Quantitative Aptitude - Arithmetic Ability Questions

52.5% of 800 + 30.5% of 2800 = ? + 87.30

420 + 854 - 87.30 = ?

1274 - 87.30 = ?

? = 1186.70

3 pairs of gloves = 2 x 54 = 108 ; 1 pairs of gloves = 36
3 pairs of leg pads = 2 x 36 = 1 pairs of leg pads = 24
3 cricket balls = 2 x 24 = 48
cost of 1 cricket ball = 16;

Here the given series 7, 11, 19, 35, ?

follows a pattern that (x 2 - 3) i.e,

7

7 x 2 - 3 = 11

11 x 2 - 3 = 19

19 x 2 - 3 = 35

35 x 2 - 3 = 67

67 x 2 - 3 = 131

Hence the next number in the given number series is 67.

After solving we get,

62/931.

None of the options satisfy.

In each number except 383, the product of first and third digits is the middle one.

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

The given series follows the pattern that,

18, 97, 26, 91, ?, 83

=> 18 + 8 = 26

=> 97 - 8 = 91

=> 26 + 8 = 34

=> 91 - 8 = 83

Hence, the missing number is 34.

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