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...

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? --> substitutein from (1) and (2) --> is xn^2+xk^2=xn^2+xk^2? Answer is YES. Sufficient.

Answer: C.
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What is PIN? Picking numbers? I think for numbers with this many variables, it's much better to go for a conceptual approach. It's too easy to make a careless math error, or some weird math property or cancellation comes up.
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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.
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if kmn not equal to zero, is x/m(m^2+n^2+k^2)=xm+yn+zk?
(1) z/k=x/m
(2) x/m=y/m

i took too much to solve these problem. Is there any short cut way?
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Question:

If kmn<>0 is (x/m)(m²+n²+k²)=xm+yn+zk?

1) z/k=x/m

2)x/m=y/n

This took me about 5 minutes to solve, could anybody here solve this in approx. 2 minutes?

Is there another way besides just simplifying?

This is question 106 from the official quant review guide.