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GRE Math Challenge #31-hot dog costs twice as much as a soda
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07 Sep 2014, 06:47
07 Sep 2014, 06:47
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52% (02:10) correct
47% (02:31) wrong based on 42 sessions
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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. \(\frac{d}{(100s +2h)}\)
E. \(\frac{d}{100(s + 2h)}\)
A. \(\frac{100d}{(s +2h)}\)
B. \(\frac{(s + 2h)}{100d}\)
C. \(\frac{d(s +2h)}{100}\)
D. \(\frac{d}{(100s +2h)}\)
E. \(\frac{d}{100(s + 2h)}\)
Re: GRE Math Challenge #31-hot dog costs twice as much as a soda
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16 Jan 2018, 01:05
16 Jan 2018, 01:05
D. \frac{100d}{(s +2h)}?
solution, thanks!
solution, thanks!
Re: GRE Math Challenge #31-hot dog costs twice as much as a soda
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11 Mar 2018, 00:15
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. \(\frac{d}{100(s + 2h)}\)
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.
