Difference between revisions of "1987 AHSME Problems/Problem 4"
(Created page with "==Problem== <math>\frac{2^1+2^0+2^{-1}}{2^{-2}+2^{-3}+2^{-4}}</math> equals <math>\text{(A)} \ 6 \qquad \text{(B)} \ 8 \qquad \text{(C)} \ \frac{31}{2} \qquad \text{(D)} \ 2...") |
(Added a solution with explanation) |
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| Line 6: | Line 6: | ||
\text{(C)} \ \frac{31}{2} \qquad | \text{(C)} \ \frac{31}{2} \qquad | ||
\text{(D)} \ 24 \qquad | \text{(D)} \ 24 \qquad | ||
| − | \text{(E)} \ 512 </math> | + | \text{(E)} \ 512 </math> |
| + | |||
| + | == Solution == | ||
| + | We can factorise to give <math>\frac{2^1 + 2^0 + 2^{-1}}{2^{-3}(2^1 + 2^0 + 2^{-1})} = \frac{1}{2^{-3}} = 8</math>, which is <math>\boxed{\text{B}}</math>. | ||
== See also == | == See also == | ||
Latest revision as of 11:21, 1 March 2018
Problem
equals
Solution
We can factorise to give 
, which is 
.
See also
| 1987 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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. 
