Difference between revisions of "1997 PMWC Problems/Problem I10"
m |
|||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
Mary took 24 chickens to the market. In the morning she | Mary took 24 chickens to the market. In the morning she | ||
− | sold the chickens at <math> | + | sold the chickens at <math>$7 each and she only sold out less than |
half of them. In the afternoon she discounted the price of | half of them. In the afternoon she discounted the price of | ||
each chicken but the price was still an integral number in | each chicken but the price was still an integral number in | ||
dollar. In the afternoon she could sell all the chickens, and | dollar. In the afternoon she could sell all the chickens, and | ||
− | she got totally | + | she got totally </math>$132 for the whole day. How many |
chickens were sold in the morning? | chickens were sold in the morning? | ||
Line 30: | Line 30: | ||
A=6. | A=6. | ||
− | == See | + | == See Also == |
{{PMWC box|year=1997|num-b=I9|num-a=I11}} | {{PMWC box|year=1997|num-b=I9|num-a=I11}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] |
Revision as of 15:04, 15 May 2012
Problem
Mary took 24 chickens to the market. In the morning she sold the chickens at $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.
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 |