The Central Idea
A ratio compares two quantities by division:
\[ \boxed{\text{ratio}=\frac{\text{first quantity}}{\text{second quantity}}}. \]A proportion states that two such comparisons are equal:
\[ \boxed{\frac{a}{b}=\frac{c}{d}}, \qquad b\neq0,\ d\neq0. \]A percent is a ratio whose reference whole is \(100\):
\[ \boxed{p\%=\frac{p}{100}}. \]These are three views of the same underlying idea: comparing one quantity with another by multiplication or division.
Additive and Multiplicative Comparisons
Applied problems often become easier once we decide whether the comparison is additive or multiplicative.
If one quantity is \(8\) and another is \(12\), then the additive difference is
\[ 12-8=4. \]The multiplicative comparison is
\[ \frac{12}{8}=\frac{3}{2}=1.5. \]Thus \(12\) is \(4\) greater than \(8\), but it is also \(1.5\) times as large as \(8\). Ratio, proportion, and percent problems are primarily about the second type of comparison.
What Is a Ratio?
A ratio compares two quantities by division.
The ratio of \(6\) to \(15\) may be written
\[ 6:15, \]or
\[ \frac{6}{15}. \]Simplifying gives
\[ \boxed{\frac{6}{15}=\frac{2}{5}}. \]The simplified ratio \(2:5\) means that for every \(2\) units of the first quantity, there are \(5\) corresponding units of the second.
The Order of a Ratio Matters
The ratio of \(a\) to \(b\) is
\[ \frac{a}{b}, \]while the ratio of \(b\) to \(a\) is
\[ \frac{b}{a}. \]These ratios are generally not equal.
For example,
\[ \frac{2}{5}\neq\frac{5}{2}. \]The wording of a problem therefore determines which quantity belongs in the numerator and which belongs in the denominator.
Part-to-Part and Part-to-Whole Ratios
Suppose a group contains \(12\) red objects and \(18\) blue objects.
The ratio of red to blue is
\[ \frac{12}{18}=\frac{2}{3}. \]This is a part-to-part ratio.
The total number of objects is
\[ 12+18=30. \]The ratio of red objects to all objects is
\[ \frac{12}{30}=\frac{2}{5}. \]This is a part-to-whole ratio.
Confusing these two comparisons is one of the most common errors in ratio problems.
Equivalent Ratios
Multiplying or dividing both terms of a ratio by the same nonzero number produces an equivalent ratio.
\[ \frac{2}{3} = \frac{2\cdot4}{3\cdot4} = \frac{8}{12}. \]Therefore,
\[ \boxed{2:3=8:12}. \]Equivalent ratios express the same multiplicative relationship at different sizes.
Scale Factors Inside Equivalent Ratios
Consider
\[ \frac{3}{5}=\frac{12}{20}. \]The numerator \(3\) was multiplied by \(4\) to obtain \(12\), and the denominator \(5\) was multiplied by the same factor \(4\) to obtain \(20\).
That shared multiplier is the scale factor connecting the equivalent ratios.
When a scale factor is easy to see, it can solve a proportion without any formal cross multiplication.
What Is a Proportion?
A proportion is an equation asserting that two ratios are equal.
\[ \boxed{\frac{a}{b}=\frac{c}{d}}. \]For example,
\[ \frac{3}{4}=\frac{15}{20}. \]Both ratios simplify to the same value:
\[ \frac{3}{4}=0.75, \qquad \frac{15}{20}=0.75. \]A proportion is therefore an equation about equivalent multiplicative comparisons.
Solving a Proportion by Scaling
Solve
\[ \frac{4}{7}=\frac{x}{35}. \]Since
\[ 7\cdot5=35, \]multiply the numerator \(4\) by the same factor:
\[ x=4\cdot5=20. \]Thus,
\[ \boxed{x=20}. \]Scaling is often the fastest method when the relationship between corresponding entries is obvious.
Solving a Proportion Algebraically
Solve
\[ \frac{5}{8}=\frac{x}{28}. \]Multiply both sides by \(28\):
\[ 28\left(\frac{5}{8}\right)=x. \]Simplify:
\[ x=\frac{28\cdot5}{8} =\frac{35}{2} =17.5. \]Therefore,
\[ \boxed{x=17.5}. \]Cross Products
If
\[ \frac{a}{b}=\frac{c}{d}, \qquad b\neq0,\ d\neq0, \]multiplying both sides by \(bd\) gives
\[ ad=bc. \]Thus,
\[ \boxed{\frac{a}{b}=\frac{c}{d}\Longrightarrow ad=bc}. \]This equality is commonly described as the equality of the cross products.
