Difference between revisions of "1992 AHSME Problems/Problem 1"
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== Problem == | == Problem == | ||
| − | If <math>3(4x+\pi)=P</math> then <math>6(8x+10\pi)=</math> | + | If <math>3(4x+5\pi)=P</math> then <math>6(8x+10\pi)=</math> |
<math>\text{(A) } 2P\quad | <math>\text{(A) } 2P\quad | ||
| − | \text{(B) } | + | \text{(B) } 4P\quad |
\text{(C) } 6P\quad | \text{(C) } 6P\quad | ||
\text{(D) } 8P\quad | \text{(D) } 8P\quad | ||
\text{(E) } 18P</math> | \text{(E) } 18P</math> | ||
| − | |||
== Solution == | == Solution == | ||
| − | <math>\ | + | We can see that <math>8x+10\pi</math> is equal to <math>2(4x+5\pi),</math> and we know that <math>2^2 = 4,</math> so the answer is <math>\boxed{B}\, .</math> |
== See also == | == See also == | ||
Latest revision as of 16:20, 18 August 2020
Problem
If 
then 
Solution
We can see that 
is equal to 
and we know that 
so the answer is 
See also
| 1992 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 2 | |
| 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. 
