


Common Word Problems
The writers of the Math IIC love word problems. They force
you to show your range as a mathematician. They demand that you
intelligently read and comprehend the problem, set up an equation
or two, and manipulate the equations to find a solution. Luckily,
the Math IIC uses only a few types of word problems, and we have
the nittygritty on all of them.
Rates
Rates are ratios that relate quantities with different
units. For example, speed is a rate that relates the two quantities
of distance and time. In general, all rates on the Math IIC have
the same structure. There will usually be a total quantity, an interval,
and the rate of quantity/interval.
The key to solving a rate problem rests in correctly identifying
the total quantity, the interval, and the rate. You can then substitute
the values into the specific problem. The Math IIC generally asks
three different types of rate problems: speed, work, and price.
Speed
In speed rate problems, time (in units of seconds, minutes,
or hours, etc.) is the interval, and distance (in units of inches,
meters, or miles, etc.) is the total quantity. For example, if you
traveled for 4 hours at 25 miles per hour, then:
Note that the hour units cancel out, since the hour in
the rate is the denominator of the fraction. But you can be sure
that the Math IIC test won’t simply give you one of the quantities
and the rate and ask you to plug it into the rate formula. Since
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
direct manner.
Here’s an example:

As with many word problems, this question provides more
information than we need to solve the problem. 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: x hours rollerblading
 Rate: 6 miles per hour
 Total quantity: 60 miles traveled
So, we can write:
Jim was rollerblading for 10 hours. Here’s a slightly
more difficult rate problem:

This problem gives the rate as 528,000 feet per hour.
It also tells you that a total quantity of 480 miles are traveled
in the race. A complicating factor in this problem is that the problem contains
inconsistent units. In the rate, the distance is measured in feet
while the length of the race is given in miles. Always read over
the problem carefully and don’t forget to adjust the units—you can
bet that the answer you would come to if you had forgotten to correct for
units will be one of the answer choices.
For this question, since we know there are 5280 feet in
a mile, we can find the rate for miles per hour:
We can now plug the information into the rate formula
to determine how many intervals passed to achieve the total quantity
at the given rate:
 Time: x hours cycling
 Rate of Speed: 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 interval
measured in units of time, the total quantity measured in units
of work, and the rate measured in work per time. For example, if
you knitted for 8 hours (think of this as 8 onehour intervals)
and produced 2 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 IIC:

First, let’s examine what the problem says: four men can
dig a 40foot well in 4 days. We are given the total quantity of
work of 40 feet and a time interval of 4 days. We need to create
a rate, using whichever units are most convenient, to carry over
to the 8man problem. Let’s try to figure out how much work each
man gets done in one day. The group of 4 men gets 40 feet of work
done in 4 days, so it is easy to see that the group gets 10 feet
of work done each day. By simply dividing 104 = 2.5, it’s clear to see that each
person gets 2.5 feet of work done per day.
The question asks us to determine how long it would take
8 men to dig a 60foot well. The total quantity is 60 feet, and
the rate is 2.5 feet per man per day. The 8man group gets 8 2.5 = 20 feet of work done per day.
 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 total quantity measured in a number of items, the interval
measured in price (usually dollars or cents), and a rate measured
in price per item. For example, if you had 8 basketballs, and you
knew that each basketball cost twentyfive dollars, you could determine
the total cost of the 8 basketballs:
Percent Change
Percentchange questions ask you to determine how a percent
increase or decrease affects the values given in the question. There
are two variations of percentchange questions: one where you are
given the percent change and asked to find either the original value
or new value; other times you will be given two of the values and
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 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, which allows you to combine
the 20% increase and the original score in one step. Since you know
that the golfer’s score went up by 20% 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 change by the original quantity. In this case,
the shirt’s price was reduced by 20 – 14 = 6 dollars. So, 620 = 0.3, 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 that would be wrong.
The key to solving double percentchange questions is realizing
that the second percentage change is dependent on the first. For
example, in the problem we just looked at, the second percent decrease
is 20% of a new, lower price—not the original. Let’s work through
the problem carefully and see. After the first sale, the price of
the bike drops 30%:
The second reduction in price knocks off an additional
20% of the sale price, not the original price:
The trickiest of the tricky percentage problems go a little
something like this:

If the answer 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%, the 20% reduction in price
was a 20% reduction of the 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 are not as simple as they seem.
But once you know their tricks, they can be easily solved.
Exponential Growth and Decay
These types of word problems take the concept of percent
change even further. For questions involving populations growing
in size or the diminishing price of a car over time, you’ll need
to perform a percent change over and over again. For such a repeated
operations, exponents can be used to simplify the calculations.
Here’s an example problem:

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:
In general, quantities like the one described in this
problem are said to be growing exponentially. The formula for calculating
exponential growth over a specific number of years is:
Exponential decay is mathematically equivalent to negative
exponential growth. 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 in exponential decay is less than 1, which
should make sense because during each unit of time, the original
amount is reduced by a fixed percentage. You will find that exponential
decay is often used to model population decreases and the loss of
physical mass over time.
Let’s work through a few example questions to get a feel
for both exponential growth and decay problems.
Simple Exponential Growth Problem

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 1000, 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 of air
is 4000 cubic centimeters, and the length of decay is 10 days. So
plug 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 the question refers to, the interest will compound
2 12 = 24 times. As long as you remember
to use 24 as the length of time, the answer is straightforward:
Here’s another compounding problem:

Sam’s account will have $2000 1.05^{15} ≈
$4157.85 in it after 15 years. Justin’s account will have $2500 1.04^{15} ≈
$4502.36 in it. So, Justin’s account will still have more money
in it after 15 years.
Logarithms
Logarithms have important uses in solving problems with
complicated exponential equations. Consider the following example:

This question is like the exponential growth problems
we’ve just seen, but with a twist. Here, we’re given the growth
rate, the initial quantity, and the ending quantity.
We need to find the length of time over which the growth rate must
be applied to the initial quantity to yield the ending quantity.
Logarithms provide the tool to solve this type of problem.
It will take roughly twenty and a half years for the town’s
population to exceed 1500. So about halfway through the year 2022,
the population will first exceed 1500.
In general, to solve a problem such as the example problem
in which the exponent is unknown, the first step is to isolate the
exponential term. Once the exponential term has been isolated, the
next step is to isolate the variable itself by taking the logarithm
of both sides and then using the power rule of logarithms to bring
the variable out of the exponent. It does not matter which base
you use in your logarithm; you could choose a base10 logarithm
or a natural logarithm for ease of calculation.
Here’s a simple example to illustrate this process:

This problem would be vastly more difficult if we didn’t
have logarithms. How would you possibly calculate 5^{1000}?
And how do you solve for x when it’s the exponent
of a number? Fortunately, we can simply take the logarithm of each
side of the equation and then apply the power rule of logarithms.
From here, the solution becomes much more clear:
