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# The average (arithmetic mean) of 3 different positive integers is 100

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The average (arithmetic mean) of 3 different positive integers is 100  [#permalink]

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20 Nov 2018, 09:19
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The average (arithmetic mean) of 3 different positive integers is 100 and the largest of these 3 integers is 120. What is the least possible value of the smallest of these 3 integers?

(A) 1
(B) 10
(C) 61
(D) 71
(E) 80

Project PS Butler : Question #29

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Re: The average (arithmetic mean) of 3 different positive integers is 100  [#permalink]

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20 Nov 2018, 11:24
4
$$\frac{(2x+120)}{3} =100$$

$$2x+120=300$$

$$2x=180$$

$$x =90$$

so 120 +90+90 to minimize one value maximize secnd one 120+119+61

IMO: C
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Re: The average (arithmetic mean) of 3 different positive integers is 100  [#permalink]

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20 Nov 2018, 14:32
1
(a+b+c)/3 = 100
a+b+120 = 300
=> a+b = 180 =>a = 180 - b
b<120 => a>60
smallest a is 61 => C
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Re: The average (arithmetic mean) of 3 different positive integers is 100  [#permalink]

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20 Nov 2018, 15:25
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HKD1710 wrote:
The average (arithmetic mean) of 3 different positive integers is 100 and the largest of these 3 integers is 120. What is the least possible value of the smallest of these 3 integers?

(A) 1
(B) 10
(C) 61
(D) 71
(E) 80

Project PS Butler : Question #29

second largest integer is 120-1=119
so least value of smallest must be (300-120)-119=61
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Re: The average (arithmetic mean) of 3 different positive integers is 100  [#permalink]

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20 Nov 2018, 16:26
Average=100={120+x+y}/3
X+y=180
Assume x to be smallest and start plugging in the answers to get y.
Now neither x nor y can be larger than 120. So a and b makes y greater than 120. Therefore are incorrect. From the rest 61 is the smallest.

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Re: The average (arithmetic mean) of 3 different positive integers is 100  [#permalink]

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27 Nov 2018, 13:43
1
Avg of 3 nos will be -
(x+ 2nd largest + largest)/3
=> (x+ 119 +120) / 3 =100
=> x= 61 (C)
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Re: The average (arithmetic mean) of 3 different positive integers is 100  [#permalink]

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17 Dec 2018, 14:49
What if the other 2 numbers are 109 and 71 or 100 and 80? It is no where mentioned that the 2nd number has to be immediate smaller number than 120. So how we can eliminate 109/100? Please explain.
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Re: The average (arithmetic mean) of 3 different positive integers is 100  [#permalink]

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22 Dec 2018, 02:42
1
tannumunu wrote:
What if the other 2 numbers are 109 and 71 or 100 and 80? It is no where mentioned that the 2nd number has to be immediate smaller number than 120. So how we can eliminate 109/100? Please explain.

Hello Tannu,

We need to find the smallest number out of the 3 and we are given that largest number is 120.
We also know that we have 3 different positive integers, i.e, the remaining 2 numbers are not equal.

Consider a parallel example: x + y = 5. (where x and y are positive integers)

If we want to find the smallest out of x and y, the OTHER number HAS TO BE AS LARGE AS POSSIBLE, that is 4.
Why?
Consider y = 3:
x+3=5. Therefore x = 2. But is this the smallest POSSIBLE value that 'x' can take? 'x' will be as small as possible only when y will be as large as possible. In this case, if we take x + 4 = 5, then x will be equal to 1 (this can be the smallest possible positive integer. Therefore y has to be 4 (or 2nd number immediately smaller to 5, as you have pointed out).

With this understanding, you can re-read the solutions provided above and hopefully understand it better.
Hope this helped.
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Re: The average (arithmetic mean) of 3 different positive integers is 100  [#permalink]

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22 Dec 2018, 18:23
Darshi04 wrote:
tannumunu wrote:
What if the other 2 numbers are 109 and 71 or 100 and 80? It is no where mentioned that the 2nd number has to be immediate smaller number than 120. So how we can eliminate 109/100? Please explain.

Hello Tannu,

We need to find the smallest number out of the 3 and we are given that largest number is 120.
We also know that we have 3 different positive integers, i.e, the remaining 2 numbers are not equal.

Consider a parallel example: x + y = 5. (where x and y are positive integers)

If we want to find the smallest out of x and y, the OTHER number HAS TO BE AS LARGE AS POSSIBLE, that is 4.
Why?
Consider y = 3:
x+3=5. Therefore x = 2. But is this the smallest POSSIBLE value that 'x' can take? 'x' will be as small as possible only when y will be as large as possible. In this case, if we take x + 4 = 5, then x will be equal to 1 (this can be the smallest possible positive integer. Therefore y has to be 4 (or 2nd number immediately smaller to 5, as you have pointed out).

With this understanding, you can re-read the solutions provided above and hopefully understand it better.
Hope this helped.

Thank you..
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Re: The average (arithmetic mean) of 3 different positive integers is 100  [#permalink]

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01 Aug 2019, 00:54
1
The average (arithmetic mean) of 3 different positive integers is 100 and the largest of these 3 integers is 120. What is the least possible value of the smallest of these 3 integers?

(A) 1
(B) 10
(C) 61
(D) 71
(E) 80

Ans= The 2nd largest no can be 119 only otherwise is be > than 120 and violate the statement. Hence 3rd no will be Total sum - sum of 2 nos = 3rd no, i.e.... 300-(120+119) = 61.

KUDOS!!
Re: The average (arithmetic mean) of 3 different positive integers is 100   [#permalink] 01 Aug 2019, 00:54
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