Quantitative Aptitude - Arithmetic Ability Questions

Go on multiplying the given numbers by 2, 3, 4, 5, 6.

So, the correct next number is 1440.

Let the numbers be 3x, 4x and 5x.

Then, their L.C.M. = 60x.

So, 60x = 2400 or x = 40.

 The numbers are (3 x 40), (4 x 40) and (5 x 40).

Hence, required H.C.F. = 40.

Given 92613 x 62742241 = ? + 89

Using unit digit method,

On LHS,

1 x 1 x 1 = 1

sqrt of last digit 1 => it must end with either 9 or 1

so now

1 x 1 = 1

1 x 9 = 9

On RHS,

? + 89

=> ? = ...1 - 89 = last digit is 11- 9 = 2

or

=> ? = ...9 - 89 = last digit is 9 - 9 = 0

Here in the options last digit 2 is option C.

There are two series (8, 11, 14, 17, 20) and (7, 12, 17, 22) increasing by 3 and 5 respectively.

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.

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