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A vendor sells h hot dogs and s sodas. If a hot dog costs twice as muc

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A vendor sells h hot dogs and s sodas. If a hot dog costs twice as muc  [#permalink]

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New post 05 Jul 2018, 20:43
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Question Stats:

67% (02:04) correct 33% (01:53) wrong based on 35 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. \(\frac{d}{100(s + 2h)}\)

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Re: A vendor sells h hot dogs and s sodas. If a hot dog costs twice as muc  [#permalink]

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New post 05 Jul 2018, 21:49
Bunuel 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. \(\frac{d}{(100s +2h)}\)

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


d $=d*100 cent

Let the unit cost of soda be p cents. So, unit cost of hot dog=2p cents

Total cost=cost of hot dog+cost of soda
or, 100d=h*2p+s*p=p(2h+s)
Or, \(p=\frac{100d}{(s+2h)}\)

Ans. (A)
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Re: A vendor sells h hot dogs and s sodas. If a hot dog costs twice as muc  [#permalink]

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New post 05 Jul 2018, 22:28
Bunuel 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. \(\frac{d}{(100s +2h)}\)

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



a quick method...
d is in dollars but we are asking in cents so d dollar = 100d in cents
so keep the choices that have 100d together
Only A and B
Now we are looking for amount so amount that is 100d should be in numerator
B gives us something per cent, not what we are looking for
A

algebraic method
let soda cost x, then hot dog costs 2x
total cost \(= sx+2hx=x(s+2h)=100d...............x=\frac{100d}{s+2h}\)
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Re: A vendor sells h hot dogs and s sodas. If a hot dog costs twice as muc  [#permalink]

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New post 13 Jul 2018, 13:20
Bunuel 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. \(\frac{d}{(100s +2h)}\)

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



We can let the cost of a soda = x cents; thus, the cost of a hot dog = 2x cents. Since d dollars = 100d cents, we can create the following equation:

2xh + xs = 100d

x(2h + s) = 100d

x = 100d/(2h + s)

Answer: A
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Re: A vendor sells h hot dogs and s sodas. If a hot dog costs twice as muc &nbs [#permalink] 13 Jul 2018, 13:20
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