Difference between revisions of "1997 AJHSME Problems/Problem 22"
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==Problem== | ==Problem== | ||
| − | A two-inch cube <math>(2\times 2\times 2)</math> of silver weighs 3 pounds and is worth < | + | A two-inch cube <math>(2\times 2\times 2)</math> of silver weighs 3 pounds and is worth <math>200. How much is a three-inch cube of silver worth? |
| − | <math>\text{(A)}\ 300\text{ dollars} \qquad \text{(B)}\ 375\text{ dollars} \qquad \text{(C)}\ 450\text{ dollars} \qquad \text{(D)}\ 560\text{ dollars} \qquad \text{(E)}\ 675\text{ dollars} | + | </math>\text{(A)}\ 300\text{ dollars} \qquad \text{(B)}\ 375\text{ dollars} \qquad \text{(C)}\ 450\text{ dollars} \qquad \text{(D)}\ 560\text{ dollars} \qquad \text{(E)}\ 675\text{ dollars}$ |
==Solution 1== | ==Solution 1== | ||
Revision as of 15:28, 5 April 2021
Contents
Problem
A two-inch cube 
of silver weighs 3 pounds and is worth 
\text{(A)}\ 300\text{ dollars} \qquad \text{(B)}\ 375\text{ dollars} \qquad \text{(C)}\ 450\text{ dollars} \qquad \text{(D)}\ 560\text{ dollars} \qquad \text{(E)}\ 675\text{ dollars}$
Solution 1
The 2x2x2 cube of silver can be divided into 
equal cubes that are 1x1x1. Each smaller cube is worth 
dollars.
To create a 3x3x3 cube of silver, you need 
of those 1x1x1 cubes. The cost of those cubes is 
dollars, which is answer 
Solution 2
Since price is proportional to the amount (or volume) of silver, and volume is proportional to the cube of the side, the price ought to be proportional to the cube of the side.
Setting up a proportion:
, which is answer
See also
| 1997 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