Why Cross Multiplication Works
Cross multiplication is not a separate algebraic law. It is a shortcut for multiplying both sides of a proportion by the common denominator.
Starting from
\[ \frac{a}{b}=\frac{c}{d}, \]multiply both sides by \(bd\):
\[ bd\left(\frac{a}{b}\right) = bd\left(\frac{c}{d}\right). \]Cancel the nonzero denominators:
\[ ad=bc. \]Understanding this derivation prevents cross multiplication from becoming an isolated memorized trick.
Example: Cross Products
Solve
\[ \frac{7}{12}=\frac{x}{30}. \]Set the cross products equal:
\[ 7(30)=12x. \]Then
\[ 210=12x, \]so
\[ x=\frac{210}{12}=17.5. \]Therefore,
\[ \boxed{x=17.5}. \]Keep Corresponding Quantities in Corresponding Positions
Suppose \(4\) notebooks cost \(\$10\), and \(7\) notebooks cost \(x\) dollars.
One consistent proportion is
\[ \frac{4\text{ notebooks}}{10\text{ dollars}} = \frac{7\text{ notebooks}}{x\text{ dollars}}. \]Another equally valid setup is
\[ \frac{10\text{ dollars}}{4\text{ notebooks}} = \frac{x\text{ dollars}}{7\text{ notebooks}}. \]What matters is consistency: notebooks must correspond with notebooks and dollars with dollars.
Units Help Organize Proportions
Units can function as structural labels.
If
\[ \frac{4\text{ notebooks}}{10\text{ dollars}} = \frac{7\text{ notebooks}}{x\text{ dollars}}, \]then the same kind of quantity appears in each horizontal position.
Writing units during setup is especially useful when several numbers appear in the problem and it is not immediately obvious which quantities should be paired.
Rates Are Ratios with Different Units
A rate is a ratio comparing quantities measured in different units.
Examples include
\[ \frac{60\text{ miles}}{1\text{ hour}}, \] \[ \frac{12\text{ dollars}}{3\text{ kilograms}}, \]and
\[ \frac{500\text{ words}}{4\text{ minutes}}. \]Rate problems use the same ratio logic developed here. Later lessons will apply that logic specifically to distance, motion, work, and other contexts.
Unit Rates
A unit rate compares a quantity with one unit of another quantity.
Suppose \(6\) notebooks cost \(\$15\). The cost per notebook is
\[ \frac{15}{6}=2.50. \]Thus the unit rate is
\[ \boxed{\$2.50\text{ per notebook}}. \]Once a unit rate is known, many proportion problems become simple multiplication problems.
Example: Solve with a Unit Rate
If \(6\) notebooks cost \(\$15\), what do \(14\) notebooks cost at the same rate?
First find the cost per notebook:
\[ \frac{15}{6}=2.50. \]Then multiply:
\[ 14(2.50)=35. \]Therefore,
\[ \boxed{\$35}. \]Ratios Can Divide a Total
Ratios are often used to divide a fixed total into proportional parts.
Suppose \(\$420\) is divided in the ratio
\[ 2:5. \]The ratio contains
\[ 2+5=7 \]total ratio parts.
Each ratio part is worth
\[ \frac{420}{7}=60. \]Therefore, the two shares are
\[ 2(60)=120 \]and
\[ 5(60)=300. \]Thus,
\[ \boxed{\$120\text{ and }\$300}. \]Example: Three-Part Ratio
Divide \(540\) in the ratio
\[ 2:3:4. \]The total number of ratio parts is
\[ 2+3+4=9. \]Each ratio part is
\[ \frac{540}{9}=60. \]Therefore, the three shares are
\[ 2(60)=120, \qquad 3(60)=180, \qquad 4(60)=240. \]Check:
\[ 120+180+240=540. \]What Is a Percent?
The word percent means “per hundred.”
\[ p\%=\frac{p}{100}. \]For example,
\[ 37\%=\frac{37}{100}=0.37. \]Percent notation is therefore another way to write a ratio.
