John is choosing a number n randomly from all integers from 56 to 150

Question

John is choosing a number n randomly from all integers from 56 to 150 inclusive. What is the probability that the number he chooses will be one where n(n + 1) is divisible by 5?

(A) 1/5

(B) 19/95

(C) 2/5

(D) 19/94

(E) 3/5

Aks111 · Answer

Total number of Integers from 56 to 150 inclusive = 150 - 56 +1 = 95

n(n+1) will be divisible by 5 if either 'n' or 'n+1' is divisible by 5. i.e., n should be multiple of 5 or 4.

We need to find how many numbers from 56 to 150 (inclusive) are multiple of 4, multiple of 5, multiple of both.

Total numbers divisible by 4 = (last multiple in the list/4) - (first multiple in the list/4) + 1 = 148/4 - 56/4 + 1 = 37-14+1 = 24

Similarly, Total numbers divisible by 5 = (last multiple in the list/5) - (first multiple in the list/5) + 1 = 150/5 - 60/5 + 1 = 30-12+1 = 19

So, in the list of integers from 56 to 150, we have 24 numbers divisible by 4 and 19 numbers divisible by 5.

However, we also have numbers that are divisible by both 4 and 5 and they need to be subtracted from the total.
i.e. Total Numbers that are divisible by 20 = 140/20 - 60/20 + 1 = 7-3+1 = 5

Hence, total numbers divisible by 4 or 5 = 24 + 19 - 5 = 38

Probability that the chosen number he will be one where n(n + 1) is divisible by 5 = 38 / 95 = 2/5

Answer: C

suu007 · Answer

total number of integers from 56 to 150= (150-56)+1=95
all sets in this range where n(n+1) are (7x8),(8x9),(9X10),(10x11),(11x22). so total number of sets are =5.
n(n+1) divided by 5 satisfies for following sets (9X10),(10x11). Similarly total number of sets divided by 5 =2.
Probability that the chosen number he will be one where n(n + 1) is divisible by 5 = 2/5

Regor60 · Answer

Write out the first five:

56 57 58 59 60

Observing this series it's clear that a multiple of 5 occurs once in five terms.

To achieve a multiple of 5 with n(n+1) it is necessary to either select the multiple of 5 or the number immediately preceding.

So, there are two ways to achieve the desired result out of every five numbers:

2/5

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samarpan.g28 · Answer

Total ways of selection 150-56+1=95.

Now, if n(n+1) is divisible by 5 then either n is div by 5 or (n+1) is div by 5.

Within that given set of numbers if n is div by 5 then n can be 60,65,70,....,150 i.e. total 19 numbers.
If (n+1) is div by 5 then n can be 59,64,69,....,149 i.e. total 19 numbers.

So total 19+19=38 ways out of 95 ways therefore probability is 38/95=2/5. Option (C) is correct.

John is choosing a number n randomly from all integers from 56 to 150

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John is choosing a number n randomly from all integers from 56 to 150

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