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MATHS

FOUNDATION LEVEL

SAMPLE QUESTIONS

PRACTISE FOR THE 2017 GCSE MATHS PAPERS

1

Question 1

Question 2

FREE GCSE SAMPLE QUESTIONS: PRACTISE FOR THE

2017 GCSE MATHS PAPERS

Below you will find sample questions and answers replicating some of the major content

areas that appear on the GCSE foundation tier Maths exam. For more information about the

GCSEs, visit our GCSE Past Papers page.

GCSE MATHS FOUNDATION LEVEL: NON-CALCULATOR

QUESTIONS

The following questions are examples of a questions likely to appear on a GCSE Maths

foundation level exam. Note, that the use of calculators would NOT be permitted for such

questions.

1. Ethan says,

"If you halve a whole number that ends in a 6, the answer always ends in a 3."

a) Give an example to show that Ethan is wrong. (1)

Su says,

"Because 3 and 13 are both prime numbers, all whole numbers ending in 3 are prime."

b) Is Su correct? You must give a reason with your answer.

Solve the simultaneous equations

3 𝑥 + 2 𝑦 = 11

𝑥 − 4 𝑦 = 13

3

Answers & Explanations

Question Answer Notes

1. a)

b)

Example

Example

C1 for appropriate example shown, eg 16.

C1 for conclusion and appropriate example shown, eg 63.

a) If you halve 6, you will get 3, but with larger numbers this is not always the case.

Work backwards and think if there are other numbers that will end in a 6 when

they are doubled. 3 will always end in a 6, but so will 8, as 8 × 2 = 16.

Therefore, 16 is one example of a number that will not end in a 3 when it is halved. There

are other examples here you could choose, for example 36, 56, etc.

b) For a number to be prime, it must have only two factors: itself and 1.

For example, the factors of 13 are 1 and 13. Thus, it is prime.

The factors of 15 are 1, 3, 5, and 15. As it has more than two factors, it is not prime.

Su thinks that all numbers which end in 3 are prime. To prove if she is correct or

not, try out some numbers first.

23 : This has only two factors, 1 and 23. Thus, it is prime.

33 : This has four factors, 1, 3, 11, and 33. So, it is not a prime number.

Therefore, you can conclude that she is wrong, as you have found an example

where a number ends in 3 but is not prime.

There are, in fact, an infinite number of examples to prove she is wrong. For

example: 63, 93, 123, etc.

4

Question Answer Notes

2

𝑥 = 5

𝑦 = − 2

M1 for correct process to eliminate one variable (condone one

arithmetic error)

M1 (dep) for substituting found value in one of the equations or

appropriate method after starting again (condone one arithmetic

error)

A1 for both correct solutions.

Linear simultaneous equations such as these can be solved most easily using the

elimination or substitution methods. Both methods aim to remove one of the variables in

an equation so it can be solved.

Elimination Method

In this method, you must match up the numerical coefficients of either 𝑥 or 𝑦 by

multiplying equations. The negative or positive signs need not match. If you choose to

match up the 𝑥 values, then multiply the second equation by 3 so that both equations

contain 3𝑥.

Equation 1: 𝟑𝒙 + 2 𝑦 = 11

Equation 2: 3(𝑥 − 4 𝑦 = 13 ) → 𝟑𝒙 − 12 𝑦 = 39.

Now the 𝑥 coefficients match. As you have multiplied every value in the equation by 3, you

have not changed the meaning of the equation. The values of 𝑥 and 𝑦 remain the same.

Writing the equations one on top of the other, it is possible to then see that if you subtract

one from the other the x terms will disappear.

Equation 1: 𝟑𝒙 + 2 𝑦 = 11

Equation 2: 𝟑𝒙 − 12 𝑦 = 39.

Subtract: 0 + 14 𝑦 = − 28

Note: It is possible to subtract the equations in either order.

Be careful when subtracting the 𝑦 terms. As you are subtracting an already negative

number, the two negatives become a positive and the calculation becomes: 2 𝑦 + 12 𝑦.

You have now eliminated the 𝑥 terms and are just left with an equation in terms of 𝑦,

which can be solved more easily, to find your first variable.