Converting Percent to Decimal
Divide by \(100\), which moves the decimal point two places to the left:
\[ 62\%=0.62, \] \[ 8\%=0.08, \]and
\[ 125\%=1.25. \]A percent may be greater than \(100\%\). That simply means the compared quantity is greater than the chosen reference whole.
Converting Decimal to Percent
Multiply by \(100\%\), which moves the decimal point two places to the right:
\[ 0.43=43\%, \] \[ 0.075=7.5\%, \]and
\[ 1.6=160\%. \]Converting Fractions to Percent
A fraction can be converted to a percent by dividing to obtain a decimal, then multiplying by \(100\%\).
For example,
\[ \frac{3}{8}=0.375. \]Therefore,
\[ 0.375(100\%)=37.5\%. \]So
\[ \boxed{\frac{3}{8}=37.5\%}. \]The Basic Percent Relationship
Most elementary percent problems can be organized around
\[ \boxed{\text{part}=\text{rate}\cdot\text{whole}}. \]If we use \(P\) for the part, \(r\) for the decimal rate, and \(W\) for the whole, then
\[ \boxed{P=rW}. \]This one equation can be rearranged to solve for any of the three quantities.
Percent Language and Algebra
The sentence
“\(18\) is \(30\%\) of \(60\)”
translates directly into
\[ 18=0.30(60). \]In many textbook problems:
- “is” becomes an equals sign;
- “of” usually indicates multiplication;
- the percent is converted to a decimal before calculation.
These language cues are useful, but the underlying relationship \(P=rW\) is more reliable than memorizing isolated wording rules.
Finding the Part
What is \(35\%\) of \(240\)?
Here,
\[ r=0.35, \qquad W=240. \]Use
\[ P=rW. \]Then
\[ P=0.35(240)=84. \]Therefore,
\[ \boxed{35\%\text{ of }240\text{ is }84}. \]Finding the Whole
\(54\) is \(30\%\) of what number?
Here,
\[ P=54, \qquad r=0.30. \]Solve
\[ 54=0.30W. \]Divide by \(0.30\):
\[ W=\frac{54}{0.30}=180. \]Thus,
\[ \boxed{54\text{ is }30\%\text{ of }180}. \]Finding the Percent
\(42\) is what percent of \(120\)?
Use
\[ 42=r(120). \]Then
\[ r=\frac{42}{120}=0.35. \]Convert to percent:
\[ 0.35=35\%. \]Therefore,
\[ \boxed{42\text{ is }35\%\text{ of }120}. \]The Percent Proportion
Percent problems can also be written as proportions.
\[ \boxed{\frac{\text{part}}{\text{whole}}=\frac{\text{percent}}{100}}. \]For example, \(42\) is what percent of \(120\)?
\[ \frac{42}{120}=\frac{p}{100}. \]Cross multiply:
\[ 42(100)=120p. \]Then
\[ p=35. \]Therefore, the percent is
\[ \boxed{35\%}. \]Equation Method versus Proportion Method
The equation
\[ P=rW \]and the proportion
\[ \frac{P}{W}=\frac{p}{100} \]are equivalent when
\[ r=\frac{p}{100}. \]Use whichever form makes the relationship easiest to see. The important issue is not the format but the correctness of the comparison.
Percent Increase
Suppose a quantity increases from \(80\) to \(92\).
The amount of increase is
\[ 92-80=12. \]Percent increase compares that change with the original value:
\[ \frac{12}{80}=0.15. \]Therefore,
\[ \boxed{15\%\text{ increase}}. \]Percent Decrease
Suppose a quantity decreases from \(250\) to \(200\).
The decrease is
\[ 250-200=50. \]Compare with the original value:
\[ \frac{50}{250}=0.20. \]Therefore,
\[ \boxed{20\%\text{ decrease}}. \]The General Percent-Change Formula
If \(O\) is the original value and \(N\) is the new value, then the signed relative change is
\[ \frac{N-O}{O}, \qquad O\neq0. \]Multiplying by \(100\%\) gives
\[ \boxed{ \text{percent change} = \frac{N-O}{O}\cdot100\% }. \]A positive result indicates an increase. A negative result indicates a decrease.
Why the Original Value Is the Reference
Percent change answers the question:
“How large was the change compared with where we started?”
