# Math problem in the SAT?

<p>Is there a fast way to answer these questions without using too much algebra? it took me 10 minutes to solve these.</p>

<p>At a certain college, the number of freshmen is three times the number of seniors. If 1/4 of the freshmen and 1/3 of the seniors attend a football game, what fraction of the total number of freshmen and seniors attends the game?
A.) 5/24
B.) 13/48
C.) 17/48
D.) 11/24
E.) 23/48

<p>At Jones College, there are a total of 100 students. If 30 of the students have cars on campus, and 50 have bicycles, and 20 have both cars and bicycles, then how many students have neither a car nor a bicycle on campus?
A.) 80
B.) 60
C.) 40
D.) 20
E.) 0</p>

<p>For the first one:</p>

<p>(Without much algebra) Say there are 100 students at the school. Because Freshman=3Seniors OR 1F = 3S
1F + 1S = 100
3S + 1S = 100
S=25
from the other equation we find that there are 75 freshman and 25 seniors
1/4 (freshman) = 1/4 (75) = 18.75
1/3 (senior) = 1/3 (25) = 8.3333333333...
[(18.75 + 8.333)/100] = 27.0833333/100 = 13/48
When you work it out on paper it takes like 1-2 minutes
I always use my own numbers when I can't think of a straightforward algebra equation</p>

<p>36 freshman and 12 seniors (assumption)</p>

<p>1/4 freshman = 9, 1/3 seniors = 4, 4+9 = 13, 48 students = 13/48</p>

<p>For the second one, if you think about it like this, 30 people who have cars minus 20 who have both equals 10 who have only cars, and 50 people who have bikes minus the 20 who have both equals 30 people who have only bikes, so 10 plus 30 plus the 20 with both equals 60 people with at least one, and 100-60=40</p>

<p>pick numbers!
first one:
let's say there are 20 seniors. you know there would have to be 60 freshmen! (80 total)
now if 1/4 of freshmen went, thats 15 of them
if 1/3 of seniors went, thats 6.6 of them
(15+6.6)/80 = 13/48</p>

<p>second one:
30 have bikes. but of those, 20 also have a car. so you add the ones that only have a car, which is 50-20 = 30. 30+30=60 with a vehicle. 100-60=40 without.</p>