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# 1/(1 + 1/(2 + 1/3))

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1/(1 + 1/(2 + 1/3))  [#permalink]

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Updated on: 25 May 2017, 15:33
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Difficulty:

5% (low)

Question Stats:

89% (00:34) correct 11% (00:35) wrong based on 123 sessions

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$$\frac{1}{1 + \frac{1}{2 + \frac{1}{3}}}$$

A)$$\frac{3}{10}$$

B) $$\frac{7}{10}$$

C) $$\frac{6}{7}$$

D) $$\frac{10}{7}$$

E) $$\frac{10}{3}$$

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Originally posted by carcass on 25 May 2017, 14:57.
Last edited by Bunuel on 25 May 2017, 15:33, edited 1 time in total.
Renamed the topic and edited the question.
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Re: 1/(1 + 1/(2 + 1/3))  [#permalink]

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25 May 2017, 15:12

Calculations in the attached image.
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Untitled.png [ 42.01 KiB | Viewed 1503 times ]

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Re: 1/(1 + 1/(2 + 1/3))  [#permalink]

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25 May 2017, 15:34
carcass wrote:
$$\frac{1}{1 + \frac{1}{2 + \frac{1}{3}}}$$

A)$$\frac{3}{10}$$

B) $$\frac{7}{10}$$

C) $$\frac{6}{7}$$

D) $$\frac{10}{7}$$

E) $$\frac{10}{3}$$

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Re: 1/(1 + 1/(2 + 1/3))  [#permalink]

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25 May 2017, 22:56
2 + 1/3 = 7/3. And 1/(7/3) = 3/7
Now, 1 + 3/7 = 10/7 and 1/(10/7) = 7/10

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Re: Quantitative :: Problem solving:: 14017  [#permalink]

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16 Nov 2019, 13:59
brandonag12 wrote:
$$\frac{1}{1+\frac{1}{2+\frac{1}{3}}$$

A. $$\frac{3}{10}$$
B. $$\frac{7}{10}$$
C. $$\frac{6}{7}$$
D. $$\frac{10}{7}$$
E. $$\frac{10}{3}$$

Quantitative :: Problem solving:: 14017

Multiply the top and bottom of $$\frac{1}{2+\frac{1}{3}}$$ by 3 to get an easier start (and it does not change the value of the fraction so this is valid).

We get $$\frac{1}{1+\frac{3}{7}}$$, we can repeat the same process and multiply top and bottom by 7 to get 7 / (7+3) = 7/10.

Ans: B
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Re: Quantitative :: Problem solving:: 14017   [#permalink] 16 Nov 2019, 13:59
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