beater wrote:If the average (arithmetic mean) of x, y, and 20 is 10 greater than the average of x, y, 20, and 30, what is the average of x and y?
A. 40
B. 45
C. 60
D. 75
E. 95
I'm curious why you'd look for a shortcut here if you find the problem straightforward? The algebra only takes three or four steps:
(x+y+20)/3 - 10 = (x+y+20+30)/4
4x + 4y + 80 - 120 = 3x + 3y + 150
x + y = 190
(x+y)/2 = 95
I did the problem a different way at first, but I'd recommend the algebraic approach above- guaranteed to work in under two minutes, and not too difficult to set up.
Still, you could also do the following, which takes a while to explain, but took about fifteen seconds to actually do:
-we know from the answer choices that x+y is equal to one of the following:
80
90
120
150
190
-Let S = sum of the first set, and T = sum of the second. We know S and T are integers, and that
S/3 - T/4 = 10
The only way this could be true is if S is divisible by 3 and T is divisible by 4 -- if S/3 is 1/3 or 2/3 greater than an integer, you'll never get an integer by subtracting T/4, that is, by subtracting some number of quarters. If that's not convincing, you can multiply by 12 on both sides and rearrange:
4S = 120 + 3T
and since the right side is divisible by 3, so must be the left side- i.e. so must be S, since 4 is not divisible by 3.
So you're looking for a value of x+y which makes x+y+20 divisible by 3, and E is the only option.
Still, this approach requires a bit of number theoretic analysis that I wouldn't expect everyone to see quickly, and since the algebra here is quite easy to work with, that seems a better approach to me.
