Math problems from the *free* official practice tests and
problems from mba.com
Nov1907
 
 

Another from my Gprep 2

by Nov1907 Thu Nov 15, 2007 2:00 am

Again a time consuming. Short method would be appreciated.
Image
shaji
 
 

Re: Another from my Gprep 2

by shaji Fri Nov 16, 2007 1:16 am

Nov1907 Wrote:Again a time consuming. Short method would be appreciated.
Image


a,y,z,b are in ascending order.
Is 2y<a+b.... is the ??
Statement 1: 2z<a+b.....Suff
Statement2: y+z<a+b....Suff

Correct answer is D.
RonPurewal
Students
 
Posts: 19744
Joined: Tue Aug 14, 2007 8:23 am
 

by RonPurewal Fri Nov 16, 2007 7:15 am

The solution posted here is correct, but it requires a rather sophisticated understanding of the absolute value. For instance, to arrive at shaji's restatement of the question prompt, you need to understand that |y - a| is rewritten as y - a, because y is greater than a, but also that |y - b| is rewritten as b - y, because y is less than b.

If you want to kill that noise, then you can appeal to the INTUITIVE DEFINITION of the absolute value, which really goes hand in glove with this particular problem. (Warning: It doesn't go well with EVERY absolute value problem, but it works admirably here.) Namely, |a - b| is the DISTANCE BETWEEN a and b on the number line.

Therefore, this question says:
a, y, z, b are arranged from left to right on a line. Is the distance between y and a greater than the distance between y and b ?
(1) z and a are closer together than are z and b
(2) y and a are closer together than are z and b

If you draw pictures satisfying these statements, you'll see that either of them is sufficient to establish a 'yes' answer to the prompt question.