1954 AHSME Problems/Problem 28

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.

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

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1954 AHSME Problems/Problem 28

Problem 28

Solution 1

Solution 2