14 𝑦 = − 28 // divide both sides by 14.

𝒚 = −𝟐.

Finally, substitute this value into either equation 1 or equation 2 to find 𝑥.

6

Question Answer Notes

3a) Red = 65

Black = 14

Gold = 21

P1 for process to start to solve the problem

Eg 400 ÷ 20 , or 4 × 2. 5

P1 for a complete process to find the total number of tiles (=100)

P1 for 65% x 100 (= 65)

P1 for 100 − 65 ÷ 5

A1 for correct answer only

b) Correct

statement

C1 for eg "fewer tiles may be needed"

Questions such as these are multi-step problems and can look daunting. But, it is always

worth attempting to get at least a few marks, even if you cannot see how to solve the

problem to completion.

Break down the process into logical steps. These steps are often how the mark scheme

awards marks.

Step 1:

Calculate how many tiles are needed in total.

Step 2:

Calculate how many red tiles are needed.

Step 3:

Calculate how many tiles are remaining.

Step 4:

Calculate how many black and gold tiles there will be.

Step 1:

To calculate how many tiles are needed, you can divide the area of the floor by the area of

each tile. Take care to convert everything into either centimetres or metres first.

Area of floor: 400 cm × 250 cm = 100 , 000 cm

<

Area of tile: 20 cm × 50 cm = 1 , 000 cm

2

Number of tiles: 100 , 000 ÷ 1 , 000 = 𝟏𝟎𝟎 𝐭𝐢𝐥𝐞𝐬

Alternatively, work out how many tiles would fit along and across the floor. If the length of

the floor is 4m = 400cm, and the length of each tile is 20cm, then divide to calculate how

many tiles you can tile along.

7

400 ÷ 20 = 20 tiles

If the width of the floor is 2.5 m = 250 cm, and the length of each tile is 50cm, then divide

to calculate how many tiles you can tile across.

250 ÷ 50 = 50 tiles

So, if you could fit 20 tiles along the length and 50 tiles across the width, then altogether

you could fit 100 tiles on the floor, as 50 × 20 = 100 tiles.

Step 2:

65% of the tiles are red, so calculate 65% of 100. As percent means out of 100, then 65%

of 100 is simply 65 tiles.

Step 3:

To find the number of remaining tiles, subtract the number of red tiles from the total.

100 − 65 = 35 tiles.

Step 4:

Those 35 tiles are split in the ratio 2: 3. For every two black tiles, there are three gold tiles.

One way to look at it is to see each section as five tiles and work out how many sections

would fit.

As there are 35 tiles, 35 ÷ 5 = 7.

So, this pattern would repeat seven times.

That would give 7 × 2 = 14 black tiles.

7 × 3 = 21 gold tiles.

In general, divide the total by the total parts of the ratio and then multiply by each

individual ratio.

So, altogether, there are:

65 red tiles, 14 black tiles, and 21 gold tiles.

Solving Tip:

Learn the conversion rules between metric units. There are 10 mm in 1 cm. There are 100

cm in 1 m. There are 1,000 m in 1 km.

mm cm m km

÷ (^10) ÷ 100 ÷ 1 , 000

× (^10) × (^100) × 1 , 000

9

Question 6

Chelsea says that when the output is 24, the input is 0.

Here is her working.

24 ÷ 3 = 8

Chelsea is wrong.

b) Explain what she has done wrong.

A flowerbed is in the shape of a trapezium, ABCD, and a semicircle.

AB is the diameter of the semicircle.

Jim is going to cover the flowerbed in seeds.

A packet of seeds costs £2.99.

Jim has been told that one packet of seeds will cover 1.5 m

2 of soil.

a) Work out the cost of covering the whole flowerbed in seeds. (5)

Jim discovers that one packet will cover less than 1.5 m^2 of soil.

b) Explain how this might affect the number of packets of seeds he will need to buy. (1)

3.6 m

1.2 m

4.1 m

A

B

C

D

10

Answers & Explanations

Question Answer Notes

4. 625g butter

5 tsp ginger

500g sugar

750g flour

2.5 eggs

M1 for ÷ 12 × 30 oe or 30 ÷ 12 (= 2. 5 )

A1 for 2 or 3 correct

A1 cao.

