Difference between revisions of "1997 PMWC Problems/Problem I10"
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==Solution== | ==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. | + | Let <math>A</math> be the number of chickens she sold before the discount and <math>B</math> be the number of chickens sold after the discount. Let <math>c</math> be the price of one chicken after the discount. |
<math>A+B=24</math> | <math>A+B=24</math> | ||
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<math>(7-c)(A)=132-24c</math> | <math>(7-c)(A)=132-24c</math> | ||
− | So c is 5 | + | So <math>c</math> is less than <math>5</math>. We make a table of <math>A</math> and <math>c</math>: |
− | c|A | + | <math>c</math> | <math>A</math> |
− | 5|6 | + | <math>5</math> | <math>6</math> |
− | 4|18 | + | <math>4</math> | <math>18</math> |
− | So c must equal 5, since when c decreases, A increases. | + | So <math>c</math> must equal <math>5</math>, since when <math>c</math> decreases, <math>A</math> increases. |
− | A=6. | + | <math>A=6</math>. |
== See Also == | == See Also == |
Latest revision as of 18:14, 7 April 2016
Problem
Mary took chickens to the market. In the morning she sold the chickens at 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 for the whole day. How many chickens were sold in the morning?
Solution
Let be the number of chickens she sold before the discount and be the number of chickens sold after the discount. Let be the price of one chicken after the discount.
So is less than . We make a table of and :
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So must equal , since when decreases, increases.
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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 |