<p>20 percent of 70 percent of an integer is equal to x percent of 40 percent of that same integer. What is the value of x?</p>
<p>.14(integer) = .4 of x% of an integer
the integer doesn’t even matter</p>
<p>divide both sides by .4</p>
<p>7/20 = x%
to get rid of the %, multiply by both sides by 100</p>
<p>so 35 = x</p>
<p>the first line of this solution isn’t really the conventional way of setting up an algebraic equation but hey, it worked.</p>
<p>alternatively, once you’ve figured out the integer remains constant (taken out of the problem), you can notice that since all the relevant numbers in play are percents, you can ignore the %'s on all of them.</p>
<p>then it just becomes 20X70 = 40x, and since 40 is twice as large as 20, x must be twice as small (one-half as large) as 70.</p>
<p>let a = integer</p>
<p>0.2(0.7a) = x(0.4a)
0.14a = 0.4xa
0.14 = 0.4x
x = 0.35 or 35%</p>
<p>Therefore, x = 35.</p>
<p>Edit: Oops, beaten to it XD;</p>
<p>It’s all about interpreting words and then substituting in expressions:
(.2)(.7)i = (x/100)(.4)i –> i’s cancel, solve for x –> x = (100(.2)(.7))/(.4).</p>
<p>I just did what constance did, but accounted for x being a percent earlier.</p>
<p>The above solutions are correct and they would work no matter what the numbers in the problem were. But as is often the case on the SAT, there’s a lazier, non-algebraic way:</p>
<p>You are ending up at the same answer having taken two percents. The first time, it was 20, then 70. The next time it was 40 and we-don’t-know. But they didn’t choose the numbers randomly to put in the problem…40 is twice as much as 20 – so the other percentage taken must be half as much as 70 to end up on the same result. It must be 35. </p>
<p>I know that this only worked because the numbers were nice – but so what?</p>