1968 AHSME Problems/Problem 8

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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

$\fbox{B}$

See also

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