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by bblast » Mon May 16, 2011 8:01 am
If a, b, and c are greater than 0 and a is twice as large as b percent of c, then in
terms of band c, what is a percent of c?

A) 2bc/100
b> 2bc^2/1000
c> bc^2/5000
d> b^2c/5000
e> 5000b/c^2

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by Stuart@KaplanGMAT » Mon May 16, 2011 8:28 am
bblast wrote:If a, b, and c are greater than 0 and a is twice as large as b percent of c, then in
terms of band c, what is a percent of c?

A) 2bc/100
b> 2bc^2/1000
c> bc^2/5000
d> b^2c/5000
e> 5000b/c^2

oa-C
Took me 4 reads to figure out what the question actually meant! Curse question writers who use "a" as a variable in word problems. Rewriting the question using caps (and adding a space so that "band" reads "b and" - that one really confused me too!):
If A, B, and C are greater than 0 and A is twice as large as B percent of C, then in
terms of B and C, what is A percent of C?

A) 2bc/100
b> 2bc^2/1000
c> bc^2/5000
d> b^2c/5000
e> 5000b/c^2
We see percents with unknown values and variables in the choices - this is a job for Picking Numbers Man (starring in a 3-D movie this summer, coming to a theatre near you)!

Our constraint: A = 2 * (B% of C)

To make the numbers simple, let's choose B=25 and C=100, giving us:

A = 2 * (25% of 100)

A = 2 * 25

A = 50

Our question: what is 50% of 100?

Answer: 50.

Plugging into the choices:

A) 2*25*100/100 = 50... works
B) 2*25*100*100/1000 = 2*25*10 = more than 50... eliminate
C) 25*100*100/5000 = 25*10/5 = 25*2 = 50... works
D) 25*25*100/5000 = 25*25/50 = not an integer... eliminate
E) 5000*25/100*100 = 5*25/10 = not an integer... eliminate

As is sometimes the case when you pick numbers, two answer choices work. So, we need to try one more set of numbers. Let's try:

B=20, C=40

A = 2*(20% of 40)
A = 2*8 = 16

Our question: what is 16% of 40?

Our answer: 6.4

A) 2*20*40/100 = 2*2*4 = 16... doesn't work... eliminate (A).
C) 20*40*40/5000 = 2*4*4/5 = 32/5 = 6.4... choose (C)!

edited for poor subject-verb agreement!
Last edited by Stuart@KaplanGMAT on Mon May 16, 2011 8:36 am, edited 1 time in total.
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by Stuart@KaplanGMAT » Mon May 16, 2011 8:33 am
For the algebraphiles in the audience:

A = 2*(B/100)*C

A = 2BC/100 (which is trap answer (A)).

Our actual question:

what is A% of C?

What is 2BC/100 % of C?

2BC/100 % of C = (2BC/100)/100 * C

= 2BC/10000 * C
= BC/5000 * C
= B(C^2)/5000
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by manpsingh87 » Mon May 16, 2011 8:34 am
bblast wrote:If a, b, and c are greater than 0 and a is twice as large as b percent of c, then in
terms of band c, what is a percent of c?

A) 2bc/100
b> 2bc^2/1000
c> bc^2/5000
d> b^2c/5000
e> 5000b/c^2

oa-C
a=2(b%ofc)
=2*(b/100)*c;
=b*c/50;-----1)
now a% of c is; a/100*c;-------2) from 1) we have a=bc/50;
substitute it in 2) we have; (b*c/50)*(1/100)*c=bc^2/5000; hence answer should be C
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