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by arzanr » Mon Feb 22, 2010 8:13 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

OA: E

I can't figure this one out for the life of me!
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by kstv » Mon Feb 22, 2010 8:40 am
Seems something is wrong with the Q
cos
H + C = 3.59
H + F = 4.4 but F = 2C. so H + 2 C = 4.4
subsract eq 1 from 2

C = .81 F = 2 X .81 = 1.62 H = 2.78
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by arzanr » Mon Feb 22, 2010 9:59 am
yup I agree.. I dont see how it can possibly make sense either
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by harsh.champ » Tue Feb 23, 2010 5:50 am
arzanr 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?

(A) $0.30
(B) $0.45
(C) $0.60
(D) $0.75
(E) $0.90

OA: E

I can't figure this one out for the life of me!
H + C = 3.59
H + F = 4.40
F = 2C
H + 2C =4.40
Subtracting, we get
C = 0.81
H = 2.78[Since, H + C = 3.59]
F = 2 x C = 1.62

Dude,whats the source??
Just look at the method -I dont think anything is wrong wid it???
It takes time and effort to explain, so if my comment helped you please press Thanks button :)



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