Percent Word Problems SAT Notes

Applied percent problems require more than one isolated calculation. A strong model identifies what each percentage acts on, preserves quantities such as pure component or total principal, and combines sequential operations in the stated order.

Mixtures conserve the pure component

Two-solution mixture

\[c_1A_1+c_2A_2=c_f(A_1+A_2)\]

Each concentration $$c$$ is a decimal, each $$A$$ is an amount, and the total pure component before mixing equals the pure component in the final mixture.

Mixture setup for two solutions
Solution	Amount	Concentration	Pure component amount
Solution A	$$x$$ liters	$30\%=0.30$	$$0.30x$$
Solution B	$$40-x$$ liters	$55\%=0.55$	$$0.55(40-x)$$
Final mixture	$$40$$ liters	$40\%=0.40$	$$0.40(40)$$

Mixture workflow

Define the unknown
Name the amount of one input solution.
Express the other amount
Use the known final total when appropriate.
Convert rates
Write every concentration as a decimal.
Count pure component
Multiply each input amount by its concentration.
Conserve
Set total input pure component equal to final pure component.
Check
The final concentration should lie between the input concentrations.

Worked example

Blend two concentrations

How many liters of a $30\%$ solution should be combined with a $55\%$ solution to make $$40$$ liters of a $40\%$ solution?

Represent
Use $$x$$ liters at $30\%$ and $$40-x$$ liters at $55\%$ .
Conserve
$$0.30x+0.55(40-x)=0.40(40)$$ .
Solve
$$0.30x+22-0.55x=16$$ , so $$x=24$$ .
Check
$40\%$ lies between $30\%$ and $55\%$ .

Use

\(24\)

liters of the

30\%

solution and

\(16\)

liters of the

55\%

solution.

Simple-interest allocation

Annual simple interest

\[I=Pr\]

For one year, interest $$I$$ equals principal $$P$$ multiplied by decimal rate $$r$$ .

Two-investment allocation model
Investment	Principal	Rate	Annual interest
Account A	$$x$$ dollars	$4\%=0.04$	$$0.04x$$
Account B	$$12{,}000-x$$ dollars	$6.5\%=0.065$	$$0.065(12{,}000-x)$$

Worked example

Split a principal between two rates

A total of $\$12{,}000$ earns $\$630$ in one year. Part earns $4\%$ , and the rest earns $6.5\%$ . How much earns $4\%$ ?

Represent
Use $$x$$ at $4\%$ and $$12{,}000-x$$ at $6.5\%$ .
Add interest
$$0.04x+0.065(12{,}000-x)=630$$ .
Solve
$$0.04x+780-0.065x=630$$ , so $$x=6{,}000$$ .
Check
$$240+390=630$$ .

\$6{,}000

earns

4\%

Discounts, markup, tax, and reverse prices

Common price-operation multipliers
Operation	Multiplier	Example
Discount by $$r$$	$$1-r$$	$35\%$ off uses $$0.65$$
Tax or markup by $$r$$	$$1+r$$	$8\%$ tax uses $$1.08$$
Recover original	$\text{final}/\text{combined multiplier}$	Reverse multiplication with division

Successive price operations

\[F=O(1-d_1)(1-d_2)(1+t)\]

Apply each discount to the current price, then apply tax as the context directs. Do not combine the rates by simple addition.

Worked example

Successive discounts

An item receives discounts of $20\%$ and then $10\%$ . What single reduction is equivalent?

Multiply factors
$$0.80(0.90)=0.72$$ .
Interpret final factor
The customer pays $72\%$ of the original.
Find reduction
$100\%-72\%=28\%$ .

The equivalent reduction is

28\%

, not

30\%

Subgroup percentages multiply

When $a\%$ of a total belongs to a subgroup and $b\%$ of that subgroup meets another condition, the combined share is the product of the decimal rates. For example, $40\%$ followed by $25\%$ gives $$0.40(0.25)=0.10$$ , or $10\%$ of the total.

Common applied-percent mistakes

Adding successive percent changes instead of multiplying change factors.
Computing a second discount from the original price rather than the current price.
Using $$35$$ instead of $$0.35$$ in an equation.
Averaging mixture percentages without weighting by their amounts.
Forgetting that the total mixture amount changes when material is added.
Applying an interest rate to the entire principal instead of its assigned portion.
Ignoring the operation order stated for discount and tax.
Confusing a percent of a subgroup with a percent of the total.

Mini check

Check your understanding

A price receives $25\%$ off and then $8\%$ tax. Which expression gives the final price from original price $$P$$ ?

$$P(0.75)(1.08)$$
$$P(1-0.25+0.08)$$
$$P(0.25)(0.08)$$
$$P/0.75+0.08$$

Show answer and explanation

Answer: $$P(0.75)(1.08)$$

The discount leaves $75\%$ of the price, and the tax multiplies that current price by $$1.08$$ .

Key takeaways

What to remember

Mixture equations conserve the amount of pure component, not merely total volume.
Two-investment problems require both a principal-total equation and an interest-total equation.
Discounts use $$1-r$$ ; tax and markup use $$1+r$$ .
Successive operations multiply, and reverse problems divide by the combined multiplier.
Percent-of-subgroup chains multiply their decimal rates.

Continue learning

Put these notes into practice

Apply the ideas with SAT-style questions, then reinforce key details with flashcards.

Go to Practice Review Flashcards

Solution	Amount	Concentration	Pure component amount
Solution A	$\(x\)$ liters	$30\%=0.30$	$\(0.30x\)$
Solution B	$\(40-x\)$ liters	$55\%=0.55$	$\(0.55(40-x)\)$
Final mixture	$\(40\)$ liters	$40\%=0.40$	$\(0.40(40)\)$

Investment	Principal	Rate	Annual interest
Account A	$\(x\)$ dollars	$4\%=0.04$	$\(0.04x\)$
Account B	$\(12{,}000-x\)$ dollars	$6.5\%=0.065$	$\(0.065(12{,}000-x)\)$

Operation	Multiplier	Example
Discount by $\(r\)$	$\(1-r\)$	$35\%$ off uses $\(0.65\)$
Tax or markup by $\(r\)$	$\(1+r\)$	$8\%$ tax uses $\(1.08\)$
Recover original	$\text{final}/\text{combined multiplier}$	Reverse multiplication with division