M is what percent of X?

[deltasquare]

If m > 0, y > 0, and x is m percent of 2y, then, in terms of y, m is what percent of x?
1. y/200
2. 2y
3. 50y
4. 50/y
5. 5000/y

Since this is a variables in choice, I assumed numbers.
Assumption 1:
Say y=10 and m = 50,
Then x = 50% of 20 = 10.
Since m = 50 and x = 10, m is 500% of x.
In terms of y, m is 50y percent of x. Hence choice is (3)

Assumption 2:
Say y = 100 and m = 40,
Then x = 40% of 200 = 80.
Since m = 40 and x = 80, m is 50% of x.
In terms of y, m is 5000/y% (5000/100=50%) of x. Hence choice is (5)

Assumption 3:
Say y =50 and m=50,
Then x = 50% of 100 = 50.
Since m = 50 and x = 50, m is 100% of x.
In terms of y, m is 2y (2*50) percent of x. Hence choice is (2)

Obviously assuming different numbers give difference answer choices. Why does this happen?
I know the correct answer is choice (5) if I use normal algebraic deduction.

Given, x = m% of 2y = m2y/100.
Rewriting, m = 100x/2y = 50x/y.
Converting to percent m = 50x/y * 100 = 5000x/y = (5000/y)*x

[tmmyc]

This issue is the fact that the numbers chosen for Assumption 1 and Assumption 3 allow for two answers to work.

Assumption 1:
3. 50y
5. 5000/y

y=10

50*10 = 5000/10 = 500


Assumption 3:
2. 2y
5. 5000/y

y=50

2*50 = 5000/50 = 100


You must test all answer choices when solving by picking numbers. When picking numbers results in more than one possible answer choice, you must select another set of numbers to see which one is actually correct. In this example, if you combine Assumption 1 and 3, you can see Choice 5 must be the answer.

[StaceyKoprince]

Well put, tmmyc. This is also why we need to follow the rules when picking numbers!

- don't pick zero or one
- don't pick the same number for two different variables (case 3 above)
- don't pick numbers that show up frequently in the answers (case 1 above)

Following the above rules won't guarantee that you won't get two answer choices that work, but it will greatly reduce your chances.

If you do get unlucky and pick something that works with more than one choice, do the problem again with different numbers (or, just pick one of those two and move on, if you think you've spent too much time already - after all, you've narrowed it down to two!). And it shouldn't generally take too long to do the problem again with different numbers because you've already done it one - you just have to do the exact same work with two different numbers.