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Grade 9 Maths Book An Easy Way to Learn Maths

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Section 1.4 Compound proportions and Rates of work

Any \(4\) numbers \(a, b, c\) and \(d\) are said to be proportional if their ratios are equal, that is:
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\begin{equation*} \frac{a}{b} = \frac{c}{d} \end{equation*}
\(\quad\quad\)or
\begin{equation*} \frac{a}{c} = \frac{b}{d} \end{equation*}
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For example, consider the numbers \(4, 10, 20\) and \(50\text{.}\) Since \(4:10 = 2:5\) and \(20:50 = 2:5\text{,}\) the two ratios are equal \((4:10 = 20:50)\text{.}\) Therefore, these four numbers are proportional.
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Proportions appear everywhere in our daily life. For instance, when preparing a meal for many people you would need to increase the amount of ingredients. To keep the same taste, you must maintain the proportions between them. Similarly, proportions help us compare relationships in areas such as speed, scale drawings, and work problems.
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\(\textbf{Rate of work}\) is the amount of work done per unit of time, often expressed as work divided by time
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Subsection 1.4.1 Proportional Parts

Activity 1.4.1.

Work in Groups
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Instructions
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Masomo School has a total of \(500\) Mathematics textbooks, \(100\) storybooks, and \(300\) Science textbooks.
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  1. Find the ratio of Mathematics textbooks to storybooks to Science textbooks.
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  2. Express each type of book as a fraction of the total number of books in the school.
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  3. Explain the steps you used to obtain these fractions.
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  4. The textbooks are to be shared among \(90\) students in the ratio \(5:1:3\) for Mathematics, storybooks, and Science respectively. Work out how many books of each type each student would get.
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  5. Share your work with your fellow learners.
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Definition 1.4.1.

\(\textbf{Proportion}\) is an equation in which two ratios are set equal to each other. For example, if \(a\) and \(b\) are in the same ratio as \(c\) and \(d\text{,}\) we can write this as:
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\begin{equation*} a:b = c:d\text{.} \end{equation*}
This means that the fraction \(\frac{a}{b}\) is equal to the fraction \(\frac{c}{d}\text{.}\)
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Key Takeaway
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If an amount \(m\) is to be divided in the ratio \(a:b:c\text{,}\) then the proportional parts will be divided as follows:
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  1. \(a\) will get \(\frac{a}{a+b+c} \times m\)
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  2. \(b\) will get \(\frac{b}{a+b+c} \times m\)
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  3. \(c\) will get \(\frac{c}{a+b+c} \times m\)
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Example 1.4.2.

A basket of fruits contains mangoes, oranges, and bananas in the ratio of \(3:2:5\text{.}\) If the basket has a total of \(100\) fruits, find the number of each type of fruit.
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Solution.
To find the number of each type of fruit, we first add the parts in the ratio: \(3 + 2 + 5 = 10\text{.}\)
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\begin{align*} \text{Number of mangoes} =\amp \frac{3}{10} \times \text{number of fruits}. \\ =\amp \frac{3}{10} \times 100 \\ =\amp 30 \\ \text{Number of oranges} =\amp \frac{2}{10} \times \text{number of fruits}. \\ =\amp \frac{2}{10} \times 100 \\ =\amp 20 \\ \text{Number of bananas} =\amp \frac{5}{10} \times \text{number of fruits}. \\ =\amp \frac{5}{10} \times 100 \\ =\amp 50 \end{align*}
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Example 1.4.3.

Find the value of \(z\) in the proportion \(12 : z = 4 : 5\text{.}\)
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Solution.
Step 1: Recall that if \(a:b = c:d\text{,}\) it can be written as a fraction: \(\frac{a}{b} = \frac{c}{d}\text{.}\)
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So,
\begin{equation*} \frac{12}{z} = \frac{4}{5}\text{.} \end{equation*}
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Step 2: Cross-multiply to eliminate the fractions: \(12 \times 5 = 4 \times z\text{.}\)
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Step 3: Simplify the equation: \(60 = 4z\text{.}\)
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Step 4: Divide both sides by \(4\text{:}\) \(z = \frac{60}{4} = 15\text{.}\)
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Therefore, the value of \(z\) is \(15\text{.}\)
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Subsection 1.4.2 Relating different ratios in real life

Activity 1.4.2.