If a value rises from \(40\) to \(50\), the increase is \(10\). Relative to the original \(40\),
\[ \frac{10}{40}=25\%. \]Comparing the same increase with the new value would give
\[ \frac{10}{50}=20\%, \]which answers a different question.
Increase by a Percent
To increase an original value \(V\) by a rate \(r\), add \(rV\):
\[ V+rV. \]Factor:
\[ \boxed{V(1+r)}. \]For a \(12\%\) increase,
\[ 1+r=1.12. \]So increasing \(250\) by \(12\%\) gives
\[ 250(1.12)=280. \]Decrease by a Percent
To decrease an original value \(V\) by a rate \(r\), subtract \(rV\):
\[ V-rV. \]Factor:
\[ \boxed{V(1-r)}. \]For a \(30\%\) decrease,
\[ 1-r=0.70. \]Therefore, decreasing \(180\) by \(30\%\) gives
\[ 180(0.70)=126. \]Percent Multipliers
Percent increases and decreases can be handled efficiently with multiplicative factors.
- Increase by \(8\%\): multiply by \(1.08\).
- Increase by \(25\%\): multiply by \(1.25\).
- Decrease by \(8\%\): multiply by \(0.92\).
- Decrease by \(25\%\): multiply by \(0.75\).
In general,
\[ \boxed{\text{new value}=\text{original value}(1\pm r)}. \]Reverse Percent Problems
Sometimes the final value and the percent change are known, but the original value is unknown.
After a quantity decreases by \(20\%\), its new value is \(72\). What was the original value?
A \(20\%\) decrease leaves \(80\%\) of the original value:
\[ 72=0.80P. \]Therefore,
\[ P=\frac{72}{0.80}=90. \]So the original value was
\[ \boxed{\$90}. \]Successive Percent Changes
Successive percent changes are applied one after another, so their multipliers are multiplied.
Suppose a value increases by \(20\%\) and then decreases by \(20\%\).
The combined multiplier is
\[ 1.20(0.80)=0.96. \]Therefore, the final value is \(96\%\) of the original:
\[ \boxed{\text{net change}=-4\%}. \]Equal percentage increases and decreases do not generally cancel because the second change is applied to a different base.
Percentage Points versus Percent Change
Suppose a rate rises from \(30\%\) to \(36\%\).
The difference is
\[ 36\%-30\%=6 \]percentage points.
But the percent increase in the rate itself is
\[ \frac{36-30}{30}=0.20=20\%. \]Therefore, the rate increased by \(6\) percentage points, which is a \(20\%\) relative increase.
Percent Greater Than \(100\%\)
A percent greater than \(100\%\) is completely valid.
If \(90\) is compared with \(60\), then
\[ \frac{90}{60}=1.5. \]As a percent,
\[ 1.5(100\%)=150\%. \]Thus \(90\) is
\[ \boxed{150\%\text{ of }60}. \]Percent Less Than \(1\%\)
Very small ratios may be expressed with percentages below \(1\%\).
For example,
\[ \frac{3}{800}=0.00375. \]Therefore,
\[ 0.00375(100\%)=0.375\%. \]So
\[ \boxed{\frac{3}{800}=0.375\%}. \]Example: Ratio from a Context
A class has \(18\) students who chose option A and \(12\) who chose option B.
The ratio of A to B is
\[ 18:12. \]Divide both terms by \(6\):
\[ \boxed{3:2}. \]This means that for every \(3\) students choosing A, there are \(2\) choosing B, assuming the same ratio.
Example: Find a Missing Quantity from a Ratio
The ratio of blue tiles to white tiles is \(3:5\). If there are \(27\) blue tiles, how many white tiles are there?
Write
\[ \frac{3}{5}=\frac{27}{w}. \]Since
\[ 3\cdot9=27, \]the corresponding white quantity is
\[ 5\cdot9=45. \]Therefore,
\[ \boxed{45\text{ white tiles}}. \]Example: Divide a Total in a Given Ratio
A total of \(960\) units must be divided in the ratio \(5:3\).
There are
\[ 5+3=8 \]ratio parts.
