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S97-17

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S97-17  [#permalink]

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New post 16 Sep 2014, 01:52
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A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

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

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Re S97-17  [#permalink]

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New post 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 \(\frac{4a + 6b}{a + 3b}\).


Answer: B
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Re: S97-17  [#permalink]

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