Topic 1

Multi-Step Algebra and Percent Word Problems

Many SAT math questions are not about a single skill in isolation. Instead, they layer two or more concepts into a single problem, requiring you to recognize which tools to use and in what order. This topic focuses on problems that combine algebraic manipulation (solving equations, simplifying expressions, working with inequalities) with percent and ratio reasoning (percent change, markups, discounts, tax, tips, and successive percent changes).

A percent change can be expressed as multiplication by a factor. A

p\%

increase multiplies by

(1 + p/100)

, and a

p\%

decrease multiplies by

(1 - p/100)

. For example, a

20\%

increase multiplies by

1.20

, and a

15\%

decrease multiplies by

0.85

. When multiple percent changes happen in sequence, multiply the factors together. A

10\%

increase followed by a

10\%

decrease is

1.10 \times 0.90 = 0.99

, which is a net

1\%

decrease, not a return to the original value. This is a classic SAT trap.

Percent-algebra hybrids typically look like word problems: a store marks up an item, then applies a discount; a population grows by a certain percent per year and you need to find when it reaches a target; a price after tax equals a given total and you need to find the pre-tax price. The algebraic component usually involves setting up and solving an equation or expression that models the situation.

Another common pattern is ``percent of'' combined with systems of equations. For instance, a mixture problem might tell you that a solution is

30\%

acid and you mix it with a solution that is

50\%

acid to get a certain volume at

40\%

acid. This requires both percent reasoning and a system of two equations (one for total volume, one for total acid).

The key to these problems is translating the words carefully into algebra before you start solving. Write down what each variable represents, set up equations that match each sentence of the problem, and then solve.

Core Skills

Translate successive percent changes into multiplication factors and combine them.
Set up and solve equations involving markup, discount, tax, and tip in sequence.
Model mixture and concentration problems with systems of equations.
Work backward from a final value through percent changes to find an original value.
Combine ratio reasoning with algebraic solving in multi-step word problems.

Example 1: Successive Percent Changes

A store increases the price of a jacket by

25\%

, then offers a

20\%

discount on the new price. If the original price was

\$80

, what is the final price? Step 1: Apply the

25\%

increase.

80 \times 1.25 = 100

Step 2: Apply the

20\%

discount to the new price.

100 \times 0.80 = 80

The final price is

\boxed{\$80}

. (Note: even though the percent increase and decrease are close in value, the final price returns to exactly

\$80

here only because

1.25 \times 0.80 = 1.00

. This is not typical. A

20\%

increase followed by a

20\%

decrease gives

1.20 \times 0.80 = 0.96

, a net

4\%

loss.)

Example 2: Working Backward from a Final Price

After a

15\%

discount and then

8\%

sales tax on the discounted price, a customer pays

\$91.80

for a pair of shoes. What was the original price before the discount? Step 1: Let the original price be

p

. The discounted price is

0.85p

. After

8\%

tax, the customer pays:

0.85p \times 1.08 = 0.918p

Step 2: Set this equal to the amount paid.

0.918p = 91.80 \implies p = \frac{91.80}{0.918} = 100

The original price was

\boxed{\$100}

Example 3: Mixture Problem

A chemist has a

40\%

acid solution and a

70\%

acid solution. How many liters of each should be mixed to obtain 12 liters of a

50\%

acid solution? Step 1: Let

x

= liters of the

40\%

solution and

y

= liters of the

70\%

solution.

x + y = 12 \quad \text{(total volume)}

0.40x + 0.70y = 0.50(12) = 6 \quad \text{(total acid)}

Step 2: From the first equation,

x = 12 - y

. Substitute:

0.40(12 - y) + 0.70y = 6 \implies 4.8 - 0.40y + 0.70y = 6

0.30y = 1.2 \implies y = 4

x = 12 - 4 = 8

Use

\boxed{8}

liters of the

40\%

solution and

\boxed{4}

liters of the

70\%

solution.

Example 4: Percent Increase with Algebraic Setup

The price of a stock increased by

20\%

in January and then decreased by

k\%

in February. The stock's price at the end of February was

8\%

more than its price at the beginning of January. What is the value of

k

? Step 1: Let the initial price be

P

. After the January increase:

1.20P

. After the February decrease:

1.20P \times \left(1 - \frac{k}{100}\right)

Step 2: The final price is

1.08P

1.20\left(1 - \frac{k}{100}\right) = 1.08

1 - \frac{k}{100} = \frac{1.08}{1.20} = 0.90

\frac{k}{100} = 0.10 \implies k = 10

The value of

k

\boxed{10}

Example 5: Ratio and Percent Combined

In a class of 40 students, the ratio of boys to girls is

3:5

. If 4 more boys join the class, what percent of the new class is boys? Step 1: Original boys:

\dfrac{3}{8} \times 40 = 15

. Original girls:

\dfrac{5}{8} \times 40 = 25

. Step 2: After 4 boys join: 19 boys, 25 girls, 44 students total.

\frac{19}{44} \times 100 \approx 43.18\%

Rounding to the nearest tenth, the answer is

\boxed{43.2}\%

Key Takeaways

Successive percent changes multiply as factors. Never add or subtract percent changes directly. A $30\%$ increase followed by a $30\%$ decrease is $1.30 \times 0.70 = 0.91$ , not $1.00$ .
To reverse a percent change, divide by the factor rather than subtracting the percent. If a price after a $20\%$ increase is $\$60$ , the original is $60 \div 1.20 = \$50$ , not $60 - 12 = \$48$ .
Mixture problems always have two equations: one for the total quantity and one for the total amount of the substance (acid, salt, juice, etc.).
When a problem layers multiple operations (markup, then discount, then tax), write the entire chain as a single expression before computing. This reduces errors and makes it easier to solve for an unknown in the chain.