1954 AHSME Problems/Problem 28

Revision as of 00:31, 28 February 2020 by Rayfish (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 (ProblemsAnswer KeyResources)
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