SAT Ratios, Proportions, and Percents

To solve problems about ratios, proportions, and percents on the SAT, you need to know:

the difference between ratios, proportions, and percents

A ratio is is a comparison between two quantities of the same kind. It can be expressed as $a$ to $b$ , $a:b,$ $a/b$ , or $\frac{a}{b}.$

A proportion is an equation stating that two ratios are equivalent.

A percent is a fraction of a hundred. It can be expressed as $\frac{a}{100},$ $a\%,$ or $a$ percent.

the properties of proportion

If $a, b, c,$ and $d$ are nonzero and $\frac{a}{b}=\frac{c}{d},$ then

\[\begin{array}{r c l} \frac{b}{a} &=& \frac{d}{c}\\ \\ \frac{a}{c} &=& \frac{b}{d}\\ \\ ad&=&bc \\ \\ \frac{a}{a+b} &=& \frac{c}{c+d} \\ \\ \frac{a+c}{b+d}&=&\frac{a}{b}=\frac{c}{d}
\end{array}\]

how to work with fractions and decimals
how to simplify ratios
how to find unknown ratios
how to convert percentages, fractions, and decimals
how to find percentage change
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Examples

If $x$ is positive and $\frac{x}{6} =\frac{24}{x},$ what is the value of $x$ ?

(A) $\ \ 144$
(B) $\ \ 12$
(C) $\ \ 2\sqrt{6}$
(D) $\ \ 4$
(E) $\ \ 2\sqrt{3}$

Correct Answer: B

Solution 1:

Tip: Know the Properties of Proportions.
We solve for $x$ by cross multiplying:

$\begin{array}{r c l l} \frac{x}{6} &=& \frac{24}{x} &\quad \text{given}\\ x^{2} &=& 144 &\quad \text{cross-multiply}\\ x &=& \pm 12 &\quad \text{square root both sides}\\ x &=& 12 &\quad \text{select the positive answer} \end{array}$

Solution 2:

Tip: Plug and check.
We plug each of the answer into the proportion and select the one that does not yield a contradiction.

(A) If $x=144:$

$\begin{aligned} \frac{x}{6} &=\frac{24}{x}\\ \frac{144}{6} &=\frac{24}{144}\\ 24 &\neq \frac{1}{6} \end{aligned}$

Wrong choice.

(B) If $x=12:$

$\begin{aligned} \frac{x}{6} &=\frac{24}{x}\\ \frac{12}{6} &=\frac{24}{12}\\ 2 &=2 \end{aligned}$

This is the correct answer.

(C) If $x=2\sqrt{6}$ :

$\begin{aligned} \frac{x}{6} &=\frac{24}{x}\\ \frac{2\sqrt{6}}{6} &=\frac{24}{2\sqrt{6}}\\ 24 &\neq 144 \end{aligned}$

Wrong choice.

(D) If $x=4$ :

$\begin{aligned} \frac{x}{6} &=\frac{24}{x}\\ \frac{4}{6} &=\frac{24}{4}\\ \frac{2}{3} &\neq 6 \end{aligned}$

Wrong choice.

(E) If $x=2\sqrt{3}$ :

$\begin{aligned} \frac{x}{6} &=\frac{24}{x}\\ \frac{2\sqrt{3}}{6} &=\frac{24}{2\sqrt{3}}\\ 12 &\neq 144 \end{aligned}$

Wrong choice.

Incorrect Choices:

(A)
If you solve $\frac{x}{6} =24,$ you will get this wrong answer.

(C)
If you solve $x =\frac{24}{x},$ you will get this wrong answer.

(D)
If you solve $6x=24,$ you will get this wrong answer.

(E)
If you solve for $\sqrt{x},$ you will get this wrong answer.

How many pens can Alice buy with $m$ dollars, if $k$ pens cost $n$ cents?

(A) $\ \ \frac{n}{100km}$

(B) $\ \ \frac{km}{n}$

(C) $\ \ \frac{nk}{100m}$

(D) $\ \ \frac{100km}{n}$

(E) $\ \ 100kmn$

Correct Answer: D

Solution 1:

Tip: Pay attention to units.
$m$ dollars $= 100m$ cents.

