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

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.

Apply a3 + b3 + c3 - 3abca2 + b2 + c2 - ab - bc - ca = a + b + c

Hence the answer is  1.5 + 4.7 + 3.8 = 10.

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

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

Given number series is 12  9  13.5  30.3  ?  341.7

Here the pattern followed by the series is

12 

12 x 0.75 = 9

9 x 1.5 = 13.5

13.5 x 2.25 = 30.3

30.37 x 3 = 91.12

91.12 x 3.75 = 341.71

Hence, the missing number in the sequence is 91.12.

Given 373+353+283-3x37x35x28372+352+282-37x35-35x28-37x28

It is in the form of

 a3+b3+c3-3abca2+b2+c2-ab-bc-ca= a + b + c

Here a = 37, b = 35 & c = 28 

=> a + b + c = 37 + 35 + 28 = 100

Therefore,   = 100