Edexcel – Foundation – June 2018 – Paper 1

GCSE Maths Foundation Paper 1 (May 2018) – Interactive Practice

GCSE Maths Foundation Paper 1 (May 2018)

Mark Scheme Legend

  • M1 = Method mark (correct method applied)
  • A1 = Accuracy mark (correct answer)
  • B1 = Independent mark (correct statement or value)
  • P1 = Process mark (completing a process)
  • C1 = Communication mark (clear explanation)

Question 1 (1 mark)

Write 6324 correct to the nearest thousand.

Worked Solution

Step 1: Identify the digit to check

To round to the nearest thousand, we look at the digit in the hundreds column.

In \(6\underline{3}24\), the digit in the hundreds column is 3.

Rule: If the digit is 5 or more, round up. If it is less than 5, round down (keep the thousands digit the same).

Since \(3 < 5\), we round down.

6000

(B1)

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Question 2 (2 marks)

(a) Write the following numbers in order of size. Start with the smallest number.

\(-6 \quad 6 \quad -5 \quad 0 \quad 12\)


(b) Write the following numbers in order of size. Start with the smallest number.

\(0.078 \quad 0.78 \quad 0.87 \quad 0.708\)

Worked Solution

Part (a): Negative Numbers

Why we do this: Negative numbers with larger digits are actually smaller (colder).

Comparing the numbers: \(-6\) is the smallest.

\(-5\) is next.

Then \(0, 6, 12\).

\(-6, -5, 0, 6, 12\)

(B1)

Part (b): Ordering Decimals

Strategy: To compare decimals easily, make them all have the same number of decimal places by adding zeros.

Original: \(0.078, \quad 0.78, \quad 0.87, \quad 0.708\)

Padded (3 d.p.):

  • \(0.078\)
  • \(0.78\mathbf{0}\)
  • \(0.87\mathbf{0}\)
  • \(0.708\)

Ordering them:

  1. \(0.078\) (Smallest)
  2. \(0.708\)
  3. \(0.780\)
  4. \(0.870\)

\(0.078, 0.708, 0.78, 0.87\)

(B1)

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Question 3 (1 mark)

Write 20% as a fraction.

Worked Solution

Step 1: Convert Percentage to Fraction

Why we do this: “Percent” means “out of 100”.

\[ 20\% = \frac{20}{100} \]

Simplify by dividing numerator and denominator by 20:

\[ \frac{20 \div 20}{100 \div 20} = \frac{1}{5} \]

\(\frac{1}{5}\) (or \(\frac{20}{100}\))

(B1)

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Question 4 (1 mark)

Here is a list of four fractions.

\[ \frac{4}{16} \quad \frac{2}{8} \quad \frac{15}{60} \quad \frac{3}{9} \]

One of these fractions is not equivalent to \(\frac{1}{4}\).

Write down this fraction.

Worked Solution

Step 1: Simplify each fraction

\(\frac{4}{16}\) divides by 4 \(\rightarrow \frac{1}{4}\)

\(\frac{2}{8}\) divides by 2 \(\rightarrow \frac{1}{4}\)

\(\frac{15}{60}\) divides by 15 \(\rightarrow \frac{1}{4}\) (because \(15 \times 4 = 60\))

\(\frac{3}{9}\) divides by 3 \(\rightarrow \frac{1}{3}\)

We can see that \(\frac{3}{9}\) is the odd one out.

\(\frac{3}{9}\)

(B1)

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Question 5 (1 mark)

Write down the first even multiple of 7.

Worked Solution

Step 1: List multiples of 7

\(7 \times 1 = 7\) (Odd)

\(7 \times 2 = 14\) (Even)

14

(B1)

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Question 6 (2 marks)

(a) Simplify \(3 \times 4t\)


(b) Simplify \(8a – 3a + 2a\)

Worked Solution

Part (a): Multiplication

Multiply the numbers first:

\[ 3 \times 4 = 12 \] \[ 12 \times t = 12t \]

(B1)

Part (b): Collecting Like Terms

Note: Order of operations applies. Work from left to right.

First: \(8a – 3a = 5a\)

Then: \(5a + 2a = 7a\)

\(7a\)

(B1)

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Question 7 (5 marks)

Here is a probability scale showing events A, B, C and D.

0 1/2 1 × × × × A B C D

(a) Write down the letter of the event that is certain.

(b) Write down the letter of the event that is unlikely.


There are 12 counters in a bag.

  • 3 of the counters are red.
  • 1 of the counters is blue.
  • 2 of the counters are yellow.
  • The rest of the counters are green.

Caitlin takes at random a counter from the bag.

(c) Show that the probability that this counter is yellow or green is \(\frac{2}{3}\).

