Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

 It is currently 20 Jul 2019, 18:00

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

Intern
Joined: 18 Feb 2011
Posts: 38
GPA: 3.91
If the average of 5 positive integers is 40 and the  [#permalink]

### Show Tags

Updated on: 24 May 2013, 04:37
3
14
00:00

Difficulty:

45% (medium)

Question Stats:

70% (02:14) correct 30% (02:32) wrong based on 501 sessions

### HideShow timer Statistics

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, 04:15.
Last edited by Bunuel on 24 May 2013, 04:37, edited 2 times in total.
Math Expert
Joined: 02 Sep 2009
Posts: 56304
Re: Averages – Descriptive Statistics problem solving  [#permalink]

### Show Tags

04 Feb 2012, 04:42
5
6
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.

Hope it's clear.
_________________
##### General Discussion
SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1787
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
If the average of 5 positive integers is 40 and the  [#permalink]

### Show Tags

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

5a + 10 = 200

$$a = \frac{190}{5} = 38$$

a+10 = 48 (Maximum Value)

_________________
Kindly press "+1 Kudos" to appreciate
Intern
Joined: 18 Mar 2014
Posts: 5
Re: If the average of 5 positive integers is 40 and the  [#permalink]

### Show Tags

12 Aug 2014, 19:59
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.
CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2960
Location: India
GMAT: INSIGHT
Schools: Darden '21
WE: Education (Education)
Re: If the average of 5 positive integers is 40 and the  [#permalink]

### Show Tags

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

_________________
Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION
Manager
Joined: 09 Oct 2015
Posts: 237
Re: If the average of 5 positive integers is 40 and the  [#permalink]

### Show Tags

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

why cant we solve taking a=e-10
CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2960
Location: India
GMAT: INSIGHT
Schools: Darden '21
WE: Education (Education)
Re: If the average of 5 positive integers is 40 and the  [#permalink]

### Show Tags

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

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
_________________
Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION
Board of Directors
Joined: 17 Jul 2014
Posts: 2539
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30
GPA: 3.92
WE: General Management (Transportation)
Re: If the average of 5 positive integers is 40 and the  [#permalink]

### Show Tags

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.
_________________
Intern
Joined: 27 Sep 2016
Posts: 3
Re: If the average of 5 positive integers is 40 and the  [#permalink]

### Show Tags

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
Current Student
Joined: 12 Aug 2015
Posts: 2609
Schools: Boston U '20 (M)
GRE 1: Q169 V154
If the average of 5 positive integers is 40 and the  [#permalink]

### Show Tags

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

_________________
Director
Joined: 14 Dec 2017
Posts: 517
Location: India
Re: If the average of 5 positive integers is 40 and the  [#permalink]

### Show Tags

14 Jun 2018, 13:37
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

Thanks,
GyM
_________________
Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2822
Re: If the average of 5 positive integers is 40 and the  [#permalink]

### Show Tags

17 Jun 2018, 19:37
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.

_________________

# Jeffrey Miller

Jeff@TargetTestPrep.com
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Intern
Joined: 12 Sep 2016
Posts: 3
Re: If the average of 5 positive integers is 40 and the  [#permalink]

### Show Tags

18 Aug 2018, 09:10
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

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.
Re: If the average of 5 positive integers is 40 and the   [#permalink] 18 Aug 2018, 09:10
Display posts from previous: Sort by