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Percentages with Variables

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by pmcch » Thu Feb 21, 2013 7:18 pm
For any positive x, y and z if x is n% of y and z is m% of y then what percentage must x be of z?
A. (n/m)%
B. (m × n)%
C. (100 / [m × n]) %
D. (100 × m/n)%
E. (100 × n/m)%



Here's what I did:

The statement x is n%: x = (n / 100) * y

The statement z is m%: z = (m / 100) * y

Then divided x / z to find percentage.

(n / 100) * y =
(m / 100) * y

= n / m

Therefore x = (n / m) × z I chose answer A.


Why is answer E the correct one?

Thank you so much in advance :)
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by Anurag@Gurome » Thu Feb 21, 2013 9:26 pm
pmcch wrote:The statement x is n%: x = (n / 100) * y
The statement z is m%: z = (m / 100) * y
Then divided x / z to find percentage.
...
Therefore x = (n / m) × z I chose answer A.
What you found is ratio of x and z. Ratio is not same as percentage.
I hope you understand your mistake now. If not, then look at your first line : "x is n%: x = (n / 100) * y"
x is n% of y but x/y = n/100

Here x/z = n/m but x is 100*(n/m)% of z
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by GMATGuruNY » Thu Feb 21, 2013 9:32 pm
pmcch wrote:For any positive x, y and z if x is n% of y and z is m% of y then what percentage must x be of z?
A. (n/m)%
B. (m × n)%
C. (100 / [m × n]) %
D. (100 × m/n)%
E. (100 × n/m)%
Let y=100.
Let n=200.
Since x is n% of y, x = 200% of 100 = 200.
Let m=400.
Since z is m% of y, z = 400% of 100 = 400.
Since x=200 and z=400, x is 50% of z.
Thus, our target is 50%.

Now we plug n=200 and m=400 into the answers to see which yields our target of 50%.
Only E works:
(100 * n/m)% = ( 100 * 200/400 )% = 50%.

The correct answer is E.

Algebraically:

Let y=100.
Since x is n% of y, x = (n/100) * 100 = n.
Since z is m% of y, z = (m/100) * 100 = m.

Question: x is what percent of z?
x/z * 100 = n/m * 100.
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by Anurag@Gurome » Thu Feb 21, 2013 9:40 pm
pmcch wrote:For any positive x, y and z if x is n% of y and z is m% of y then what percentage must x be of z?
Let us assume x = 2, y = 100, and z = 100

Hence, x is 2% of y ---> n = 2
And, z is 100% of y ---> m = 100
And, x is 2% of z

Hence, the correct option must be equal to 2 when n = 2 and m = 100
  • A. (n/m) = 2/100 ---> NO
    B. (m X n) = 200 ---> NO
    C. (100 / [m X n]) 100/200 ---> NO
    D. (100 X m/n) = 100*100/2 ---> NO
    E. (100 X n/m) = 100*2/100 = 2 ---> YES
The correct answer is E.
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by pmcch » Fri Feb 22, 2013 12:30 pm
Thank you!
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by Jeff@TargetTestPrep » Thu Dec 14, 2017 4:16 pm
pmcch wrote:For any positive x, y and z if x is n% of y and z is m% of y then what percentage must x be of z?
A. (n/m)%
B. (m × n)%
C. (100 / [m × n]) %
D. (100 × m/n)%
E. (100 × n/m)%
We are given:

x = (n/100)y

x/(n/100) = y

100x/n = y

AND

z = (m/100)y

z/(m/100) = y

100z/m = y

We can equate our two equations and then express x as a percentage of z:

100x/n = 100z/m

100x/100z = not n/m

x/z = n/m

Thus, x/z * 100 = n/m * 100.

Answer: E

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