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

Answer: (d)

It is given that last 3 digits are randomly dialled. Then each of the digits can be selected out of 10 digits (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) in 10 ways. Hence the required probability = $(1/{10})^3 = 1/{1000}$

Answer: (d)

Each box can be filled in 2 ways.

Hence total no. of ways = $2^5$ = 32

Blue balls can not be filled in adjacent boxes

Total no. of such cases in which blue is filled in 2 adjacent boxes is

2 blue + 3 blue + 4 blue + 5 blue

= 4 ways ( 12, 23, 34, 45) + 3 ways ( 123, 234, 345) + 2 ways (1234, 2345)+ 1 way

= 10 ways

Hence total cases in which blue balls can not be filled in adjacent boxes = 32 - 10 = 22

Answer: (d)

Probability that the trouser is not black $2/3$

Probability that the shirt is not black = $3/4$

Since, picking of a shirt and a trouser are independent, required probability = $2/3 × 3/4 = 1/2$

Answer: (c)

If a number is divisible by 3, the sum of the digits in it must be a multiple of 3. The sum of the given six numerals is 0 + 1 + 2 + 3 + 4 + 5 = 15. So to make a five digit number divisible by 3 we can either exclude 0 or 3. If 0 is left out, then 5! = 120 number of ways are possible. If 3 is left out, then the number of ways of making a five digit numbers is 4 × 4! = 96, because 0 cannot be placed in the first place from left, as it will give a number of four digits.

Hence, total number = 120 + 96 = 216.

Answer: (c)

Let there be n teams participating in the championship. Then, total no. of matches = $^nC_2$ = 153

or ${n!}/{2!(n - 2)!} = {1/2}(n - 1)$ × n = 153

or $n^2$ – n – 306 = 0

or (n – 18) (n + 17) = 0

or n = 18

[n ≠ -ve]