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 "1954 AHSME Problems/Problem 28"

(Created page with "== Problem 28== If <math>\frac{m}{n}=\frac{4}{3}</math> and <math>\frac{r}{t}=\frac{9}{14}</math>, the value of <math>\frac{3mr-nt}{4nt-7mr}</math> is: <math> \textbf{(A)}\...")
 
 
(One intermediate revision by one other user not shown)
Line 15: Line 15:
 
Because the ratio works for any set of integers satisfying <math>\frac{m}{n}=\frac{4}{3}</math> and <math>\frac{r}{t}=\frac{9}{14}</math>, it has to satisfy <math>m=4</math>, <math>n=3</math>, <math>r=9</math>, and <math>t=14</math>. From here it is just simple arithmetic.
 
Because the ratio works for any set of integers satisfying <math>\frac{m}{n}=\frac{4}{3}</math> and <math>\frac{r}{t}=\frac{9}{14}</math>, it has to satisfy <math>m=4</math>, <math>n=3</math>, <math>r=9</math>, and <math>t=14</math>. From here it is just simple arithmetic.
  
−
<math>\frac{3\cdot4\cdot9-3\cdot14}{4\cdot3\cdot14-7\cdot4\cdot9}\implies \frac{3(36-14)}{4(42-63)}\implies \frac{3(22)}{4(-21)}\implies \boxed{\frac{-11}{14} (\textbf{B})}</math>
+
<math>\frac{3mr-nt}{4nt-7mr}\implies\frac{3\cdot4\cdot9-3\cdot14}{4\cdot3\cdot14-7\cdot4\cdot9}\implies \frac{3(36-14)}{4(42-63)}\implies \frac{3(22)}{4(-21)}\implies \boxed{\frac{-11}{14} (\textbf{B})}</math>
 +
 
 +
==See Also==
 +
 
 +
{{AHSME 50p box|year=1954|num-b=27|num-a=29}}
 +
 
 +
{{MAA Notice}}

Latest revision as of 00:31, 28 February 2020

Problem 28

If $\frac{m}{n}=\frac{4}{3}$ and $\frac{r}{t}=\frac{9}{14}$, the value of $\frac{3mr-nt}{4nt-7mr}$ is:

$\textbf{(A)}\ -5\frac{1}{2}\qquad\textbf{(B)}\ -\frac{11}{14}\qquad\textbf{(C)}\ -1\frac{1}{4}\qquad\textbf{(D)}\ \frac{11}{14}\qquad\textbf{(E)}\ -\frac{2}{3}$

Solution 1

From $\frac{m}{n}=\frac{4}{3}$, we have $3m=4n$. From $\frac{r}{t}=\frac{9}{14}$, we have $14r=9t\implies 7r=4.5t$

This simplifies the fraction to $\frac{4nr-nt}{4nt-7r\cdot m}\implies \frac{4nr-nt}{4nt-4.5mt}\implies \frac{4nr-nt}{4nt-1.5t\cdot3m}\implies \frac{4nr-nt}{4nr-1.5\cdot t\cdot 4n}\implies \frac{\frac{4\cdot7r}{7}t-nt}{4nt-6nt}\implies \frac{\frac{4\cdot9t}{7\cdot2}n-nt}{-2nt}\implies \frac{nt(\frac{36}{14}-1)}{-2(nt)}\implies\frac{\frac{22}{14}}{-2}\implies \boxed{\frac{-11}{14} (\textbf{B})}$

Solution 2

Because the ratio works for any set of integers satisfying $\frac{m}{n}=\frac{4}{3}$ and $\frac{r}{t}=\frac{9}{14}$, it has to satisfy $m=4$, $n=3$, $r=9$, and $t=14$. From here it is just simple arithmetic.

$\frac{3mr-nt}{4nt-7mr}\implies\frac{3\cdot4\cdot9-3\cdot14}{4\cdot3\cdot14-7\cdot4\cdot9}\implies \frac{3(36-14)}{4(42-63)}\implies \frac{3(22)}{4(-21)}\implies \boxed{\frac{-11}{14} (\textbf{B})}$

See Also

1954 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 27 		Followed by
Problem 29
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions


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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1954_AHSME_Problems/Problem_28&oldid=118541"