Common Word Problems
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 nitty-gritty on all of them.
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.
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:
Jim rollerblades 6 miles per hour. One morning, Jim starts rollerblading and doesn’t stop until he has gone 60 miles. For how many hours did he rollerblade?
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:
At a cycling race, there are 50 cyclists in all, each representing a state. The cyclist from California can cycle 528,000 feet per hour. If the race is 480 miles long, how long will it take him to finish the race? (Note that 1 mile = 5280 feet)
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.
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 one-hour 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:
Four men can dig a 40-foot well in 4 days. How long would it take for 8 men to dig a 60-foot well? Assume that each of these eight men works at the same pace as each of the four men.
First, let’s examine what the problem says: four men can dig a 40-foot 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 8-man 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 60-foot well. The total quantity is 60 feet, and the rate is 2.5 feet per man per day. The 8-man 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.
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 twenty-five dollars, you could determine the total cost of the 8 basketballs:
Percent Change
Percent-change questions ask you to determine how a percent increase or decrease affects the values given in the question. There are two variations of percent-change 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:
A professional golfer usually has an average score of 72 but recently went through a major slump. His new average is 20 percent worse (higher) than it used to be. What is his new average?
This is a percent-change 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 percent-change problem:
A shirt whose original price was 20 dollars has now been put on sale for 14 dollars. By what percentage did its price drop?
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 percent-change question asks you to determine the cumulative effect of two percent changes in the same problem. For example:
A bike has an original price of 300 dollars. Its price is reduced by 30%. Then, two weeks later, its price is reduced by an additional 20%. What is the final sale price of the bike?
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 percent-change 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:
A computer has a price of 1400 dollars. Its price is raised 20% and then lowered 20%. What is the final selling price of the computer?
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:
If a population of 100 grows by 5% per year, how great will the population be in 50 years?
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
A population of bacteria grows by 35% every hour. If the population begins with 100 specimens, how many are there after 6 hours?
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
A fully inflated beach ball loses 6% of its air every day. If the beach ball originally contains 4000 cubic centimeters of air, how many cubic centimeters does it hold after 10 days?
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
A bank offers a 4.7% interest rate on all savings accounts, compounded monthly. If 1000 dollars are initially put into a savings account, how much money will the account hold two years later?
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 puts 2000 dollars into a savings account that pays 5% interest compounded annually. Justin puts 2500 dollars into a different savings account that pays 4% annually. After 15 years, whose account will have more money in it, if no more money is added or subtracted from the principal?
Sam’s account will have $2000 1.0515 ≈ $4157.85 in it after 15 years. Justin’s account will have $2500 1.0415 ≈ $4502.36 in it. So, Justin’s account will still have more money in it after 15 years.
Logarithms have important uses in solving problems with complicated exponential equations. Consider the following example:
The population of a small town is 1000 on January 1, 2002. It grows at a constant rate of 2% per year. In what year does the population of the town first exceed 1500?
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 base-10 logarithm or a natural logarithm for ease of calculation.
Here’s a simple example to illustrate this process:
If 6x = 51000, then find the value of x.
This problem would be vastly more difficult if we didn’t have logarithms. How would you possibly calculate 51000? 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:
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