  # Math questions

<p>21) What is the value of y when 2^y = 15 ?</p>

<p>23) In 1980, ratio of male students to female students at Williams College was 2 males to 3 females. Since then, the enrollment of male students in the college has increased by 400 and the enrollment of female students has remained the same. The ratio of males to females is currently 1 to 1. How many students are currently enrolled at Williams College? (Answer is 2400).</p>

<p>21) Take the natural log (ln) of both sides:</p>

<p>ln (2^y) = ln 15</p>

<p>You can bring the y down as a coefficient (it's a property of logarithms)</p>

<p>y*ln 2 = ln 15</p>

<p>Divide the ln 2 over so you get y.</p>

<p>23) You know the ratio of M:F in 1980 and the present. You also know what number of males changed the ratio from 2:3 to 3:3 (I changed 1:1 to 3:3). The male side of the proportion increased from 2 to 3 with the addition of 400 students. So, "1" in the proportion means 400 students. Now, there are 6 groups of 400 students (3 on the M side, 3 on the F side). 6*400=2400 students.</p>

<p>You won't see a question like 21 on the regular SAT: logarithms are not tested.</p>

<p>You <em>wlll</em> get exponent questions that can be solved without logs.</p>

<p>E.g.: 3^y + 3^y + 3^y = 81, y=?</p>

<p>Is there any easier way to break #23?</p>

<p>I vaguely remember a question like #23 on one of Barron's practice tests. What if it didn't end up to be 1:1? What if it ended up to be 4:5 or some weird ratio?</p>

<p>If someone could come up with a template to deal with this kind of problems, it would be very helpful.</p>

<p>
[quote]
I vaguely remember a question like #23 on one of Barron's practice tests. What if it didn't end up to be 1:1? What if it ended up to be 4:5 or some weird ratio?</p>

<p>If someone could come up with a template to deal with this kind of problems, it would be very helpful.

[/quote]

I don't recommend memorizing a template. It is best to try to understand how to do the problem with reasoning.</p>

<p>Ratios represent fractions.</p>

<p>Let's say we keep the same question but change the ratios to 2:3 to 4:5.</p>

<p>2:3 implies that, for every 2 boys, there are 3 girls. So, for every 5 people, uniformly, there are 2 boys and 3 girls.</p>

<p>2:3 --> 2/(2+3) = 2/5 of the old group are boys
4:5 --> 4/(4+5) = 4/9 of the new group are boys</p>

<p>Let X = total number of people initially
Let X+400 = total number of people after 400 boys enter the school</p>

<p>number of boys initially + 400 = number of boys now
(2/5 of the initial population) + 400 = 4/9 of the population now
(2/5)<em>X + 400 = (4/9)</em>(X+400)</p>

<p>Solve for X, then, to answer the question, find X+400 (the total number of students now).</p>

<hr>

<p>Here is similar work applied to the original problem:
2:3 to 1:1</p>

<p>2/(2+3) = 2/5 of original group are boys
1/(1+1) = 1/2 of new group are boys</p>

<p>(2/5)<em>X + 400 = (1/2)</em>(X+400)
(2/5)<em>X + 400 = (1/2)</em>X + 200
(2/5)<em>X - (1/2)</em>X = -200
(4/10)<em>X - (5/10)</em>X = -200
(-1/10)<em>X = -200
X = (-200)</em>(-10) = 2000
X+400=2400</p>