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# If (a + 2b)/(17b - 2a) = 1, which of the following must be true about

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Joined: 02 Sep 2009
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If (a + 2b)/(17b - 2a) = 1, which of the following must be true about  [#permalink]

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28 Aug 2018, 05:25
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Difficulty:

15% (low)

Question Stats:

89% (00:44) correct 11% (01:14) wrong based on 54 sessions

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If $$\frac{(a+2b)}{(17b-2a)}=1$$, which of the following must be true about the relationship between a and b?

A. a is 3 more than b.

B. b is 3 more than a.

C. a is $$\frac{1}{5}$$ of b.

D. a is 5 times b.

E. a is $$\frac{3}{5}$$ of b.

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Re: If (a + 2b)/(17b - 2a) = 1, which of the following must be true about  [#permalink]

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28 Aug 2018, 05:40
Bunuel wrote:
If $$\frac{(a+2b)}{(17b-2a)}=1$$, which of the following must be true about the relationship between a and b?

A. a is 3 more than b.

B. b is 3 more than a.

C. a is $$\frac{1}{5}$$ of b.

D. a is 5 times b.

E. a is $$\frac{3}{5}$$ of b.

Simplifying the given expression,
$$\frac{(a+2b)}{(17b-2a)}=1$$
Or, a+2b=17b-2a
Or, 3a=15b
Or, a=5*b
Or, a is 5 times b.

Ans. (D)
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Re: If (a + 2b)/(17b - 2a) = 1, which of the following must be true about  [#permalink]

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28 Aug 2018, 06:06
Bunuel wrote:
If $$\frac{(a+2b)}{(17b-2a)}=1$$, which of the following must be true about the relationship between a and b?

A. a is 3 more than b.

B. b is 3 more than a.

C. a is $$\frac{1}{5}$$ of b.

D. a is 5 times b.

E. a is $$\frac{3}{5}$$ of b.

$$\frac{(a+2b)}{(17b-2a)}=1$$

Or, $$a + 2b = 17b - 2a$$

Or, $$3a = 15b$$

Or, $$a = 5b$$, Answer must be (D)
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Re: If (a + 2b)/(17b - 2a) = 1, which of the following must be true about  [#permalink]

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30 Aug 2018, 17:04
Bunuel wrote:
If $$\frac{(a+2b)}{(17b-2a)}=1$$, which of the following must be true about the relationship between a and b?

A. a is 3 more than b.

B. b is 3 more than a.

C. a is $$\frac{1}{5}$$ of b.

D. a is 5 times b.

E. a is $$\frac{3}{5}$$ of b.

Simplifying, we have:

a + 2b = 17b - 2a

3a = 15b

a = 5b

So a is 5 times b.

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Re: If (a + 2b)/(17b - 2a) = 1, which of the following must be true about &nbs [#permalink] 30 Aug 2018, 17:04
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