Work in Groups
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Instructions
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Mr Kamboi, a Grade \(9\) teacher, organized for a short distance race. The race had \(5\) stations that is station \(A\text{,}\) \(B\text{,}\) \(C\text{,}\) \(D\) and \(E\text{.}\)
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The distance between the stations is as follows:
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Station Position in (m)
A 0
B 20
C 40
D 80
E 160
  1. Use the information in the table to make a sketch showing the stations and the distances between them.
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  2. If the distance between the stations is in continued proportion, find the distance between station \(A\) and \(B\text{,}\) \(B\) and \(C\text{,}\) \(C\) and \(D\text{,}\) and \(D\) and \(E\text{.}\)
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  3. Identify any proportional relationships between the distance \(AB\) and \(CD\) and the distance \(BC\) and \(DE\text{.}\)
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  4. Work out the ratio of \(AB:BC\) and \(CD:DE\) in simplest form.
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  5. Share your work with your fellow learners.
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Definition 1.4.4.

If \(4\) numbers, \(a, b, c\) and \(d\) are in continued proportion, then:
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\begin{equation*} \frac{a}{b} = \frac{b}{c} = \frac{c}{d} \end{equation*}
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\(\quad\quad\)Since
\begin{equation*} \frac{a}{b} = \frac{b}{c}, \end{equation*}
\(\quad\quad\)then
\begin{equation*} b^2 = ac \end{equation*}
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\(\quad\quad\)Therefore,
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\begin{equation*} b = \pm \sqrt{ac} \end{equation*}
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Key Takeaway
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\(\bullet\) When numbers are in continued proportion, they form a geometric progression because each term is obtained by multiplying the previous term by the same common ratio.
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\(\bullet\) For example, consider the numbers \(3, 6, 12, 24\text{.}\) Here, each number is equal to the previous number multiplied by \(2\text{.}\) Since the common ratio is \(2\text{,}\) the four numbers are said to be in continued proportion.
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\(\bullet\) In real life, ratios help us compare quantities. For example, if James takes \(5\) minutes to solve a math problem and Esther takes \(4\) minutes for the same task, we can compare their times using the ratio \(5:4\text{.}\) Ratios like this allow us to see how one quantity relates to another in everyday situations.
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Example 1.4.5.

If the ratio of the ages of John and Mary is \(3:4\text{,}\) and the ratio of their ages in \(5\) years will be \(4:5\text{,}\) what are their current ages?
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Solution.
Let John’s current age be \(3x\) and Mary’s current age be \(4x\text{.}\)
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In \(5\) years, their ages will be \(3x + 5\) and \(4x + 5\) respectively.
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Since the ratio of their ages at that time is given as \(4:5\text{,}\) we can write:
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\begin{align*} \frac{3x + 5}{4x + 5} =\amp \frac{4}{5} \end{align*}
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Cross-multiplying:
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\begin{align*} 5(3x + 5) =\amp 4(4x + 5) \end{align*}
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Expanding both sides:
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\begin{align*} 15x + 25 =\amp 16x + 20 \end{align*}
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Simplifying gives:
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\begin{align*} x =\amp 5 \end{align*}
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Substituting back:
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John’s current age =
\begin{align*} 3 \times 5 =\amp 15 \end{align*}
years.
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Mary’s current age =
\begin{align*} 4 \times 5 =\amp 20 \end{align*}
years.
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Therefore, John is \(15\) years old and Mary is \(20\) years old.
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Example 1.4.6.

Given that the ratio \(x:y=2:3\text{,}\) find the ratio \((5x - 2y):(x+y)\)
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Solution.
Since \(x:y = 2:3\text{,}\) we can let:
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\begin{align*} x =\amp 2k\\ y =\amp 3k \end{align*}
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Substituting into \((5x - 2y):(x+y)\text{:}\)
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\begin{align*} 5x - 2y =\amp 5(2k) - 2(3k)\\ =\amp 10k - 6k \\ =\amp 4k\\ x + y =\amp 2k + 3k\\ =\amp 5k \end{align*}
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Therefore:
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\begin{align*} (5x - 2y):(x+y) =\amp 4k:5k \end{align*}
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Simplify the ratio by cancelling \(k\) (where \(k ≠ 0\)):
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\begin{gather*} 4:5 \end{gather*}
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Therefore, the ratio \((5x-2y):(x+y)\) simplifies to \(4:5\text{.}\)
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Example 1.4.7.

