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 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> |
==Solution 1== | ==Solution 1== |
Latest revision as of 16:28, 5 April 2021
Contents
Problem
A two-inch cube of silver weighs 3 pounds and is worth 200. How much is a three-inch cube of silver worth?
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 |
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