If kmn ≠ 0, is x/m*(m^2 + n^2 + k^2) = xm + yn + zk?

Question

If  kmn
eq{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}

Bunuel · Accepted Answer

SOLUTION

If  kmn
eq{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.

MulanQ · Answer

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

Manofsteel · Answer

Simplifying the question statement given:-
x/y*(m^2+n^2+k^2)=xm+yn+zk
x/y*m^2+x/y*n^2+x/y*k^2 = xm+yn+zk

Thus simplified equation:-
xm+x/y*n^2+x/y*k^2 = xm+yn+zk

Now Statement 1:- z/k = x/m
Substituting this in the simplified equation above we get - 
xm+x/y*n^2+zk = xm+yn+zk
Since no other info is provided  this statement is insufficient.

Note -   Looking at this itself u can figure out that the only info needed to prove both sides of the above equation equal is x/m = y/n. This can be achieved by combining both the statements. So from here itself u can directly conclude that Option (C) is the answer

Statement 2:- x/m = y/n
Substituting this in the simplified equation above we get - 
xm+yn+x/y*k^2 = xm+yn+zk
Since no other info is provided  this statement is insufficient.

Combining both the statements we get x/m = y/n = z/k
Substituting this in the simplified equation we get 
xm+yn+zk = xm+yn+zk

Thus both sides are proved to be equal.
 Hence Option (C) is the answer.

This question can be solved much faster as explained earlier since both statements are exactly the same. i.e they provide exactly half of the solution.

charronecrom · Answer

Can someone please check my work to see if this is a valid approach:

x/m (m^2 + n^2 + k^2) = xm + yn + zk --> n^2x + k^2x = mny + kmz --> n^2x - mny = kmz - k^2y -->  n(nx - my) = k (mz - kx)

1) mz = kx --> mz - kx = 0 
n(nx - my) = 0 --> Not Sufficient

2) nx = my --> nx - my = 0
0 = k (mz - kx) --> Not Sufficient

T) 0 = 0 --> Sufficient

satishov · Answer

Bunuel  why kmn#0 is given in the question stem ?

Bunuel · Answer

k, m, and n are in the denominators. Since we cannot divide by 0, then we should rule out this case because otherwise the fractions won't be defined.

ScottTargetTestPrep · Answer

Solution:

Question Stem Analysis:
 
We need to determine: Is x/m * (m^2 + n^2 + k^2) = xm + yn + zk ? If it is, then we have:

xm + x/m * n^2 + x/m * k^2 = xm + yn + zk

x/m * n^2 + x/m * k^2 = yn + zk

So the question becomes: Is x/m * n^2 + x/m * k^2 = yn + zk ?

Statement One Alone:

Since z/k = x/m, we have:

z/k * n^2 + z/k * k^2 = yn + zk ?

z/k * n^2 + zk = yn + zk ?

z/k * n^2 = yn ?

However, since there is no way we can determine whether z/k * n^2 = yn, statement one alone is not sufficient.

Statement Two Alone: 
 
Since x/m = y/n, we have:

y/n * n^2 + y/n * k^2 = yn + zk ?

yn + y/n * k^2 = yn + zk ?

y/n * k^2 = zk ?

However, since there is no way we can determine whether y/n * k^2 = zk, statement two alone is not sufficient.

Statements One and Two Together: 
 
Since z/k = x/m and x/m = y/n, in the equation x/m * n^2 + x/m * k^2 = yn + zk, we can substitute the first x/m with y/n and the second x/m with z/k to obtain:

y/n * n^2 + z/k * k^2 = yn + zk ?

yn + zk = yn + zk ?

We see that the answer is yes. Both statements together are sufficient.

Answer: C

KanikaJain · Answer

If kmn≠0, is (x/m)∗(m^2+n^2+k^2)=xm+yn+zk

(1) z/k=x/m
(2) x/m=y/n

Equation: xm + (xn^2/m) + (xk^2/m)

Statement 1:
x/m = z/k thus: 
xm + (xn^2/m) + (xk^2/m) = xm + (xn^2/m)+zk

Statement 2:
x/m=y/n thus: 
xm + (xn^2/m) + zk = xm + yn +zk

thus C

If kmn ≠ 0, is x/m*(m^2 + n^2 + k^2) = xm + yn + zk?

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If kmn ≠ 0, is x/m*(m^2 + n^2 + k^2) = xm + yn + zk?

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