Difference between revisions of "1950 AHSME Problems/Problem 31"
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| − | ==Problem== | + | == Problem == |
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John ordered <math>4</math> pairs of black socks and some additional pairs of blue socks. The price of the black socks per pair was twice that of the blue. When the order was filled, it was found that the number of pairs of the two colors had been interchanged. This increased the bill by <math> 50\%</math>. The ratio of the number of pairs of black socks to the number of pairs of blue socks in the original order was: | John ordered <math>4</math> pairs of black socks and some additional pairs of blue socks. The price of the black socks per pair was twice that of the blue. When the order was filled, it was found that the number of pairs of the two colors had been interchanged. This increased the bill by <math> 50\%</math>. The ratio of the number of pairs of black socks to the number of pairs of blue socks in the original order was: | ||
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\textbf{(E)}\ 1:8</math> | \textbf{(E)}\ 1:8</math> | ||
| − | ==Solution== | + | == Solution == |
| − | Let the number of blue socks be represented as <math>b</math>. We are informed that the price of the black sock is twice the price of a blue sock; let us assume that the price of one pair of | + | Let the number of blue socks be represented as <math>b</math>. We are informed that the price of the black sock is twice the price of a blue sock; let us assume that the price of one pair of blue socks is <math>1</math>. That means the price of one pair of black socks is <math>2</math>. |
| − | Now from the third and fourth sentence, we see that <math>1.5( | + | Now from the third and fourth sentence, we see that <math>1.5(2(4)+1(b))=1(4)+2(b)</math>. Simplifying gives <math>b=16</math>. This means the ratio of the number of pairs of black socks and the number of pairs of blue socks is <math>\boxed{\textbf{(C)}\ 1:4}</math>. |
| − | ==See Also== | + | == See Also == |
{{AHSME 50p box|year=1950|num-b=30|num-a=32}} | {{AHSME 50p box|year=1950|num-b=30|num-a=32}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 00:01, 12 October 2020
Problem
John ordered 
pairs of black socks and some additional pairs of blue socks. The price of the black socks per pair was twice that of the blue. When the order was filled, it was found that the number of pairs of the two colors had been interchanged. This increased the bill by 
. The ratio of the number of pairs of black socks to the number of pairs of blue socks in the original order was:
Solution
Let the number of blue socks be represented as 
. We are informed that the price of the black sock is twice the price of a blue sock; let us assume that the price of one pair of blue socks is 
. That means the price of one pair of black socks is 
.
Now from the third and fourth sentence, we see that 
. Simplifying gives 
. This means the ratio of the number of pairs of black socks and the number of pairs of blue socks is 
.
See Also
| 1950 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 30 |
Followed by Problem 32 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
