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25 Apr 2018, 21:49
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
71% (02:43) correct
29%
(02:43)
wrong
based on 408
sessions
History
Date
Time
Result
Not Attempted Yet
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
(A) 42
(B) 49
(C) 70
(D) 52
(E) 91
25 Apr 2018, 22:34
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Bunuel
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
ruchirS
Joined: 19 Dec 2017
Last visit: 16 Jul 2024
Posts: 4
Given Kudos: 37
Location: Kuwait
Schools: ISB '20 (A) Rotman '21 (A$)
GMAT 1: 640 Q48 V31
GMAT 2: 710 Q49 V37
GPA: 3.64
26 Apr 2018, 00:08
Kudos
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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
(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
