percent problem

[HPengineer]

If w is 40% less than x, x is 40% less then y, and z is 46% less then y, then z is greater than w by what percent of w?

smart number method??

[beat_gmat_09]

If w is 40% less than x, x is 40% less then y, and z is 46% less then y, then z is greater than w by what percent of w?

smart number method??

Pick x = 100
W - 40% less than x = 0.6*100 = 60
X - 100 = (40% less =) 0.6y = 100. y = 1000/6 approx = 166.66 (no need to calculate)
Z - (46% less) = 0.54 * (1000/6) = 0.09 * 1000 = 90

To find answer, (90-60/60) * 100 = 50%

[Brian@VeritasPrep]

Hey guys,

It may not be any easier (but maybe not any harder, either), but you can also do this one using fractions (remember - "percent" just means "over 100" so each percent gives you a fraction!).

If w is 40% less than x, then that means that w is 60% of x, so w = 3/5 x
If x is 40% less than y, then that means that x is 60% of y, so x = 3/5 y
If z is 46% less than y, then that means that z is 54% of y, so z = 27/50 y

NOTE: as ugly as 27/50 is, you should recognize that you have some 3s and 5s in the other fractions...this will ultimately reduce!

They want to get w in terms of z, so let's start to combine:

w = 3/5 x, and x = 3/5 y, so we can say that w = (3/5)(3/5)y

w = 9/25 y

Now to get Y in terms of z, we know that z = 27/50 y, so solving for y we'd have y = 50/27 z.

If w = 9/25 y, and y = 50/27 z, then:

w = (9/25)(50/27)z

Here is where that reduction gets quick - the 50 divides over the 25 and the 9 divides over the 27 leaving

w = 2/3 z

Finally, they want to know "z is greater than w by what percent of w".

So, we need to get z in terms of w. If w = 2/3 z then z = 3/2 w. Then we can phrase the question:

3/2 w is greater than w by what percent of w?

3/2 w - w = x/100 (w)

1/2 w = x/100 w

x = 50%



Now, like I said, this may not be any faster (and could be in fact slower) than picking numbers, but I bring it up because I always emphasize to my students that:

You should pick numbers because you see that it will be an easier/faster way to solve the problem, not because you're scared of the algebra!

If I were writing for GMAC (hint, hint, GMAC...I'd love to), the way I'd turn a 65th percentile percents problem into an 85th percentile percents problem is to structure it so that the "obvious" numbers to pick will lead to awkward math. For example if everyone should pick 100, I'd make it so that you'd have to divide by 3 a few times.

So...I highly recommend being able to use algebra as well as being able to pick numbers, particularly as you study. On test day, you'll probably tend toward whichever is easier for you, but I hope that no one relies only on one strategy in practice as a way to avoid the discomfort of algebra, as there may well be cases in which a question will require algebra so you might as well get used to it in practice.