# Work with Direct Proportion Algebraically

In this worksheet, students will find the values of unknown variables which are directly proportional to each other, using appropriate notation to express proportional relationships algebraically.

Key stage:  KS 4

GCSE Subjects:   Maths

GCSE Boards:   AQA, Eduqas, Pearson Edexcel, OCR

Curriculum topic:   Ratio, Proportion and Rates of Change

Curriculum subtopic:   Ratio, Proportion and Rates of Change, Direct and Inverse Proportion

Difficulty level:

### QUESTION 1 of 10

Questions involving proportion can be confusing due to the symbols and letters used.

Quite often the wording can confuse too.

Here are some key explanations that may help.

The word variable is used to define an element which changes.

Direct proportion exists between two variables when one is a simple multiple of the other.

This means that their ratio is a constant.

e.g. 1 kilogram = 2.2 pounds

There is a multiplying factor of 2.2 between the number of kilograms compared to pounds.

So if one doubles, the other doubles.

If one is multiplied by 10, the other is multiplied by 10.

If one is halved, the other is halved.

The two variables are linked by a constant (k) which, as the word implies, does not change.

(We use the letter 'k' in proportion questions to represent an unknown constant - that makes a change from x and y, doesn't it?)

We need to find the value of k each time, then substitute this into our formula to solve any problem involving these variables.

The symbol for direct proportion is:

So to write 'pay is directly proportional to time' in this way we would write:

Pay ∝ Time

This is where the letter 'k' can help us too.

If we want to express 'Pay' in relation to 'Time', we could say that:

Pay = k × Time

Let's look at these notations in action now using an example.

e.g. T is directly proportional to M. If T = 20 when M = 4, find:

a) T when M = 3

T ∝ M and T = k × M

Now let's substitute in what we know:

20 = k × 4

So to find k, we need to divide 20 by 4:

20 ÷ 4 = 5

Therefore, our multiplier (k) is 5.

Now we can substitute this multiplier into our expression, to find T when M = 3:

T = k × 3

T = 5 × 3

So T = 15 when M = 3.

b) M when T = 10

We can use the same expressions from above and the same multiplier for k:

T = k × M

10 = 5 × M

M = 10 ÷ 5 = 2

Okay - let's dive in!

In this activity, you will find the values of unknown variables which are directly proportional to each other, using appropriate notation to express proportional relationships algebraically.

The variable x is directly proportional to y.

If x = 45 when y = 3, find the missing values in the statements below.

The variable q is directly proportional to p.

If p = 2 when q = 100, find the missing values in the statements below.

The variable y is directly proportional to x.

If y = 20 when x = 8, what is the value of x when y = 35?

12

14

16

18

The variable y is directly proportional to x.

If y = 15 when x = 3, find the value of:

a) y when x = 10;

b) x when y = 65.

12

14

16

18

The variable y is directly proportional to x.

If y = 52 when x = 8, find the value of:

a) when x = 14;

b) x when y = 143.

12

14

16

18

y ∝ x

Consider the values in the table below:

 x 3 4 B y A 18 63

Use what you know to find the values of the missing numbers, represented as A and B in the table.

12

14

16

18

The distance d (in km) covered by an airplane is directly proportional to the time taken t (in hours).

The airplane covers a distance of 1800 km in 3.6 hours.

Find the value of d when t = 8.

y is directly proportional to the square of x.

If y = 36 when x = 3, find:

a) y when x = 5;

b) x when y = 25.

y is directly proportional to the cube of x.

If y = 28.8 when x = 2, find:

a) y when x = 4;

b) x when y = 450.

The cost of fuel per hour c (in £) is directly proportional to the cube of its speed s (in mph).

A boat travelling at 10 mph uses £50 of fuel per hour.

What is the cost of fuel per hour when the boat is travelling at 5 mph?

• Question 1

The variable x is directly proportional to y.

If x = 45 when y = 3, find the missing values in the statements below.

EDDIE SAYS
Don't get stressed with this type of question. We just need to remember how to write the relationships out, then we can substitute the information we have been given into these formulae. x ∝ y and x = k × y When x = 45, y = 3, so we know that: 45 = k × 3 45 ÷ 3 = 15 So our multiplier (k) is 15. Now let's apply this information in the two scenarios we have been given. x when y = 5: x = k × 5 x = 15 × 5 = 75 y when x = 90: x = k × y 90 = 15 × y 15 ÷ 90 = 6 y = 6 How was that first challenge? Hopefully your brain isn't hurting too much already!
• Question 2

The variable q is directly proportional to p.

If p = 2 when q = 100, find the missing values in the statements below.

EDDIE SAYS
So let's start with creating our formulae: q ∝ p and q = k × p Then let's find our multiplier (k): 100 = k × 2 100 ÷ 2 = 50 Now we can apply this multiplier to the new scenarios. q when p = 3: q = k × 3 q = 50 × 3 = 150 p when q = 300: q = k × p 300 = 50 × p 300 ÷ 50 = 6 p = 6 Which emoji are you feeling now then?
• Question 3

The variable y is directly proportional to x.

