VMHS Math Circle
VIII. The Grand Algebra Review (Part III: Applications)
I couldn’t think of a better name to tie these problems together. What this document will cover
are the types of problems that AMC applies to real-life contexts. Now, these problems are pretty
instinctive. What’s mentioned here is pretty simply stated, and here aren’t a lot of formulas to remember
here. Despite this, the best way to get better at these problems is to practice practice practice!
d = rt
It’s the good, old distance equals rate times time equation. We obviously know that if something
moves at a constant rate after a given time has elapsed, it goes a certain distance. Getting the obvious
aside, the AMC adds some tricks to this equation. Some problems might involve scenarios where one
thing goes faster than another, ones in which one thing stops and goes a certain speed, or ones in which
two objects run around a racetrack. Granted, these seem like pretty shady examples, but the AMC will
definitely throw something like that in there. Let’s see some problems.
Leo and Aaron run along the same linear trail in the same direction at a constant speed. However, Aaron
runs one hour later than Leo. If Leo runs 3 miles per hour, and Aaron runs 5 miles per hour, in how many
hours (after Leo begins) do they meet?
One thing that would help a lot is a diagram:
Leo
Aaron
The circles are there to mark each hour that has elapsed. Keep in mind that going a distance at a
constant rate involves continuity. We can’t arbitrarily match up distances and make assumptions from
there.
Also note that after the third hour, Aaron runs 10 miles while Leo runs 9 miles. Since after the second
hour, Aaron ran 5 miles and Leo ran 6, it has to be between the second and third hour when they meet.
Now that we’ve cleared that up, let’s use d = rt. Note that because Aaron and Leo meet, distance must
be equal.
d = 3t
d = 5(t – 1)
We have that t – 1 there because Aaron runs an hour less than Leo does. Substituting, we get:
3t = 5t – 5  -2t = -5  t = 2.5 hours.
Michael is riding his scooter. He notices that the time he takes to get to a local store is 10 minutes.
However, on a futuristic conveyor belt, Michael gets to the store in one-quarter of the time if he both
rides and is on the booster track. If Michael is on the booster track but not riding his scooter, in how
long will he get to the store? Assume the same starting point and finishing point for all three cases.
Let’s whip out some d = rt. The distance should be the same, so “d” will not change in the equations we
make. It’s clear that one-quarter of 10 minutes is 2.5 minutes. Carrying on, we get:
d = 10r
d = 2.5(r + b)
Note that we define “b” as the speed of the booster pad. Setting these equal to each other, we get:
10r = 2.5r + 2.5b  7.5r = 2.5b  b = 3r
Now, plugging in a dummy value for “r” in 1, we have that b = 3. If we plug in 1, and d = 10 from our first
equation, 10 = 3t because we’re going at 3 units per minute. Thus, t = 10/3.
Assume that both Aaron and Leo are running on the same linear track again in the same direction, with
constant and improved speeds of 10 miles per hour and 6 miles per hour, respectively. This time, Aaron
starts running when Leo is 5 miles ahead of him. In how long will Aaron catch up to Leo?
This one is a bit more confusing. You’d think that we’d have to set two rt’s equal to each other because
two bodies meeting is an indicator that distance is equal, and it is, in a sense.
However, when Leo and Aaron meet, starting with the point that Leo was 5 miles ahead of Aaron,
they would have ran different distances. I’ll make this clearer in my solution.
As a consequence of what I just said, time, rather than distance, should be the constant here. That is, we
make two d/r’s equal to each other. Doing so, we have:
Leo’s time = Aaron’s time  d/6 = (d + 5)/10  10d = 6d + 30  d = 7.5.
The (d + 5) is there for the expression we make for Aaron because he has to run (d + 5) meters to meet
with Leo. This is the main reason why this problem is different than the first. Again, both people meet
when they have run different distances.
To finish, since both times are equal, we can just use the expression we had for Leo to calculate time:
7.5/6 = 1.25 hours.
Of course, the AMC will come up with ways to make this harder, but that will come in our
practice problems. Next up,
Work
Every now and then, you’ll run into some problems concerning how fast people can do something
if working together or individually. If you do, make sure to keep these formulae in mind.
General Work Formula
w1t1 = w2t2 = jk, with “w” being the amount of workers, “t” being the time, and “j” being the number of
jobs.
This is primarily used when you’re dealing with how long it takes to complete a job or something related,
given some extra information. Note that “k” is basically a constant that scales “j” up to meet “wt.” Let’s
put in a sample problem.
If 20 workers can make 2 computers in 1 day, how long does it take for 10 workers to make 4
computers, assuming that each worker works at constant and identical rates?
Let the number of computers be equal to the number of jobs. Using our general work equation, we
have: (20)(1) = 2k, so “k” should be equal to 10. Plugging into our work equation again, we have:
(10)(t) = (4)(10)  t = 4 days.
