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PS - Set1

Expert replies
by RM » Sat Jun 16, 2007 11:00 pm
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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Source: — Problem Solving |

Is it right??

by moneyman » Sun Jun 17, 2007 1:31 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

With the available data,two equations can be derived..

h+c=3.59->1
h+f=4.40->2

It is stated that french fries cost twice as much as colesaw ..Therefore f=2c

Substituting the value of f in 2 we get

h+2c=4.40

Now subtracting 1 from 2,we get c=0.81 and as we know,f=2c then f=1.62..

Is the question right..??
Maxx
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by RM » Sun Jun 17, 2007 8:33 am
I am also getting the same answer, but the answer given is 0.90.
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by thumpin_termis » Mon Jun 18, 2007 7:54 pm
Either the question or the OA is wrong. I also get 1.62 for fries.

If the cost of fries were really .90, then that means:
h + f = 4.40
h + .90 = 4.40
h = 4.40 - .90
h = 3.50

If we stick this back into the coleslaw equation,
h + c = 3.59
3.50 + c = 3.59
c = 3.59 - 3.50
c = 0.09

And this means that french fries (90 cents) cost way more than twice that of coleslaw (9 cents), so something is wrong here.
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