Chapter 4 — Proportions
Topics in this chapter:
4.1 Direct and Inverse Proportions
Direct Proportion: Two quantities are directly proportional when their quotient is constant. As one increases, so does the other.
\[\frac{a}{c} = k \quad (\text{constant})\]Inverse Proportion: Two quantities are inversely proportional when their product is constant. As one increases, the other decreases.
\[a \cdot c = k \quad (\text{constant})\]The constant $k$ is called the constant of proportionality.
Notation: Ratios can be written as fractions $\frac{a}{b}$ or with colons $a : b$. For three quantities: $a : b : c$. When solving problems, always convert colons to fractions.
Example 4-1: A 10-foot pole casts an 8-foot shadow. How long is a pole that casts a 12-foot shadow?
Solution: Height is directly proportional to shadow length:
\[\frac{\text{pole}}{\text{shadow}} = \frac{10}{8} \implies \frac{\text{pole}}{12} = \frac{10}{8} \implies \text{pole} = 15 \text{ feet}.\]Example 4-2: The ratio of boys to girls in a class is $3:2$. If there are 35 students, how many girls?
Solution: Out of every 5 students, 2 are girls. So $\frac{2}{5} \times 35 = 14$ girls.
Exercise 4-1: In 4 games, Michael Jordan scores 124 points. At this rate, how many will he score in the next 6 games?
Exercise 4-2: The ratio of wins to losses of the Yankees is $15:16$. If they lost 64 games, how many games did they play?
Joint Proportion
If a quantity is directly proportional to more than one variable, it is jointly proportional to those values.
Example 4-3: If $x$ and $y$ are inversely proportional and $x = 10$ when $y = 6$, find $x$ when $y = 4$.
Solution: $xy = (10)(6) = 60$, so when $y = 4$: $x = \frac{60}{4} = 15$.
Example 4-4: Given $x$ is directly proportional to $y$ and $z$, and inversely proportional to $w$, with $x = 4$ when $(w,y,z) = (6,8,5)$, find $x$ when $(w,y,z) = (4,10,9)$.
Solution: The quantity $\frac{xw}{yz}$ is constant:
\[\frac{xw}{yz} = \frac{(4)(6)}{(8)(5)} = \frac{3}{5}\]When $(w,y,z) = (4,10,9)$:
\[x = \frac{3yz}{5w} = \frac{3(10)(9)}{5(4)} = \frac{27}{2}\]Example 4-5: It takes 3 days for 4 people to paint 5 houses. How long for 2 people to paint 6 houses?
Solution: More houses → more time (direct). More people → less time (inverse). So:
\[\frac{\text{houses}}{(\text{people})(\text{days})} = \frac{5}{(4)(3)} = \frac{5}{12}\] \[\text{days} = \frac{12 \cdot \text{houses}}{5 \cdot \text{people}} = \frac{12 \cdot 6}{5 \cdot 2} = \frac{36}{5}\]Example 4-6 (Clock Problem): It is four o’clock. How many minutes until the minute and hour hands are coincident?
Solution: After $x$ minutes, the minute hand is at the $x$-minute mark. The hour hand has moved $\frac{x}{60}$ of the way from 4 to 5. Starting at the 20-minute mark, it is at position $20 + 5!\left(\frac{x}{60}\right)$ minutes.
Setting equal: $x = 20 + \frac{x}{12}$, so $\frac{11x}{12} = 20$, giving $x = \frac{240}{11}$ minutes.
Exercise 4-3: If 5 hens can lay 24 eggs in 5 days, how many days are needed for 8 hens to lay 20 eggs?
4.2 Manipulating Proportions
Key Proportion Identities: If $\frac{a}{b} = \frac{c}{d}$, then all of the following hold:
\[\frac{a}{b} = \frac{c}{d} = \frac{a+c}{b+d} = \frac{a-c}{b-d} = \frac{a + kc}{b + kd} \quad \text{for all } k.\]Multiplying proportions: Given $\frac{x}{y} = \frac{4}{5}$ and $\frac{z}{y} = \frac{4}{3}$, find $\frac{x}{z}$:
\[\frac{x}{z} = \frac{x}{y} \cdot \frac{y}{z} = \frac{4}{5} \cdot \frac{3}{4} = \frac{3}{5}\]Note how the $y$’s cancel when we multiply the ratios.
