<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)%
b)n/(2n+75)%
c)n/(100(2n+75)%
d)100n/(n+75)%
e)100n/(2n+75)%</p>
<p>Answer is e. When done algebraically it makes perfect sense, but when you do it by plugging in for a value of n and then evaluating the answer choices, its come out to be d. Its weird. any help?</p>
<p>Men = n
Women = n +75
Men/(Women + Men) *100 = percent of men enrolled
100n/(2n+75)</p>
<p>Let men = 15. Women then equals 90.
percent men enrolled is 15/105 *100 = 150/105
e. 100(15)/(2(15)+75) = 150/(30+75) = 150/105
You probably made a mistake while you were plugging in.</p>
<p>there are 75 MORE women than men.</p>
<p>n=75
women=150</p>
<p>E works</p>
<p>Thanks, I made a dumb mistake…When pluggin in I was using the amount of women as the total instead of the women+men…Wow, I really need to stop making dumb mistakes everywhere. I know the concepts to each math problem I miss but all of the ones I do are just dumb/careless mistakes…Anyone have advice to limiting these? They mean the potential difference between an 800 and a 690(thats big) to me.</p>
<p>Actually, I end up making more mistakes if I plug in rather than using variables. But that’s just me and a few others. After I finish the section, I go over each problem again, using either a different method or retracing my steps. This is usually when I plug in.</p>
<p>the only ones that are prone to careless errors are the ones later in the section.</p>
<p>get through the first half or first 3/4 of the section as quickly yet efficiently as possible. then use the rest of your time to carefully and slowly read the “hard” problems.</p>