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If the average of 5 positive integers is 40 and the

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If the average of 5 positive integers is 40 and the [#permalink]

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If the average of 5 positive integers is 40 and the difference between the largest and the smallest of these 5 numbers is 10, what is the maximum value possible for the largest of these 5 integers?

A. 50
B. 52
C. 49
D. 48
E. 44
[Reveal] Spoiler: OA

Last edited by Bunuel on 24 May 2013, 04:37, edited 2 times in total.
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Re: Averages – Descriptive Statistics problem solving [#permalink]

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nafishasan60 wrote:
If the average of 5 positive integers is 40 and the difference between the largest and the smallest of these 5 numbers is 10, what is the maximum value possible for the largest of these 5 integers?
A. 50
B. 52
C. 49
D. 48
E. 44


The average of 5 positive integers is 40 --> the sum of these 5 integers is 5*40=200;
The difference between the largest and the smallest of these 5 numbers is 10 --> let the smallest integer be x and the largest x+10;

Now, we want to maximize x+10.

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

Hence, in order to maximize x+10 we should minimize all other terms, thus make them equal to the smallest integer x. We would have: x+x+x+x+(x+10)=200 --> x=38 --> x+10=48.

Answer: D.

Hope it's clear.
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Re: If the average of 5 positive integers is 40 and the [#permalink]

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New post 30 Jun 2014, 05:39
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If the average of 5 positive integers is 40 and the [#permalink]

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New post 01 Jul 2014, 03:00
Let the 5 integers be as follows:

To have maximum value of a+10, let all others be "a"

a+10 ............ a ............... a .............. a .............. a

Addition = 200

5a + 10 = 200

\(a = \frac{190}{5} = 38\)

a+10 = 48 (Maximum Value)

Answer = D
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Re: If the average of 5 positive integers is 40 and the [#permalink]

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New post 12 Aug 2014, 19:59
My method was,

x + (x+1) + (x+2) + (x+3) + (x+4) / 5 = 40

Thus 5x + 10 = 200
and x = 38

Since we know the range is 10, that is ( x + 4 ) - x = 10

so, ( x + 4) - 38 = 10

(x + 4) = 48

** Take (x + 4) as a whole number.
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Re: If the average of 5 positive integers is 40 and the [#permalink]

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Re: If the average of 5 positive integers is 40 and the [#permalink]

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New post 02 Dec 2015, 07:25
nafishasan60 wrote:
If the average of 5 positive integers is 40 and the difference between the largest and the smallest of these 5 numbers is 10, what is the maximum value possible for the largest of these 5 integers?

A. 50
B. 52
C. 49
D. 48
E. 44


Sum of 5 Integer (a, b, c, d, e) = 5*40 = 200

e - a = 10
i.e. e = a+10

For e to be maximum remaining 4 MUST be as small as possible
Since smallest of 5 numbers is a so to minimize other numbers we can take them equal to the smallest of 5 numbers


i.e. a+a+a+a+(a+10) = 200
i.e. 5a = 190
i.e. a = 38

i.e. Largest e = 38+10 = 48

Answer: option D
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Re: If the average of 5 positive integers is 40 and the [#permalink]

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New post 02 Dec 2015, 08:49
GMATinsight wrote:
nafishasan60 wrote:
If the average of 5 positive integers is 40 and the difference between the largest and the smallest of these 5 numbers is 10, what is the maximum value possible for the largest of these 5 integers?

A. 50
B. 52
C. 49
D. 48
E. 44


Sum of 5 Integer (a, b, c, d, e) = 5*40 = 200

e - a = 10
i.e. e = a+10

For e to be maximum remaining 4 MUST be as small as possible
Since smallest of 5 numbers is a so to minimize other numbers we can take them equal to the smallest of 5 numbers


i.e. a+a+a+a+(a+10) = 200
i.e. 5a = 190
i.e. a = 38

i.e. Largest e = 38+10 = 48

Answer: option D



why cant we solve taking a=e-10
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Re: If the average of 5 positive integers is 40 and the [#permalink]

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New post 02 Dec 2015, 09:02
rahulkashyap wrote:
GMATinsight wrote:
nafishasan60 wrote:
If the average of 5 positive integers is 40 and the difference between the largest and the smallest of these 5 numbers is 10, what is the maximum value possible for the largest of these 5 integers?

A. 50
B. 52
C. 49
D. 48
E. 44


Sum of 5 Integer (a, b, c, d, e) = 5*40 = 200

e - a = 10
i.e. e = a+10

For e to be maximum remaining 4 MUST be as small as possible
Since smallest of 5 numbers is a so to minimize other numbers we can take them equal to the smallest of 5 numbers


i.e. a+a+a+a+(a+10) = 200
i.e. 5a = 190
i.e. a = 38

i.e. Largest e = 38+10 = 48

Answer: option D



why cant we solve taking a=e-10


We can solve it like that too

in that case the equation will look like

(e-10)+(e-10)+(e-10)+(e-10)+e = 200
i.e. 5e - 40 = 200
i.e. 5e = 240
i.e. e = 48

I hope this helps
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Re: If the average of 5 positive integers is 40 and the [#permalink]

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New post 05 Oct 2016, 07:37
I solved it the same way bunuel explained.
sum is 200.
x is the smallest
x+10 is the greatest
to maximize x+10, suppose all numbers are the same
x+x+x+x+x+10=200
5x=190
x=190/5
x=38
x+10=48.
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Re: If the average of 5 positive integers is 40 and the [#permalink]

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New post 05 Oct 2016, 07:59
nafishasan60 wrote:
If the average of 5 positive integers is 40 and the difference between the largest and the smallest of these 5 numbers is 10, what is the maximum value possible for the largest of these 5 integers?

A. 50
B. 52
C. 49
D. 48
E. 44


It average of 5 integers is 40, then their sum is \(40 *5=200\)
Let max value be \(x\) .
Therefore smallest value is \(x -10\)
for one value to be maximum (i.e. x ) , all others have to be minimum (i.e. x- 10 ) .
So, \(4(x -10) + x = 200\)
\(x =48\)
Ans. D
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If the average of 5 positive integers is 40 and the [#permalink]

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New post 17 Dec 2016, 21:23
Excellent Question
Here is what i did in this Question =>

Using =>

\(Mean=\frac{Sum}{#}\)


Sum(5)=40*5=200


Now to maximise the largest quantity,we must minimise all other quantities.
Let a be the smallest element
The series would be -> a,a,a,a,a+10

Hence 5a+10=200
a=38

Helen the largest element = 38+10 = 48

Hence D

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If the average of 5 positive integers is 40 and the   [#permalink] 17 Dec 2016, 21:23
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