# 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

ratiois 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

proportionis an equation stating that two ratios are equivalent.A

percentis 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
- how to translate words into math

## 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 \(x=144:\)

\[\begin{align} \frac{x}{6} &=\frac{24}{x}\\ \frac{144}{6} &=\frac{24}{144}\\ 24 &\neq \frac{1}{6} \end{align}\]

Wrong choice.

(B) If \(x=12:\)

\[\begin{align} \frac{x}{6} &=\frac{24}{x}\\ \frac{12}{6} &=\frac{24}{12}\\ 2 &=2 \end{align}\]

This is the correct answer.

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

\[\begin{align} \frac{x}{6} &=\frac{24}{x}\\ \frac{2\sqrt{6}}{6} &=\frac{24}{2\sqrt{6}}\\ 24 &\neq 144 \end{align}\]

Wrong choice.

(D) If \(x=4\):

\[\begin{align} \frac{x}{6} &=\frac{24}{x}\\ \frac{4}{6} &=\frac{24}{4}\\ \frac{2}{3} &\neq 6 \end{align}\]

Wrong choice.

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

\[\begin{align} \frac{x}{6} &=\frac{24}{x}\\ \frac{2\sqrt{3}}{6} &=\frac{24}{2\sqrt{3}}\\ 12 &\neq 144 \end{align}\]

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

**Cite as:**SAT Ratios, Proportions, and Percents.

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