The writers of the Math IC love word problems. These problems force youto show your range as a mathematician. They demand that you read andcomprehend the problem, set up an equation or two, and manipulate theequations to find the solution. Luckily, the Math IC uses only a fewtypes of word problems, and we have the nitty-gritty on all of them.
Rates
A rate is a ratio of related qualitiesthat have different units. For example, speed is a rate that relatesthe two quantities of distance and time. Here is the general rateformula:
No matter the specifics, the key to a rateproblem is correctly placing the given information in the threecategories. Then, you can substitute the values into the rate formula.We’ll look at the three most common types of rate: speed, work, andprice.
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, sincethe hour in the rate is at the bottom of the fraction. But you can besure that the Math IC test won’t simply give you one of the quantitiesand the rate and ask you to plug it into the rate formula. Because ratequestions are in the form of word problems, the information that you’llneed to solve the problem will often be given in a less straightforwardmanner.
Here’s an example:
Jim rollerblades 6 miles per hour. Onemorning, Jim starts rollerblading and doesn’t stop until he has gone 60miles. How many hours did he rollerblade?
This question provides more information thansimply the speed and one of the quantities. We know unnecessary factssuch as how Jim is traveling (by rollerblades) and when he started (inthe morning). Ignore them and focus on the facts you need to solve theproblem.
- Time a: x hours rollerblading
- Rate: 6 miles per hour
- Quantity b: 60 miles
So, we can write:
Jim was rollerblading for 10 hours. Thisproblem requires a little analysis, but basically we plugged somenumbers into the rate equation and got our answer. Here’s a slightlymore difficult rate problem:
At a cycling race, there are 50 cyclistsin all, each representing astate. The cyclist from California cancumulatively cycle 528,000 feetper hour. If the race is 480 mileslong, how long will it take him tofinish the race?
Immediately, you should pick out the givenrate of 528,000 feet per hour and notice that 480 miles are traveled.You should also notice that the question presents a units problem: thegiven rate is in feet cycled per hour, and the distance traveled is inmiles.
Sometimes a question will give youinconsistent units, like in this example. Always read over the problemcarefully and don’t forget to adjust the units—the answer choices arebound to include non-adjusted 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 thefirst quantity measured in time, the second quantity measured in workdone, and the rate measured in work done per time. For example, if youknitted 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:
Four men can dig a 40 foot well in 4 days.How long would it take for 8men to dig a 60 foot well? Assume thatthese 8 men work at the same paceas the 4 men.
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 workof 40 feet and a time of 4 days. We need to create our own rate, usingwhichever units might be most convenient, to carry over to the 8-menproblem. The group of 4 men dig 40 feet in 3 days. Dividing 40 feet by4 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 a60 foot well. The work done by the 8 men is 60 feet, and they work at arate of 10 feet per day per 4 men. Can we use this information toanswer the question? Yes. The rate of 10 feet per day per 4 menconverts to 20 feet per day per 8 men, which is the size of the newcrew. 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 ofcreativity—but nothing you can’t handle. Just remember the classic rateformula and use it wisely.
Price
In rate questions dealing with price, youwill usually find the first quantity measured in numbers of items, thesecond measured in price, and the rate in price per item. Let’s say youhad 8 basketballs, and you knew that each basketball cost $25 each:
Percent Change
In percent-change questions, you will need todetermine how a percent increase or decrease affects the values givenin the question. Sometimes you will be given the percent change, andyou will have to find either the original value or new value. Othertimes, you will be given one of the values and be asked to find thepercent change. Take a look at this sample problem:
A professional golfer usually has anaverage score of 72, but herecently went through a major slump. Hisnew average is 20 percent worse(higher) than it used to be. What ishis new average?
