 It is currently 26 Mar 2019, 00:48 ### 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

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. # The average (arithmetic mean) of 4 different integers is 75.  Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:
Intern Joined: 03 Jun 2016
Posts: 37
Followers: 0

Kudos [?]: 27 , given: 4

The average (arithmetic mean) of 4 different integers is 75. [#permalink] 00:00

Question Stats: 0% (00:00) correct 100% (00:00) wrong based on 1 sessions
The average (arithmetic mean) of 4 different integers is 75. If the largest integer is 90, what is the least possible value of the smallest integer?

A)1
B)19
C)29
D)30
E)33
[Reveal] Spoiler: OA
Manager Joined: 13 Aug 2016
Posts: 77
GRE 1: Q158 V154 Followers: 0

Kudos [?]: 38 , given: 21

Re: The average (arithmetic mean) of 4 [#permalink]
Suppose four different integers are : a,b,c & d.The average of these four different integers is :

$$Avg$$ = (a+b+c+d)/4 = 75

=>

Sum of these four different integers is : sum = a+b+c+d = 300

The difference between the largest integer and the average is : 90-75 = 15

if largest integer is 90 (suppose d=90) then a+b+c = 300-90 = 210
=> Average of a,b & c is : (a+b+c)/3 = 210/3 = 70

How have you calculated that the least integer should be 33 (option E)?I have tried to calculate but finding it difficult;could you please share the detailed solution? GRE Prep Club Legend  Joined: 07 Jun 2014
Posts: 4857
GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
Followers: 105

Kudos [?]: 1783  , given: 397

Re: The average (arithmetic mean) of 4 [#permalink]
2
KUDOS
Expert's post
Hi phoenixio,

Well the mean is given to be 75. Let the integers be x,y,z, 90.

So we know that:

$$\frac{(x+y+z+90)}{4}$$= 75

$$x+y+z= 210$$

Lets say z is the smallest. So in order for z to be smallest possible x and y should be largest values possible.

Now we also know that $$x,y,z < 90$$ (since 90 is the largest integer) and x,y,z, 90 are "different".

So x = 89 and y= 88 satisfy this criteria:

Hence $$z = 210 - 89 -88 =33$$.

Cheers

PS: Source of the Que ?
_________________

Sandy
If you found this post useful, please let me know by pressing the Kudos Button

Try our free Online GRE Test Intern Joined: 11 Nov 2017
Posts: 4
Followers: 0

Kudos [?]: 6  , given: 0

Re: The average (arithmetic mean) of 4 [#permalink]
2
KUDOS
a + b + c + d = 75*4
let a < b < c < d, so d = 90

a + b + c = 210
b+c = 210 - a => 1
we know that b < 90 and c < 90

b+c < 180

210 - a < 180 substitute from 1

a > 30
GRE Prep Club Legend  Joined: 07 Jun 2014
Posts: 4857
GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
Followers: 105

Kudos [?]: 1783 , given: 397

Re: The average (arithmetic mean) of 4 [#permalink]
Expert's post
181charan wrote:
a + b + c + d = 75*4
let a < b < c < d, so d = 90

a + b + c = 210
b+c = 210 - a => 1
we know that b < 90 and c < 90

b+c < 180

210 - a < 180 substitute from 1

a > 30

Nice!!
Well done!
_________________

Sandy
If you found this post useful, please let me know by pressing the Kudos Button

Try our free Online GRE Test Re: The average (arithmetic mean) of 4   [#permalink] 11 Nov 2017, 11:42
Display posts from previous: Sort by

# The average (arithmetic mean) of 4 different integers is 75.  Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group Kindly note that the GRE® test is a registered trademark of the Educational Testing Service®, and this site has neither been reviewed nor endorsed by ETS®.