1982 AHSME Problems/Problem 2

Problem

If a number eight times as large as $x$ is increased by two, then one fourth of the result equals

$\text{(A)} \ 2x + \frac{1}{2} \qquad  \text{(B)} \ x + \frac{1}{2} \qquad  \text{(C)} \ 2x+2 \qquad  \text{(D)}\ 2x+4 \qquad \text{(E)}\ 2x+16$

Solution

The number $8$ times as large as $x$ will be $8x$, and $8x$ increased by two will give $8x+2$. Hence finally, the answer is $\frac{1}{4}(8x+2) = \boxed{\text{(A)}\ 2x + \frac{1}{2}}$.

See Also

1982 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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