Quantitative Aptitude - Arithmetic Ability Questions

Here,P(A)=4/9=p/(p+q)

P(odd against event A)=q/p=5/4

Total coins 30

In that,

1 rupee coins 20

50 paise coins 10

Probability of total 1 rupee coins =  20C11

Probability that 11 coins are picked = 30C11

Required probability of a coin now picked from the box is 1 rupee = 20C11/30C11 = 2/3.

Number of shares = 14400120=120.

Face value = Rs.(100 x 120) = Rs.12000.

Annual income = Rs (5/100*12000) = Rs. 600

One-quarter of one-seventh = 1/4 x 1/7
= 1/28
Now, 1/28 of a land costs = Rs. 30,000
Thus, 8/35 of the land will cost = 30,000 x 28 x 4/15
= Rs. 2,24,000.

Total of results = 50 x 9 = 450

Total of first four results = 52 x 4 = 208

Total of Last four results = 49 x 4 = 196

Hence the fifth result = 450 - (208 + 196) = 46

Given Five boxes of bananas sell for Rs. 30.

=> 1 Box of Bananas for = 30/5 = Rs. 6

Then, for Rs. 9

=> 9/6 = 3/2 = 1.5

Hence, for Rs. 9, 1.5 box of bananas can buy.

B runs around the track in 10 min.

 i.e ,    Speed of B = 10 min per round

 Therefore,  A beats B by 1 round

  Time taken  by A to complete 4 rounds 

  = Time taken by B to complete 3 rounds  

 = 30 min 

Therefore, A's speed = 30/4 min per round = 7.5 min per round

 Hence, if the race is only of one round A's time over the course = 7 min 30 sec

A person can be chosen out of 18 people in 18 ways to be seated between Musharraf and Manmohan. Now consider Musharraf, Manmohan, and the third person, sitting between them, as a single personality, we can arrange them in 17! ways but Musharraf and Manmohan can also be arranged in 2 ways. 

Required number of permutations = 18 x (17!) x 2 = 2 x 18!