# SAT Algebra Word Problems

To solve SAT algebra word problems, you need to know:

- how to translate words into math
- manipulate algebraic expressions
- work with fractions and decimals
- how to work with ratios, proportions, and percents

## Examples

A fruit basket contains the same number of apples and pears. If Eric eats 5 apples and 1 pear, there will be twice as many pears as apples. How many pears remain in the basket?(A) \(\ \ 4\)

(B) \(\ \ 8\)

(C) \(\ \ 9\)

(D) \(\ \ 10\)

(E) \(\ \ 11\)

Correct Answer: B

Solution 1:Let \(x\) be the number of apples and the number of pears in the basket. Eric eats 5 apples, and therefore there are \(x-5\) apples remaining. He also eats 1 pear, and therefore there are \(x-1\) pears remaining. We're told that after Eric eats there are twice as many pears as apples in the basket. We set up an equation.

\[\begin{array}{l c l l} 2(x-5) &=& x-1 &\quad \text{create equation} &(1)\\ 2x-10 &=& x-1 &\quad \text{use distributive property} &(2)\\ x-10 &=& -1 &\quad \text{subtract}\ x\ \text{from both sides} &(3)\\ x &=& 9 &\quad \text{add}\ 10\ \text{to both sides} &(4)\\ \end{array}\]

We're looking for the number of pears remaining in the basket, but \(x=9\) is the original number of pears. Since he ate \(1\) pear, there are \(8\) pears remaining.

Solution 2:Here, for each answer choice, we work our way backwards. We compute, based on how many pears are left, how many pears and apples there were to start with. Remember that there was the same number of apples and pears before Eric ate.

(A) If 4 pears are left, then there were 4+1=5 pears to start with.

Since there are twice as many pears remaining as apples, then there are \(\frac{4}{2}=2\) apples left, and there were 2+5=7 apples to start with. But \(5\neq 7.\) This is the wrong choice.(B) If 8 pears are left, then there were 8+1=9 pears to start with.

Since there are twice as many pears remaining as apples, then there are \(\frac{8}{2}=4\) apples left, and there were 4+5=9 apples to start with. \(9=9\) and therefore this is the correct answer.(C) If 9 pears are left, then there were 9+1=10 pears to start with.

Since there are twice as many pears remaining as apples, then there are \(\frac{9}{2}=4.5\) apples left, and there were 4.5+5=9.5 apples to start with. But \(9 \neq 9.5.\) This is the wrong choice.(D) If 10 pears are left, then there were 10+1=11 pears to start with.

Since there are twice as many pears remaining as apples, then there are \(\frac{10}{2}=5\) apples left, and there were 5+5=10 apples to start with. But \(10\neq 11\). This is the wrong choice.(E) If 11 pears are left, then there were 11+1=12 pears to start with.

Since there are twice as many pears remaining as apples, then there are \(\frac{11}{2}=5.5\) apples left, and there were 5.5+5=10.5 apples to start with. \(10.5\neq 12.\) This is the wrong choice.

Incorrect Choices:

If you got this problem wrong, you should review SAT Translating Word Problems.

(A)

This wrong choice is the number of apples that remain in the basket.

(C)

If you solve for the original number of pears, you will get this wrong answer.

(D)

When accounting for the pear Eric ate, if instead of subtracting 1 from the original number of pears, you add 1, you will get this wrong answer.

(E)When finding how many apples remain in the basket after Eric consumed 5 of them, if you add 5 to \(x\) instead of subtract 5 from it, you will get\[\begin{array}{l c l l} 2(x\fbox{+}5) &=& x-1 &\quad \text{mistake: added}\ 5\ \text{to}\ x &(1)\\ \end{array}\]

Or, you may get this wrong answer if in step (3) of the solution you subtract 10 from both sides and you ignore the negative signs.

Katie and Beth had the same number of marbles on January 1st. By the end of the year, Katie's collection of marbles increased by \(25\%\) and Beth's decreased by \(75\%\). On December 31st, the number of marbles Beth owned was what percent of the number of marbles Katie owned?

(A) \(\ \ 18.75\%\)

(B) \(\ \ 20\%\)

(C) \(\ \ 33.33\%\)

(D) \(\ \ 75\%\)

(E) \(\ \ 100\%\)

At 2:00 P.M. a train A leaves a station traveling at 100 mi/hr. At 4:30 P.M. train B leaves the station on a parallel route and traveling at 150 mi/hr. How far away from the station will the second train overtake the first train?

(A) \(\ \ 250\) mi

(B) \(\ \ 500\) mi

(C) \(\ \ 750\) mi

(D) \(\ \ 800\) mi

(E) \(\ \ 1500\) mi

Correct Answer: C

Solution:

Tip: Distance = Rate \(\times\) Time.

Between 2:00 P.M. and 4:30 P.M. there are 2.5 hours. This means that train A travels 2.5 hrs more than train B. If \(t\) denotes the hours train B travels, then train A travels \((t+2.5)\) hrs. Let the distance the two trains travel be denoted by \(d\). Using \(d=rt\) we create an equation representing the distance each train traveled.\[\begin{array}{l l c l l} d&=&100(t+2.5) &\quad \text{train A} &(1)\\ d&=&150t &\quad \text{train B} &(2)\\ \end{array}\]

Since the distance the two trains traveled is the same, we set \((1)\) equal to \((2).\)

\[\begin{array}{r c l} 150t&=&100(t+2.5) &\quad (1) = (2) \\ 150t &=& 100t + 250 &\quad \text{use distributive property}\\ 50t &=& 250 &\quad \text{subtract}\ 100t\ \text{from both sides}\\ t&=&5\ \text{hrs} &\quad \text{divide both sides by}\ 25\\ \end{array}\]

So, train B travel 5 hours before it catches up with train A. But we are looking for the distance the trains have traveled when train B takes over train A. Plugging \(t=5\) into \((2)\), we get

\(5 \times 150 = 750\) miles.

Incorrect Choices:

(A)

This is just the sum of the speeds of the two trains.

(B)

If you find the distance train A travels from 4:30 P.M. till train B overtakes it, you will get this wrong answer.

(D)

If you think the difference between 2:00 P.M. and 4:30 P.M. is 2 hrs, you will get this wrong answer.

(E)

If you find the combined distance the trains traveled, you will get this wrong answer.

## Review

If you thought these examples difficult and you need to review the material, these links will help:

- Fractions-Word Problems
- Decimals-Word Problems
- Percents-Word Problems
- Ratio and Proportion Word Problem
- Systems of Linear Equations Word Problems-Easy
- Linear Equations Word Problems
- Linear Inequalities Word Problems
- Convert Percentages, Fractions, and Decimals
- Percentage Change
- SAT Translating Word Problems
- SAT Fractions and Decimals
- SAT Direct and Inverse Variation

## SAT Tips for Algebra Word Problems

- Read the entire question carefully.
- Distance = Rate \(\times\) Time.
- Know the Properties of Proportions.
- Replace variables with numbers.
- Pay attention to units.
- SAT General Tips

**Cite as:**SAT Algebra Word Problems.

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