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GmatKiss: Legendary Member; Posts: 2789; Joined: Tue Jul 26, 2011 12:19 amFollowed by: 43 members; Location: Chennai, India; Thanked: 206 times; GMAT Score:640

by GmatKiss » Mon May 21, 2012 7:41 am

The average of three positive integers is 44 and the median is 42. What is the least possible value of the greatest of the three numbers?

A) 6
B) 42
C) 48
D) 89
E) 128

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Source: Beat The GMAT — Problem Solving | Mon May 21, 2012 7:41 am

hey_thr67: Master | Next Rank: 500 Posts; Posts: 299; Joined: Tue Feb 15, 2011 10:27 amFollowed by: 2 members; Thanked: 9 times

by hey_thr67 » Mon May 21, 2012 8:08 am

IMO: C

least possible maximum number will be 42,42,x Now Average is 44 So, x has to 2(44-42) + 44 So as to get the 44 as average. The answer is 48.

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Mon May 21, 2012 8:10 am

GmatKiss wrote:The average of three positive integers is 44 and the median is 42. What is the least possible value of the greatest of the three numbers?

A) 6
B) 42
C) 48
D) 89
E) 128

Let the numbers be x, y, z, where x < y < z (so, the median = y, and we're trying to find the least possible value of z)

Average = 44, so x+y+z = 3(44) = 132

If the median (y) is 42, then x+z = 90 (since x+y+z = 132 )

Now let's check the answer choices:
A) Eliminate it since z > 42
B) If z = 42, then x = 48 but x is supposed to be the smallest number. Eliminate it.
C) If z = 48, then x = 42. Perfect! This satisfies the condition that x < y < z

So, we get x=42, y=42 and z=48

So, the answer is C

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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