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The decimal r=2.666666 continues forever
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10 Nov 2025, 01:04
10 Nov 2025, 01:04
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82% (01:04) correct
17% (01:06) wrong based on 35 sessions
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The decimal $\(r=2.666666\)$ continues forever in that repeating decimal pattern. When written as a fraction in lowest terms,\( r=\frac{a}{b} \), where a and b are positive numbers.
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
\(a+b\) |
\(10\) |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
- Part of the project: GRE Daily Quant Challenge - Topic-Wise 2025!!!
- Also replying to the unanswered questions
Re: The decimal r=2.666666 continues forever
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10 Dec 2025, 10:08
10 Dec 2025, 10:08
Expert Reply
$\(r=2.666666 \ldots\)$ as a repeating decimal equals $\(\frac{8}{3}\)$, so $\(a=8\)$ and $\(b=3\)$, giving $\(a+b=11\)$, which is greater than 10 , so Quantity $A$ is greater.
Converting to a fraction
Let $\(r=2 . \overline{6}\)$, meaning the digit 6 repeats forever.
Then $\(r=2+0 . \overline{6}=2+\frac{2}{3}=\frac{6}{3}+\frac{2}{3}=\frac{8}{3}\)$.
Thus $\(a=8, b=3\)$, and $\(a+b=8+3=11\)$, which is greater than 10 , so Quantity $\(\mathrm{A}>\)$ Quantity B .
Converting to a fraction
Let $\(r=2 . \overline{6}\)$, meaning the digit 6 repeats forever.
Then $\(r=2+0 . \overline{6}=2+\frac{2}{3}=\frac{6}{3}+\frac{2}{3}=\frac{8}{3}\)$.
Thus $\(a=8, b=3\)$, and $\(a+b=8+3=11\)$, which is greater than 10 , so Quantity $\(\mathrm{A}>\)$ Quantity B .
gmatclubot
Re: The decimal r=2.666666 continues forever [#permalink]
10 Dec 2025, 10:08
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