Difference between revisions of "1997 PMWC Problems/Problem I10"

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A=6.
 
A=6.
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== See also ==
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{{PMWC box|year=1997|num-b=I9|num-a=I11}}
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[[Category:Introductory Algebra Problems]]

Revision as of 12:16, 15 January 2008

Problem

Mary took 24 chickens to the market. In the morning she sold the chickens at $$7 each and she only sold out less than half of them. In the afternoon she discounted the price of each chicken but the price was still an integral number in dollar. In the afternoon she could sell all the chickens, and she got totally $$132 for the whole day. How many chickens were sold in the morning?

Solution

Let A be the number of chickens she sold before the discount and B be the number of chickens sold after the discount. Let c be the price of one chicken after the discount.

$A+B=24$

$7A+cB=132$

$(7-c)(A)=132-24c$

So c is 5 or less. We make a table of A and c:

c|A

5|6

4|18

So c must equal 5, since when c decreases, A increases.

A=6.

See also

1997 PMWC (Problems)
Preceded by
Problem I9
Followed by
Problem I11
I: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
T: 1 2 3 4 5 6 7 8 9 10