1997 AJHSME Problems/Problem 22

Revision as of 15:28, 5 April 2021 by Jayrenfu (talk | contribs) (Problem)

Problem

A two-inch cube $(2\times 2\times 2)$ of silver weighs 3 pounds and is worth $200.  How much is a three-inch cube of silver worth?$\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 $8$ equal cubes that are 1x1x1. Each smaller cube is worth $\frac{200}{8} = 25$ dollars.

To create a 3x3x3 cube of silver, you need $27$ of those 1x1x1 cubes. The cost of those $27$ cubes is $27 \cdot 25 = 675$ dollars, which is answer $\boxed{E}$

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:

$\frac{200}{2^3} = \frac{x}{3^3}$

$x = 200 \cdot \frac{3^3}{2^3} = 675$, which is answer $\boxed{E}$

See also

1997 AJHSME (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png