Difference between revisions of "1997 AHSME Problems/Problem 4"
Talkinaway (talk | contribs) |
|||
| (4 intermediate revisions by 3 users not shown) | |||
| Line 1: | Line 1: | ||
| + | ==Problem== | ||
| + | |||
| + | If <math>a</math> is <math>50\%</math> larger than <math>c</math>, and <math>b</math> is <math>25\%</math> larger than <math>c</math>, then <math>a</math> is what percent larger than <math>b</math>? | ||
| + | |||
| + | <math> \mathrm{(A)\ } 20\% \qquad \mathrm{(B) \ }25\% \qquad \mathrm{(C) \ } 50\% \qquad \mathrm{(D) \ } 100\% \qquad \mathrm{(E) \ }200\% </math> | ||
| + | |||
| + | __TOC__ | ||
| + | == Solution == | ||
| + | ===Solution 1=== | ||
| + | |||
| + | Translating each sentence into an equation, <math>a = 1.5c</math> and <math>b = 1.25c</math>. | ||
| + | |||
| + | We want a relationship between <math>a</math> and <math>b</math>. Dividing the second equation into the first will cancel the <math>c</math>, so we try that and get: | ||
| + | |||
| + | <math>\frac{a}{b} = \frac{1.5}{1.25}</math> | ||
| + | |||
| + | <math>\frac{a}{b} = \frac{150}{125}</math> | ||
| + | |||
| + | <math>\frac{a}{b} = \frac{6}{5}</math> | ||
| + | |||
| + | <math>a = 1.2b</math> | ||
| + | |||
| + | In this case, <math>a</math> is <math>1.2 - 1 = 0.2 = 20\%</math> bigger than <math>b</math>, and the answer is <math>\boxed{A}</math>. | ||
| + | |||
| + | ===Solution 2=== | ||
| + | |||
| + | Arbitrarily assign a value to one of the variables. Since <math>c</math> is the smallest variable, let <math>c = 100</math>. | ||
| + | |||
| + | If <math>a</math> is <math>50\%</math> larger than <math>c</math>, then <math>a = 150</math>. | ||
| + | |||
| + | If <math>b</math> is <math>25\%</math> larger than <math>c</math>, then <math>b = 125</math>. | ||
| + | |||
| + | We see that <math>\frac{a}{b} = \frac{150}{125} = 1.2</math> So, <math>a</math> is <math>20\%</math> bigger than <math>b</math>, and the answer is <math>\boxed{A}</math>. | ||
| + | |||
== See also == | == See also == | ||
{{AHSME box|year=1997|num-b=3|num-a=5}} | {{AHSME box|year=1997|num-b=3|num-a=5}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 13:12, 5 July 2013
Problem
If 
is 
larger than 
, and 
is 
larger than , then is what percent larger than ?
Solution
Solution 1
Translating each sentence into an equation, 
and 
.
We want a relationship between and . Dividing the second equation into the first will cancel the , so we try that and get:
In this case, is 
bigger than , and the answer is 
.
Solution 2
Arbitrarily assign a value to one of the variables. Since is the smallest variable, let 
.
If is larger than , then 
.
If is larger than , then 
.
We see that 
So, is 
bigger than , and the answer is .
See also
| 1997 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. 
