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# S97-17

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Math Expert
Joined: 02 Sep 2009
Posts: 54371

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16 Sep 2014, 01:52
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Difficulty:

75% (hard)

Question Stats:

32% (02:17) correct 68% (01:38) wrong based on 50 sessions

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If $$ab \neq 0$$ and $$a \neq -3b$$, what is the value of $$\frac{(4a + 6b)}{(a + 3b)}$$?

(1) $$a - 3b = 6$$

(2) $$\frac{2a}{(a + 3b)} = 4$$

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Math Expert
Joined: 02 Sep 2009
Posts: 54371

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16 Sep 2014, 01:52
Official Solution:

We cannot simplify the given expression very much, because the denominator (which is a sum, $$a + 3b$$) is not a factor of the numerator. If we really wanted to, we could split the numerator and write the expression as a sum:

$$\frac{4a}{(a + 3b)} + \frac{6b}{(a + 3b)} =$$ ?

Or we could leave the question as is. Either way, be sure not to cancel any of the coefficients, because the denominator is a sum - we can’t simply cancel the 6 in the numerator with the 3 in the denominator, for instance.

Statement 1: INSUFFICIENT. This gives us a relationship between $$a$$ and $$b$$. However, if we use it to solve for one of the variables and then we substitute that expression into the question, we'll quickly see that we will not get a single number:

From the statement: $$a = 6 + 3b$$.

Substitute into the original question:

$$\frac{4(6 + 3b) + 3b}{6 + 3b + 3b} =$$ ?

We can stop here if we see that the denominator is $$6 + 6b$$, which will not cancel with the numerator of the combined fraction (which equals $$24 + 15b$$).

Statement 2: SUFFICIENT. We can get a constant ratio between $$a$$ and $$b$$, which will actually cancel in the question.

From the statement:
$$\frac{2a}{(a + 3b)} = 4$$
$$2a = 4a + 12b$$
$$-2a = 12b$$
$$a = -6b$$

Substitute into the question:
$$\frac{4(-6b) + 6b}{-6b + 3b} =$$
$$= \frac{-24b + 6b}{-3b}$$
$$= \frac{-18b}{-3b}$$
$$= 6$$

Note that it is okay to cancel out the $$b$$'s, since $$ab \neq 0$$ and thus neither variable equals 0.

As long as we have a constant ratio between $$a$$ and $$b$$, we will get a number out of an expression such as $$\frac{4a + 6b}{a + 3b}$$.

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Joined: 24 Feb 2017
Posts: 35
Schools: CBS '20 (S)
GMAT 1: 760 Q50 V42

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13 Jul 2017, 23:34
1
Rephrasing the stem:

$$\frac{(4a+6b)}{(a+3b)}$$

= $$\frac{2*(2a+3b)}{(a+3B)}$$

= $$\frac{2*(a + a + 3b)}{(a+3b)}$$

= $$\frac{2a}{(a+3b)} + \frac{2(a+3b)}{(a+3b)}$$

= $$\frac{2a}{(a+3b)} + 2$$

STATEMENT 1 Only gives the difference, so the sum a+3b can be any value. So, INSUFFICIENT
STATEMENT 2 Directly gives the value of the fraction $$\frac{a}{(a+3b)}$$ So, SUFFICIENT

(B)
Re: S97-17   [#permalink] 13 Jul 2017, 23:34
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# S97-17

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