Cooperative Work Formula
(Together Time)/(First Individual’s Time) + (Together Time)/(Second Individual’s Time) = 1
There exist two types of problems: input and output. There’s no complicated explanation for how to solve
either of these, but let’s start off by explaining which involves what. Input problems involve an input,
obviously, such as time, or resources. Similarly, output problems tell you the outputs, or what is being
produced or accomplished. The AMC only focuses in the scope of time. By saying this, I mean that any
work problem that the AMC throws at you only has to do with the time it takes for individuals to
complete a task. A couple problems might help.
Worker A can finish a job in 3 hours. When working at the same time as Worker B, they can finish the
job in 2 hours. How long does it take for Worker B to finish the job if he works alone?
Simply enough, just plug in your knowns into the equation I gave you. Let “x” denote the time it takes
for worker B to finish his job alone.
2/3 + 2/x = 1  2/x = 1/3  x = 6 hours.
A 10,000 liter pool is filled by two pipes: A and B. Pipe A delivers 1,000 liters per hour. When pipe A and
B are both on, they can fill this pool in 4 hours. How many liters per hour can pipe B deliver?
This is a perfect example of an output problem, since you’re given the rate at which Pipe A fills the pool.
The perfect thing to do with output problems is to convert them into input problems.
If Pipe A delivers 1,000 L per hour, how long does it take for a 10,000 L pool to be filled? That’s right, 10
hours. We know our together time: 4 hours. We can also convert the 10,000 L pool into 1 job. Let’s plug
in our knowns and let “t” be equal to the time it takes for pipe B to transfer water.
4/10 + 4/t = 1  4t + 40 = 10t  40 = 6t  t = 20/3
Notice what we didn’t find what we were looking for! We’re looking for pipe B’s rate, and to do so, we
divide 10,000 L by 20/3 hours to get our rate in liters per hour. Doing so, we get 1,500 L/hr.
Alternatively, you could do this in a much simpler way. What’s our rate if we fill a 10,000 L pool in 4
hours? 2,500 L/hr. If one pipe fills at a rate of 1,000 L/hr, at what rate must the other pipe fill the pool
for us to get 2,500 L/hr? 2,500 L/hr – 1,000 L/hr = 1,500 L/hr.
One thing to note here: “application” problems will comprise some of the first 10 or 15 problems.
You definitely won’t see them in the final five problems. One more category to go!
Percents/Ratios
There are some problems on the AMC that deal with fractions and percents. It is clear that such
percent/ratio problems deal with a real-life context, such as money or other scenarios. Sometimes, the
AMC tells us to express a certain quantity with a variable expression. To clarify, let’s get to our usual
drill of some practice problems:
Arlen has a fund of $1,500. 10% of the initial fund is spent on groceries while 25% of what is left is spent
on bills. How much of the fund is left after these two payments?
Well, we know that 10% of 1,500 is 150. 1350 is left from this if Arlen pays 150 bucks. 25% of 1350 is
1350/4 = 337.5. 1350 – 337.5 = $1012.50.
“x” workers can produce “y” computers at a constant rate, in “z” days. What expression shows the time
it takes for “a” workers to produce “b” computers at this rate, using only a, b, x, y, and z?
If you’re stuck, one great thing to do is use dummy values that are easy to work with.
Let’s say that 5 workers can produce 10 computers in 2 days. This rate is 1 computer a day. At this rate,
6 workers can produce 18 computers in 3 days. However, we can only use a, b, x, y, and z, so we make
these variables correspond to the numbers we have in the following way:
a=6
b = 18
x=5
y = 10
z=2
We want to rearrange to get 3, so after some experimenting, we get:
bxz/ay.
In a class, 20% of the students are girls. After 4 girls leave and are replaced with boys, 10% of the
students are girls. How many students are in the initial class?
Let “I” be the initial amount of people. From the first sentence, we know that 0.2I = g. If four girls leave
and are replaced with boys, the initial population remains the same. Hence, we can form a relation from
our second sentence. This is 0.1I = g – 4. Substituting for g, we have that 0.1I = 0.2I – 4. Solving for I, we
get that -0.1I = 4  I = 40.
TIPS:
 Please remember that an increase of a certain rate over a certain period of time compounds. For
example, if the number 100 was increased by 10% twice, the final answer would not be 120! It
would be 121. Why? A 10% increase of 100 is 110, and a 10% increase of 110 is 121. Always
assume that this is the case unless the AMC tells you that a certain quantity increases by a
specific amount throughout a certain period of time rather than a specific percent.
 There are some cases in which you can and cannot add speed.
 Times in which you can add speed or rate include instances in which someone is sped up/slowed
by something. For example, if a kayaker was rowing at 2 mph along a 5 mph current, that
kayaker’s speed would increase to 7 mph.
 However, if you’re trying to find average speed, do not add speeds together! Take the total
distance over the total time to find average speed. Examples of this would include instances in
which you’re trying to find average speed or gas mileage given data for different trips. Why do
we need to do this? Speed affects the time it takes to go places. If we go faster, then we spend less
time with our faster speed to get to a certain place. We have to adjust for that.
 In a work problem, if two things oppose one another’s work, subtract the slower/less efficient
force from the faster/more efficient force! Say, if a pipe took one hour to fill a tank and a pipe
took two hours to drain a tank, take x – x/2 = 1 to find the time it takes for the tank to be filled if
both pipes were open.