Worked Solution

Part (a): Certainty

A probability of 1 means an event is Certain.

D

(B1)

Part (b): Unlikely

Unlikely means a probability between 0 and 0.5.

B

(B1)

Part (c): Calculating Probability

Total counters = 12

Green = \(12 – (3 + 1 + 2) = 12 – 6 = 6\)

Yellow + Green = \(2 + 6 = 8\)

Probability = \(\frac{8}{12}\)

Simplify: \(\frac{8 \div 4}{12 \div 4} = \frac{2}{3}\)

Shown as \(\frac{2}{3}\)

(M1) Green counters, ✓ (M1) Probability setup, ✓ (C1) Conclusion.

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Question 8 (2 marks)

3 kg of meat costs £54.

Nina buys 2 kg of the meat.

Work out how much Nina pays.

Worked Solution

Step 1: Find cost of 1 kg
\[ 54 \div 3 = 18 \]

1 kg costs £18.

Step 2: Find cost of 2 kg
\[ 18 \times 2 = 36 \]

£36

(M1) for ÷3 or ×2, ✓ (A1) for answer.

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Question 9 (2 marks)

The centre of this circle is marked with a cross (×).

(a) Write down the mathematical name of the straight line shown in the circle.


(b) Write down the mathematical name of the straight line that is touching the circle.

Worked Solution

Part (a)

Radius

(B1)

Part (b)

Tangent

(B1)

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Question 10 (3 marks)

Tim and three friends go on holiday together for a week.

The 4 friends will share the costs of the holiday equally.

Here are the costs of the holiday:

  • £1280 for 4 return plane tickets
  • £640 for the villa
  • £220 for hire of a car for the week

Work out how much Tim has to pay for his share of the costs.

Worked Solution

Step 1: Find Total Cost

Sum the costs:

\[ 1280 + 640 + 220 = 2140 \]

(P1)

Step 2: Divide by 4

There are 4 friends (Tim + 3 friends).

\[ 2140 \div 4 = 535 \]

(P1)

£535

(A1)

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Question 11 (2 marks)

Write down an example to show that each of the following two statements is not correct.

(a) The factors of an even number are always even.


(b) All the digits in odd numbers are odd.

Worked Solution

Part (a)

We need an even number with an odd factor.

Example: 6 is even. Factors of 6 are 1, 2, 3, 6.

3 is a factor and it is odd.

e.g., 6 has factor 3 (or 12 has factor 3)

(C1)

Part (b)

We need an odd number that contains an even digit.

Example: 23 is odd, but the digit 2 is even.

e.g., 21, 23, 45

(C1)

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Question 12 (7 marks)

A shop sells desktop computers, laptops and tablets.

The composite bar chart shows information about sales over the last three years.

0 100 200 300 400 500 600 700 800 900 Number sold 2015 2016 2017 Desktop Laptops Tablets

(a) Write down the number of desktop computers sold in 2015.

(b) Work out the total number of laptops sold in the 3 years.

(c) State the item that had the greatest increase in sales over the 3 years. Give a reason for your answer.

Alex says, “In 2017, more tablets were sold than desktop computers. This means the shop makes more profit from the sale of tablets than from the sale of desktop computers.”

(d) Is Alex correct? You must justify your answer.

Worked Solution

Part (a)

100

(B1)

Part (b)

2015: \(260 – 100 = 160\)

2016: \(340 – 120 = 220\)

2017: \(440 – 160 = 280\)

Total: \(160 + 220 + 280 = 660\)

660

(A1)

Part (c)

Tablets increased the most.

Tablet increase: \(340 – 60 = 280\).

(Laptops: \(120\), Desktops: \(60\))

Tablets (with reason)

(C1)

Part (d)

No, because we do not know the profit per item.

(C1)

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Question 13 (3 marks)

A piece of wire is 240 cm long.

Peter cuts two 45 cm lengths off the wire.

He then cuts the rest of the wire into as many 40 cm lengths as possible.

Work out how many 40 cm lengths of wire Peter cuts.

Worked Solution

Step 1: Calculate used wire
\[ 2 \times 45 = 90 \text{ cm} \]
Step 2: Calculate remaining wire
\[ 240 – 90 = 150 \text{ cm} \]

(P1)

Step 3: Divide by 40
\[ 150 \div 40 = 3.75 \]

He can cut 3 full lengths.

(P1)

3

(A1)

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Question 14 (4 marks)

Gavin, Harry and Isabel each earn the same monthly salary.

Each month:

  • Gavin saves 28% of his salary.
  • Harry spends \(\frac{3}{4}\) of his salary and saves the rest.
  • The amount Isabel saves : the amount she spends = 3 : 7.

Work out who saves the most of their salary each month.

