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Simple Q

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Senior Manager
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Simple Q [#permalink]

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New post 15 May 2010, 11:50
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A
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D
E

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Question Stats:

67% (00:53) correct 33% (00:41) wrong based on 1 sessions

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The average of 4 different positive integers is 125. If the largest of these integers is 150, what is the least possible value of the smallest integer?

* 1
* 2
* 12
* 53
* 100

Kudos [?]: 171 [0], given: 0

Manager
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Posts: 151

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Re: Simple Q [#permalink]

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New post 17 May 2010, 01:36
i also think the ans is D... whats OA..?

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Senior Manager
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Joined: 25 Jun 2009
Posts: 298

Kudos [?]: 151 [0], given: 6

Re: Simple Q [#permalink]

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New post 17 May 2010, 02:45
nverma wrote:
The average of 4 different positive integers is 125. If the largest of these integers is 150, what is the least possible value of the smallest integer?

* 1
* 2
* 12
* 53
* 100

D,

Let the numbers be a,b,c and d

then \(\frac{a+b+c+d}{4}\) \(= 125\)

a+ b+c +d = 500

d = 150

a+b+c= 350

now 150 is the largest among the lot, then let c= 149 and b = 148 then equation becomes

a+ 148+ 149 = 350 or a = 53

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Re: Simple Q [#permalink]

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New post 17 May 2010, 02:46
D should be the ans...

Lets take other numbers to be 148 and 146. Adding these 2 numbers to 150 and then subtracting with 125*4 , ans comes to 56, so only D is closest to that.

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Posts: 179

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Re: Simple Q [#permalink]

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New post 17 May 2010, 07:59
another way to solve
in order to minimize one no we have to maximize all the other no.s to get the same average , the max. no. is 150 so the maximum limit of other no.s is 149 & 148
NOW (148 +149+150 +X)/4 = 125 so x=53 ans option D

Kudos [?]: 34 [0], given: 17

Re: Simple Q   [#permalink] 17 May 2010, 07:59
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