bekkilyn wrote:I agree with mpaudena. Dissect everything you do in practice! That's how you're going to learn exactly why the 1/100 is taken into account in that final equation. Learn exactly what's going on in all the algebra steps for each problem even if you end up plugging in numbers on an exam for more speed. If you don't see what's going on in the algebra, then that may be a sign to go back and brush up on some of those concepts too since they may be needed on a different question that you can't plug in.

It's possible you may be jumping ahead just a bit and that's why the difference between using 25 or 25/100 isn't getting clearer, but you're the one who will know if that's true or not.

Nevertheless, it does seem that you would always just use the 25 without the 1/100 when plugging into the answers for these types of equations. I'm very reluctant to use "always" though, considering that the GMAT likes to be tricky, so still take that remark with a grain of salt!

Where does this VIC strategy actually come from? Is it from one of the testing companies? And why do I keep wanting to call it "Variable Induced Coma"?

Stacey Koprince wrote:I received a PM asking me to comment. I like the "variable induced coma" acronym a lot better.

If you pick numbers for a VIC, and the problem says something like "x percent of blah blah blah" then you literally pick something for x and x alone. If I say x=25, then every place in the problem where it says x, I now read that as 25 instead - literally pretend the problem no longer says x anywhere, in the text or in the answers. Instead, it says 25.

So I'd re-read the problem and it would actually say "25 percent of blah blah blah" - I'm ONLY substituting for the variable itself, not for the other words / concepts in the problem. When you do the math for that problem, you will actually use 25 as a percent, because that's what the problem tells you to do (since the word "percent" is in the problem after the number 25).

Eg, on the original VIC posted here:

z = 10
x = 40
y = 50

Go back and re-read the problem right now, substituting in the above values for the variables ONLY - notice that the word "percent" is still in the problem. Now do the math and notice that you will actually use those numbers above as percentages, once you do the math itself!

Started out costing $10. Then price raised 40 percent, so 10+ 10*(40/100) = 14. Then price decreased 50 percent, so 14 - (14)(50/100) = 7. Final price is 7. Go to my answers and plug in the values for the variables ONLY again:
A: [10,000(10) + (100)(10)(-10) - (10)(40)(50)]/10000 let's get rid of three zeroes first: [(10)(10) + (1)(1)(-10) - (1)(4)(5)]/10 = (100 - 10 - 20)/10 = 70/10 = 7. Bingo.
Now, use your knowledge that A did work to eliminate some other choices.
B is identical to A, except that it says y-x instead of x-y. Is that going to give you the exact same answer? No. B is wrong.
C is identical to A, except that the opening term (10,000z) is missing. That's also not going to give you the same answer, then.
D combines the two errors introduced in B and C. Unlikely to work, though you can check it if you want.
E: (10,000) / [(100)(50)(10) + (40)(50)] Denominator is bigger than numerator, so can't be 7.

As some people have noted above, if you don't have variables in the answers, then you don't pick numbers anyway - so the above doesn't matter. There, you either need to write out an equation and solve (and, yes, you'd write out the equation using the /100) or you'd use the answer choices to work backwards (in which case you may or may not have to write out an equation with something over 100).

Oh, and that problem about Jones elementary - it's not a typo. Part of the trick to the problem is that the number of total students is the same number used to describe the "x percent of boys" part of the problem.
90 = (x/100)B
B = 0.4x
two equations, two variables - plug and chug!

Stacey Koprince wrote:In this case, I do think the VIC method is the easiest, despite the annoying form of the answers.

Notice my discussion of the answer choices: I didn't have to do anything besides look at B, C, and D (and compare them to A's set-up) in order to eliminate them. And then, for E, all I had to do was write it out but not actually simplify it - I just thought about it before continuing to solve and realized it couldn't equal 7. So, I fully solved only one of the answer choices.

It's often the case, with VIC, that you do not have to solve every answer choice fully. (In fact, you should NOT do this.) Don't think of it as "I need to find out what number each answer choice equals." Instead, think of it as "I need to find the answer choice that equals 7." If, as you're looking at or working through a choice, you can tell that it's not going to equal 7 (or whatever you want for the problem you're doing), then cross off that choice immediately and move on to the next one. You can almost always avoid fully solving for at least two choices, and often three (or, if we get lucky, as in this case, four!).

Stacey Koprince wrote:I still would've saved a lot of time after solving A, yes, because B, C, and D share significant parts. Re-use what you've already done.

So I'd go back and look at what I'd written out for A and say, ok, B is the same except that instead of x-y, we have y-x. So that one number, and only that one number, changes to +10 instead of -10. How does that change the outcome of the calculation?

Look at the second line of your work - okay, that one -10 is still intact and it should now say 10 instead. Everything else is the same. So move onto the third line - now we change that -10 to +10: (100 + 10 - 20)/10

And that's all that you really have to do here - just find the one small change and carry that through the problem. The only real math I have to do is right at the end.

Go look at some of these in OG. Whenever they give you ridiculous answer choices (as you have for this problem), there will be significant similarities between multiple choices, so you can re-use some of the work you already did. For others, you'll be able to estimate (as in E).

The whole point is that the test has to build in some shortcuts or there really wouldn't be an easier way to do the problem!

Stacey Koprince wrote:I generally do recommend that someone be scoring in the 80th+ percentile in order to get the most out of the quest for 750 workshops because these workshops touch on only a subset of difficult problems. I personally don't think you're as likely to see these problems in the first place unless you're scoring at the 80+ percentile already.

At the same time - I don't have actual research to back up this particular threshold. I just chose the 80th percentile based on a gut feeling, having worked with this stuff for a while. So maybe it will be helpful at the 70th percentile. Or maybe we're both wrong and it should be more like 90th percentile - I don't know for sure.

All of that is to say: I wouldn't necessarily argue with Mike, because there's room for reasonable disagreement here. But I'm still going to stick with my own general 80th percentile threshold when making recommendations about the quests.

If you do want to do a quest, you may want to look at the SC quest instead; your percentiles there look good. That could help raise your verbal score and get you closer to (if not over) that 700 threshold.

Stacey Koprince wrote:If you're scoring 60%+ with average difficulty levels in the high 600s to low 700s, then yes, you'd likely benefit from the SC Quest. Especially if your numbers have been getting better as you study more - so your numbers are better on your later CATs (and I'm assuming this is the case in making this assessment, but let me know if that's not the case).

Just a note - I also see that you're struggling a bit with NP and algebra. Those two areas are common on the test (especially NP), so definitely do some more work in those areas too.