We set up the proportion:

$\begin{array}{r c l l} \frac{k}{n} &=& \frac{x}{100m} &\quad \frac{\mbox{pens}}{\mbox{cents}}\\ \frac{100km}{n} &=& x &\quad \text{multiply both sides by}\ m\\ \end{array}$

Alice can purchase $\frac{100km}{n}$ pens with $m$ dollars.

Solution 2:

Tip: Replace variables with numbers.
Let the price of $k=1$ pen be $n=100$ cents, and let Alice have $m=2$ dollars. Since $2$ dollars $=200$ cents, with $200$ cents she can buy $2$ pens.

We check which of the choices yields $2$ pens.

(A) $\ \ \frac{n}{100km} =\frac{100}{100\cdot 1 \cdot 2} =\frac{1}{2} \neq 2.$ Wrong choice.

(B) $\ \ \frac{km}{n} = \frac{1 \cdot 2}{100} = \frac{1}{50} \neq 2.$ Wrong choice.

(C) $\ \ \frac{nk}{100m} = \frac{100 \cdot 1}{100 \cdot 2} = \frac{1}{2} \neq 2.$ Wrong choice.

(D) $\ \ \frac{100km}{n} = \frac{100\cdot 1 \cdot 2}{100}= 2.$ Correct answer.

(E) $\ \ 100kmn = 100\cdot 1 \cdot 2 \cdot 100 = 20,000.$ Wrong choice.

Incorrect Choices:

(A), (B), (C), and (E)
Solution 2 eliminates these choices by replacing the variables with numbers.

Which of the following statements is always true?

$\begin{array}{l l l c l} &\text{I.} &a \%\ \text{of}\ b &=& b\%\ \text{of}\ a\\ &\text{II.} &(a+b)\% &=& a\% + b\%\\ &\text{III.} &(ab)\% &=& a\% \cdot b\%\\ \end{array}$

(A) I only
(B) II only
(C) I and II only
(D) II and III only
(E) I, II and III

Correct Answer: C

Solution 1:

We analyze each of the options.

I. $a\%\ \text{of}\ b = \frac{a}{100} \cdot b = \frac{ab}{100} = \frac{b}{100} \cdot a = b\%\ \text{of}\ a.$ This is true.

II. $(a+b)\% = \frac{a+b}{100}=\frac{a}{100} + \frac{b}{100}=a \% + b \%.$ This is true.

III. Starting with the right hand side,

$a\% \cdot b\% = \frac{a}{100} \cdot \frac{b}{100} = \frac{ab}{100} \cdot \frac{1}{100} = (ab)\%\ \text{of}\ \frac{1}{100} \neq (ab)\%.$

Only options I and II are true statements, and therefore answer (C) is correct.

Solution 2:

Tip: Replace variables with numbers.
Let $a=10$ and $b=20.$ For each option, we plug these in and check if the two sides of the equation yield the same answer.

I. Left Hand Side = $a\%\ \text{of}\ b = 10\%\ \text{of}\ 20 = \frac{10}{100} \cdot 20 = 2$

and

Right Hand Side = $b\%\ \text{of}\ a = 20\%\ \text{of}\ 10 = \frac{20}{100} \cdot 10 = 2.$

LHS = RHS and therefore this option is true.

II. LHS = $(a+b)\% = (10+20)\% = \frac{10+20}{100}=\frac{30}{100}$

and

RHS = $a\% + b\% = 10\% + 20\% = \frac{10}{100} + \frac{20}{100} = \frac{30}{100}.$

LHS = RHS and therefore this options is true.

III. LHS = $(ab)\% = (10\cdot 20)\% =30\% = \frac{30}{100}$

and

RHS = $a\% \cdot b\% = \frac{10}{100} \cdot \frac{20}{100} = \frac{2}{100}.$

But, $\frac{30}{100} \neq \frac{2}{100},$ and therefore option III is false. The correct answer is choice (C).

Incorrect Choices:

(A), (B), (D), and (E)
Solution 1 shows why these choices are wrong. Solution 2 eliminates them by plugging and checking.

Review

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

SAT Tips for Ratios, Proportions, and Percents

Know the Properties of Proportions.

Replace variables with numbers.

Pay attention to units.

SAT General Tips

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