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

Redeem

grockit tough PS

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Legendary Member
Posts: 1077
Joined: Mon Dec 13, 2010 1:44 am
Thanked: 118 times
Followed by:33 members
GMAT Score:710

grockit tough PS

by bblast » Tue May 31, 2011 9:15 am
The ratio of the area of a square mirror to its frame is 16 to 33. If the frame has a uniform width (c) around the mirror, which of the following could be the value, in inches, of c ?

I. 2
II. 7/2
III. 5

https://s3.amazonaws.com/production.groc ... S422Q0.png


ans 1,2 and 3
Cheers !!

Quant 47-Striving for 50
Verbal 34-Striving for 40

My gmat journey :
https://www.beatthegmat.com/710-bblast-s ... 90735.html
My take on the GMAT RC :
https://www.beatthegmat.com/ways-to-bbla ... 90808.html
How to prepare before your MBA:
https://www.youtube.com/watch?v=upz46D7 ... TWBZF14TKW_
Top
Source: — Problem Solving |
User avatar
Legendary Member
Posts: 1309
Joined: Mon Apr 04, 2011 5:34 am
Location: India
Thanked: 310 times
Followed by:123 members
GMAT Score:750

by cans » Tue May 31, 2011 9:24 am
let side of square = a, thus area of square mirror = a^2
Also frame area = total area - mirror area = (a+2c)^2 - a^2 = 4c^2+4ac
thus ration = a^2/(4c^2+4ac) = 16/33 (this is given)
cross multiply:- 33a^2 = 64c^2+64ac
or 33a^2 -64ac - 64c^2 = 0
if we solve this, we will get c in terms of a.
As a is not given, c can take any value..
Top
Legendary Member
Posts: 1448
Joined: Tue May 17, 2011 9:55 am
Location: India
Thanked: 375 times
Followed by:53 members

by Frankenstein » Tue May 31, 2011 9:24 am
Hi,
ar(inner square) /ar(frame) = 16/33
So, ar(inner square) /[ar(outer square) - ar(inner square)] = 16/33
So ar(inner square) /ar(outer square) = 16/16+33 = 16/49
So side length of inner square/side length of outer square = 4/7
If 4x is inner length, c = 3x.

Cheers!
Top
Senior | Next Rank: 100 Posts
Posts: 97
Joined: Sun May 15, 2011 9:19 am
Thanked: 18 times
Followed by:1 members

by SoCan » Tue May 31, 2011 9:41 am
While it's good to think about the equations while you're studying, you'll save a lot of time during the test if you think about questions like this one conceptually.

There are no constraints on what the areas of the mirror or its frame could be besides their ratio. Therefore, you can set c to whatever value you'd like. Once you choose c, the rest of the dimensions are obviously constrained. It's like asking, "if a rectangle's longer side is twice as long as its shorter side, how long can the shorter side be?" Anything, but once you choose, the longer side is constrained. Thinking about the question like this should allow you to solve it in 10-20 seconds, freeing up valuable time for questions that require more work.

Of course, the question would be trickier and would require more work if there was another constraint placed - e.g., the area of the mirror must be an integer, the side of the frame must be a multiple of 3, etc.
Top
Post new topic Post Reply
• Page 1 of 1