If \(22, X\) and \(88\) are in continued proportion, find the value of \(X\text{.}\)
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Solution.
If \(22, X\) and \(88\) are in continued proportion, then we can express this relationship as:
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\begin{align*} \frac{22}{X} =\amp \frac{X}{88} \end{align*}
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Cross-multiplying gives us:
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\begin{align*} 22 \times 88 =\amp X^2 \end{align*}
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Simplifying this, we have:
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\begin{align*} 1936 =\amp X^2 \end{align*}
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Hence, \(X = \sqrt{1936}\)
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\begin{equation*} = 44 \end{equation*}
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Subsection 1.4.3 Working out compound proportions using the ratio method

Activity 1.4.3.

Work in Groups
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What you need
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Instructions
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1. Measure the length and width of the following trianngles, A and B, using a ruler.
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2. Record the measurements of the lengths and widths of triangle A and triangle B.
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3. Calculate the ratio of the lengths and widths of triangle A to triangle B.
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Example: length ratio = \(\frac{\text{Length of B}}{\text{Length of A}}\text{.}\)
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4. Compare the two ratios. What do you notice about the relationship between the two triangles?
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5. Share your work with your fellow learners.
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Key Takeaway
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\(\bullet\) Two quantities in compound proportion increases or decreases in the same ratio.
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For example, if you have the ratios \(3:4\) and \(6:8\text{,}\) you can set up the proportion as follows:
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\(\frac{3}{4} = \frac{6}{8}\)
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Cross-multiplying gives you \(3 \times 8 = 4 \times 6\text{,}\) which simplifies to \(24\text{.}\) This confirms that the two ratios are equivalent.
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\(\bullet\) How to work out compound proportions using the ratio method:
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  1. Identify the two ratios you want to compare or relate.
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  2. Set up a proportion by equating the two ratios.
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  3. Cross-multiply to eliminate the fractions.
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  4. Solve for the unknown variable, if applicable.
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  5. Check your solution by substituting back into the original ratios to ensure they are equal.
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Example 1.4.8.

A class has \(8\) girls and \(6\) boys. More students are admitted to the class so that the ratio of girls to boys remains the same. If the new number of boys in the class is \(12\text{,}\) determine the new number of girls using the ratio method.
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Solution.
The initial ratio of girls to boys is:
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\begin{equation*} 8:6 \end{equation*}
which simplifies to
\begin{equation*} 4:3 \end{equation*}
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Let the number of girls after admission be \(y\text{.}\)
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Since the ratio must remain the same:
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\begin{align*} \frac{y}{12} =\amp \frac{4}{3} \end{align*}
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Cross-multiplying gives:
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\begin{align*} 3y =\amp 4 \times 12\\ 3y =\amp 48 \end{align*}
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Dividing both sides by 3:
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\begin{align*} y =\amp \frac{48}{3} = 16 \end{align*}
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Therefore, the new number of girls in the class is \(16\text{.}\)
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Example 1.4.9.

Kipngetich measured the heights of a mango tree and a person in a photo. Their heights in the photo were \(10\) cm and \(5\) cm respectively. Since the ratio of their actual heights is the same as in the photo, determine the actual height of the mango tree if the person’s actual height is \(2.5\) meters.
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Solution.
The photo ratio is \(10:5 = 2:1\text{.}\)
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Let the actual height of the tree be \(h\text{.}\) Then:
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\begin{align*} \frac{h}{2.5} =\amp \frac{2}{1}\\ h = 2 \times 2.5 =\amp 5 \end{align*}
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Therefore, the actual height of the mango tree is \(5\) meters.
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Subsection 1.4.4 Rates of work

Activity 1.4.4.

