Quantitative Aptitude - Arithmetic Ability Questions

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

4       30       ?     1440       7195      71940

Here the given series follows a pattern of

4

4 x 10 - 10 = 30

30 x 5 - 5  = 145

145 x 10 - 10 = 1440

1440 x 5 - 5 = 7195

7195 x 10 - 10 = 71940

Hence, the missing number is 145

The terms are alternatively multiplied by 1.5 and divided by 3. However, 18.5 does not satisfy it.

The Sum of the digits in each number, Except 324 is 10.

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

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

 22 = 2 x 11

 54 = 2×33

108 = 22×33

135 = 33×5 

198 = 2×32×11

 ∴ L.C.M = 22×33×5×11=5940

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