Quantitative Aptitude - Arithmetic Ability Questions

The given number series follows the pattern that,

24×0 + 4  = 4  

4×1  + 9   = 13

13×2 + 16 = 42

42×3 + 25  = 151

151×4 + 36 =  640

Therefore, the odd number in the given series is 41

The series follows the following pattern

5 X 5 – 5 = 20

20 X 4 – 4 = 76

76 X 3 – 3 = 225

225 X 2 – 2 = 448

Hence, the term 225 denoted by (d) completes the series

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.

With close observation, you will note that each number in the list is in the middle of two prime numbers. Thus:

4 is in the middle of 3 and 5, 6 is in the middle of 5 and 7, 12 is in the middle of 11 and 13, 18 is in the middle of 17 and 19, 30 is in the middle of 29 and 31. 42 is in the middle of 41 and 43, 60 is in the middle of 59 and 61.

Therefore, the next number would be the one that is in the middle of the next two prime numbers, which is 72 (which is in the middle of 71 and 73).

At 2 o'clock, the hour hand is at 2 and the minute hand is at 12, i.e. they are 10 min spaces apart. 

To be together, the minute hand must gain 10 minutes over the hour hand. 

Now, 55 minutes are gained by it in 60 min.

  10 minutes will be gained in 6055×10 min. = 101011 min. 

The hands will coincide at 101011 min. past 2.

Let the numbers be x, x + 1, x + 2,  x + 3, x + 4, x + 5 and x + 6,

Then   (x + (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) + (x + 6)) / 7 = 20.

or  7x + 21 = 140 or 7x = 119 or x =17.

Latest number = x + 6 = 23.

 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

Clearly, n(S) = (6 x 6) = 36.

Let E = Event that the sum is a prime number.

Then E= { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4,3),(5, 2), (5, 6), (6, 1), (6, 5) }

n(E) = 15.

P(E) = n(E)/n(S) = 15/36 = 5/12.