Ratio, Rate And Speed

In this chapter, we will be discussing the below mentioned topics in detail:

• Finding and simplifying ratios of quantities with units
• Finding unknowns in a ratio
• Problems involving ratios of two quantities
• Problems involving ratios of three quantities

A) Finding Ratio

A ratio compares two quantities of the same kind that either have no units or are measured in the same units.

The ratio \(a : b\), where \(a\) and \(b\) are positive integers, has no units.

Let’s understand this with the help of some examples:

Question 1:

A map has a scale where \(1 \;cm\) represents \(1 \;km\).  Express the map scale as a ratio.

Solution: 

\(\begin{align*} 1\;cm &: 1\;km \\ 1\;cm &: 1000\;m & &\text{(Converting kilometres into metres)}\\ 1\;cm &: 100\;000\;cm & &\text{(Converting metres into centimetres)}\\ 1 &: 100\;000 \\ \end{align*}\)

Hence, the ratio would be \(1 : 100 000\).

B) Equivalent Ratio

Equivalent ratios are ratios that remain the same when compared.

Let’s understand this with the help of some examples:

Question 2:

Without using a calculator, simplify each of the following ratios.

1. \(\begin{align*} \frac{3}{8} : 3\frac{3}{4} \end{align*}\)
2. \(\begin{align*} 0.24 : 0.08 \end{align*}\)

Solution: 

\(\begin{align*} \frac38 &: 3\frac34\\ \frac38 &: \frac{15}{4}\\ \frac38 × 8 &: \frac{15}4 × 8\\ 3 &: 30\\ \frac33 &: \frac{30}3\\ 1 &: 10 \end{align*}\)

\(\begin{align*} 0.24 &: 0.08\\ 0.24 × 100 &: 0.08 × 100\\ 24 &: 8\\ \frac{24}8 &: \frac88\\ 3&:1 \end{align*}\)

C) Finding unknowns in a ratio

Let’s understand this with the help of some examples:

Question 3:

1. Given that \(\begin{align*} 3x - 5 : 8 = x : 2 \end{align*}\), find the value of \(x\). 
2.  Given that \(\begin{align*} \frac{5a}{9} = \frac{2b}{15} \end{align*}\), find the ratio of \(\begin{align*} a : b \end{align*}\).

Solutions:

\(\begin{align*} \frac{3x-5}{8} &= \frac{x}{2}\\ 2 \;(3x – 5) &= 8x\\ 6x – 10 &= 8x\\ –10 &= 2x\\ x &= \;–5 \end{align*}\)

\(\begin{align*} \frac{5a}9 &= \frac{2b}{15}\\ 5a × 15 &= 9 × 2b\\ 75a &= 18b\\ \frac{a}b &= \frac{18}{75}\\ &= \frac{6}{25}\\ a : b &= 6 : 25 \end{align*}\)

D) Problems involving ratios of two quantities

Let’s understand this with the help of some examples:

Question 4:

The ratio of the number of children to the number of adults at an event is \(3 : 7\).If there are \(32\) fewer children than adults, calculate the total number of people at the event.

Solution: 

Let the number of children \(= 3x\)

Then, number of adults \(= 7x\)

\(\begin{align*} 7x – 3x &= 32\\ 4x &= 32\\ x &= 8 \end{align*}\)

Total number of people

\(\begin{align*} &= 3x + 7x\\ &= 10x\\ &= 10 × 8\\ &= 80 \end{align*}\)

E) Problems involving ratios of three quantities

Let’s understand this with the help of some examples:

Question 5:

Patrick, Rachel and Quincy shared a sum of money in the ratio \(3 : 4 : 7\). If Patrick and Quincy each receive \($10\) from Rachel, the ratio becomes \(7 : 6 : 15\). Calculate the total amount of money all three of them had at first.

Solution: 

Let Patrick have \($3x\) at first. 

Then Rachel had \($4x\) and Quincy had \($7x\) at first.

\(\begin{align*} && P&:  &  Q&:   &   &R \\ \text{Before} &&         3x&: &   4x &: &&7x\\ \text{After} && 3x + 10 &: & 4x – 20 &: &&7x + 15 \end{align*}\)

\(\begin{align*} \frac{3x+10}{4x-20} &= \frac76\\ 6(3x +10) &= 7(4x \;– 20)\\ 18x + 60 &= 28x \;– 140\\ 200 &= 10x\\ x &= 20 \end{align*}\)

Total money at first

\(\begin{align*} &= 14x\\ &= 14(20)\\ &= $280 \end{align*}\)