VIC Help Needed: Choosing Smarter Values for Variables

[josephgreer]

I've noticed several VIC problems I've gotten wrong simply because of the values I chose for the variables in the question. Sometimes, my chosen values result in more than one of the answers being correct.

What is the best strategy for value selection? Here is an example from the third MGMAT sample test:

if x is m percent of 2y, what percent of x is m?

(a) y/200
(b) 2y
(c) 50y
(d) 50/y
(e) 5000/y

Here's how I approached the problem:
Assign values to X and Y as 5 and 10 respectively, then solve for m.

x = (m/100)(2y) ==> 5 = (m/100)*20 ==> 5 = 20m/100 ==> 5 = m/5 ==> m =25

This makes sense. 5 is 25% of 20. Now, what percent of 5 is 25? We'll call the mystery percentage z.

m/x = z/100 ==> 25/5 = z/100 simplify ==> 5/1 = z/100 now cross-multiply ==> 500 = z

Therefore m is 500% of x. Since y is 10, 500 = 50*10 = 50y. The answer is C.

Upon reviewing the problem, noting I got it wrong, I realized that answer E also worked. 5000/y = 5000/10=500. This has happened a few times on tests. I've chosen some values, worked through the math, then noticed that the solution set included duplicates. Going back and choosing new numbers burns significant time.

What can you recommend to help me select smarter values for my variables??

Thanks

[Guest]

What's the official answer? I keep getting E.

[Guest]

E is the official answer. Basically, if I had chosen some different numbers for X and Y, I would have gotten an answer that only was true for E. In other words, there are numbers you can choose for your variables that result in multiple right answers.

What numbers did you use for X, Y, and M?

[Guest]

I solved it using algebra, because when I pick numbers I always run into the same problem that you stated. My strategy on these problems is to try the algebra first, but if I'm getting stuck in the mud, I will go with picking numbers.

Here's how I solved it:

The trick to this problem I would say is knowing what percent means. If you keep in your mind that percent always means "per hundred" or "out of 100" it helps.

if x is m percent of 2y, what percent of x is m?

x is m percent of 2y : x = (m/100) * 2y = 2ym/100 = ym/50

what percent of x is m?

we're looking for what number divided by 100 multiplied by x is equal to m. this number is not x or y, so I called it n. now I can write an equation.

what percent of x is m? : m = (n/100) * x = nx/100

using the two equations we can solve for n in terms of y.

x = ym/50
m = nx/100

substitue x in the second equation

m = [n * (ym/50)] / 100 = (nym/50)/100 = nym/5000

5000m = nym

n = 5000 / y

[josephgreer]

Thanks for the algebraic solution. I shy away from some of these in favor of the number picking approach. But you are right. The number approach isn't always the best.

One thing I noticed was that the numbers I chose were multiples of 5. Numbers in the answer were also multiples of 5. Perhaps if I had looked at the answer choices and selected numbers that were not multiples of 5, I'd have been better.

While I like the algebraic approach, I find the number picking approach faster. Any feedback from MGMAT STAFF on this one?

[StaceyKoprince]

There are drawbacks to both algebra and picking numbers - it's important to know both techniques and to know how to decide when to use which one (that last is a bit variable from person to person). The biggest drawback to algebra is that the most common careless mistakes for that problem are built into the answer choices, so if you make a mistake, you aren't likely to figure that out b/c you'll get an answer that is actually in the answer choices.

Follow these rules when picking numbers for VIC:
- don't pick zero or one
- don't pick numbers that show up everywhere in the answer choices
- follow any constraints given by the problem (eg, if they tell you it's a positive number, don't pick a negative one)
- use any other logic you can to avoid picking numbers that might do weird things in the answer choices (eg, on this one, 4 and 8 might have been better given that there are lots of multiples of 5 and 10 in the choices)

And test ALL of the answer choices. Don't stop when one works. If you get two that work, do the problem again with different numbers.

Now, that last part always scares my students into thinking "I should never use this method b/c I might have to do it twice and I have to test all of the answers? No way."

So, several more things:
- when testing the answers, STOP whenever you can tell that this choice is not going to be the number you're trying to find; you only care if the number is what you want. You don't actually have to find all 5 numerical values; you just need to know whether it's right (or close to right, in which case, do finish the calculation). For example, on this problem, I don't need to know what y/200 is; I just need to know that it's not going to be 500 (if I used 5 and 10).
- if you follow the above rules for picking numbers, it's unlikely that you'll hit the "two choices work" situation much.
- if you do hit that situation anyway, you will be able to do the problem a second time VERY quickly b/c you've already done it once. All you have to do is run two different numbers through the exact same set-up, and it's extremely unlikely that the second set would cause the same problem.
- finally, if you are testing a second set of numbers, only test them on the two choices you have left; you've already eliminated the other three.