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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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Updated on: 24 May 2013, 03:37
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Question Stats:

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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

Originally posted by nafishasan60 on 04 Feb 2012, 03:15.
Last edited by Bunuel on 24 May 2013, 03:37, edited 2 times in total.
Added the OA
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Posts: 64102
Re: Averages – Descriptive Statistics problem solving  [#permalink]

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04 Feb 2012, 03:42
8
1
7
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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If the average of 5 positive integers is 40 and the  [#permalink]

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01 Jul 2014, 02:00
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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12 Aug 2014, 18:59
1
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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02 Dec 2015, 06:25
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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02 Dec 2015, 07:49
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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02 Dec 2015, 08:02
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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05 Oct 2016, 06:37
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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05 Oct 2016, 06:59
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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17 Dec 2016, 20:23
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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Re: If the average of 5 positive integers is 40 and the  [#permalink]

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14 Jun 2018, 12:37
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

Let a, b, c, d, e be the 5 positive integers, where a is smallest & e is largest.

Given average = 40, hence a+b+c+d+e = 5*40 = 200......(i)

& e - a = 10......(ii)

for e to be maximum, we need all other numbers to be as small as possible, hence a=b=c=d

We get from (i) 4a + e = 200.......(iii)

Solving (ii) & (iii), we get a = 38 & hence Maximum value of e = 48

Answer D.

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

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17 Jun 2018, 18:37
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

We are given that we have a sum of 40 x 5 = 200 and a range of 10.

Let’s strategically check our answer choices:

If the largest integer is 50, then the 4 other numbers could be 40. However, 50 + 40 x 4 = 210, which is greater than 200. Thus, we see 50 (and also 52) cannot be correct.

Let’s check 49:

49 + 4 x 39 = 49 + 156 = 205, which is also greater than 200.

Let’s check 48:

48 + 38 x 4 = 48 + 152 = 200, so 48 is the largest number in the set.
Alternate Solution:

Let us denote the greatest element in this set by M and the smallest element in this set by m.

To maximize the greatest integer in the set, we should keep the remaining integers as small as possible. For that purpose, let’s assume all the elements in the set besides M are equal to m. Then, according to the information given in the question, we have

(4m + M)/5 = 40

4m + M = 200

and

M - m = 10

Let’s multiply the last equation by -1 and add to the preceding equation:

5m = 190

m = 38

Substituting m = 38 in M - m = 10, we see that the greatest value of M is 48.

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

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18 Aug 2018, 08:10
In a questions like this one, I like to assign variables to the value the question asks.
So, for example in this question we are looking for "the maximum value possible for the largest of these 5 integers".
Let it be "x".
Then, to maximize "x", we need to minimize all other unknowns. Since the question stem does not say anything about integers being different (which is some times a case), we can minimize 4 other numbers to be (x-10) (Inferred from difference the largest and the smallest being "10").

the first 4 numbers = (x-10)
the fifth number = x
Now, from this formula: total number of items * avg = sum of items
5*40 = 200 = (x-10)*4 + x
200 = 4x-40+x
5x = 240
x = 48
Hence, the answer is D

The main take a way, at least for me, is assigning the variable ("x") to the thing we need to solve for at the end. This way, we can avoid mistakes solving for something different than the question actually asks.
On some of the trickier questions there would be an answer choice 38, that would be intended to punish those who assign x to the smallest number in the set and by the time he/she finishes the calculations, forgets to add extra 10 to find the largest number is the set.
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Re: If the average of 5 positive integers is 40 and the  [#permalink]

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25 Aug 2019, 00:20
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

Given:
1. The average of 5 positive integers is 40.
2. The difference between the largest and the smallest of these 5 numbers is 10.

Asked: What is the maximum value possible for the largest of these 5 integers?

Let the numbers be {s, s, s, s, s+10} since to maximise largest number, all other numbers need to be minimised
4s + s + 10 = 5*40 = 200
5s = 200 - 10 = 190
s = 190/5 = 38
Largest number = s + 10 = 48

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

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25 Aug 2019, 10:46
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 nos ; 5*40 ; 200
range os no ; 10
let smallest be x so largest no ; 10+x
5x+10=200
x=38
largest ; 48
IMO D
Re: If the average of 5 positive integers is 40 and the   [#permalink] 25 Aug 2019, 10:46

# If the average of 5 positive integers is 40 and the

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