This question is about keeping the same proportional relationship throughout. This means

multiplying all the ingredients by the same number, or otherwise the recipe will taste

different.

Currently, the ingredients make 12 gingerbread men. If Nikisha wants to make 30, one

option is to first calculate the ingredients needed for one gingerbread man and then

multiply up to make 30.

Option 1:

To make one gingerbread man: ÷ by 12.

To make 30 gingerbread men: × answer by 30.

For example: 300 g flour ÷ 𝟏𝟐 = 25 g flour.

25 g flour × 𝟑𝟎 = 750 g flour.

Option 2:

The second option is to find out how to get from 12 to 30 in one step. Do this by dividing

30 by 12.

30 ÷ 12 = 2. 5

This means, as 30 is 2.5 times bigger than 12, you need to multiply all the ingredients by

For example: 300 g flour × 𝟐. 𝟓 = 750 g flour.

The result should be the same, whether you use method 1 or 2.

Therefore, the ingredients Nikisha should use are:

625g butter

5 tsp ginger

500g sugar

750g flour

2.5 eggs

12

The correct process would have been:

16 × 3 = 48

If the output was 24, the input should have been 48. Therefore, Chelsea was

wrong.

Question Answer Notes

6 a) £20.

P1 process to find area of circle or semicircle π × 0. 6

< (÷ 2 )

P1 process to find area of trapezium (= 9. 84 m^2 )

P1 process to find number of packets " 10. 41 " ÷ 1. 5

P1 process to find cost " 7 " × 2. 99

A1 cao

b) Correct

statement

C1 eg. He might need to buy more boxes.

a) This question should be solved in logical stages.

(i) Find the area of the trapezium.

(ii) Find the area of the semicircle.

(iii) Find the total area of the flowerbed.

(iv) Work out the number of packets needed.

(v) Work out the total cost.

First, recall the formula for the area of a trapezium:

In this formula, a and b are the parallel sides and h is the perpendicular height.

Therefore, 𝐴 =

N

<

1. 2 + 3. 6 × 4. 1 = 𝟗. 𝟖𝟒 m

2 .

Next, find the area of a semicircle. This will be half the area of a circle:

Here, r is the radius of the circle. If the diameter is 1.2 m, then the radius is half the

diameter, which is 0.6 m. You must also halve the formula above as you only want a

semicircle.

So, the area of the semicircle is: 𝐴 =

N

<

𝜋 × 0. 6

< = 𝟎. 𝟓𝟕 m^2 (2 dp).

𝐴 =

N

<

(a + b) h

𝐴 = 𝜋 𝑟

<

13

Therefore, the total area of the flowerbed is 9. 84 + 0. 57 = 𝟏𝟎. 𝟒𝟏 m^2.

Jim must cover an area of 10.41 m^2 , and each packet of seeds will cover 1.5 m^2 , so to

determine how many packets he will need, you must calculate how many lots of 1.5 m

2 are

in 10.41 m

2

. This can be done by dividing.

10. 41 ÷ 1. 5 = 6. 94 packets

It is not possible to buy part of a packet of seeds. Therefore, Jim must buy seven packets

to ensure he has enough to cover the flowerbed.

The final step is to calculate the total cost of buying seven packets of seeds. Each packet

costs £2.99, so to find the cost of seven packets, you must multiply:

7 × 2. 99 = 20. 93.

Therefore, Jim must pay £20.93.

Solving Tip:

In the new GCSE, you are no longer provided with a formula booklet. Therefore, you must

learn the formulae for areas and volumes of shapes. However, if you forget a formula, for

example the area of a trapezium, you can split the trapezium up into simpler shapes like

triangles and rectangles and find its area that way.

b) If Jim discovers that each packet of seeds covers less than 1.5 m

2 of soil, then he

might have to buy more packets to cover the whole flowerbed. It is not certain

that he will need to buy more as he already has slightly more seeds than he

needed. This was because he was not able to buy part of a pack, so he bought

seven packets when he needed 6.94.

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