If the average (arithmetic mean) of five distinct positive integers is

Question

If the average (arithmetic mean) of five distinct positive integers is 10, what is the least possible value of the greatest of the five numbers?

(A) 11
(B) 12
(C) 24
(D) 40
(E) 46

Bunuel · Accepted Answer

If the average (arithmetic mean) of five  distinct positive integers  is 10, what is the least possible value of the greatest of the five numbers?

(A) 11
(B) 12
(C) 24
(D) 40
(E) 46

Given:  0<a < b < c< d< e  and  a + b + c + d + e = 10*5 .

We want to minimise e.

General rule for such kind of problems, when the sum is fixed:  
to maximize one quantity, minimize the others; 
 to minimize one quantity, maximize the others.

So, here we should maximize a, b, c, and d. Since the integers are distinct, then the max value of d is e - 1, the max value of c is e - 2, the max value of b is e - 3 and the max value of a is e - 4.

Thus,  (e - 4) + (e - 3) + (e - 2) + (e - 1) + e = 50 ;
e = 12.

Answer: B.

14. Min/Max Problems 
 
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Abhishek009 · Answer

a + ( a + 1 ) + ( a + 2 ) + ( a + 3 ) + (a + 4) = 50    ( As the numbers are distinct )

Or,  5a + 10 = 50

Or,  5a = 40

Or,  a = 8

So, The least value of the greatest of the five numbers is 8 + 4 = 12, answer , must be (B) 12

pushpitkc · Answer

Since the average of 5 numbers is 10, the total must be 50.
The numbers, must be consecutive for the greatest number to be least (8,9,10,11,12)

The least possible value of the greatest number(as it has been given that the numbers are distinct) has to be  12(Option B)

ThatDudeKnows · Answer

We can also test answer choices on questions like this.

A) If the greatest is 11, how are we going to get to an average of 10? The highest we could have is 11, 10, 9, 8, 7, which doesn't work. We are close, but (spoiler alert!) need something just a little bigger. Eliminate.
B) Adjusting what we just did for A, now we have 12, 11, 10, 9, 8. Yay!!

Answer choice B.

CruelForm · Answer

Can someone tell me what mistake I'm making?  Bunuel

Let the numbers be a,b,c,d,e where e is the largest of all.
Given a+b+c+d+e = 50 
Also since e is largest we can write : 
a<e
b<e
c<e
d<e
adding inequalities 
 a+b+c+d <4e
50-e<4e
10<e
e=11

Bunuel · Answer

Your inequality only proves that e > 10, so the smallest possible integer candidate is 11. But you still need to check whether e = 11 is actually possible.

If e = 11, the four largest possible distinct integers below it are 10, 9, 8, and 7. Their total with 11 is:

7 + 8 + 9 + 10 + 11 = 45

So you cannot reach 50. Thus,  e = 11 is impossible .

For e = 12:

8 + 9 + 10 + 11 + 12 = 50

So the least possible value is 12.

If the average (arithmetic mean) of five distinct positive integers is

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If the average (arithmetic mean) of five distinct positive integers is

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