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21 Mar 2022, 05:17
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C
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Difficulty:
95%
(hard)
Question Stats:
43% (02:44) correct
57%
(02:50)
wrong
based on 84
sessions
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Not Attempted Yet
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
(A) 1/5
(B) 19/95
(C) 2/5
(D) 19/94
(E) 3/5
21 Mar 2022, 15:43
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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
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
06 Nov 2023, 00:00
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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
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
