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..??
PS - Set1
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thumpin_termis
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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.
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.