You must show how you get your answer.

Worked Solution

Step 1: Gavin

Gavin saves 28%.

Step 2: Harry

Harry spends \(\frac{3}{4}\), so he saves \(\frac{1}{4}\).

Convert to percentage: \(\frac{1}{4} = 25\%\).

(P1)

Step 3: Isabel

Ratio Saves : Spends = 3 : 7.

Total parts = \(3 + 7 = 10\).

Fraction saved = \(\frac{3}{10}\).

Convert to percentage: \(30\%\).

(P1)

Step 4: Compare

Gavin: 28%

Harry: 25%

Isabel: 30%

Isabel

(C1) (with supported working)

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Question 15 (2 marks)

Work out 15% of 160 grams.

Worked Solution

Method: Split into 10% and 5%

10% of 160 = 16

5% of 160 = 8 (half of 16)

15% = 16 + 8 = 24

(M1)

24 grams

(A1)

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Question 16 (6 marks)

P = 4x + 3y

x = 5

y = -2

(a) Work out the value of P.


(b) Expand \(4e(e + 2)\)


(c) Solve \(3(m – 4) = 21\)

Worked Solution

Part (a)
\[ P = 4(5) + 3(-2) \] \[ P = 20 – 6 \] \[ P = 14 \]

14

(A1)

Part (b)
\[ 4e \times e = 4e^2 \] \[ 4e \times 2 = 8e \]

\(4e^2 + 8e\)

(B2)

Part (c)

Divide by 3:

\[ m – 4 = 7 \]

Add 4 to both sides:

\[ m = 11 \]

m = 11

(A1)

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Question 17 (2 marks)

There are some chocolates in a box.

\(\frac{1}{4}\) of the chocolates contain nuts.

The rest of the chocolates do not contain nuts.

Write down the ratio of the number of chocolates that contain nuts to the number of chocolates that do not contain nuts.

Give your answer in the form \(1 : n\)

Worked Solution

Step 1: Identify Fractions

Nuts: \(\frac{1}{4}\)

No Nuts: \(1 – \frac{1}{4} = \frac{3}{4}\)

Step 2: Form Ratio

Nuts : No Nuts

1 : 3

1 : 3

(A1)

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Question 18 (3 marks)

\(A = \{\text{multiples of 5 between 14 and 26}\}\)

\(B = \{\text{odd numbers between 14 and 26}\}\)

(a) List the members of \(A \cup B\)


(b) Describe the members of \(A \cap B\)

Worked Solution

Step 1: List Sets

\(A = \{15, 20, 25\}\)

\(B = \{15, 17, 19, 21, 23, 25\}\)

Part (a): Union (A ∪ B)

Union means everything in A OR B (no repeats).

15, 17, 19, 20, 21, 23, 25

(A1)

Part (b): Intersection (A ∩ B)

Intersection means in BOTH A AND B.

\(\{15, 25\}\)

Odd multiples of 5 (between 14 and 26)

(C1)

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Question 19 (4 marks)

(a) Work out \(2\frac{1}{7} + 1\frac{1}{4}\)


(b) Work out \(1\frac{1}{5} \div \frac{3}{4}\)

Give your answer as a mixed number in its simplest form.

Worked Solution

Part (a)

Convert to improper fractions:

\[ \frac{15}{7} + \frac{5}{4} \]

Common denominator (28):

\[ \frac{60}{28} + \frac{35}{28} = \frac{95}{28} \]

Convert back to mixed:

\[ 3\frac{11}{28} \]

\(3\frac{11}{28}\) (or \(\frac{95}{28}\))

(A1)

Part (b)

Convert to improper fraction:

\[ \frac{6}{5} \div \frac{3}{4} \]

Flip and multiply (KFC):

\[ \frac{6}{5} \times \frac{4}{3} = \frac{24}{15} \]

Simplify (divide by 3):

\[ \frac{8}{5} \]

Convert to mixed:

\[ 1\frac{3}{5} \]

\(1\frac{3}{5}\)

(A1)

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Question 20 (3 marks)

In a village,

  • the number of houses and the number of flats are in the ratio 7 : 4
  • the number of flats and the number of bungalows are in the ratio 8 : 5

There are 50 bungalows in the village.

How many houses are there in the village?

Worked Solution

Step 1: Combine Ratio

Houses : Flats = 7 : 4

Flats : Bungalows = 8 : 5

To match the “Flats” part (4 and 8), multiply the first ratio by 2.

H : F = 14 : 8

Combined: H : F : B = 14 : 8 : 5

(P1)

Step 2: Calculate Value of One Part

5 parts = 50 bungalows.

1 part = 10.

