Difference between revisions of "2013 AMC 8 Problems/Problem 2"
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==Problem== | ==Problem== | ||
| − | A sign at the fish market says, "50% off, today only: half-pound packages for just \$3 per package." What is the regular price for a full pound of fish, in dollars? | + | A sign at the fish market says, "50<math>\%</math> off, today only: half-pound packages for just \$3 per package." What is the regular price for a full pound of fish, in dollars? |
<math>\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 15</math> | <math>\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 15</math> | ||
==Solution== | ==Solution== | ||
| − | + | 50% off the price of half a pound of fish is \$3, so 100%, the regular price, of a half pound of fish is \$6. If half a pound of fish costs \$6, then a whole pound of fish is <math>\boxed{\textbf{(D)}\ 12}</math> dollars. | |
| + | |||
| + | ==Solution 2== | ||
| + | Suppose a full pound at normal price costs <math>x</math> dollars. Then, with the 50% off deal, the full pound would cost <math>x/2</math> dollars. A half pound of this would cost <math>x/4</math> dollars (including the 50% off). 50% off half a pound is 3, so we form the equation <math>x/4=3\Rightarrow x=\boxed{\textbf{(D)}\ 12}</math>. | ||
==Video Solution== | ==Video Solution== | ||
Latest revision as of 07:58, 8 April 2023
Problem
A sign at the fish market says, "50
off, today only: half-pound packages for just $3 per package." What is the regular price for a full pound of fish, in dollars?
Solution
50% off the price of half a pound of fish is $3, so 100%, the regular price, of a half pound of fish is $6. If half a pound of fish costs $6, then a whole pound of fish is 
dollars.
Solution 2
Suppose a full pound at normal price costs 
dollars. Then, with the 50% off deal, the full pound would cost 
dollars. A half pound of this would cost 
dollars (including the 50% off). 50% off half a pound is 3, so we form the equation 
.
Video Solution
https://youtu.be/om14Kv2jCck ~savannahsolver
See Also
| 2013 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
