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# If a, b, and c are positive numbers, a+b+c=?

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 4877
GPA: 3.82
If a, b, and c are positive numbers, a+b+c=? [#permalink]

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21 Feb 2017, 00:08
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If a, b, and c are positive numbers, a+b+c=?

$$1) a^2+b^2+c^2=56$$
$$2) ab+bc+ca=10$$
[Reveal] Spoiler: OA

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If a, b, and c are positive numbers, a+b+c=? [#permalink]

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21 Feb 2017, 00:39
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MathRevolution wrote:
If a, b, and c are positive numbers, a+b+c=?

$$1) a^2+b^2+c^2=56$$
$$2) ab+bc+ca=10$$

$$(a+b+c)^2 = a^2+b^2+c^2 + 2*(ab+bc+ca)$$

St I
a^2+b^2+c^2=56

St II
ab+bc+ca=10

Combining St I and II, we have both the terms which are enough to find $$(a+b+c)^2$$ which in turn will give (a+b+c)
$$(a+b+c)^2 = a^2+b^2+c^2 + 2*(ab+bc+ca)$$
$$(a+b+c)^2 = 56 + 20 = 76$$
$$(a+b+c) = 76^{0.5}$$

Hence Option C is correct
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Math Revolution GMAT Instructor
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Re: If a, b, and c are positive numbers, a+b+c=? [#permalink]

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23 Feb 2017, 00:06
==> In the original condition, there are 3 variables (x, y, z) and in order to match the number of variables to the number of equations, there must be 3 equations. Since there is 1 for con 1) and 1 for con 2), E is most likely to be the answer. By solving con 1) and con 2), you get $$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)=56+2(10)=76$$, which becomes $$a+b+c=\sqrt{76}$$, hence it is unique and sufficient.

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Re: If a, b, and c are positive numbers, a+b+c=? [#permalink]

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23 Feb 2017, 01:41
MathRevolution wrote:
If a, b, and c are positive numbers, a+b+c=?

$$1) a^2+b^2+c^2=56$$
$$2) ab+bc+ca=10$$

Hi,

Here there is no necessity to calculate the value if we know the formula of $$(a+b+c)^2$$ then it is more than enough

So $$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$$

$$1) a^2+b^2+c^2=56$$ We don't the value of ab+bc+ca--Not Suff

$$2) ab+bc+ca=10$$ We don't the value of $$a^2+b^2+c^2=56$$--Not Suff

By Combining 1 and 2 it is suff

Hence C
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Intern
Joined: 27 Feb 2017
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Location: United States
GMAT 1: 720 Q42 V47
GPA: 3.1
Re: If a, b, and c are positive numbers, a+b+c=? [#permalink]

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01 Mar 2017, 19:38
MathRevolution wrote:
If a, b, and c are positive numbers, a+b+c=?

$$1) a^2+b^2+c^2=56$$
$$2) ab+bc+ca=10$$

In the first condition, why can't we just take the square root of both sides and get a+b+c=sqrt(56) ?

Thanks for the help.
Senior Manager
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Re: If a, b, and c are positive numbers, a+b+c=? [#permalink]

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01 Mar 2017, 20:06
josephlassen wrote:
MathRevolution wrote:
If a, b, and c are positive numbers, a+b+c=?

$$1) a^2+b^2+c^2=56$$
$$2) ab+bc+ca=10$$

In the first condition, why can't we just take the square root of both sides and get a+b+c=sqrt(56) ?

Thanks for the help.

Hi josephlassen,

$$\sqrt{}(a^2+b^2+c^2)=56==>a+b+c=\sqrt{}(56)$$ This is not right,However $$\sqrt{}(a+b+c)^2=a+b+c$$ is right

$$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$$
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Resource: GMATPrep RCs With Solution

Re: If a, b, and c are positive numbers, a+b+c=?   [#permalink] 01 Mar 2017, 20:06
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