If the average of 10, 12, n, and n is greater than 25, what is the lea

Question

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

PKN · Answer

Given 22+2n>100
 Or,2n>78
 Or,n>39
So the least possible value of n for which the average of 4 numbers is greater than 25 is 40.
Ans.(C)

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sudarshan22 · Answer

+1 for   C  .

(10+12+N+N)/4 > 25
22 + 2N > 100
2N > 78
N > 39

Since, the least, therefore, N = 40

Hence,   C  .

KSBGC · Answer

Given

(10 + 12 + n+ n) / 4 >25

22 + 2n > 100

2n> 78

n> 39

Here only possible conclusion is that the least value of n is 40.

The best answer is C.

GMATinsight · Answer

Average of 10, 12, n and n = (10+12+n+n)/4 > 25

i.e. 22+2n > 100

i.e. n > 39

i.e. Least possible value of n = 40 (Integer)

Answer: Option C

pandeyashwin · Answer

Avg of 4 no is greater than 25
Sum is >100
10+12 + 2n > 100
N>39

C

MathRevolution · Answer

For the average of '4 numbers to be greater than 25, the sum should be greater than 4 * 25 = 100.

=> 10 + 12 = 22, so remaining 100 - 22 = 78 will make n = 39.

For it to be greater than 100, n = 40

Answer C

BrushMyQuant · Answer

Theory: Mean is the average of the all the numbers in the set. 
 => Mean =  \frac{Sum Of All The Numbers In The Set }{ Total Number Of Numbers In The Set}

We are given that the average of 10, 12, n, and n is greater than 25 and we need to find the least possible value of n

Mean =  \frac{10+12+n+n}{4}  =  \frac{22+2n}{4}  =  \frac{2(11+n)}{4}  =  \frac{11+n}{2} 
Ans we know that Mean > 25
=>  \frac{11+n}{2}  > 25
=> 11+n > 25*2
=> 11+n > 50
=> n > 50-11
=> n > 39
=> Least possible value of n = 40

So, Answer will be C
Hope it helps!

Watch the following video to Learn the Basics of Statistics

https://www.youtube.com/watch?v=Cvy_HHw4KIs

bumpbot · Answer

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If the average of 10, 12, n, and n is greater than 25, what is the lea

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If the average of 10, 12, n, and n is greater than 25, what is the lea

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