This is a percent-change question in whichyou need to find how the original value is affected by a percentincrease. First, to answer this question, you should multiply 72 by .20to see what the change in score was:
Once you know the score change, then youshould add it to his original average, since his new average is higherthan it used to be:
It is also possible to solve this problem bymultiplying the golfer’s original score by 1.2. Since you know that thegolfer’s score went up by twenty percent over his original score, youknow that his new score is 120% higher than his old score. If you seethis 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 priceand the sale price and need to determine the percent decrease. All youneed to do is divide the amount by which the quantity changed by theoriginal quantity. In this case, the shirt’s price was reduced by 20 –14 = 6 dollars. So, 6
20 = .3, a 30% drop in the price of the shirt.
Double Percent Change
A slightly trickier version of thepercent-change question asks you to determine the cumulative effect oftwo percent changes in the same problem. For example:
A bike has an original price of 300dollars. Its price is reduced by30%. Then, two weeks later, its priceis reduced by an additional 20%.What is the final sale price of thebike?
One might be tempted to say that the bike’sprice is discounted 30% + 20% = 50% from its original price, but thekey to solving double percent-change questions is to realize that eachpercentage change is dependent on the last. For example, in the problemwe just looked at, the second percent decrease is 20 percent of a new,lower price—not the original amount. Let’s work through the problemcarefully and see. After the first sale, the price of the bike drops 30percent:
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:
A computer has a price of 1400 dollars.Its price is raised 20%, andthen lowered 20%. What is the finalselling price of the computer?
If this question sounds too simple to betrue; it probably is. The final price is not the same as the original.Why? Because after the price was increased by 20 percent, the reductionin price was a reduction of 20 percent of a new, higher price.Therefore, the final price will be lower than the original. Watch andlearn:
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 theconcept of percent change even further. In questions involvingpopulations growing in size or the diminishing price of a car overtime, you need to perform percent-change operations repeatedly. Solvingthese problems would be time-consuming without exponents. Here’s anexample:
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. Youmay already see the shortcut you can use to avoid having to do, in thiscase, 50 separate calculations. The final answer is simply:
In general, quantities like the onedescribed in this problem are said to be growing exponentially. Theformula for calculating how much an exponential quantity will grow in aspecific number of years is:
Exponential decay is mathematicallyequivalent to negative exponential growth. But instead of a quantitygrowing at a constant percentage, the quantity shrinks at a constantpercentage. Exponential decay is a repeated percent decrease. That iswhy the formulas that model these two situations are so similar. Tocalculate exponential decay:
The only difference between the twoequations is that the base of the exponent is less than 1, becauseduring each unit of time the original amount is reduced by a fixedpercentage. Exponential decay is often used to model populationdecreases, 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
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 ofbacteria, makes it quite clear that this is an exponential growthproblem. To solve the problem, you just need to plug the appropriatevalues 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
A fully inflated beach ball loses 6% ofits air every day. If the beachball originally contains 4000 cubiccentimeters of air, how many cubiccentimeters does it hold after 10days?
Since the beach ball loses air, we know thisis an exponential decay problem. The decay rate is .06, the originalamount is 4000 cubic centimeters of air, and the time is 10. Pluggingthe information into the formula:
More Complicated Exponential Growth Problem
A bank offers a 4.7% interest rate on allsavings accounts, compoundedmonthly. If 1000 dollars is initially putinto a savings account, howmuch money will the account hold two yearslater?
This problem is a bit tricky for the simplereason that the interest on the account is compounded monthly. Thismeans that in the 2 years that question refers to, there will be 2
12 = 24 compoundings of interest. The time variable in the equation isaffected by these monthly compoundings: it will be 24 instead of 2.Thus, our answer is:
Here’s another compounding problem:
Sam puts 2000 dollars into a savingsaccount that pays 5% interestcompounded annually. Chris puts 2500dollars into a different savingsaccount that pays 4% annually. After15 years, whose account will havemore money in it, if no more money isadded or subtracted from theprincipal?
Sam’s account will have $2000
1.0515 ≈ $4157.85 in it after 15 years. Chris’s account will have $2500
1.0415 ≈ $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.
