advance math mcq pdf Model Questions & Answers, Practice Test for ibps rrb so officer sclae 2 3 single exam 2024

Question :26

A bag contains 2 red, 3 green and 2 blue balls. 2 balls are to be drawn randomly. What is probability that the balls drawn contain no blue ball ?

Answer: (a)

2 balls can be drawn in the following ways 1 red and 1 green or 2 red or 2 green

Required probability = ${^2C_1 × ^3C_1}/{^7C_2} + {^2C_2}/{^7C_2} + {^3C_2}/{^7C_2}$

= $6/{21} + 1/{21} + 3/{21} = {10}/{21}$

Question :27

Total number of four digit odd numbers that can be formed using 0, 1, 2, 3, 5, 7 (using repetition allowed) are

Answer: (d)

Required number of numbers = 5 × 6 × 6 × 4

= 36 × 20 = 720.

Question :28

A fair coin is tossed repeatedly. If the tail appears on first four tosses, then the probability of the head appearing on the fifth toss equals

Answer: (c)

The event that the fifth toss results a head is independent of the event that the first four tosses results tails.

∴ Probability of the required event = 1/2

Question :29

There are three parallel straight lines. Two points, 'A' and 'B', are marked on the first line, points, 'C' and 'D' are marked on the second line; and points, 'E' and 'F', are marked on the third line. Each of these 6 points can move to any position on its respective straight line. Consider the following statements:
1. The maximum number of triangles that can be drawn by joining these points is 18.
2. The minimum number of triangles that can be drawn by joining these points is zero. Which of the statement(s) given above is/are correct?

Answer: (a)

Maximum number of triangles can be formed by selecting 3, 4 or 5 points out of 6 at a time.

So, maximum no. of triangles = $^6C_3 + ^6C_4 + ^6C_5$ which is clearly more than 18. Now, triangles formed will be minimum i.e., zero, when the points will overlap on the same line and all the points are along the same vertical line.

Question :30

An examination paper contains 8 questions of which 4 have 3 possible answers each, 3 have 2 possible answers each and the remaining one question has 5 possible answers. The total number of possible answers to all the questions is:

Answer: (d)

4 questions having 3 answers, each can be answered in $3^4$ ways.

Similarly, we have $2^3 \text"and" 5^1$ ways.

i.e. total possible answers = $3^4 × 2^3× 5^1$ = 3240

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