ABC26GN0723 · Decimal Fractions
Subject: General Aptitude · Chapter: Decimal Fractions · Exam: 2005 · Marks: · Difficulty:
$(8.3 \overline{1}+0 . \overline{6}+0.00 \overline{2})$ is equal to
(a)$8.9 \overline{12}$
(b)$8 . \overline{912}$
(c)$8.9 \overline{79}$
(d)$8.97 \overline{9}$
Answer
Explanation
$$\begin{aligned} (8.3 \overline{1}+0 . \overline{6}+0.00 \overline{2}) & =8+\frac{31-3}{90}+\frac{6}{9}+\frac{2}{900} \\ & =\frac{7200+280+600+2}{900} \\ & =\frac{8082}{900}=8 \frac{882}{900}=8+\frac{979-97}{900}=8.97 \overline{9} . \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.
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