- Question-1
- Explanation

Jeeva has a story book of 2047 pages. He read 1 page on the first day, 2 pages on the second a to complete the book?

a) 10 days

b) 12 days

c) 11 days

d) 13 days

**Answer**: c) 11 days

Total number fo pages = 2047

On the 1^{st} day, jeeva read 1 page.

On the 2^{nd} day, he read 2 pages.

On the 3^{rd} day, he read 4 pages and so on…….

Let x denote the number of pages read on the xth day.

Therefore, 1 + 2 + 4 + …..+ X = 2047

Then by using the formula, the sum of x numbers in G.P = a(r^{x} – 1)/(r – 1) where a – first term & r – common ratio, we get

1 (2^{x} – 1)/ (2 – 1) = 2047

2^{x} – 1 = 2047

2^{x} = 2048

2^{x} = 2^{11}

X = 11.

So jeeva took 11 days to complete the book.

- Question-2
- Explanation

Find the total number of distinct vehicle numbers that can be formed using two letters followed by two numbers. Letters need to be distinct.

**Answer:** This question comes under permutations and combinations section. Out of 26 alphabets two distinct letters can be chosen in 26p2 ways. Coming to numbers part, there are 10 ways ( any number from 0 to 9 can be chosen) to choose the first digit and similarly another 10 ways to choose the second digit. Hence there are totally 10 * 10 = 100 ways. Combined with letters there are 26p2 * 100 ways = 65000 ways to choose vehicle numbers.

- Question-3
- Explanation

A bacteria doubles itself each single day. It totally takes 15 days for the bacteria to fill a test tube. Find an approximate number of days for the bacteria to fill 1/3 of the jar.

**Answer: ** Though the question looks like a tough one, it is actually a simple question. If it takes 15 days for the bacteria to fill the entire test tube, on the 14^{th} day it would had filled half of the tube, on the 13^{th} day it would have filled a quarter of the tube and so on. Now 1/3^{rd} comes somewhere in middle between quarter and half filling of the tube. Hence the bacteria would need somewhere between 13 to 14 days to fill 1/3^{rd} of the jar.

- Question-4
- Explanation

A leather box contains 8 black balls and 6 white balls. Two draws of three balls each are made, the balls being replaced after the first draw. What is the probability that the balls were black in the first draw and white in the second draw?

a) 70 / 8281

b) 140 / 20449

c) 25/ 5445

d) 35/ 5448

**Answer: **a) 70 / 8281

Total number of balls = 8 + 6 = 14

Total ways of drawing 3 balls, N (S) = 14C3 = 364

No of ways to draw 3 black balls = N (E1) for black balls = 8C3 = 56

Probability of all balls being black = P(E1) = N(E1)/N(S) = 56/ 364 = 14 /91

No of ways to draw 3 white balls = N(E2) for white balls = 6C3 = 20

Probability of all balls being white = P(E2) = 20 / 364 = 5/ 91

Probability of drawing three black balls in first draw and drawing 3 white balls in the second draw = P (E1) * P (E2)

Therefore P(E) = 14 / 91 * 5 / 91 = 70 / 8281

- Question-5
- Explanation

Fourteen persons are sitting around a circular table facing the center. What is the probability that three particular persons sit together?

a) 2/ 9

b) 1 / 13

c) 2 / 13

d) 1 / 26

**Answer**: d) 1 / 26

In a circle of n different persons, the total number of arrangements possible = (n – 1)!

Total number of arrangements = n(S) = (14 – 1)! = 13!

Taking three persons as a unit, total persons = 12 (in 4 units)

Therefore number of ways for these 12 persons to around the circular table = (12 – 1)! = 11!

In any given unit, 3 particular person can sit in 3!. Hence total number of ways that any three person can sit = n(E) = 11! * 3!

Therefore p ( E) = probability of three persons sitting together = n(E) / n(S) = 11! * 3! Divided by 13!

3 * 2 divided by 13 * 12 = 1/26.