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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: 6826
GMAT 1: 760 Q51 V42
GPA: 3.82
If a, b, and c are positive numbers, a+b+c=?  [#permalink]

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21 Feb 2017, 00:08
00:00

Difficulty:

45% (medium)

Question Stats:

53% (01:01) correct 47% (01:33) wrong based on 58 sessions

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

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

_________________

MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only $149 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" ##### Most Helpful Community Reply Senior Manager Joined: 19 Apr 2016 Posts: 274 Location: India GMAT 1: 570 Q48 V22 GMAT 2: 640 Q49 V28 GPA: 3.5 WE: Web Development (Computer Software) If a, b, and c are positive numbers, a+b+c=? [#permalink] ### Show Tags 21 Feb 2017, 00:39 5 1 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 no info about ab+bc+ca ------------Insufficient St II ab+bc+ca=10 no info about a^2+b^2+c^2 ------------Insufficient 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 Hit Kudos if you liked it ##### General Discussion Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6826 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: If a, b, and c are positive numbers, a+b+c=? [#permalink] ### Show Tags 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. Therefore, the answer is C. Answer: C _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$149 for 3 month Online Course"
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Director
Joined: 06 Jan 2015
Posts: 570
Location: India
Concentration: Operations, Finance
GPA: 3.35
WE: Information Technology (Computer Software)
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
Posts: 2
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.
Director
Joined: 06 Jan 2015
Posts: 570
Location: India
Concentration: Operations, Finance
GPA: 3.35
WE: Information Technology (Computer Software)
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

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

आत्मनॊ मोक्षार्थम् जगद्धिताय च

Resource: GMATPrep RCs With Solution

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