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abc≠0 and 3/a + 3/b = 3/c. What is the sum a+b?

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abc≠0 and 3/a + 3/b = 3/c. What is the sum a+b?  [#permalink]

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New post 17 May 2017, 23:29
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abc≠0 and 3/a + 3/b = 3/c. What is the sum a+b?

(1) ab=c

(2) ab=ac+bc
Spoiler: OA

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Re: abc≠0 and 3/a + 3/b = 3/c. What is the sum a+b?  [#permalink]

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New post 17 May 2017, 23:55
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Answer A.

3/a + 3/b = 3/c
1/a + 1/b = 1/c
(a+b)/ab = 1/c
a+b = ab/c
St 1: ab = c ; i.e a+b=1----- sufficient
St 2: ab=ac+bc; ab=c(a+b); a+b=ab/c---- cannot find a+b----- Insufficient.
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Re: abc≠0 and 3/a + 3/b = 3/c. What is the sum a+b?  [#permalink]

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New post 20 May 2017, 06:37
obliraj wrote:
Answer A.

3/a + 3/b = 3/c
1/a + 1/b = 1/c
(a+b)/ab = 1/c
a+b = ab/c
St 1: ab = c ; i.e a+b=1----- sufficient
St 2: ab=ac+bc; ab=c(a+b); a+b=ab/c---- cannot find a+b----- Insufficient.


It's because if the we know the product of a and b equals the numerator of 1/c or c then the top we can just set equal to a + b= 1

Hence

A
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Re: abc≠0 and 3/a + 3/b = 3/c. What is the sum a+b?  [#permalink]

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New post 05 Feb 2019, 10:49
abc≠0 and 3/a + 3/b = 3/c. What is the sum a+b?

(1) ab=c

(2) ab=ac+bc

solution:
3/a+3/b=3/c =3(a+b)/ab=3/c.

1.ab=c ,substitute in the above equation we get a+b=1.
A is sufficient

2.ab=c(a+b), substitute in equation above, we get 1/c so it provies the same info that we already know. so not sufficient

A is the correct answer.
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Re: abc≠0 and 3/a + 3/b = 3/c. What is the sum a+b?   [#permalink] 05 Feb 2019, 10:49
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