Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1997 AHSME Problems/Problem 4"
m
Line 5: Line 5:
 
<math> \mathrm{(A)\ } 20\% \qquad \mathrm{(B) \ }25\% \qquad \mathrm{(C) \  } 50\% \qquad \mathrm{(D) \  } 100\% \qquad \mathrm{(E) \  }200\%  </math>
 
<math> \mathrm{(A)\ } 20\% \qquad \mathrm{(B) \ }25\% \qquad \mathrm{(C) \  } 50\% \qquad \mathrm{(D) \  } 100\% \qquad \mathrm{(E) \  }200\%  </math>
  
−
==Solution 1==
+
__TOC__
 +
== Solution ==
 +
===Solution 1===
  
 
Translating each sentence into an equation, <math>a = 1.5c</math> and <math>b = 1.25c</math>.
 
Translating each sentence into an equation, <math>a = 1.5c</math> and <math>b = 1.25c</math>.
Line 21: Line 23:
 
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{B}</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{B}</math>.
  
−
==Solution 2==
+
===Solution 2===
  
 
Arbitrarily assign a value to one of the variables.  Since <math>c</math> is the smallest variable, let <math>c = 100</math>.
 
Arbitrarily assign a value to one of the variables.  Since <math>c</math> is the smallest variable, let <math>c = 100</math>.
Line 34: Line 36:
 
== 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]]

Revision as of 18:27, 9 August 2011

Problem

If $a$ is $50\%$ larger than $c$, and $b$ is $25\%$ larger than $c$, then $a$ is what percent larger than $b$?

$\mathrm{(A)\ } 20\% \qquad \mathrm{(B) \ }25\% \qquad \mathrm{(C) \ } 50\% \qquad \mathrm{(D) \ } 100\% \qquad \mathrm{(E) \ }200\%$

Solution

Solution 1

Translating each sentence into an equation, $a = 1.5c$ and $b = 1.25c$.

We want a relationship between $a$ and $b$. Dividing the second equation into the first will cancel the $c$, so we try that and get:

$\frac{a}{b} = \frac{1.5}{1.25}$

$\frac{a}{b} = \frac{150}{125}$

$\frac{a}{b} = \frac{6}{5}$

$a = 1.2b$

In this case, $a$ is $1.2 - 1 = 0.2 = 20\%$ bigger than $b$, and the answer is $\boxed{B}$.

Solution 2

Arbitrarily assign a value to one of the variables. Since $c$ is the smallest variable, let $c = 100$.

If $a$ is $50\%$ larger than $c$, then $a = 150$.

If $b$ is $25\%$ larger than $c$, then $b = 125$.

We see that $\frac{a}{b} = \frac{150}{125} = 1.2$ So, $a$ is $20\%$ bigger than $b$, and the answer is $\boxed{B}$.


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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1997_AHSME_Problems/Problem_4&oldid=41377"