Quantitative Aptitude - Arithmetic Ability Questions

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

given

 48?+32?= 320 

=> 6?+4?=40

Now squaring on both sides

=> 36 x ? + 16 x ? + 48 x ? = 1600

=> 100 x ? = 1600

=>  ? = 16

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

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).

Here the given series is 1, 3, 6, 11, 20, 37, ??,

and the logic behind the series is

1 x 2 + 1 = 3

3 x 2 + 0 = 6

6 x 2 + (-1) = 11

11 x 2 + (-2) = 20

20 x 2 + (-3) = 37

37 x 2 + (-4) = 70

Hence, the next number in the given series is 70.

 As per the BODMAS rule, the priority in which the operations should be done is:

Note: Addition and subtraction can be treated on same priority (from left to right) when they are in consecutive order.

(115/5) + 12 × 6 = ? + (64/4) – 35 23 + 72 = ? + 16 – 35 95 = ? - 19 95+19 = ? ? = 114

Let S be the sample space.

Then, n(S) = number of ways of drawing 3 balls out of 15 = 15C3  =15*14*133*2*1= 455.  

Let E = event of getting all the 3 red balls.

 n(E) = 5C3 = 5*42*1 = 10.

 => P(E) = n(E)/n(S) = 10/455 = 2/91.