One part is
\[ \frac{960}{8}=120. \]Therefore, the two allocations are
\[ 5(120)=600 \]and
\[ 3(120)=360. \]Thus,
\[ \boxed{600\text{ and }360}. \]Example: Percent of a Total
A collection contains \(480\) items. If \(37.5\%\) belong to one category, how many items are in that category?
Convert the percent:
\[ 37.5\%=0.375. \]Multiply:
\[ 0.375(480)=180. \]Therefore,
\[ \boxed{180\text{ items}}. \]Example: Find the Percent from Counts
A group contains \(75\) successful outcomes out of \(120\) trials.
The success ratio is
\[ \frac{75}{120}=0.625. \]Convert to percent:
\[ 0.625(100\%)=62.5\%. \]Thus,
\[ \boxed{62.5\%} \]of the trials were successful.
Example: Find the Original Whole
\(84\) represents \(35\%\) of a total. Find the total.
Let \(W\) be the whole:
\[ 84=0.35W. \]Solve:
\[ W=\frac{84}{0.35}=240. \]Therefore,
\[ \boxed{W=240}. \]Example: Percent Increase in Context
A quantity rises from \(640\) to \(760\).
The increase is
\[ 760-640=120. \]Relative to the original value,
\[ \frac{120}{640}=0.1875. \]Therefore,
\[ \boxed{18.75\%\text{ increase}}. \]Example: Percent Decrease in Context
A quantity falls from \(900\) to \(765\).
The decrease is
\[ 900-765=135. \]Relative to the original value,
\[ \frac{135}{900}=0.15. \]Therefore,
\[ \boxed{15\%\text{ decrease}}. \]Translating Ratio Statements into Equations
Suppose the ratio of \(x\) to \(y\) is \(4:7\).
This can be written
\[ \frac{x}{y}=\frac{4}{7}. \]Equivalently,
\[ 7x=4y. \]If \(y\) is known, then
\[ x=\frac{4}{7}y. \]The verbal ratio therefore translates naturally into an algebraic relationship.
Translating “Times As Large”
If \(A\) is \(1.6\) times as large as \(B\), then
\[ \boxed{A=1.6B}. \]The corresponding ratio is
\[ \frac{A}{B}=1.6. \]As a percent,
\[ A=160\%\text{ of }B. \]These are three equivalent descriptions of the same multiplicative comparison.
“Percent Of” and “Percent More Than” Are Different
If \(A\) is \(120\%\) of \(B\), then
\[ A=1.20B. \]This means \(A\) is \(20\%\) more than \(B\), not \(120\%\) more than \(B\).
In general,
\[ (100+r)\%\text{ of a quantity} \]means
\[ r\%\text{ more than that quantity}. \]Checking with Estimation
Exact algebra should be accompanied by a reasonableness check.
Suppose \(19\%\) of \(410\) is calculated as \(77.9\).
Since \(20\%\) of \(410\) is
\[ 82, \]a result slightly below \(82\) is reasonable.
Estimation can quickly expose misplaced decimals or incorrect percent conversions.
Checking a Proportion by Substitution
Suppose
\[ \frac{5}{8}=\frac{x}{24} \]gives \(x=15\).
Substitute:
\[ \frac{15}{24}=\frac{5}{8}. \]Since
\[ \frac{15}{24}=\frac{5}{8}, \]the solution checks.
Checking Ratio Allocations
When a total is divided in a ratio, use two checks.
- Verify that the shares add to the original total.
- Verify that the shares preserve the required ratio.
If \(420\) is divided into \(120\) and \(300\), then
\[ 120+300=420 \]and
\[ 120:300=2:5. \]Both conditions are satisfied.
Common Error: Reversing a Ratio
If the problem asks for the ratio of red to blue and there are \(8\) red and \(12\) blue, then
\[ \frac{\text{red}}{\text{blue}} = \frac{8}{12} = \frac{2}{3}. \]Writing
\[ \frac{12}{8} \]answers the different question “blue to red.”
Common Error: Mixing Part-to-Part and Part-to-Whole
If a group has \(4\) cats and \(6\) dogs, then cats to dogs is
\[ \frac{4}{6}=\frac{2}{3}. \]Cats to all animals is
\[ \frac{4}{10}=\frac{2}{5}. \]These are different comparisons. Identify the reference quantity before calculating.
Common Error: Inconsistent Proportion Setup
Suppose \(3\) units correspond to \(12\) dollars and \(5\) units correspond to \(x\) dollars.
