BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

manhatan 5

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Legendary Member
Posts: 789
Joined: Sun May 06, 2007 1:25 am
Location: Southern California, USA
Thanked: 15 times
Followed by:6 members

manhatan 5

by resilient » Sun Feb 17, 2008 11:21 pm
x, y, and z are consecutive integers, and x < y < z. What is the average of x, y, and z?

(1) x = 11

(2) The average of y and z is 12.5.

qa is d i chose c
Appetite for 700 and I scraped my plate!
Top
Source: — Data Sufficiency |
Senior | Next Rank: 100 Posts
Posts: 74
Joined: Tue Sep 25, 2007 7:58 am
Thanked: 4 times

Re: manhatan 5

by joshi.komal » Tue Feb 19, 2008 2:13 am
Enginpasa1 wrote:x, y, and z are consecutive integers, and x < y <z> [x + (x+1) +(x+2)]/3
=> [3x + 3]/3
=> x+1
NOw your rephrased question is:
What is x?

Any stmt. answering the this question is sufficient

(1) x = 11 SUFFICIENT :ELIMINATE choices B,C,E

(2) The average of y and z is 12.5. Solve it: y = x+1 and Z = x+2
(y+z)/2 = 12.5
=>[(x+1)+(x+2)]/2 = 12.5
SUFFICIENT you can get the value of x

Hence the answer is D



Hope it helps

Thanks
Komal
Top
Post new topic Post Reply
• Page 1 of 1