What's the fastest way to do these math problems?

xiggi
Tue, 04 Aug 2009 02:13:36 +0000

If a and b are integers such that a + b < 1000 and a/b = .625, what is the greatest possible value of b?

a = .625b, so .625b + b < 1000 —> b < 615

.625 = 5/8, a = 5/8 b, so b must be divisible by 8. We test down from 615: 614, 612, 610, 608 and find that 608 is the answer.



The largest integer <1000 is 999. Once you establish that .625 is also 5/8, all you need to do is 999/13 (which is 76.xx) and multiply the largest integer (76) by 8 to find 76*8 or 608.