Difference between revisions of "1983 AHSME Problems/Problem 7"
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== Problem== | == Problem== | ||
| − | Alice sells an item at | + | Alice sells an item at <math>\$10</math> less than the list price and receives <math>10\%</math> of her selling price as her commission. |
| − | Bob sells the same item at | + | Bob sells the same item at <math>\$20</math> less than the list price and receives <math>20\%</math> of his selling price as his commission. |
| − | If they both get the same commission, then the list price | + | If they both get the same commission, then the list price is |
| − | <math>\textbf{(A) } 20\qquad | + | <math>\textbf{(A) } \$20\qquad |
| − | \textbf{(B) } 30\qquad | + | \textbf{(B) } \$30\qquad |
| − | \textbf{(C) } 50\qquad | + | \textbf{(C) } \$50\qquad |
| − | \textbf{(D) } 70\qquad | + | \textbf{(D) } \$70\qquad |
| − | \textbf{(E) } 100 </math> | + | \textbf{(E) } \$100 </math> |
==Solution== | ==Solution== | ||
| − | If <math>x</math> is the list price, then <math>10\%(x-10)=20\%(x-20)</math>. Solving this | + | If <math>x</math> is the list price, then <math>10\%(x-10)=20\%(x-20)</math>. Solving this equation gives <math>x=30</math>, so the answer is <math>\boxed{\textbf{(B) }\$30}</math>. |
| + | |||
| + | ==See Also== | ||
| + | {{AHSME box|year=1983|num-b=6|num-a=8}} | ||
| + | |||
| + | {{MAA Notice}} | ||
Latest revision as of 23:44, 19 February 2019
Problem
Alice sells an item at 
less than the list price and receives 
of her selling price as her commission.
Bob sells the same item at 
less than the list price and receives 
of his selling price as his commission.
If they both get the same commission, then the list price is
Solution
If 
is the list price, then 
. Solving this equation gives 
, so the answer is 
.
See Also
| 1983 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
