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Math Revolution GMAT Instructor
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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=? \(1) a^2+b^2+c^2=56\) \(2) ab+bc+ca=10\)
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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 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



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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. Therefore, the answer is C. Answer: C
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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+caNot 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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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.



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