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# If abc ≠ 0, then (ab + bc + ac)/(1/a + 1/b + 1/c)

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Math Expert
Joined: 02 Sep 2009
Posts: 50009
If abc ≠ 0, then (ab + bc + ac)/(1/a + 1/b + 1/c)  [#permalink]

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15 Mar 2018, 00:17
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Difficulty:

15% (low)

Question Stats:

91% (00:55) correct 9% (01:48) wrong based on 45 sessions

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If abc ≠ 0, then $$\frac{ab + bc + ac}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$$

A. a + b + c

B. abc

C. 1/(a + b + c)

D. 1/(ab) + 1/(ac) + 1/(bc)

E. ab + ac + bc

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If abc ≠ 0, then (ab + bc + ac)/(1/a + 1/b + 1/c)  [#permalink]

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15 Mar 2018, 01:15
Bunuel wrote:
If abc ≠ 0, then $$\frac{ab + bc + ac}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$$

A. a + b + c

B. abc

C. 1/(a + b + c)

D. 1/(ab) + 1/(ac) + 1/(bc)

E. ab + ac + bc

The denominator of the expression can be simplified as follows:
$${\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$$ = $$\frac{bc + ac + ab}{abc}$$

Therefore, the expression $$\frac{ab + bc + ac}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$$ = $$ab + bc + ac*\frac{abc}{bc + ac + ab} = abc$$(Option B)
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Re: If abc ≠ 0, then (ab + bc + ac)/(1/a + 1/b + 1/c)  [#permalink]

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15 Mar 2018, 02:55
1
Bunuel wrote:
If abc ≠ 0, then $$\frac{ab + bc + ac}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$$

A. a + b + c

B. abc

C. 1/(a + b + c)

D. 1/(ab) + 1/(ac) + 1/(bc)

E. ab + ac + bc

Since both question and answer have only variables, we'll pick easy numbers.
This can help avoid simplification errors and is an Alternative approach.

Say a=b=c=1. Then our answers are
A. 3
B. 1
C. 1/3
D. 3
E. 3
Since our expression is (1+1+1)/(1+1+1) = 1, then (B) is our answer.
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Re: If abc ≠ 0, then (ab + bc + ac)/(1/a + 1/b + 1/c) &nbs [#permalink] 15 Mar 2018, 02:55
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