰⁰C₅₁ (p)⁵¹ (1 - p)⁴⁹ = ¹⁰⁰C₅₀ (p)⁵⁰ (1 - p)⁵⁰   

50p = 51 - 51p
101p = 51
p = 51/101

Number of Nestle chocolates = 5
Number of Dairy Milk chocolates = 18
Total number of chocolates = 18 + 5 = 23
Probability that a picked up chocolate is Nestle = 5/23

Suppose wholesale price = Rs. 100
Since the trader buys at Rs. 90 and sells at Rs. 110,
Therefore, the required profit percentage = [(110 - 90)/90] × 100 = 22.22%

Five boys and three girls can be seated on 8 chairs in 8! different ways.
_ B _ B _ B _ B _ B _
As shown above, 5 boys can be seated at 5 places in 5! ways. After this, 3 girls can be seated in the remaining 6 places in ⁶P₃ ways.
Favourable outcomes = 5! × ⁶P₃
Probability = (5!/3!) × (6!/8!) = 5/14

P(E) = 30/100 = 3/10, P(H) = 20/100 = 1/5, P(E ∩ H) = 10/100 = 1/10.
P(English or Hindi) = P(E ∪ H) = P(E) + P(H) – P(E ∩ H) = 4/10 = 2/5.

Sample space is 21.
Even numbers between 0 and 20 are {2, 4, 6, 8, 10, 12, 14, 16, 18}
Number of favourable cases = 9
Hence, the probability is 9/21 = 3/7.

S = sample space. n(S) = 36.
E = event of getting two numbers whose sum is odd = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5)}.
n(E) = 18.

P(E) = n(E)/n(S) = 18/36 = 1/2.