# Word Problems Study Guide for the Math Basics

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## How to Prepare for the Word Problems on a Math Test

### General Information

Pure math is straightforward: Isolate the variable, get the answer. But math is nearly pointless without application. Most test makers know this. Enter word problems, those pesky things that seem to confuse even the best math student. More and more questions on tests are in this format, and are no longer simply equations, numbers, and symbols. Fear not! This study guide will help you translate those word problems into manageable calculations.

### Identifying the Question

Imagine you’re taking the reading comprehension section of a standardized test (ACT, SAT, etc.). Any tutor will tell you the best thing to do is to read at least the first few questions before reading the passage. The same goes for mathematical word problems. You can’t do anything unless you know what the question is. The first step to solving a word problem is to identify the question. While identifying the question, you should assign a variable to the unknown. It’s good practice to use a letter or symbol that is meaningful to the situation (like the first letter of the word).

Take this example:

Jim decided to start a business. He has $4000 saved up. If it costs$2500 to rent a space and $1.25 to produce each product, how many products can he produce upon startup? Identify the question: How many products can he produce? Assign the variable: Let $$p =$$ products. Here’s one that’s a little trickier: Kristyn went to the store and spotted the perfect dress on sale for 20% off the original$65 price tag. She had a $50 bill and is wondering if it’s enough to purchase the dress. She decides to ask an associate to ring up the order anyway. What amount does the cash register show after the discount and a 6% sales tax? Identify the question: What is the final price of the dress? Assign the variable: Let $$p =$$ final price of the dress ### Identifying Useful Data Test makers will take any chance they can to throw a curveball to make sure you really know what you’re doing. Sometimes this will be in the form of giving too much information, like this example: A quarter mile away from school, Eleanor and Daisy are on their way home when they start feeling raindrops. How long does it take them to get home if they’re a half mile away and they can run at 8 fps? Identifying the necessary data first will be helpful. Identify the question: How long does it take them to get home? Assign the variable: Let $$t =$$ time to get home. Useful data: distance from home, speed of running. Unneeded data: distance from school Another Example: You rolled a 6-sided die 4 times already and rolled two 2’s, a 3, and a 6. What is the probability that you roll a 5 on your next roll? Identify the question: What is the probability of rolling a 5? Assign the variable: Let $$p =$$ probability of rolling a 5 Useful information: 6 sided die Unneeded data: all the previous rolls—they don’t affect the probability of any other roll Of course, always look out for key words like not or except. Usually, on a multiple choice test, one of the options will be the answer to the problem if you didn’t read that word. Here’s an example: John had 16 pieces of candy on Sunday. He ate $$\frac{1}{8}$$ of them on Monday, 4 more on Tuesday, and $$\frac{1}{5}$$ of the remainder on Wednesday. How many pieces did he not eat? A) 7 pieces B) 8 pieces C) 12 pieces D) 10 pieces The correct answer is B) 8 pieces, but if you didn’t read the not part of the question, you’ll think the answer was C) 12 pieces because that’s how much candy he ate. Some tests use bold print or italics to highlight tricky words like not, but many don’t, so be careful. ### Setting Up Equations After identifying the question and weeding out the unneeded data it’s time to set up the equation: the final step to translating your word problem into a purely mathematical problem. Here’s some common words which translate into mathematical operations. $\begin{array}{|c|c|} \hline \textbf{Words} & \textbf{Operation}\\ \hline \text{sum, increased by, added to, total} & +\\ \hline \text{difference, decreased by, subtracted from} & - \\ \hline \text{product, times, of} & \times \\ \hline \text{quotient, ratio, divided by} & \div \\ \hline \text{equals, totals to, is, are, equivalent} & = \\ \hline \text{is less than} & \lt \\ \hline \text{is greater than, is more than} & \gt \\ \hline \text{is less than or equal to, no more than, at most} & \le \\ \hline \text{is greater than or equal to, no less than, at least} & \ge \\ \hline \end{array}$ You also might want to know some helpful formulas: $\text{Distance} = \text{speed} \cdot \text{time}$ $\text{Percent of change} = \dfrac{\text{new amount} - \text{original amount}}{\text{original amount}} \cdot 100$ $\text{Probability} = \dfrac{\text{number of ways to succeed}}{\text{total number of possible outcomes}}$ $\text{Average} = \dfrac{ \text{sum of amounts}}{\text{total number of objects}}$ Use the information above to try this example: A farmer wants to sell a third of his brood of hens to his neighbor. If he had 60 hens in the beginning, how many hens does he want to sell? Identify the question: How many hens did he sell? Assign a variable: Let $$h =$$ hens sold Key words: of means multiplication Equation: $$\frac{1}{3} \cdot 60 = h$$ Solve: $$20 \text{ hens} = h$$ Example: Lucy and Karen are on a cross country team. Lucy runs at a steady pace of 7.5 mph, which is 0.5 mph faster than Karen. How long does it take Lucy to run the 8 mile practice course? Identify the question: How long does it take Lucy to 8 miles? Assign a variable: Let $$t =$$ the time it takes Lucy to run 8 miles. Unnecessary information: Lucy runs 0.5 mph faster than Karen. Important formulas: $$\text{Distance} = \text{speed} \cdot \text{time}$$ Equation: $$8 = 7.5 \cdot t$$ Solve: $$8\div 7.5 = 1.0666 \text{ hours} = t$$ ### Checking Your Answer A good multiple choice testmaker will assume that the testtaker makes a common mistake or two. And they will usually use the answer derived from making that mistake as one of the choices. So if you think to yourself “My answer is there, I must have done it right!” you probably should think again. #### Plug it back in Most problems require an equation to solve. If you feel it was a pretty easy equation to come up with, plug your answer back into the equation. It’s the fastest way to check and can easily check for algebraic mistakes. Here’s an example with a common mistake. Word Problem: The difference of 20 and the product of 2 and the quantity (x+1) is 10. What is x? A) $$6$$ B) $$-6$$ C) $$4$$ D) $$-2$$ Equation: $20-2(x+1)=10$ Your work and solution: $20-2x+2=10$ $22 - 2x =10$ $-2x=-12$ $x=\dfrac{-12}{-2}$ $x=6$ Your answer is one of the options (A). But let’s quickly check: $20-2(6+1)\;?\;10$ $20 - 2(7) \;?\; 10$ $20-14 \;?\; 10$ $6 \ne 10$ Now, look for the error. It happened right at the beginning. It was a pretty straightforward word problem to convert into an equation, but you forgot what your algebra teacher has been urging you to remember since day 1, “Don’t forget to distribute the sign!”. You didn’t distribute the negative sign when you distributed the $$2$$. Finally, fix the error: $20-2(x+1)=10$ $20\;-2x\;-2=10$ $18 \;-2x=10$ $-2x=-8$ $x = \dfrac{-8}{-2} = 4$ which was answer (C) Quickly, check again: $20-2(4+1)\;?\;10$ $20-2(5)\;?\;10$ $20-10 \;?\; 10$ $10=10$ Nice job! Great catch! #### Estimation Sometimes, you just plain forget how to perform some sort of mathematical operation without a calculator. When this happens, don’t panic. If you just use a little estimation and common sense, you can probably find the answer choice that is correct. Take long division of decimals, for instance. Look at this example: Candy bars cost$0.97 a piece at the register. If I paid $11.64 for candy bars for my friends and I, how many candy bars did I buy? A) 10 B) 12 C) 15 D) 11 Clearly, the equation written should look like this: $0.97b=11.64$ where $$b=$$ the number of candy bars bought. Also clear is the method to solve this: $b=\dfrac{11.64}{0.97}$ But your test doesn’t allow calculators and you forgot how to do this. So you should probably estimate and see if your answer is there. Round$0.97 to $1 and$11.64 to $12. Now, $1b=12$ $b= \dfrac{12}{1}=12$ Which is answer B). #### Number Sense Of course, not all tests are multiple choice, so you might have to rely on your number sense to see if you did things right. Let’s look at this problem: You went out to dinner and the bill came out to$56.32. You want to tip 20%. What’s the amount you should pay your server?

Maybe you did the problem this way:

$56.32 \cdot 20\%= \text{tip}$ $56.32 \cdot 0.20 = \text{tip}$ $112.64 = \text{tip}$ $\text{ final amount} = \text{ bill amount} + \text{ tip}$ $\text{ final amount} = 56.32 + 112.64$ $\text{final amount} = \168. 96$

Wait a minute! Wasn’t your original bill under \$60? You’ve been out to a restaurant before… no reasonable tip almost triples the bill. So what did you do wrong? It was probably a careless keystroke on the calculator. Punch the keys again:

$56.32 \cdot 20\%= \text{tip}$ $56.32 \cdot 0.20 = \text{tip}$ $11.26 = \text{tip}$

Here was your mistake. Carry on:

$\text{ final amount} = \text{ bill amount} + \text{ tip}$ $\text{ final amount} = 56.32 + 11.26$ $\text{ final amount} = \67.58$

That’s more like it.

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