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M03-16

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Bunuel
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M03-16 [#permalink] New post   16 Sep 2014, 00:20
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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
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D
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Bunuel
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Re M03-16 [#permalink] New post   16 Sep 2014, 00:20
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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.

Given that the average of 4 different positive integers is 125, 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 all integers are distinct, then \(y_{\text{max=148\) and \(z_{\text{max=149\). Therefore, \(x+148+149=350\) and \(x=53\).


Answer: D
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Bunuel
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Re: M03-16 [#permalink] New post   22 May 2015, 05:24
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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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Re: M03-16 [#permalink] New post   22 May 2015, 05:19
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Why cant be 1 ;

1+2+347=350
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Re: M03-16 [#permalink] New post   08 Jan 2017, 05:25
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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)
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Re: M03-16 [#permalink] New post   27 Jul 2017, 11:16
I think this is a high-quality question and I agree with explanation.
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Re: M03-16 [#permalink] New post   15 May 2019, 16:59
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What made this easier to understand how to solve in review is putting each variable up against the constraints:
- that each number is distinct
- that each number is less than 150

Thus
X + Y + Z + 150 = 500
X + Y + Z = 350

We can then treat the variables as x < y < z < 150 to maximise each

Test values as per Bunuel's solution: x< 148 < 149 for x<y<z
148 + 149 = 297, so is this a possible value? (stepping through my thinking)

if x + 297 = 350, then x = 53

This is an answer choice.
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Re: M03-16 [#permalink] New post   07 Mar 2022, 21:13
I think this is a high-quality question and I agree with explanation.
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Re: M03-16 [#permalink] New post   04 Mar 2023, 12:09
I think this is a high-quality question and I agree with explanation.
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Re: M03-16 [#permalink] New post   22 Jul 2023, 04:54
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re M03-16 [#permalink] New post   12 Aug 2023, 10:55
I think this is a high-quality question and I agree with explanation.
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Re: M03-16 [#permalink] New post   28 Dec 2023, 11:18
I used the following approach -

By definition, all numbers in a set can be represented with the help of mean. For example, in this case we can have the numbers as 125,125,125, and 125

But since one number is 150, then one of the 125 increases +25
Similarly, we can increase two other 125 as +24 and +23 (since we want to find the least possible value of the remaining)

Total increase = 25+24+23 = 72
So the remaining 125 will go down by 72 to maintain the same sum
Hence number = 125-72 = 53
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Re: M03-16 [#permalink]
28 Dec 2023, 11:18
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