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Seven positive integers have an arithmetic average of 34 and a median

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Seven positive integers have an arithmetic average of 34 and a median  [#permalink]

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New post 25 Apr 2018, 21:49
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Seven positive integers have an arithmetic average of 34 and a median of 42. If the greatest of these integers is 7 more than 2 times the least of the integers, what is the maximum possible value of the greatest integer?

(A) 42
(B) 49
(C) 70
(D) 52
(E) 91

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Re: Seven positive integers have an arithmetic average of 34 and a median  [#permalink]

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New post 25 Apr 2018, 22:34
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Bunuel wrote:
Seven positive integers have an arithmetic average of 34 and a median of 42. If the greatest of these integers is 7 more than 2 times the least of the integers, what is the maximum possible value of the greatest integer?

(A) 42
(B) 49
(C) 70
(D) 52
(E) 91


Let the least integer be x. So the greatest integer = 2x+7. Knowing that x is an integer, 2x will be even and thus 2x+7 will be odd. There are only 2 odd numbers in options - 49 and 91. Since question asks for max possible value, lets try with the bigger option first, 91. If it works, well thats the answer otherwise answer has to be 49.

So lets put 2x+7 = 91. This gives x = 42. But if smallest is 42, then average cannot be 34. Rejected.

So only one option left, 49 which should be the answer. Hence B answer
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Re: Seven positive integers have an arithmetic average of 34 and a median  [#permalink]

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New post 26 Apr 2018, 00:08
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Seven positive integers have an arithmetic average of 34 and a median of 42. If the greatest of these integers is 7 more than 2 times the least of the integers, what is the maximum possible value of the greatest integer?

(A) 42
(B) 49
(C) 70
(D) 52
(E) 91

the answer by amanvermagmat is great way to do the question and saves time,

I solved it in more conventional way,

let the seven numbers in increasing order be a,b,c,d,e,f,h

given that median is 42, so d = 42

largest no. is seven more than two times the smallest, h = 2a+7

Average of seven no.s is given as 34,

a + b + c + 42 + e + f + 2a+7 = 238

we need to maximise 2a+7, so we have to minimise all other terms

the question does not mention that numbers are distinct so we can repeat the numbers,
so lowest value for b and c is a (the smallest number itself)

also the lowest value for e and f will be at least 42. (equal to the median)

plugging it all in,

a + a + a + 42 + 42 + 42 + (2a+7) = 238
a = 21

So the largest no is (21 x 2) + 7 = 49
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Re: Seven positive integers have an arithmetic average of 34 and a median  [#permalink]

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New post 13 May 2018, 10:02
As per question

Let smallest number be x and in order to calculate largest integer 2nd and 3rd numbers are also x

42 is median, in order for 7th to be greatest 5th and 6th need to be equal to median

x x x 42 42 42 2x+7
_ _ _ __ __ __ ____

5x+7 +42*3=34*7

x=21

Greatest= 2*21+7=49

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Re: Seven positive integers have an arithmetic average of 34 and a median &nbs [#permalink] 13 May 2018, 10:02
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