Word Problems Study Guide for the Math Basics

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Proportion Magic

About 10-25% of problems on the ACT can be at least partially solved using the magic of proportions. So, it’s good to have a solid grasp on solving proportions. In general, use this technique:

cross multiply and then divide

Here’s a mathematical example:

\[\dfrac{3}{x}=\dfrac{9}{6}\] \[3 \cdot 6 = 9 \cdot x\] \[18 = 9x\] \[18 \div 9 = x\] \[2 = x\]

The key is to realize when to use proportions for a word problem. If there are three pieces of given information (numbers) and the question is looking for a fourth, chances are the problem can be solved using a proportion.

Assign a variable for the missing amount, set up a proportion and then solve. Let’s look at this problem and the possible proportions that can be used to solve it.

If Harry can paint 2 paintings in 7 minutes, how many can he paint in 28 minutes?

Notice, there are three pieces of information given (2 paintings, 7 minutes, and 28 minutes) and one piece of information being asked for (let \(p\) be the missing amount of paintings). Now, set up your proportion any way you want, as long as items in the same row and column are related.

\[\begin{array}{|c|c|} \hline \textbf{paintings} & \textbf{minutes}\\ \hline 2 & 7\\ \hline p & 28 \\ \hline \end{array}\] \[\begin{array}{|c|c|c|} \hline \textbf{paintings} & 2 & p \\ \hline \textbf{minutes} & 7 & 28\\ \hline \end{array}\] \[\begin{array}{|c|c|} \hline \textbf{minutes} & \textbf{paintings}\\ \hline 28 & p\\ \hline 7 & 2 \\ \hline \end{array}\]

But this one won’t work:

\[\begin{array}{|c|c|c|} \hline \textbf{paintings} & 2 & p \\ \hline \textbf{minutes} & 28 & 7\\ \hline \end{array}\]

Notice that, in the first column, “2 paintings” is grouped with “28 minutes,” which isn’t the case according to the problem.

Now, just pick one and solve. Let’s do the first one:

\[\dfrac{2}{p}=\dfrac{7}{28}\] \[2 \cdot 28 = 7 \cdot p\] \[56 = 7 \cdot p\] \[56 \div 7 = p\] \[8 = p\]

So, Harry can paint 8 paintings in 28 minutes.

Multi-Step Problems

Not all problems can be easily written as linear equations. Sometimes you have to go slowly and sometimes you need to use a more advanced approach.

Step by Step

Let’s go back to the problem from the very beginning:

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

Don’t be a hero, though. While there is a way to set up one equation for the problem (here’s one: \(65-20\%\cdot65 + 6\% \cdot (65-20\%\cdot65) = p\)), it might lead you into trouble. Perhaps before we go for the gold, we should assign a different variable for the first step.

Assign an intermediate variable: Let \(d =\) the discounted price of the dress

First equation

\[65-20\% \cdot 65 = d\] \[65 - 0.2 \cdot 65= d\] \[65 - 13 = d\] \[52 = d\]

So the dress was $52 after the discount. Now let’s add the tax: Let \(p =\) the final price of the dress:

Second Equation

\[52 + 6\% \cdot 52 = p\] \[52 + 0.06 \cdot 52 = p\] \[52 + 3.12 = p\] \[\$55.12= p\]

So the dress is $55.12, and no, Kristyn doesn’t have enough money for it.

Don’t Forget Systems of Equations

No word problem section of a math test is complete without one that requires a system of equations to solve it. You know the kind:

Julie and Layla are selling flower bulbs for a school fundraiser. Customers can buy packages of tulip bulbs and packages of crocus bulbs. Julie sold 2 packages of tulip bulbs and 2 packages of crocus bulbs for a total of $34. Layla sold 2 packages of tulip bulbs and 6 packages of crocus bulbs for a total of $82. Find the cost each of one package of tulips bulbs and one package of crocus bulbs.

First assign the variables: Let t = the cost of a package of tulip bulbs and c = the cost of one package of crocus bulbs.

Now, write the system of equations:

\[2t+2c =34\] \[2t + 6 c = 82\]

There are two main ways to solve. One way is the substitution method.

Step 1: Isolate either variable in either equation. Let’s isolate \(t\) in the first equation.

\[2t + 2c = 34\] \[2t = 34 - 2c\] \[t = \dfrac{34-2c}{2}\] \[t = 17 - c\]

Step 2: Substitute that value (\(17-c\)) in for the variable (\(t\)) in the other equation.

\[2t + 6c = 82\] \[2(17-c) + 6 c = 82\]

Step 3: Isolate that variable.

\[34 - 2c + 6c = 82\] \[34+4c = 82\] \[4c = 48\] \[c = 12\]

Step 4: Substitute that value into either original equation. Let’s plug \(c=12\) into the first equation.

\[2t+2c=34\] \[2t + 2(12) = 34\] \[2t + 24 = 34\] \[2t = 10\] \[t= 5\]

So, tulip bulbs cost $5 per package and crocus bulbs cost $12 per package.

The other way to solve the equation is the elimination method. Here is the system again:

\[2t+2c =34\] \[2t + 6 c = 82\]

Step 1: Ensure that one variable of the first equation has the opposite (meaning negative) coefficient of the same variable of the second equation. That is, if the first equation has \(5x\) the second should have \(-5x\). In this case, we’re pretty close, but not quite there. Both equations have \(2t\), but one needs to be negative. So, multiply the second equation by \(-1\) and our new system will be:

\[2t+2c =34\] \[-2t + -6 c = -82\]

Step 2: As long as everything is lined up, add the two equations vertically:

\[\;\;2t+\;\;2c =34\] \[\underline{+(-2t + -6 c = -82)}\] \[\quad\;0t + -4c = -48\]

Or, equivalently:

\[-4c = -48\]

The \(t\) is eliminated, hence the name elimination method.

Step 3: Isolate the leftover variable.

\[-4c = -48\] \[c = 12\]

Step 4: Substitute that back into either of the original equations and solve.

\[2t+2c=34\] \[2t + 2(12) = 34\] \[2t + 24 = 34\] \[2t = 10\] \[t= 5\]

And we get the same answer using either method.

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