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If the average of 10, 12, n, and n is greater than 25, what
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24 Nov 2017, 20:48
24 Nov 2017, 20:48
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If the average of 10, 12, n, and n is greater than 25, what is the least possible value of integer n ?
A. 38
B. 39
C. 40
D. 41
E. 42
A. 38
B. 39
C. 40
D. 41
E. 42
Drill 1
Question: 12
Page: 523
Question: 12
Page: 523
Re: If the average of 10, 12, n, and n is greater than 25, what
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02 Dec 2017, 11:04
02 Dec 2017, 11:04
sandy wrote:
If the average of 10, 12, n, and n is greater than 25, what is the least possible value of integer n ?
A. 38
B. 39
C. 40
D. 41
E. 42
A. 38
B. 39
C. 40
D. 41
E. 42
Drill 1
Question: 12
Page: 523
Question: 12
Page: 523
Now we need to find the least value of n
let us take the average =
\(\frac{(10 + 12 + n + n)}{4}\) = 25
or 22 + 2n = 100
or n = 39 . Now n cannot be 39 because the average of all these 4 nos > 25
so the least value should be 40
Re: If the average of 10, 12, n, and n is greater than 25, what
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10 Dec 2017, 17:14
10 Dec 2017, 17:14
Expert Reply
Explaination
To find the average, divide the total by the number of values. So,\(\frac{10+12+n+n}{4}> 25.\)
Multiply both sides of the inequality by 4 and then subtract 22 (10 + 12) to find 2n > 78. Divide by 2 to find n > 39. So, n is NOT 39; it is the least integer greater than 39, which is 40.
Alternatively, you could plug in the answers starting with choice (A) because the question asks for the least possible value. If n = 38 or 39, then the average is not greater than 25. If n = 40, the average is greater than 25.
The answer is choice C.
To find the average, divide the total by the number of values. So,\(\frac{10+12+n+n}{4}> 25.\)
Multiply both sides of the inequality by 4 and then subtract 22 (10 + 12) to find 2n > 78. Divide by 2 to find n > 39. So, n is NOT 39; it is the least integer greater than 39, which is 40.
Alternatively, you could plug in the answers starting with choice (A) because the question asks for the least possible value. If n = 38 or 39, then the average is not greater than 25. If n = 40, the average is greater than 25.
The answer is choice C.
