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by Captchar » Sun Dec 23, 2012 10:40 am
At a certain diner, a hamburger and coleslaw cost $3.59, and a hamburger and french fries cost $4.40. If french fries cost twice as much as coleslaw, how much do french fries cost?

(A) $0.30
(B) $0.45
(C) $0.60
(D) $0.75
(E) $0.90
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by Anurag@Gurome » Sun Dec 23, 2012 10:47 am
Captchar wrote:At a certain diner, a hamburger and coleslaw cost $3.59, and a hamburger and french fries cost $4.40. If french fries cost twice as much as coleslaw, how much do french fries cost?
Fries = 2*Cloeslaw

Hamburger + Coleslaw = Hamburger + Fries/2 = $3.59 ........... (1)
Hamburger + Fries = $4.40 ...................................... (2)

Subtracting (1) from (2), Fries/2 = $(4.40 - 3.59) = $0.81

Hence, price of french fries = $(2*0.81) = $1.62

Check your source. None of the options are correct.
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by tabsang » Mon Dec 24, 2012 1:26 am
Yep. I've got the same thing:

Hamburger (H)
Coleslaw (C)
French fries (F)

Given:
H+C=3.59...(1)
H+F=4.40...(2)

Also, F=2C
Substitute the above in (2),
H+2C=4.40...(3)

Subtract (1) from (3)

We get: C=0.81
Thus, F=2C=2*0.81=$1.62=Wrong Answer choices in the original question :P :D
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by The Iceman » Mon Dec 24, 2012 2:14 am
With keen observation you can eliminate the need of writing equations in such problems.

Clearly the difference in the price of the items purchased two at a time (with one common item) is 81 cents and one item being twice of the other implies the cost of the cheaper item itself is 81 cents. Hence the cost of french fries = 2*0.82=$1.62
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