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# M03-16

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Math Expert
Joined: 02 Sep 2009
Posts: 43830

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15 Sep 2014, 23:20
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Difficulty:

45% (medium)

Question Stats:

70% (01:01) correct 30% (01:04) wrong based on 233 sessions

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The average (arithmetic mean) of 4 different positive integers is 125 and the largest of these integers is 150, what is the least possible value of the smallest of the 4 integers?

A. 1
B. 2
C. 12
D. 53
E. 100
[Reveal] Spoiler: OA

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Math Expert
Joined: 02 Sep 2009
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15 Sep 2014, 23:20
Expert's post
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Official Solution:

The average (arithmetic mean) of 4 different positive integers is 125 and the largest of these integers is 150, what is the least possible value of the smallest of the 4 integers?

A. 1
B. 2
C. 12
D. 53
E. 100

Suppose the 4 integers in ascending order are $$x$$, $$y$$, $$z$$ and 150.

Since the average of 4 different positive integers is 125 then their sum is $$125*4=500$$. So, $$x+y+z+150=500$$, which simplifies to $$x+y+z=350$$.

We want to minimize $$x$$, so we should maximize $$y$$ and $$z$$, since given that all integers are distinct then $$y_{\text{max}}=148$$ and $$z_{\text{max}}=149$$. Therefore $$x+148+149=350$$ and $$x=53$$.

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Joined: 21 Sep 2014
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22 May 2015, 04:19
Why cant be 1 ;

1+2+347=350
Math Expert
Joined: 02 Sep 2009
Posts: 43830

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22 May 2015, 04:24
1
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Expert's post
vikramd wrote:
Why cant be 1 ;

1+2+347=350

Because we are told that the largest of these integers is 150.
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Intern
Joined: 17 Jun 2016
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03 Sep 2016, 12:14
Yes exactly my question.
If we are asked to find the least possible value, then why we are not considering 1?
Manager
Joined: 02 Jun 2015
Posts: 192
Location: Ghana

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03 Sep 2016, 14:02
Notice that each element/integer of the set must be DIFFERENT.

So if we have the largest to be150, we have to maximise the other two elements to 49 & 48 to determine the least possible value of the smallest, which will result in 53 (D) as indicated in official solution given by Bunuel.

Hope this helps.

Sent from my Infinix X509 using GMAT Club Forum mobile app
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Kindly press kudos if you find my post helpful

Director
Joined: 02 Sep 2016
Posts: 786

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08 Jan 2017, 04:25
X is not 1 because:

1) All the integers are different so x,y, and z cannot take 150 as their value.
2) Simple rule to minimize a number is to maximize other numbers. Simple!!

Let's suppose x=1
y=148 (as given)
z=149 (as given in the solution)

1+148+149+150= 448 (But we found x+y+z+150= 500)

Therefore x=53 (as given in the solution)
Intern
Joined: 17 Dec 2016
Posts: 14
Location: United States (NY)
Concentration: Sustainability
GPA: 3.76
WE: Other (Military & Defense)

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27 Jul 2017, 10:16
I think this is a high-quality question and I agree with explanation.
Intern
Joined: 17 Feb 2015
Posts: 1

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31 Aug 2017, 20:43
Yes, but lets take the example wherein the other numbers are 118, 116,115 and 1? In this case, we have the sum of the other four numbers to be 350 and the least value of the smallest integer to be 1?

What am I missing here?
Math Expert
Joined: 02 Sep 2009
Posts: 43830

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31 Aug 2017, 20:58
ttvenkat15 wrote:
Yes, but lets take the example wherein the other numbers are 118, 116,115 and 1? In this case, we have the sum of the other four numbers to be 350 and the least value of the smallest integer to be 1?

What am I missing here?

What is your set? {150, 118, 116, 115, 1}? But there are FOUR numbers in the set including 150, not 5!
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Re: M03-16   [#permalink] 31 Aug 2017, 20:58
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# M03-16

Moderators: chetan2u, Bunuel

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