Step 3: Calculate Houses

Houses = 14 parts.

\[ 14 \times 10 = 140 \]

(P1)

140

(A1)

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Question 21 (4 marks)

Renee buys 5 kg of sweets to sell.

She pays £10 for the sweets.

Renee puts all the sweets into bags.

She puts 250 g of sweets into each bag.

She sells each bag of sweets for 65p.

Work out her percentage profit.

Worked Solution

Step 1: Find number of bags

5 kg = 5000 g

\[ 5000 \div 250 = 20 \text{ bags} \]

(P1)

Step 2: Find Total Revenue
\[ 20 \times 65\text{p} = 1300\text{p} = £13.00 \]

(P1)

Step 3: Calculate Percentage Profit

Profit = \(13 – 10 = £3\)

\[ \frac{\text{Profit}}{\text{Original Cost}} \times 100 \] \[ \frac{3}{10} \times 100 = 30\% \]

(P1)

30%

(A1)

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Question 22 (4 marks)

A cycle race across America is 3069.25 miles in length.

Juan knows his average speed for his previous races is 15.12 miles per hour.

For the next race across America he will cycle for 8 hours per day.

(a) Estimate how many days Juan will take to complete the race.


Juan trains for the race.

The average speed he can cycle at increases.

It is now 16.27 miles per hour.

(b) How does this affect your answer to part (a)?

Worked Solution

Part (a): Estimation

Round the numbers:

Distance \(\approx 3000\)

Speed \(\approx 15\)

Miles per day = \(15 \times 8 = 120\)

Days = \(3000 \div 120\)

\(300 \div 12 = 25\)

25 days

(A1)

Part (b)

Fewer days required (because he is faster).

(C1)

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Question 23 (6 marks)

Here is a solid square-based pyramid, VABCD.

5 cm M V A B C D 6 cm 4 cm

The base of the pyramid is a square of side 6 cm.

The height of the pyramid is 4 cm.

\(M\) is the midpoint of \(BC\) and \(VM = 5\) cm.

(a) Draw an accurate front elevation of the pyramid from the direction of the arrow.

(b) Work out the total surface area of the pyramid.

Worked Solution

Part (a): Front Elevation

Looking from the front, the pyramid looks like a triangle.

Isosceles triangle with Base = 6cm and Height = 4cm.

(A1)

Part (b): Surface Area

Base Area: \(6 \times 6 = 36\)

Triangle Face Area: \(\frac{1}{2} \times 6 \times 5 = 15\)

Total Area: \(36 + 4(15) = 36 + 60 = 96\)

96 cm\(^2\)

(A1)

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Question 24 (5 marks)

A pattern is made from four identical squares.

The sides of the squares are parallel to the axes.

x y O A(6, 7) B(38, 36) C

Point A has coordinates (6, 7).

Point B has coordinates (38, 36).

Point C is marked on the diagram.

Work out the coordinates of C.

Worked Solution

Step 1: Find side length

Horizontal distance = \(38 – 6 = 32\).

There are 4 squares across.

Side length = \(32 \div 4 = 8\).

(P1)

Step 2: Find C coordinates

C is at the top-right of the second square.

x-coordinate: \(6 + 8 + 8 = 22\)

y-coordinate: Start at top B (36) and go down 2 squares.

\(36 – 8 – 8 = 20\).

(P1)

(22, 20)

(A1)

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Question 25 (3 marks)

On the grid below, draw the graph of \(y = 1 – 4x\) for values of \(x\) from \(-3\) to \(3\).

Worked Solution

Step 1: Create Table of Values
  • If \(x = -3, y = 1 – 4(-3) = 13\)
  • If \(x = 0, y = 1\)
  • If \(x = 3, y = 1 – 4(3) = -11\)

(B2) for points

Straight line drawn from (-3, 13) to (3, -11).

(B1)

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Question 26 (2 marks)

\(\mathbf{a} = \begin{pmatrix} 5 \\ 2 \end{pmatrix} \quad \mathbf{b} = \begin{pmatrix} -1 \\ 7 \end{pmatrix}\)

Work out \(2\mathbf{a} + \mathbf{b}\) as a column vector.

Worked Solution

Step 1: Multiply vector a
\[ 2\mathbf{a} = 2 \begin{pmatrix} 5 \\ 2 \end{pmatrix} = \begin{pmatrix} 10 \\ 4 \end{pmatrix} \]
Step 2: Add vector b
\[ \begin{pmatrix} 10 \\ 4 \end{pmatrix} + \begin{pmatrix} -1 \\ 7 \end{pmatrix} = \begin{pmatrix} 10 + (-1) \\ 4 + 7 \end{pmatrix} \]

\(\begin{pmatrix} 9 \\ 11 \end{pmatrix}\)

(A1)

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