


Common Word Problems
The writers of the Math IC love word problems. These problems
force you to show your range as a mathematician. They demand that
you read and comprehend the problem, set up an equation or two,
and manipulate the equations to find the solution. Luckily, the
Math IC uses only a few types of word problems, and we have the
nittygritty on all of them.
Rates
A rate is a ratio of related qualities that have different
units. For example, speed is a rate that relates the two quantities
of distance and time. Here is the general rate formula:
No matter the specifics, the key to a rate problem is
correctly placing the given information in the three categories.
Then, you can substitute the values into the rate formula. We’ll look
at the three most common types of rate: speed, work, and price.
Speed
In the case of speed, time is quantity a and
distance is quantity b. For example, if you traveled
for 4 hours at 25 miles per hour, then:
Note that the hour units canceled out, since the hour
in the rate is at the bottom of the fraction. But you can be sure
that the Math IC test won’t simply give you one of the quantities and
the rate and ask you to plug it into the rate formula. Because rate
questions are in the form of word problems, the information that
you’ll need to solve the problem will often be given in a less straightforward
manner.
Here’s an example:

This question provides more information than
simply the speed and one of the quantities. We know unnecessary
facts such as how Jim is traveling (by rollerblades) and when he
started (in the morning). Ignore them and focus on the facts you
need to solve the problem.
 Time a: x hours rollerblading
 Rate: 6 miles per hour
 Quantity b: 60 miles
So, we can write:
Jim was rollerblading for 10 hours. This problem requires
a little analysis, but basically we plugged some numbers into the
rate equation and got our answer. Here’s a slightly more difficult
rate problem:

Immediately, you should pick out the given rate of 528,000
feet per hour and notice that 480 miles are traveled. You should
also notice that the question presents a units problem: the given
rate is in feet cycled per hour, and the distance traveled is in
miles.
Sometimes a question will give you inconsistent units,
like in this example. Always read over the problem carefully and
don’t forget to adjust the units—the answer choices are bound to
include nonadjusted options, just to throw you off.
For this question, since we know there are 5,280 feet
in a mile, we can find the rate for miles per hour:
We can now plug the information into the rate formula:
 Time: x hours cycling
 Rate: 100 miles per hour
 Distance: 480 miles
So it takes the cyclist 4.8 hours to finish the race.
Work
In work questions, you will usually find the first quantity
measured in time, the second quantity measured in work done, and
the rate measured in work done per time. For example, if you knitted
for 8 hours and produced two sweaters per hour, then:
Here is a sample work problem. It is one of the harder
rate questions you might come across on the Math IC:

First, let’s examine what that problem says: 4 men can
dig a 40 foot well in 4 days. We are given a quantity of work of
40 feet and a time of 4 days. We need to create our own rate, using
whichever units might be most convenient, to carry over to the 8men
problem. The group of 4 men dig 40 feet in 3 days. Dividing 40 feet
by 4 days, you find that the group of 4 digs at a pace of 10 feet
per day.
From the question, we know that 8 men dig a 60
foot well. The work done by the 8 men is 60 feet, and they work
at a rate of 10 feet per day per 4 men. Can we use this information
to answer the question? Yes. The rate of 10 feet per day per 4 men
converts to 20 feet per day per 8 men, which is the size of the
new crew. Now we use the rate formula:
 Time: x days of work
 Rate: 20 feet per day per eight men
 Total Quantity: 60 feet
This last problem required a little bit of creativity—but
nothing you can’t handle. Just remember the classic rate formula
and use it wisely.
Price
In rate questions dealing with price, you will usually
find the first quantity measured in numbers of items, the second
measured in price, and the rate in price per item. Let’s say you
had 8 basketballs, and you knew that each basketball cost $25 each:
Percent Change
In percentchange questions, you will need to determine
how a percent increase or decrease affects the values given in the
question. Sometimes you will be given the percent change, and you
will have to find either the original value or new value. Other
times, you will be given one of the values and be asked to find
the percent change. Take a look at this sample problem:

