<p>There are 75 more women than men enrolled in Linden College. If there are n men enrolled, then, in terms of n, what percent of those enrolled are men?</p>
<p>(A) (n / n + 75)%</p>
<p>(B) (n / 2n + 75)%</p>
<p>(C) (n / 100(2n + 75))%</p>
<p>(D) (100n / n + 75)%</p>
<p>(E) (100n / 2n + 75)%</p>
<p>I keep getting B, which is the wrong answer.</p>
<p>Does anyone know how to do this? This the last question in section 3 of the math section in the blue book test 2.</p>
<p>The answer is E. You’re doing it right, except for it asks for the answers in terms of percent, so you have to multiply by 100. Thus, getting 100n instead of (1)n.</p>
<p>Isn’t it D? </p>
<p>Say there are 25 men, 75 women–100 overall. 25% of the enrolled students, therefore, are men.</p>
<p>(100n / n + 75)%
(100*25 / 100) = 25%</p>
<p>I think it is D also…</p>
<p>
There’s 75 more women than men, not 50. You would want 25 men, 100 women. In that case men are 20% of the total, as E would predict.</p>
<p>There are 75 more women then men so your senario of 25 men and 75 women cannot be possible. It would have to be 25 men and 100 women with n=25. women would be n+ 75. </p>
<p>So % men would be n/n+ (n+75) * 100. </p>
<p>100n/ 2n+75 %.</p>
<p>^Correct. OP, you only needed to multiply by 100, because it asks for a percentage answer. That is the “trick” part of the question.</p>
<p>The answer page says E,</p>
<p>but I’m still somewhat confused. If the next SAT had a question like this, I would still get it wrong.</p>
<p>So from a big picture view, is it:</p>
<p>Men/Men+Women
= n / n+(n+ 75)
= (n / 2n + 75) X 100%
= 100n/2n+75</p>
<p>Is that the process?</p>
<p>To find percent #of boys/total students * 100. N=number of boys n+75=number of girls total=2n+75. </p>
<p>Number of boys= n
number of students= 2n+75</p>
<p>percent= n/2n+75 ( this gives a decimal) * 100 ( this gives a percentage) = 100n/2n+75 E is correct</p>