Quantitative Aptitude - Arithmetic Ability Questions

In given Series, all the numbers except 54 are multiples of 5.

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

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

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.

From the beginning, the next term comes by adding prime numbers in a sequence of 2, 3, 5, 7, 9, 11, 13... to its previous term. But 165 will not be in the series as it must be replaced by 166 since 153+13 = 166.

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.

1, 1 + 1² = 2, 2 + 2² = 6, 6 + 3² = 15, 15 + 4² = 31, 31 + 5² = 56, 56 + 6² = 92

Last number of given series must be 92 not 91.

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.