Difference between revisions of "1997 PMWC Problems/Problem I10"
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| − | 6 | + | ==Problem== |
| + | Mary took 24 chickens to the market. In the morning she | ||
| + | sold the chickens at <math>\</math>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 <math>\</math>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. | ||
| + | |||
| + | <math>A+B=24</math> | ||
| + | |||
| + | <math>7A+cB=132</math> | ||
| + | |||
| + | <math>(7-c)(A)=132-24c</math> | ||
| + | |||
| + | 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. | ||
Revision as of 17:13, 8 October 2007
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.
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.
