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Senior | Next Rank: 100 Posts
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by smushkas » Fri Aug 03, 2007 9:06 pm
Hey guys,

Can anyone help me with this simple question?

5. 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

Correct answer is (E)

I did solve it this way:
x + y = 3.59
x + 2y = 4.40
but it was wrong.
Thanks in advance for your attention! :D
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Re: Please help

by gabriel » Sat Aug 04, 2007 12:20 am
smushkas wrote:Hey guys,

Can anyone help me with this simple question?

5. 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

Correct answer is (E)

I did solve it this way:
x + y = 3.59
x + 2y = 4.40
but it was wrong.
Thanks in advance for your attention! :D
let hamburger cost x $..

let the coleslaw cost y $..

So, the french fries cost 2y $ ..


so the eqn are

x + y = 3.59 and x + 2y = 4.40 ...

.. nothing wrong with the eqn's .. so either the answer choices are wrong or the given numbers are wrong .. i have a feeling that it should be 3.95 $ and not 3.59 $ ..
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Senior | Next Rank: 100 Posts
Posts: 49
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Thanked: 3 times

by smushkas » Sat Aug 04, 2007 4:07 am
Thanks a bunch gabriel!

I thought i was totally wrong. Even with the correct answer hamburger price does not make any sense to me. :shock:
1) 3.59 - 0.45 = 3.14
2) 4.40 - 0.90 = 3.50

It seems 2 prices were charged for hambugers. There definitely something must be wrong...

Thanks again gabriel for your help!
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