1968 AHSME Problems/Problem 34

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Problem

With $400$ members voting the House of Representatives defeated a bill. A re-vote, with the same members voting, resulted in the passage of the bill by twice the margin by which it was originally defeated. The number voting for the bill on the revote was $\frac{12}{11}$ of the number voting against it originally. How many more members voted for the bill the second time than voted for it the first time?

$\text{(A) } 75\quad \text{(B) } 60\quad \text{(C) } 50\quad \text{(D) } 45\quad \text{(E) } 20$

Solution

Let the number of votes for the bill the first time be $a$ and the number of votes for the bill the second time be $b$. Then, from the problem, we know that $b=\frac{12}{11}(400-a)$. Furthermore, the original vote's margin of defeat was $(400-a)-a=400-2a$, and the revote's margin of victory was $b-(400-b)=2b-400$. Thus, from the problem, we know that $2(400-2a)=2b-400$. Substituting the value of $b$ we obtained earlier into this equation, we see that: 2(4002a)=21211(400a)4004002a=400121112a11200200a=4006116a11100300240011=5a115a=330024005a=900a=180 From this fact, we can conclude that $c=\frac{12}{11}*(400-180)=240$, so $c-a=\boxed{60}$, which is answer choice $\fbox{B}$.

See also

1968 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 33
Followed by
Problem 35
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