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

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

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New post 21 Feb 2017, 01:08
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  45% (medium)

Question Stats:

53% (00:46) correct 47% (01:11) wrong based on 55 sessions

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

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

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

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New post 21 Feb 2017, 01:39
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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
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Re: If a, b, and c are positive numbers, a+b+c=? [#permalink]

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New post 23 Feb 2017, 01:06
==> 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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New post 23 Feb 2017, 02:41
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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Re: If a, b, and c are positive numbers, a+b+c=? [#permalink]

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New post 01 Mar 2017, 20:38
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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New post 01 Mar 2017, 21:06
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=?   [#permalink] 01 Mar 2017, 21:06
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