A consistent proportion is
\[ \frac{3}{12}=\frac{5}{x}. \]The setup
\[ \frac{3}{12}=\frac{x}{5} \]reverses the positions in the second ratio and does not preserve correspondence.
Common Error: Using the Percent Number Without Converting
To find \(18\%\) of \(250\), use
\[ 0.18(250), \]not
\[ 18(250). \]Since
\[ 18\%=0.18, \]the correct result is
\[ \boxed{45}. \]Common Error: Dividing by \(100\) Twice
Once \(25\%\) has been converted to
\[ 0.25, \]it should not be divided by \(100\) again.
For example,
\[ 25\%\text{ of }80=0.25(80)=20. \]The decimal \(0.25\) already contains the conversion from percent notation.
Common Error: Using the New Value as the Percent-Change Base
A value rises from \(50\) to \(65\).
The increase is \(15\), but the percent increase is
\[ \frac{15}{50}=0.30=30\%, \]not
\[ \frac{15}{65}. \]Percent change is measured relative to the original value unless a problem explicitly defines a different reference.
Common Error: Assuming Equal Opposite Percent Changes Cancel
Increasing \(100\) by \(50\%\) gives
\[ 100(1.5)=150. \]Decreasing \(150\) by \(50\%\) gives
\[ 150(0.5)=75. \]The final value is not \(100\) because the increase and decrease use different reference amounts.
Common Error: Confusing Percent with Percentage Points
If a rate changes from \(40\%\) to \(50\%\), the increase is
\[ 10 \]percentage points.
Relative to the original \(40\%\), however, the percent increase is
\[ \frac{50-40}{40}=0.25=25\%. \]The two descriptions measure different things.
A Reliable Procedure for Ratio Problems
- Identify the two quantities being compared.
- Preserve the order stated in the problem.
- Attach units or labels when they clarify the comparison.
- Write the ratio as a fraction when algebra will be required.
- Simplify the ratio if the problem asks for simplest form.
- If a total is being divided, add the ratio parts first.
- Check that the final comparison matches the requested order.
A Reliable Procedure for Proportion Problems
- Identify corresponding quantities.
- Place like quantities in matching positions.
- Write two equivalent ratios.
- Look first for an obvious scale factor.
- If needed, solve algebraically or use equal cross products.
- Substitute the result back into the proportion.
- Check units and context.
A Reliable Procedure for Percent Problems
- Identify the part, rate, and whole.
- Convert percent notation to a decimal rate when using \(P=rW\).
- Write the equation \(P=rW\).
- Solve for the unknown quantity.
- For percent change, use the original value as the reference.
- For increases or decreases, consider using the multiplier \(1\pm r\).
- Estimate to verify that the result has a reasonable magnitude.
A General Modeling Workflow
Ratio, proportion, and percent problems are early examples of a broader algebraic modeling process.
- Identify the quantities.
- Determine how they are related.
- Choose variables for unknown quantities.
- Translate the relationship into an equation.
- Solve the equation.
- Attach the correct units.
- Interpret the result in the original context.
- Check the result numerically and conceptually.
The final lesson of this section will return to this workflow in a more general form.
Practice 1: Simplify a Ratio
Simplify
\[ 42:63. \]Practice 2: Part-to-Whole Ratio
A collection has \(14\) green items and \(21\) yellow items. Find the ratio of green items to the total number of items.
Practice 3: Solve a Proportion
Solve
\[ \frac{6}{11}=\frac{x}{44}. \]Practice 4: Proportion with a Noninteger Solution
Solve
\[ \frac{9}{14}=\frac{x}{35}. \]Practice 5: Divide a Total by a Ratio
Divide \(720\) in the ratio
\[ 3:5. \]Practice 6: Three-Part Ratio
Divide \(1{,}260\) in the ratio
\[ 2:3:4. \]Practice 7: Find the Part
Find \(28\%\) of \(450\).
Practice 8: Find the Whole
\(72\) is \(24\%\) of what number?
Practice 9: Find the Percent
\(54\) is what percent of \(240\)?
Practice 10: Percent Increase
A quantity rises from \(320\) to \(392\). Find the percent increase.
Practice 11: Percent Decrease
A quantity falls from \(850\) to \(697\). Find the percent decrease.
