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23 Feb 2014, 07:41
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C
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Difficulty:
25%
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Question Stats:
80% (02:07) correct
20%
(02:36)
wrong
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If \(kmn\neq{0}\), is \(\frac{x}{m}*(m^2+n^2+k^2)=xm+yn+zk\)?
(1) \(\frac{z}{k} = \frac{x}{m}\)
(2) \(\frac{x}{m} = \frac{y}{n}\)
(1) \(\frac{z}{k} = \frac{x}{m}\)
(2) \(\frac{x}{m} = \frac{y}{n}\)
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23 Feb 2014, 07:42
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SOLUTION
If \(kmn\neq{0}\), is \(\frac{x}{m}*(m^2+n^2+k^2)=xm+yn+zk\)?
Is \(\frac{x}{m}(m^2+n^2+k^2)=xm+yn+zk\)? -->multiply both part by \(m\), to get rid of fraction part and open the brackets --> \(xm^2+xn^2+xk^2=xm^2+ynm+zkm\) --> \(xm^2\) will cancel out and the question becomes is \(xn^2+xk^2=ynm+zkm\)
(1) \(\frac{z}{k}=\frac{x}{m}\) --> \(zm=kx\) --> substitute zm with kx --> is \(xn^2+xk^2=ynm+xk^2\) --> \(xk^2\) will cancel out and the question becomes is "\(xn^2=ynm\)?" Not sufficient.
(2) \(\frac{x}{m}=\frac{y}{n}\) --> \(xn=ym\) --> substitute ym with xn --> is \(xn^2+xk^2=xn^2+zkm\) --> \(xn^2\) will cancel out and the question becomes "is \(xk^2=zkm\)?" Not sufficient.
(1)+(2) is \(xn^2+xk^2=ynm+zkm\)? --> substitute in from (1) and (2) --> is \(xn^2+xk^2=xn^2+xk^2\)? Answer is YES. Sufficient.
Answer: C.
If \(kmn\neq{0}\), is \(\frac{x}{m}*(m^2+n^2+k^2)=xm+yn+zk\)?
Is \(\frac{x}{m}(m^2+n^2+k^2)=xm+yn+zk\)? -->multiply both part by \(m\), to get rid of fraction part and open the brackets --> \(xm^2+xn^2+xk^2=xm^2+ynm+zkm\) --> \(xm^2\) will cancel out and the question becomes is \(xn^2+xk^2=ynm+zkm\)
(1) \(\frac{z}{k}=\frac{x}{m}\) --> \(zm=kx\) --> substitute zm with kx --> is \(xn^2+xk^2=ynm+xk^2\) --> \(xk^2\) will cancel out and the question becomes is "\(xn^2=ynm\)?" Not sufficient.
(2) \(\frac{x}{m}=\frac{y}{n}\) --> \(xn=ym\) --> substitute ym with xn --> is \(xn^2+xk^2=xn^2+zkm\) --> \(xn^2\) will cancel out and the question becomes "is \(xk^2=zkm\)?" Not sufficient.
(1)+(2) is \(xn^2+xk^2=ynm+zkm\)? --> substitute in from (1) and (2) --> is \(xn^2+xk^2=xn^2+xk^2\)? Answer is YES. Sufficient.
Answer: C.
28 Aug 2014, 13:28
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I use plug in approach and find it easier to understand:
Does x/m(m^2 + n^2 + k^2) = xm + yn + zk ?
1. let z/k = 1/2 = x/m = 2/4, then the above becomes:
Does 1/2(16 + n^2 + 4) = 8 + yn + 2 -->insuficient
2. let x/m = 2/4 = y/n = 4/8, can guess that we will have something just like (1) ---> insuficient
1 and 2 together, using the plug in number, we have:
Does 1/2 (16 + 84 + 4) = 8 + 32 + 2 ---> Yes --> Answer is C
Does x/m(m^2 + n^2 + k^2) = xm + yn + zk ?
1. let z/k = 1/2 = x/m = 2/4, then the above becomes:
Does 1/2(16 + n^2 + 4) = 8 + yn + 2 -->insuficient
2. let x/m = 2/4 = y/n = 4/8, can guess that we will have something just like (1) ---> insuficient
1 and 2 together, using the plug in number, we have:
Does 1/2 (16 + 84 + 4) = 8 + 32 + 2 ---> Yes --> Answer is C
