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19 Aug 2010, 03:29
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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) \(\frac{x}{m}=\frac{y}{n}\)
This is just an awful arithmetic problem. Can't I just pick numbers and PIN them? The problem is: as soon as I choose numbers for all of the variables, I ALWAYS get a definite answer...
19 Aug 2010, 07:57
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heyholetsgo
Is x/m*(m^2+n^2+k^2)=xm+yn+zk?
(1) z/k=x/m
(2) x/m=y/n
This is just an awful arithmetic problem. Can't I just pick numbers and PIN them? The problem is: as soon as I choose numbers for all of the variables, I ALWAYS get a definite answer...
(1) z/k=x/m
(2) x/m=y/n
This is just an awful arithmetic problem. Can't I just pick numbers and PIN them? The problem is: as soon as I choose numbers for all of the variables, I ALWAYS get a definite answer...
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: is \(xm^2+xn^2+xk^2=xm^2+ynm+zkm\)?
\(xm^2\) 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 \(xn^2+xk^2=ynm+zkm\)? --> substitutein from (1) and (2) --> is \(xn^2+xk^2=xn^2+xk^2\)? Answer is YES. Sufficient.
Answer: C.
22 Dec 2010, 13:02
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A question with too many variables should not be done by picking numbers. Takes too much time, plus prone to errors.
then again, if you are short on time, and have barely half a minute, plug in and run.. Let me take both methods:
Proportion: This question tests your understanding of ratios and proportion.
Concept tested: If \(\frac{x}{m} = \frac{y}{n} = \frac{z}{k}\), then \(\frac{x}{m} = \frac{y}{n} = \frac{z}{k} = \frac{{x+y+z}}{{m+n+k}}\)
Question: Is \(\frac{x}{m}(m^2+n^2+k^2)=xm+yn+zk\)?
or Is \(\frac{x}{m}=\frac{xm+yn+zk}{(m^2+n^2+k^2)}\)?
1.)x/m = z/k
Since we have no information about n and y, this is not sufficient. If you are not convinced, put x = 0, z = 0.
Question becomes: Is \(0=\frac{yn}{(m^2+n^2+k^2)}\)?
We do not know since this depends on the values of y and n. Not sufficient.
2.)x/m=y/n
Similarly statement 2 alone is also not sufficient.
Using both together, we get \(\frac{x}{m} = \frac{y}{n} = \frac{z}{k}\)
which is same as \(\frac{x}{m}=\frac{xm}{m^2} = \frac{yn}{n^2} = \frac{zk}{k^2} = \frac{xm+yn+zk}{(m^2+n^2+k^2)}\)
Sufficient. Answer C.
Plugging in numbers:
1. x/m = z/k
Put x = 0, z = 0.
Question becomes: Is \(0=\frac{yn}{(m^2+n^2+k^2)}\)?
We do not know since this depends on the values of y and n. Not sufficient.
2. x/m=y/n
Put x = 0, y = 0.
Question becomes: Is \(0=\frac{zk}{(m^2+n^2+k^2)}\)?
We do not know since this depends on the values of z and k. Not sufficient.
Using both together, put x = 0, y = 0, z = 0
Question becomes: Is 0=0? Answer 'Yes'
Try putting x = 2, y = 2, z = 2, m = 1, n = 1, k = 1
It satisfies so mark the answer as (C) and move ahead. Remember, this is not foolproof. In some case, you could have taken values which satisfied the equation but not those which did not satisfy it. But if you don't have much time to spare, this gives you a decent probability of getting the answer right.
then again, if you are short on time, and have barely half a minute, plug in and run.. Let me take both methods:
Proportion: This question tests your understanding of ratios and proportion.
Concept tested: If \(\frac{x}{m} = \frac{y}{n} = \frac{z}{k}\), then \(\frac{x}{m} = \frac{y}{n} = \frac{z}{k} = \frac{{x+y+z}}{{m+n+k}}\)
Question: Is \(\frac{x}{m}(m^2+n^2+k^2)=xm+yn+zk\)?
or Is \(\frac{x}{m}=\frac{xm+yn+zk}{(m^2+n^2+k^2)}\)?
1.)x/m = z/k
Since we have no information about n and y, this is not sufficient. If you are not convinced, put x = 0, z = 0.
Question becomes: Is \(0=\frac{yn}{(m^2+n^2+k^2)}\)?
We do not know since this depends on the values of y and n. Not sufficient.
2.)x/m=y/n
Similarly statement 2 alone is also not sufficient.
Using both together, we get \(\frac{x}{m} = \frac{y}{n} = \frac{z}{k}\)
which is same as \(\frac{x}{m}=\frac{xm}{m^2} = \frac{yn}{n^2} = \frac{zk}{k^2} = \frac{xm+yn+zk}{(m^2+n^2+k^2)}\)
Sufficient. Answer C.
Plugging in numbers:
1. x/m = z/k
Put x = 0, z = 0.
Question becomes: Is \(0=\frac{yn}{(m^2+n^2+k^2)}\)?
We do not know since this depends on the values of y and n. Not sufficient.
2. x/m=y/n
Put x = 0, y = 0.
Question becomes: Is \(0=\frac{zk}{(m^2+n^2+k^2)}\)?
We do not know since this depends on the values of z and k. Not sufficient.
Using both together, put x = 0, y = 0, z = 0
Question becomes: Is 0=0? Answer 'Yes'
Try putting x = 2, y = 2, z = 2, m = 1, n = 1, k = 1
It satisfies so mark the answer as (C) and move ahead. Remember, this is not foolproof. In some case, you could have taken values which satisfied the equation but not those which did not satisfy it. But if you don't have much time to spare, this gives you a decent probability of getting the answer right.