Example 4-7: Find $\frac{x}{y}$ if $\;\dfrac{x + 2y}{x - y} = \dfrac{3}{4}$.
Solution: Cross-multiply: $4(x + 2y) = 3(x - y)$, so $4x + 8y = 3x - 3y$, giving $x + 11y = 0$, thus $x = -11y$ and:
\[\frac{x}{y} = -11\]Exercise 4-4: Find $\frac{2y}{x}$ if $\frac{x}{3z} = 3$ and $\frac{y}{4z} = 2$.
4.3 Conversion Factors
Conversion factors are fractions equal to $1$ used to change units. Since $1\text{ inch} = 2.54\text{ cm}$:
\[\frac{1\text{ inch}}{2.54\text{ cm}} = 1\]Choose the form that cancels the unwanted unit.
Example: Convert 180 cm to feet, given 1 inch = 2.54 cm and 1 foot = 12 inches.
\[180\text{ cm} \times \frac{1\text{ in}}{2.54\text{ cm}} \times \frac{1\text{ ft}}{12\text{ in}} \approx 5.9\text{ ft}\]Then $0.9\text{ ft} \times \frac{12\text{ in}}{1\text{ ft}} = 10.8\text{ in}$, so about $5’11’’$.
For squared/cubed units, apply the conversion factor multiple times:
\[5\text{ ft}^2 \times \left(\frac{0.3048\text{ m}}{1\text{ ft}}\right)^2 \approx 0.4645\text{ m}^2\]For negative exponents (e.g. $\text{kg}\cdot\text{m}^{-3}$), the conversion factor cube goes in the numerator:
\[\frac{\text{kg}}{\text{m}^3} \times \left(\frac{0.3048\text{ m}}{1\text{ ft}}\right)^3 \approx 0.02832\;\frac{\text{kg}}{\text{ft}^3}\]Example 4-8: Convert $16.2\;\text{in}^2\;\text{oz}^{-1}$ to cm and grams, given 1 in = 2.54 cm, 1 oz = 28.35 g.
Solution:
\[16.2\;\text{in}^2\;\text{oz}^{-1} \times \left(\frac{2.54\;\text{cm}}{1\;\text{in}}\right)^2 \approx 104.5\;\text{cm}^2\;\text{oz}^{-1}\] \[104.5\;\text{cm}^2\;\text{oz}^{-1} \times \frac{1\;\text{oz}}{28.35\;\text{g}} \approx 3.686\;\text{cm}^2\;\text{g}^{-1}\]Example 4-9: If 4 gleeps = 3 glops and 2 glops = 5 glips, how many glips equal 10 gleeps?
Solution:
\[10\;\text{gleeps} \times \frac{3\;\text{glops}}{4\;\text{gleeps}} = 7.5\;\text{glops}\] \[7.5\;\text{glops} \times \frac{5\;\text{glips}}{2\;\text{glops}} = 18.75\;\text{glips}\]4.4 Percent
Percent means “per hundred”: $30\% = \frac{30}{100} = 0.30$.
Key translation: “of” → multiply, “is” / “are” → equals.
$(5\%)\;(\text{all people}) = (\text{left-handed people})$
Example 4-10: Write $33\frac{1}{3}\%$ as a decimal and a fraction.
Solution: $33\frac{1}{3}\% = \frac{100/3}{100} = \frac{1}{3} = 0.\overline{3}$.
Example 4-11: Write $\frac{2}{5}$ as a percent.
Solution: $\frac{2}{5} = \frac{x}{100}$, so $x = 40$. Thus $\frac{2}{5} = 40\%$.
