Sec 1 Maths · Coordinates and linear graphs

Reading linear graphs questions with answers

Practice questions on reading linear graphs, from the basics up to exam level, each with a full worked solution. Try them on paper first, then open the solution to check your working.

Worked example

The graph shows the volume of water, $y$ litres, left in a tank $x$ minutes after it starts to drain. Use the graph to find the volume of water left after $2$ minutes.

Step 1 of 1

Go up from 2 on the horizontal axis to the line, then across to the vertical axis

Volume:

70

litres

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Questions

Question 1Read a value off the graph
The graph shows the taxi fare, $ $y$ , for a trip of $x$ km. Use the graph to find the fare for a trip of $6$ km.
Show the worked solution
1. Step 1: Go up from 6 on the horizontal axis to the line, then across to the vertical axisFare: $ $10$
Question 2Read a value off the graph
The graph shows the height, $y$ cm, of a candle $x$ hours after it is lit. Use the graph to find the height of the candle after $10$ hours.
Show the worked solution
1. Step 1: Go up from 10 on the horizontal axis to the line, then across to the vertical axisHeight: $10$ cm
Question 3The gradient as a rate
The graph shows the length, $y$ cm, of a spring when a mass of $x$ kg hangs from it. Use the gradient of the line to find the extension of the spring, in cm, for each kg.
Show the worked solution
1. Step 1: Take two points on the line where it crosses grid lines: where it starts and where it endsThe line goes from $(0, 12)$ to $(8, 36)$ .Rise: $36 - 12 = 24$ ; run: $8 - 0 = 8$
2. Step 2: Gradient = rise ÷ run: this is the rate $\text{Gradient} = \dfrac{24}{8}$ $\phantom{\text{Gradient}} = 3$ Rate: $3$ cm per kg
Question 4The gradient as a rate
The graph shows the length, $y$ cm, of a spring when a mass of $x$ kg hangs from it. Use the gradient of the line to find the extension of the spring, in cm, for each kg.
Show the worked solution
1. Step 1: Take two points on the line where it crosses grid lines: where it starts and where it endsThe line goes from $(0, 15)$ to $(8, 39)$ .Rise: $39 - 15 = 24$ ; run: $8 - 0 = 8$
2. Step 2: Gradient = rise ÷ run: this is the rate $\text{Gradient} = \dfrac{24}{8}$ $\phantom{\text{Gradient}} = 3$ Rate: $3$ cm per kg
Question 5Starting value and rate
The graph shows the length, $y$ cm, of a spring when a mass of $x$ kg hangs from it. Assuming that the line continues in the same way, find the length of the spring when a mass of $15$ kg hangs from it.
Show the worked solution
1. Step 1: The starting value is where the line meets the vertical axis $\text{Start} = 12$
2. Step 2: The rate is the gradient: rise ÷ runThe line goes from $(0, 12)$ to $(8, 36)$ .Rise: $36 - 12 = 24$ ; run: $8 - 0 = 8$ $\text{Rate} = 3$
3. Step 3: Length = starting value + rate × mass $12 + 3 \times 15 = 57$ Length: $57$ cm
Question 6Starting value and rate
The graph shows the taxi fare, $ $y$ , for a trip of $x$ km. Assuming that the line continues in the same way, find the fare for a trip of $15$ km.
Show the worked solution
1. Step 1: The starting value is where the line meets the vertical axis $\text{Start} = 4$
2. Step 2: The rate is the gradient: rise ÷ runThe line goes from $(0, 4)$ to $(10, 19)$ .Rise: $19 - 4 = 15$ ; run: $10 - 0 = 10$ $\text{Rate} = 1.5$
3. Step 3: Fare = starting value + rate × distance $4 + 1.5 \times 15 = 26.5$ Fare: $ $26.50$

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Worked example

The graph shows the volume of water,

y

litres, left in a tank

x

minutes after it starts to drain. Use the graph to find the volume of water left after

2

minutes.

Step 1 of 1

Go up from 2 on the horizontal axis to the line, then across to the vertical axis

Volume:

70

litres

Questions

Question 1Read a value off the graph

The graph shows the taxi fare, $ $y$ , for a trip of $x$ km. Use the graph to find the fare for a trip of $6$ km.

Show the worked solution

Step 1: Go up from 6 on the horizontal axis to the line, then across to the vertical axisFare: $ $10$

Question 2Read a value off the graph

The graph shows the height, $y$ cm, of a candle $x$ hours after it is lit. Use the graph to find the height of the candle after $10$ hours.

Show the worked solution

Step 1: Go up from 10 on the horizontal axis to the line, then across to the vertical axisHeight: $10$ cm

Question 3The gradient as a rate

The graph shows the length, $y$ cm, of a spring when a mass of $x$ kg hangs from it. Use the gradient of the line to find the extension of the spring, in cm, for each kg.

Show the worked solution

Step 1: Take two points on the line where it crosses grid lines: where it starts and where it endsThe line goes from $(0, 12)$ to $(8, 36)$ .Rise: $36 - 12 = 24$ ; run: $8 - 0 = 8$
Step 2: Gradient = rise ÷ run: this is the rate $\text{Gradient} = \dfrac{24}{8}$ $\phantom{\text{Gradient}} = 3$ Rate: $3$ cm per kg

Question 4The gradient as a rate

The graph shows the length, $y$ cm, of a spring when a mass of $x$ kg hangs from it. Use the gradient of the line to find the extension of the spring, in cm, for each kg.

Show the worked solution

Step 1: Take two points on the line where it crosses grid lines: where it starts and where it endsThe line goes from $(0, 15)$ to $(8, 39)$ .Rise: $39 - 15 = 24$ ; run: $8 - 0 = 8$
Step 2: Gradient = rise ÷ run: this is the rate $\text{Gradient} = \dfrac{24}{8}$ $\phantom{\text{Gradient}} = 3$ Rate: $3$ cm per kg

Question 5Starting value and rate

The graph shows the length, $y$ cm, of a spring when a mass of $x$ kg hangs from it. Assuming that the line continues in the same way, find the length of the spring when a mass of $15$ kg hangs from it.

Show the worked solution

Step 1: The starting value is where the line meets the vertical axis $\text{Start} = 12$
Step 2: The rate is the gradient: rise ÷ runThe line goes from $(0, 12)$ to $(8, 36)$ .Rise: $36 - 12 = 24$ ; run: $8 - 0 = 8$ $\text{Rate} = 3$
Step 3: Length = starting value + rate × mass $12 + 3 \times 15 = 57$ Length: $57$ cm

Question 6Starting value and rate

The graph shows the taxi fare, $ $y$ , for a trip of $x$ km. Assuming that the line continues in the same way, find the fare for a trip of $15$ km.

Show the worked solution

Step 1: The starting value is where the line meets the vertical axis $\text{Start} = 4$
Step 2: The rate is the gradient: rise ÷ runThe line goes from $(0, 4)$ to $(10, 19)$ .Rise: $19 - 4 = 15$ ; run: $10 - 0 = 10$ $\text{Rate} = 1.5$
Step 3: Fare = starting value + rate × distance $4 + 1.5 \times 15 = 26.5$ Fare: $ $26.50$