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GRE Math Challenge #31-hot dog costs twice as much as a soda [#permalink]
07 Sep 2014, 06:47

Expert's post

00:00

Question Stats:

44% (01:56) correct
55% (02:58) wrong based on 18 sessions

A vendor sells h hot dogs and s sodas. If a hot dog costs twice as much as a soda, and if the vendor takes in a total of d dollars, how many cents does soda costs?

Re: GRE Math Challenge #31-hot dog costs twice as much as a soda [#permalink]
11 Mar 2018, 00:15

sandy wrote:

A vendor sells h hot dogs and s sodas. If a hot dog costs twice as much as a soda, and if the vendor takes in a total of d dollars, how many cents does soda costs?

A. \frac{100d}{(s +2h)}

B. \frac{(s + 2h)}{100d}

C. \frac{d(s +2h)}{100}

D. \frac{100d}{(s +2h)}

E. \frac{d}{100(s + 2h)}

Given options A and D both are same, so pls correct it pls.

Re: GRE Math Challenge #31-hot dog costs twice as much as a soda [#permalink]
13 Mar 2018, 23:33

1

This post received KUDOS

The question should be more precise. It should say that how many cents does "each" soda costs.

Anyway, here's the solution by plugging in numbers. Suppose, one soda costs $1 (100 cents) so one hot dog costs $2 (200 cents) as the hot dog costs twice as much as soda.

Suppose the vendor sells 10 hot dogs and 10 sodas so in total he makes d = $30.

Now let's try A by plugging h = 10, s = 10, and d = 10. We have (100*30)/(10 + 2(10)) = 100 cents. So A is the answer as each soda costs 100 cents.

greprepclubot

Re: GRE Math Challenge #31-hot dog costs twice as much as a soda
[#permalink]
13 Mar 2018, 23:33