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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)
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Ans :A
9,10,10,10,11

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Ans :A
9,10,10,10,11

Sent from my Mi 4i using GMAT Club Forum mobile app

It cannot be 9,10,10,10,11 because it has been given that the integers are distinct.
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Hi,

I don't understand how to get the "least" possible value of the greatest. Could someone explain it please.

Thank you!
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shamsm
Hi,

I don't understand how to get the "least" possible value of the greatest. Could someone explain it please.

Thank you!

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


Why 0<a<b<c<d<e?

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I don't understand the word "distinct". In my opinion, this word does not mean "consecutive". So, why do we should calculate

Thus, (e−4)+(e−3)+(e−2)+(e−1)+e=50. ?
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I don't understand the word "distinct".

The word distinct means that the numbers must not be same,

(Say) All the numbers are 2

Or Say, the sequence be 2, 2 , 1, 3 , 2

Thus distinct means that the numbers must not be equal.

AQHER
In my opinion, this word does not mean "consecutive". So, why do we should calculate

Thus, (e−4)+(e−3)+(e−2)+(e−1)+e=50. ?

Since the question stem asks " least possible value of the greatest of the five numbers " , we have assumed that the numbers are consecutive..

Hope this helps !!
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Bunuel
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


because sum (50) is multiple of 5, and integers are distinct, assume consecutive integers
checking options,
A: 4x+6=(50-11) no, x not integer
B: 4x+6=(50-12) yes
x=8
8,9,10,11,12
B
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pushpitkc
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)

Why does the set have to be consecutive numbers? I understand everything up until the necessity to have a consecutive set since the question read distinct numbers.

Thank you in advance.
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Bunuel
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

We can use summation formula; sum= n/2 (2a+(n-1)d). 5/2(2a+ 4 x d) = 50; a+2d= 10. a has to start with 2, as all the number are integers(distinct). If a=1, d will be in fraction.

Possible combination: a=2, d=4; a=4, d=3; a=6, d=2; a=8, d=1.

possible number for the above sequence: (2,6,10,14,18), (4,7,10,13,16), (6, 8,10,12,14) & (8,9,10,11,12). So the least highest is 12.
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Bunuel
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

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.
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