Work in Groups
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What you need
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- Sticks or stones (\(60\) in number)
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- Markers or chalk to draw a start point (where the participants will start picking the sticks or stones) and a drop zone (where they will place the picked sticks or stones).
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- A watch to measure the time taken.
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- Book and pen to record the results.
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Instructions
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  1. Outside your classroom, form \(2\) groups, A and B.
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  2. Mark a start point and a drop zone for each group such that the distance between the start point and the drop zone is \(10\) metres.
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  3. One volunteer from each group will pick up the sticks or stones from the start point and place them in the drop zone. Each start point will have \(30\) sticks or stones.
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  4. The volunteer will start picking the sticks or stones at the same time as the group members starts the watch.
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  5. The volunteers will pick one stick or stone at a time and place it in the drop zone. They will continue this until all the sticks or stones are picked.
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  6. The group members will stop the watch when the volunteer has placed all the sticks or stones in the drop zone. Group members of group A and B will record the time taken by volunteer from their group to pick all the sticks or stones.
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  7. Repeat the activity with \(5\) volunteers from each group.The volunteers will take turns picking the sticks or stones.
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  8. The volunteer from each group will drop the stick or stone in the drop zone and return to tag the next volunteer in the line till the last stick is picked while the group members record the total time taken by \(5\) volunteers to complete the task.
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  9. Go back to your groups and find the average time taken by \(1\) person to complete the task and the rate of work.
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    Find also the average time taken by \(5\) volunteers to complete the task and the rate of work.
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  10. Share your findings with the other learners.
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Definition 1.4.10.

The rate of work is a measure of how quickly work is done or how much work is completed in a given amount of time. It is often expressed as the amount of work done per unit of time, such as hours, days, or weeks.
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Key Takeaway
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\(\bullet\) To solve work-rate problems, first determine how much work is completed in a single unit of time (such as one hour or one day).
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\(\bullet\) The rate of work is given by:
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\(\text{Rate of work} = \frac{\text{Work done}}{\text{Time taken}}\)
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Example 1.4.11.

A group of workers can complete a task in \(8\) hours. If they work at the same rate, how long will it take them to complete the same task if they double their effort?
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Solution.
Let the whole task be \(1\) unit of work.
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Original rate:
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\begin{equation*} \text{rate}=\frac{1}{8}\ \text{task per hour} \end{equation*}
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Doubling effort doubles the rate:
\begin{equation*} \end{equation*}
\begin{equation*} \text{new rate}=2\!\times\!\frac{1}{8}=\frac{1}{4}\ \text{task per hour} \end{equation*}
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Time = work ÷ rate:
\begin{equation*} \end{equation*}
\begin{equation*} \text{time}=\frac{1}{\frac{1}{4}}=4\ \text{hours} \end{equation*}
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Therefore, it will take \(4\) hours.
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Example 1.4.12.

A factory produces \(100\) units of a product in \(5\) hours. If the factory increases its production rate by \(20%\text{,}\) how many units will it produce in \(5\) hours?
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Solution.
First, find the current rate.
\begin{equation*} \end{equation*}
\begin{equation*} \text{Rate}=\frac{100}{5}= 20 \ \text{units/hour} \end{equation*}
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Next, increase rate by \(20\%\text{.}\)
\begin{equation*} \end{equation*}
\begin{equation*} \text{New rate}=20+(0.20\times 20)= 24 \ \text{units/hour} \end{equation*}
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Finally, find production in \(5\) hours.
\begin{equation*} \end{equation*}
\begin{equation*} \text{Units}=24\times 5= 120 \end{equation*}
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Therefore, the factory will produce \(120\) units in \(5\) hours.
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Example 1.4.13.

Mary, John and Essy did a piece of work in \(45\) hours, \(40\) hours and \(30\) hours respectively. How long can John take to complete the work when he starts after Mary and Essy have worked for \(13\) hours each?
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Solution.
The work done by each person in one hour is given as follows:
\begin{equation*} \text{Mary's rate} = \tfrac{1}{45}, \quad \text{John's rate} = \tfrac{1}{40}, \quad \text{Essy's rate} = \tfrac{1}{30} \end{equation*}
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The combined rate of Mary and Essy is:
\begin{equation*} \frac{1}{45} + \frac{1}{30} = \frac{5}{90} = \frac{1}{18} \end{equation*}
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In \(13\) hours, Mary and Essy together will complete:
\begin{equation*} 13 \times \frac{1}{18} = \frac{13}{18} \end{equation*}
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The remaining portion of work is therefore:
\begin{equation*} 1 - \frac{13}{18} = \frac{5}{18} \end{equation*}
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To finish this, John working alone will take:
\begin{align*} \text{Time taken by John} =\amp \frac{\text{Remaining work}}{\text{John's rate}}\\ \frac{\frac{5}{18}}{\frac{1}{40}} =\amp \frac{5}{18} \times 40 \\ =\amp \frac{200}{18} \\ =\amp \frac{100}{9} \end{align*}
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Hence, John will require \(11\frac{1}{9}\) hours, to complete the remaining work.
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\(\textbf{Digital corner}\)
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