Quantitative Aptitude - Arithmetic Ability Questions

Average = total runs / no.of innings = 32 

So, total = Average x no.of innings = 32 x 10 = 320.

Now increase in avg = 4runs. So, new avg = 32+4 = 36runs 

Total runs = new avg x new no. of innings = 36 x 11 = 396 

Runs made in the 11th inning = 396 - 320 = 76

With close observation, you will note that each number in the list is in the middle of two prime numbers. Thus:

4 is in the middle of 3 and 5, 6 is in the middle of 5 and 7, 12 is in the middle of 11 and 13, 18 is in the middle of 17 and 19, 30 is in the middle of 29 and 31. 42 is in the middle of 41 and 43, 60 is in the middle of 59 and 61.

Therefore, the next number would be the one that is in the middle of the next two prime numbers, which is 72 (which is in the middle of 71 and 73).

Let the son's present age be x years .Then, (38 - x) = x => x= 19. 

Son's age 5 years back = (19 - 5) = 14 years

On 31st December, 2005 it was Saturday. 

Number of odd days from 2006 to 2009 = (1 + 1 + 2 + 1) = 5 days. 

On 31st December 2009, it was Thursday.

Thus, on 1st Jan, 2010 it is Friday

We have n(s) =52C2 52 = 52*51/2*1= 1326. 

Let A = event of getting both black cards 

     B = event of getting both queens 

A∩B = event of getting queen of black cards 

n(A) = 52*512*1 = 26C2 = 325, n(B)= 26*252*1= 4*3/2*1= 6  and  n(A∩B) = 4C2 = 1 

P(A) = n(A)/n(S) = 325/1326;

P(B) = n(B)/n(S) = 6/1326 and 

P(A∩B) = n(A∩B)/n(S) = 1/1326 

P(A∪B) = P(A) + P(B) - P(A∩B) = (325+6-1) / 1326 = 330/1326 = 55/221

Given number series follows a pattern that,

2

2 x 1.5 + 1 = 4

4 x 1.5 + 1 = 7 (not equal to 8)

7 x 1.5 + 1 = 11.5

11.5 x 1.5 + 1 = 18.25

18.25 x 1.5 + 1 = 28.375

Hence, the odd number in the given number series is 8.

Terms at odd places are 5, 6, 7, 8 etc. and each term at even place is 16.

So, 9 is wrong.