1968 AHSME Problems/Problem 8

Problem

A positive number is mistakenly divided by $6$ instead of being multiplied by $6.$ Based on the correct answer, the error thus committed, to the nearest percent, is :

$\text{(A) } 100\quad \text{(B) } 97\quad \text{(C) } 83\quad \text{(D) } 17\quad \text{(E) } 3$

Solution

Let $n$ be the original number in question. The number divided by 6 is $\frac{n}{6}$, and the number multiplied by 6 is $6n$. Because $\frac{n}{6}=\frac{1}{36}*6n$, and $\frac{1}{36}=0.02 \overline{7} \approx 0.03=3$%, the error (which is our deviation from the correct value) is about $97$%. Thus, our answer is $\fbox{B}$.

See also

1968 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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
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