Exercise 4-5: Write each as a decimal and a fraction: i. $35\%$ ii. $175\%$ iii. $66\frac{2}{3}\%$ iv. $16\frac{2}{3}\%$
Exercise 4-6: Write as percents: i. $\frac{1}{8}$ ii. $2\frac{1}{2}$ iii. $0.1$ iv. $3.5$
Percent Increase and Decrease
- Percent decrease: New value $= \text{original} \times (1 - r)$, where $r$ is the rate as a decimal.
- Percent increase: New value $= \text{original} \times (1 + r)$.
- A $25\%$ discount on $22: $\;22(0.75) = $16.50$.
- A $25\%$ increase on $22: $\;22(1.25) = $27.50$.
Interest
Compound interest: A loan of $P$ at annual rate $r$ after $n$ years:
\[\text{Amount owed} = P(1 + r)^n\]Example 4-12: $75$ is $20\%$ of what number?
Solution: $(0.20)x = 75 \implies x = 375$.
Example 4-13: A car originally costs $10,000. Decreased by $25\%$, then increased by $25\%$. Result?
Solution: After decrease: $10000 \times 0.75 = $7500$. After increase: $7500 \times 1.25 = $9375$.
The result is not $10,000!
Example 4-14: Borrow $4000 at $5\%$ annual interest. How much after 4 years?
Solution:
\[4000(1.05)^4 \approx \$4862.03\]Concentration / Mixture Problems
Example 4-15: Mix 2 L of $20\%$ acid with 8 L of $50\%$ acid. What is the resulting concentration?
Solution:
\[\text{Acid} = 2(0.2) + 8(0.5) = 0.4 + 4 = 4.4\;\text{L}\] \[\text{Concentration} = \frac{4.4}{10} = 44\%\]Exercise 4-7: A ring’s price is decreased $40\%$ and the result is increased $50\%$. The final price is $360. What was the original price?
Exercise 4-8: A car was originally $8000, now on sale for $7000. What percent decrease?
Exercise 4-9: The US loans $1.5 million to France. What annual interest rate makes France owe $2 million after one year?
Exercise 4-10: A chemist has 80 mL of $20\%$ acid solution. How many mL must be removed and replaced by pure acid to get a $40\%$ solution? (MAΘ 1992)
Exercises & Problems
End-of-Chapter Problems
Problem 56. What percent of 20 is 13?
Problem 57. A town’s population increases $25\%$ in 1991. By what percent must it decrease the next year to return to its original population?
Problem 58. A number is increased by $50\%$, then decreased by $40\%$. The final number is 8 less than the original. Find the original.
Problem 59. A metric calendar divides the year (365 days) into 10 metric months, each into 10 metric weeks, each into 10 metric days. To the nearest day, how many days are in 4 metric months, 5 metric weeks, and 8 metric days? (MATHCOUNTS 1985)
Problem 60. If the ratio of $2x - y$ to $x + y$ is $2:3$, find $x:y$. (MAΘ 1992)
Problem 61. If 8 quarters stacked are $\frac{1}{2}$ inch high, how many quarters make a stack 1 foot high? (MATHCOUNTS 1991)
Problem 62. If $y^2$ varies inversely as $x^3$, and $y = 3$ when $x = 2$, find $y$ when $x = 9$ (assuming $y > 0$). (MAΘ 1990)
Problem 63. The discount on a stereo is $69, and the rate of discount is $15\%$. What was the original price? (MAΘ 1990)
Problem 64. Given $\frac{x}{y} = \frac{2}{3}$ and $\frac{y}{z} = \frac{3}{4}$, find $\frac{x}{z}$. (MATHCOUNTS 1992)
Problem 65. A test has two parts (60% and 40% of the grade). A student gets 95% on part one. What percent correct on part two gives a 90% overall average? (MATHCOUNTS 1988)
Problem 66. Jennifer gave $\frac{1}{2}$ her Gummy Bears to Jessica, $\frac{1}{3}$ to Jana, and 15 to Julie. The bag was then empty. How many were there originally? (MATHCOUNTS 1991)