If y = 20 when x = 8, what is the value of x when y = 35?

14
EDDIE SAYS
Are you ready to throw your hands in the air yet?! It is just a matter of practice. So let's start with creating our formulae: y ∝ x and y = k × x Then let's find our multiplier (k): 20 = k × 8 20 ÷ 8 = 2.5 Now we can apply this multiplier to the new scenario: y = 2.5 × x 35 = 2.5 × x x = 35 ÷ 2.5 x = 14
• Question 4

The variable y is directly proportional to x.

If y = 15 when x = 3, find the value of:

a) y when x = 10;

b) x when y = 65.

EDDIE SAYS
Let's apply our same three steps. 1. Create formulae: y ∝ x and y = k × x 2. Find multiplier (k): 15 = k × 3 15 ÷ 3 = 5 3. Apply multiplier: a) y = 5 × 10 y = 50 b) 65 = 5 × x x = 65 ÷ 5 x = 13
• Question 5

The variable y is directly proportional to x.

If y = 52 when x = 8, find the value of:

a) when x = 14;

b) x when y = 143.

EDDIE SAYS
Keep solving... 1. Create formulae: y ∝ x and y = k × x 2. Find multiplier (k): 52 = k × 8 52 ÷ 8 = 6.5 (Don't worry that this is a decimal - exactly the same rules still apply!) 3. Apply multiplier: a) y = 6.5 × 14 y = 91 b) 143 = 6.5 × x x = 143 ÷ 6.5 x = 22
• Question 6

y ∝ x

Consider the values in the table below:

 x 3 4 B y A 18 63

Use what you know to find the values of the missing numbers, represented as A and B in the table.

EDDIE SAYS
It is best to write out what we want to know, as visualising this in the same way we have been practising will make things easier. We want to find out y, when x = 3 (A). Also, we want to find x when y = 63 (B). Now we can just solve this in the same way that we have been practising. y ∝ x and y = k × x We have both variables in the middle column, so we need to use this info to find our multiplier: 18 = k × 4 18 ÷ 4 = 4.5 A) y = 4.5 × 3 y = 13.5 B) 63 = 4.5 × x x = 63 ÷ 4.5 x = 14
• Question 7

The distance d (in km) covered by an airplane is directly proportional to the time taken t (in hours).

The airplane covers a distance of 1800 km in 3.6 hours.

Find the value of d when t = 8.

4000
4000 km
EDDIE SAYS
This time, our variables are hidden within the words of a problem, so we need to break it down. Now we can just solve this in the same way that we have been practising. d ∝ t and d = k × t 1800 = k × 3.6 1800 ÷ 3.6 = 500 (our multiplier) d = 500 × 8 So when t = 8, d = 4000 You're flying now...
• Question 8

y is directly proportional to the square of x.

If y = 36 when x = 3, find:

a) y when x = 5;

b) x when y = 25.

EDDIE SAYS
Oops, this is a bit of a curve-ball! This time, we do not have a simple relationship between our variables, it is slightly more complex. The question tells us that 'y is directly proportional to the square of x', we write this as: y ∝ x2 and y = k × x2 Now we just proceed as normal by finding our multiplier: 36 = k × (3)2 36 ÷ 9 = 4 Now we can apply this multiplier as usual: a) y = 4 × (5)2 y = 4 × 25 y = 100 b) 25 = 4 × x2 x2 = 25 ÷ 4 x = √ 6.25 x = 2.5
• Question 9

y is directly proportional to the cube of x.

If y = 28.8 when x = 2, find:

a) y when x = 4;

b) x when y = 450.

EDDIE SAYS
Similarly to the last question, we have to tackle a slightly more complex relationship here. The question tells us that 'y is directly proportional to the cube of x', we write this as: y ∝ x3 and y = k × x3 Now we just proceed as normal by finding our multiplier: 28.8 = k × (2)3 28.8 ÷ 8 = 3.6 Now we can apply this multiplier as usual: a) y = 3.6 × (4)3 y = 3.6 × 64 y = 230.4 b) 450 = 3.6 × x3 x3 = 450 ÷ 3.6 x = ∛ 125 x = 5
• Question 10

The cost of fuel per hour c (in £) is directly proportional to the cube of its speed s (in mph).

A boat travelling at 10 mph uses £50 of fuel per hour.

What is the cost of fuel per hour when the boat is travelling at 5 mph?

EDDIE SAYS
Again, our variables are hidden within the words of a problem, so we need to break it down. c ∝ s3 and c = k × s3 50 = k × (10)3 50 = k × 1000 50 ÷ 1000 = 0.05 (our multiplier) c = k × s3 c = 0.05 × (5)3 c = 0.05 × 125 c = 6.25 It's expensive to own a boat, isn't it?! Great job! You can now find the values of unknown variables which are directly proportional to each other, using appropriate notation to express proportional relationships algebraically.
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