Practice 12: Reverse Percent
After a quantity decreases by \(15\%\), its new value is \(102\). Find the original value.
Practice 13: Successive Changes
A value increases by \(10\%\) and then decreases by \(10\%\). What percent of the original value remains?
Practice 14: Percentage Points
A rate rises from \(24\%\) to \(30\%\). Find both the increase in percentage points and the relative percent increase.
Practice 15: Identify the Error
A student says that a rise from \(80\) to \(100\) is a \(20\%\) increase because the amount of increase is \(20\).
Explain the error and find the correct percent increase.
Practice Answers
Practice 1
Divide both terms by \(21\):
\[ \boxed{42:63=2:3}. \]Practice 2
The total is
\[ 14+21=35. \]Therefore,
\[ \frac{14}{35}=\frac{2}{5}. \]The ratio of green items to all items is
\[ \boxed{2:5}. \]Practice 3
Since \(11\cdot4=44\),
\[ x=6\cdot4=24. \]Thus,
\[ \boxed{x=24}. \]Practice 4
\[ 9(35)=14x. \] \[ 315=14x. \] \[ x=22.5. \]Therefore,
\[ \boxed{x=22.5}. \]Practice 5
The ratio contains
\[ 3+5=8 \]parts. Each part is
\[ \frac{720}{8}=90. \]Therefore,
\[ \boxed{270\text{ and }450}. \]Practice 6
The ratio contains
\[ 2+3+4=9 \]parts. Each part is
\[ \frac{1260}{9}=140. \]Therefore,
\[ \boxed{280,\ 420,\ 560}. \]Practice 7
\[ 28\%=0.28. \] \[ 0.28(450)=126. \]Therefore,
\[ \boxed{126}. \]Practice 8
\[ 72=0.24W. \] \[ W=\frac{72}{0.24}=300. \]Therefore,
\[ \boxed{300}. \]Practice 9
\[ 54=r(240). \] \[ r=\frac{54}{240}=0.225. \] \[ 0.225=22.5\%. \]Therefore,
\[ \boxed{22.5\%}. \]Practice 10
The increase is
\[ 392-320=72. \]Relative to the original value,
\[ \frac{72}{320}=0.225. \]Therefore,
\[ \boxed{22.5\%\text{ increase}}. \]Practice 11
The decrease is
\[ 850-697=153. \]Then
\[ \frac{153}{850}=0.18. \]Therefore,
\[ \boxed{18\%\text{ decrease}}. \]Practice 12
A \(15\%\) decrease leaves \(85\%\) of the original value:
\[ 102=0.85P. \] \[ P=\frac{102}{0.85}=120. \]Therefore,
\[ \boxed{\$120}. \]Practice 13
Multiply the successive factors:
\[ 1.10(0.90)=0.99. \]Therefore,
\[ \boxed{99\%} \]of the original value remains.
Practice 14
The increase in percentage points is
\[ 30\%-24\%=6 \]percentage points.
The relative percent increase is
\[ \frac{30-24}{24}=0.25=25\%. \]Therefore,
\[ \boxed{6\text{ percentage points and }25\%\text{ relative increase}}. \]Practice 15
The student found the absolute increase but did not compare that increase with the original value.
The correct calculation is
\[ \frac{100-80}{80} = \frac{20}{80} = 0.25. \]Therefore,
\[ \boxed{25\%\text{ increase}}. \]What to Remember
A ratio compares quantities multiplicatively:
\[ \boxed{\text{ratio}=\frac{\text{first quantity}}{\text{second quantity}}}. \]A proportion states that two ratios are equal:
\[ \boxed{\frac{a}{b}=\frac{c}{d}}. \]Percent means per hundred:
\[ \boxed{p\%=\frac{p}{100}}. \]The most useful percent equation is
\[ \boxed{P=rW}. \]For percent change,
\[ \boxed{ \text{percent change} = \frac{\text{new}-\text{original}}{\text{original}}\cdot100\% }. \]The most important habit is to identify the reference quantity before calculating. Ratios, proportions, and percentages are meaningful only in relation to what is being compared.
Where We Go Next
Ratio and percent methods become especially powerful when quantities are connected by unit conversions, scale factors, and changing proportional relationships.
The next lesson, Scale, Conversion, and Variation Problems, develops those applications systematically.