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# If abc ≠ 0 and the sum of the reciprocals of a, b, and c equals the re

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If abc ≠ 0 and the sum of the reciprocals of a, b, and c equals the re [#permalink]

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16 Sep 2015, 03:56
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If abc ≠ 0 and the sum of the reciprocals of a, b, and c equals the reciprocal of the product of a, b, and c, then a =

A. (1 + bc)/(b + c)
B. (1 – bc)/(b + c)
C. (1 + b + c)/(bc)
D. (1 – b – c)/(bc)
E. (1 – b – c)/(b + c)

Kudos for a correct solution.
[Reveal] Spoiler: OA

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Re: If abc ≠ 0 and the sum of the reciprocals of a, b, and c equals the re [#permalink]

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16 Sep 2015, 05:38
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(1/a)+(1/b)+(1/c) = 1/abc

(bc+ac+ab)/abc = 1/abc ---> bc+ac+ab=1 ---> ac+ab = 1-bc ---> a(b+c)= 1-bc ---> a= (1 - bc)/(b+c)

Ans- B

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Re: If abc ≠ 0 and the sum of the reciprocals of a, b, and c equals the re [#permalink]

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16 Sep 2015, 07:03
Bunuel wrote:
If abc ≠ 0 and the sum of the reciprocals of a, b, and c equals the reciprocal of the product of a, b, and c, then a =

A. (1 + bc)/(b + c)
B. (1 – bc)/(b + c)
C. (1 + b + c)/(bc)
D. (1 – b – c)/(bc)
E. (1 – b – c)/(b + c)

Kudos for a correct solution.

(1/a)+(1/b)+(1/c) = 1/(a*b*c)

i.e. (bc+ac+ab)/abc = 1/abc

i.e. bc+ac+ab=1

i.e. ac+ab = 1-bc

i.e. a(b+c)= 1-bc

i.e. a= (1 - bc)/(b+c)

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If abc ≠ 0 and the sum of the reciprocals of a, b, and c equals the re [#permalink]

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17 Sep 2015, 14:54
Bunuel wrote:
If abc ≠ 0 and the sum of the reciprocals of a, b, and c equals the reciprocal of the product of a, b, and c, then a =

A. (1 + bc)/(b + c)
B. (1 – bc)/(b + c)
C. (1 + b + c)/(bc)
D. (1 – b – c)/(bc)
E. (1 – b – c)/(b + c)

Kudos for a correct solution.

$$\frac{1}{a}$$ + $$\frac{1}{b}$$ + $$\frac{1}{c}$$ = $$\frac{1}{abc}$$

or $$\frac{bc + ac + ab}{abc}$$ = $$\frac{1}{abc}$$

or $$ac + ab$$ = $$1 - bc$$

or $$a(c+b)$$ = $$1 - bc$$

or $$a$$ = $$\frac{(1-bc)}{(b+c)}$$

Kudos [?]: 191 [0], given: 59

Math Expert
Joined: 02 Sep 2009
Posts: 42279

Kudos [?]: 132880 [0], given: 12390

Re: If abc ≠ 0 and the sum of the reciprocals of a, b, and c equals the re [#permalink]

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20 Sep 2015, 21:06
Bunuel wrote:
If abc ≠ 0 and the sum of the reciprocals of a, b, and c equals the reciprocal of the product of a, b, and c, then a =

A. (1 + bc)/(b + c)
B. (1 – bc)/(b + c)
C. (1 + b + c)/(bc)
D. (1 – b – c)/(bc)
E. (1 – b – c)/(b + c)

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

First, translate the given information into an equation. Go phrase by phrase. The sum of the reciprocals of a, b, and c is 1/a + 1/b + 1/c. Notice that you first take reciprocals, then you add the reciprocals together.

Now, set that equal to “the reciprocal of the product of a, b, and c,” which is 1/(abc). Notice that we first take the product of a, b, and c (which is abc), and then take the reciprocal of that product.

The equation is this:
1/a + 1/b + 1/c = 1/(abc)

Now rearrange to isolate a on one side. Make a common denominator on the left side (abc), so that you can add the fractions:
1/a + 1/b + 1/c = bc/(abc) + ac/(abc) + ab/(abc) = (bc + ac + ab)/(abc)

Since the right side of the original equation is 1/(abc), which happens to have the same denominator, you can set the numerators equal:

bc + ac + ab = 1

Now solve for a:

ac + ab = 1 – bc
a(c + b) = 1 – bc
a = (1 – bc)/(c + b)

Theoretically, you can solve this problem by plugging numbers for the variables, but finding three consistent values of a, b, and c (to satisfy the complicated condition) is rather difficult. A pure algebraic approach is faster and more secure in this case.

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Re: If abc ≠ 0 and the sum of the reciprocals of a, b, and c equals the re [#permalink]

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18 Oct 2017, 06:33
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Re: If abc ≠ 0 and the sum of the reciprocals of a, b, and c equals the re   [#permalink] 18 Oct 2017, 06:33
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