Quantitative Aptitude - Arithmetic Ability Questions

Given expression = (45 x 750/100) - (25 x 480/100) = (337.50 - 120) = 217.50

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

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

The numbers are 7 x 8, 8 x 9, 9 x 10, 10 x 11, 11 x 12, 12 x 13.

So, 150 is wrong.

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.