Quantitative Aptitude - Arithmetic Ability Questions

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

From the beginning, the next term comes by adding prime numbers in a sequence of 2, 3, 5, 7, 9, 11, 13... to its previous term. But 165 will not be in the series as it must be replaced by 166 since 153+13 = 166.

The Given number series 3, 6.5, 14, 29, 64, 136 follows a pattern that

3

3 x 2 + 0.5 = 6.5

6.5 x 2 + 1 = 14

14 x 2 + 2 = 30 ≠29

30 x 2 + 4 = 64

64 x 2 + 8 = 136

Thus the wrong number in the series is 29

Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively.
Then,
(2x+4000) / (3x+4000) = 40 / 57
⇒ 57 × (2x + 4000) = 40 × (3x+4000)
⇒ 6x = 68,000
⇒ 3x = 34,000
Sumit's present salary = (3x + 4000) = Rs.(34000 + 4000) = Rs. 38,000

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.

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