This is a percentchange question in which you need to
find how the original value is affected by a percent increase. First,
to answer this question, you should multiply 72 by .20 to see what
the change in score was:
Once you know the score change, then you should add it
to his original average, since his new average is higher than it
used to be:
It is also possible to solve this problem by multiplying
the golfer’s original score by 1.2. Since you know that the golfer’s
score went up by twenty percent over his original score, you know
that his new score is 120% higher than his old score. If you see
this immediately, you can skip a step and multiply 72 1.2 = 86.4.
Here’s another example of a percentchange problem:

In this case, you have the original price and the sale
price and need to determine the percent decrease. All you need to
do is divide the amount by which the quantity changed by the original
quantity. In this case, the shirt’s price was reduced by 20 – 14
= 6 dollars. So, 620 = .3, a 30%
drop in the price of the shirt.
Double Percent Change
A slightly trickier version of the percentchange question
asks you to determine the cumulative effect of two percent changes
in the same problem. For example:

One might be tempted to say that the bike’s price is discounted
30% + 20% = 50% from its original price, but the key to solving
double percentchange questions is to realize that each percentage
change is dependent on the last. For example, in the problem we
just looked at, the second percent decrease is 20 percent of a new,
lower price—not the original amount. Let’s work through the problem
carefully and see. After the first sale, the price of the bike drops
30 percent:
The second reduction in price knocks off an additional
20 percent of the sale price, not the original
price:
The trickiest of the tricky percentage problems go a little
something like this:

If this question sounds too simple to be true; it probably
is. The final price is not the same as the original. Why? Because
after the price was increased by 20 percent, the reduction in price
was a reduction of 20 percent of a new, higher price. Therefore,
the final price will be lower than the original. Watch and learn:
Now, after the price is reduced by 20%:
Double percent problems can be more complicated than they
appear. But solve it step by step, and you’ll do fine.
Exponential Growth and Decay
These types of word problems take the concept of percent
change even further. In questions involving populations growing
in size or the diminishing price of a car over time, you need to
perform percentchange operations repeatedly. Solving these problems
would be timeconsuming without exponents. Here’s an example:

To answer this question, you might start by calculating
the population after one year:
Or use the faster method we discussed in percent increase:
After the second year, the population will have grown
to:
And so on and so on for 48 more years. You may already
see the shortcut you can use to avoid having to do, in this case,
50 separate calculations. The final answer is simply:
In general, quantities like the one described in this
problem are said to be growing exponentially. The formula for calculating
how much an exponential quantity will grow in a specific number
of years is:
Exponential decay is mathematically equivalent to negative
exponential growth. But instead of a quantity growing at a constant
percentage, the quantity shrinks at a constant percentage. Exponential
decay is a repeated percent decrease. That is why the formulas that
model these two situations are so similar. To calculate exponential
decay:
The only difference between the two equations is that
the base of the exponent is less than 1, because during each unit
of time the original amount is reduced by a fixed percentage. Exponential
decay is often used to model population decreases, as well as the
decay of physical mass.
Let’s work through a few example problems to get a feel
for both exponential growth and decay problems.
Simple Exponential Growth Problems

The question, with its growing population of bacteria,
makes it quite clear that this is an exponential growth problem.
To solve the problem, you just need to plug the appropriate values
into the formula for a repeated percent increase. The rate is .035,
the original amount is 100, and the time is 6 hours:
Simple Exponential Decay Problem

Since the beach ball loses air, we know this is an exponential
decay problem. The decay rate is .06, the original amount is 4000
cubic centimeters of air, and the time is 10. Plugging the information
into the formula:
More Complicated Exponential Growth Problem

This problem is a bit tricky for the simple reason that
the interest on the account is compounded monthly. This means that
in the 2 years that question refers to, there will be 2 12 = 24 compoundings of interest.
The time variable in the equation is affected by these monthly compoundings:
it will be 24 instead of 2. Thus, our answer is:
Here’s another compounding problem:

Sam’s account will have $2000 1.05^{15} ≈
$4157.85 in it after 15 years. Chris’s account will have $2500 1.04^{15} ≈
$4502.36 in it. So, Chris’s account will still have more
money in it after 15 years. Notice, however, that Sam’s account is gaining
on Chris’s account.