Problem 67. A man runs $x$ feet in $y$ seconds. How many yards can he run in $z$ minutes? (MAΘ 1992)
Problem 68. $x$ is directly proportional to $y$ and inversely proportional to $z$. If $x = \frac{1}{2}$ when $y = \frac{3}{4}$ and $z = \frac{2}{3}$, find $x$ when $y = \frac{7}{8}$ and $z = \frac{7}{9}$. (MATHCOUNTS 1989)
Problem 69. A woman has part of $4500 at $4\%$ and the rest at $6\%$. If annual return on each investment is the same, what is the average rate on the $4500? (AHSME 1953)
Problem 70. Wages of 3 men for 4 weeks is $108. At the same rate, how many weeks will 5 men work for $135? (MAΘ 1991)
Problem 71. Tennis coaches Lob and Smash share earnings with assistants Love and Vantage. Lob : Love = 17 : 12, Love : Smash = 3 : 4, Smash : Vantage = 32 : 15. Total earnings $3150. How much did Love receive? (MAΘ 1990)
Problem 72. In a closed bottle, pressure $\times$ volume = constant. By what percent must the volume decrease to increase pressure by $25\%$? (MATHCOUNTS 1992)
Problem 73. In an election, A got $33.\overline{3}\%$, B got $\frac{9}{20}$, C got $\frac{2}{15}$, and D got the remaining 75 votes. How many voted total? (MATHCOUNTS 1990)
Problem 74. Ms. A owns a house worth $10,000. She sells to Mr. B at $10\%$ profit. Mr. B sells it back at $10\%$ loss. How much does Ms. A make? (AHSME 1955)
Problem 75. Railroad rails are 30 ft long. A train’s speed in mph approximately equals the number of clicks heard in how many seconds? (AHSME 1953)
Problem 76. A town’s population increased by 1200, then decreased by $11\%$, leaving 32 fewer people than before the increase. What was the original population? (AHSME 1974)
Problem 77. The cost of living increases $2\%$ each quarter. To the nearest tenth of a percent, what annual percentage increase does this correspond to? (MAΘ 1990)
Problem 78. $x$ varies directly as $y$ and inversely as $z^2$. If $x = 10$ when $y = 4$ and $z = 14$, find $x$ when $y = 16$ and $z = 7$. (AHSME 1959)
Problem 79. Two joggers run around an oval in opposite directions. One completes a lap in 56 seconds. They meet every 24 seconds. How long does the second jogger take per lap? (MATHCOUNTS 1986)
Problem 80. If $x$ varies as the cube of $y$, and $y$ varies as the fifth root of $z$, then $x$ varies as the $n$th power of $z$. Find $n$. (AHSME 1954)
Problem 81. It is between 10:00 and 11:00. Six minutes from now, the minute hand will be exactly opposite the hour hand’s position 3 minutes ago. What is the exact time? (MAΘ 1991)
Problem 82. If $p$ is $50\%$ of $q$ and $r$ is $40\%$ of $q$, what percent of $r$ is $p$? (Mandelbrot #1)
Problem 83. Country A has $c\%$ of the world’s population and $d\%$ of the wealth. Country B has $e\%$ of the population and $f\%$ of the wealth. Find the ratio of wealth of a citizen of A to a citizen of B. (AHSME 1993)
Problem 84. $A, B, C, D, E$ are consecutive points on a line. $\frac{AB}{BC} = \frac{1}{3}$, $\frac{BC}{CD} = \frac{1}{4}$, $\frac{CD}{DE} = \frac{1}{2}$. Find $\frac{AC}{BE}$. (MATHCOUNTS 1992)
Problem 85. From $t = 0$ to $t = 1$ population increased by $i\%$; from $t = 1$ to $t = 2$ it increased by $j\%$. By what percent did it increase from $t = 0$ to $t = 2$, in terms of $i$ and $j$? (AHSME 1991)
Problem 86. If for three distinct positive numbers $x, y, z$:
\[\frac{y}{x-z} = \frac{x+y}{z} = \frac{x}{y},\]find the numerical value of $\frac{x}{y}$. (AHSME 1992)