Browse · harp

smc

prealgebra senior

Problem

The fraction

\frac{1}{3}

:

(A)

equals 0.33333333

(B)

is less than 0.33333333 by \frac{1}{3 \cdot 1 0 ^{8}}

(C)

is less than 0.33333333 by \frac{1}{3 \cdot 1 0 ^{9}}

(D)

is greater than 0.33333333 by \frac{1}{3 \cdot 1 0 ^{8}}

Solution

\frac{33333333}{1 0 ^{8}} - \frac{1}{3} ⟹ \frac{3 \cdot 1 0 ^{7} + 3 \cdot 1 0 ^{6} + \dots + 3 \cdot 1 0 ^{0}}{1 0 ^{8}} - \frac{1}{3} ⟹ \frac{3 ( 1 0 ^{8} - 1 )}{9 \cdot 1 0 ^{8}} - \frac{1}{3} ⟹ \frac{1 0 ^{8} - 1}{3 \cdot 1 0 ^{8}} - \frac{1}{3} ⟹ \frac{1 0 ^{8}}{3 \cdot 1 0 ^{8}} - \frac{1}{3 \cdot 1 0 ^{8}} - \frac{1}{3} ⟹ \frac{- 1}{3 \cdot 1 0 ^{8}}

Because we did

.33333333 - \frac{1}{3}

, it is

is greater than 0.33333333 by \frac{1}{3 \cdot 1 0 ^{8}}

Final answer

D