<p>If a and b are integers such that a + b < 1000 and a/b = .625, what is the greatest possible value of b?</p>
<p>a/b = 0.625 implies that a = 0.625b
a + b = 1.625b
1.625b < 1000
b < 615.38
Greatest value of b is 615</p>
<p>hmm... the answer is apparently 608...</p>
<p>the answer is 608:
if a/b is .625, you know that a::b is a ratio, 5::8 (simplify 625/1000, this can be done without a calculator, just divide by 5 twice; it goes to 5/8). so,
5x+8x must be less than 1000
13x=1000
x is about 76.9 (i dont have my calc with me)
so you round down to 76 (or else sum will be greater than 1000)
so A is 5<em>76
and B is 8</em>76, or 608</p>
<p>wait i just realized something- im supposed to round down AFTER multiplying by 8... so i think JoeBloggs is right (615 checks, after all)</p>
<p>::feels stupid:: lol</p>
<p>The problem is that if B=615, then A is not an integer. Ilovemath's method ensures that A and B are integers by rounding off x before multiplying by 5 and 8</p>
<p>no, i love math, you are correct.
joe bloggs has 615 which will not create an integer of "a" for the ratio of 5:8
76.875:615 is 5:8 but both need to integers. that's why you need to base it off the greatest integer for the 1:8 ratio portion.</p>
<p>Yeah, at the very least, one should recognize the ratio of 5:8 and know that the denominator cannot be an odd number...</p>
<p>if this were mc, i'd say look at the choices</p>
<p>ok cool lol</p>
<p>hah, i got 615 too and i thought it was easy....damn that is tricky</p>
<p>ilovemath - great solution!
Joe Bloggs' approach is more natural (probably would come fiirst to 95% of SAT'ers), but his solution lacks a final check that a+b=1000, just to make sure.
When you do that you discover that "a" is not integer. There is no time to look for the alternative solution.
So you keep going:
b < 615.38, </p>
<h1>but a=.625b gotta be integer.</h1>
<p>Part 2.
You need to know all your 8th for the SAT:
1/8=.125
2/8=?
3/8=?
...
5/8=.625
So a=5/8 b, and b has to be a multiple of 8.</p>
<h1>Multiple of 8 closest to 615 is 608 (how do you find it?)</h1>
<p>You might not see part 2 right away on a real test. In this case I'd suggest to move on and return back to this question if time allows.
Rhetorical question:
Would you rather stick with it until done because you've already invested ~1 min. in it?
Unfortunately 615 would definitely be one of the answers. :)</p>
<p>S/he means s/he would check all the answer choices to make sure s/he has the right answer.</p>
<p>Where is this problem from, anyway?</p>
<p>
True.
Plugging in the answer choices would be actually faster then solving.</p>
<p>Ya, just plug them in, multiply by .625, see if you get an integer. IF not, then toss it. If more than one answer yields an integer, add them up.</p>