Edexcel Foundation June 2018 Paper 1

Question 1 (1 mark)

Write \(6324\) correct to the nearest thousand. [cite: 173]

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Question 2 (2 marks)

(a) Write the following numbers in order of size. Start with the smallest number. [cite: 174]

\(-6\) \(6\) \(-5\) \(0\) \(12\) [cite: 175, 177, 178, 179, 180]

(b) Write the following numbers in order of size. Start with the smallest number. [cite: 181]

\(0.078\) \(0.78\) \(0.87\) \(0.708\) [cite: 182, 185, 186, 187]

Worked Solution

Step 1: Part (a) – Ordering Integers

Why we do this:

We are asked to order both positive and negative whole numbers from smallest to largest. Negative numbers are always smaller than positive numbers. The “more negative” a number is (the further left on a number line), the smaller it is.

Working:

Looking at the negatives: \(-6\) is further left on the number line than \(-5\), so \(-6\) is the smallest. Next is \(-5\).

Then comes \(0\).

Finally, we order the positive numbers: \(6\) is smaller than \(12\).

Ordering them gives: \(-6, -5, 0, 6, 12\).

✓ (B1) Correct ordering. [cite: 103]

Step 2: Part (b) – Ordering Decimals

Why we do this:

To easily compare decimals, it helps to write them so they all have the same number of decimal places by adding placeholder zeros at the end. Here, the longest decimals have 3 decimal places.

Working:

Let’s pad all the numbers to 3 decimal places:

\(0.078 \rightarrow 0.078\)
\(0.78 \rightarrow 0.780\)
\(0.87 \rightarrow 0.870\)
\(0.708 \rightarrow 0.708\)

Now, ignoring the “0.”, we can compare the parts like whole numbers: \(078, 780, 870, 708\).

Order from smallest to largest:

\(078 \rightarrow 0.078\)

\(708 \rightarrow 0.708\)

\(780 \rightarrow 0.78\)

\(870 \rightarrow 0.87\)

✓ (B1) Correct ordering. [cite: 103]

What this tells us:

Adding placeholder zeros is a fail-safe way to avoid common mistakes, like thinking \(0.78\) is smaller than \(0.708\) just because “78 is less than 708”.

Final Answers:

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

(b) \(0.078, 0.708, 0.78, 0.87\)

Total: 2 marks

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Question 3 (1 mark)

Write \(20\%\) as a fraction. [cite: 183]

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Question 4 (1 mark)

Here is a list of four fractions. [cite: 199]

\(\frac{4}{16}\) \(\frac{2}{8}\) \(\frac{15}{60}\) \(\frac{3}{9}\) [cite: 200, 201, 204, 205]

One of these fractions is not equivalent to \(\frac{1}{4}\). [cite: 202]

Write down this fraction. [cite: 203]

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Question 5 (1 mark)

Write down the first even multiple of 7. [cite: 207]

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Question 6 (2 marks)

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

(b) Simplify \(8a – 3a + 2a\)

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Question 7 (5 marks)

Here is a probability scale.

It shows the probability of each of the events A, B, C and 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

Step 1: Part (a) & (b) – Understanding the Probability Scale

What are we being asked to find?

We need to match the words “certain” and “unlikely” to the numerical probabilities on the scale.
• “Certain” means the probability is exactly \(1\).
• “Unlikely” means the probability is between \(0\) (impossible) and \(\frac{1}{2}\) (evens).

Working:

(a) The letter at \(1\) on the scale is D.

(b) The letter situated between \(0\) and \(\frac{1}{2}\) is B.

✓ (B1) for D, (B1) for B.

Step 2: Part (c) – Finding the number of Green counters

Why we do this:

Before we can find the probability of choosing yellow or green, we need to know how many green counters there are. We know the total and the amounts of all the other colors.

Working:

Total counters = \(12\)

Counters we know = \(3\) (red) \(+ 1\) (blue) \(+ 2\) (yellow) = \(6\)

Green counters = \(12 – 6 = 6\)

✓ (M1) for finding the number of green counters.

Step 3: Calculating the Probability

Why we do this:

The probability of picking a yellow OR green counter is the total number of yellow and green counters divided by the total number of counters in the bag.

Working:

Number of yellow or green counters = \(2\) (yellow) \(+ 6\) (green) = \(8\)

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

Simplify the fraction by dividing top and bottom by \(4\):

\[ \frac{8 \div 4}{12 \div 4} = \frac{2}{3} \]

✓ (M1) for adding the probabilities or totals correctly (\(\frac{8}{12}\) seen).

✓ (C1) for a correct conclusion supported by accurate figures.

What this tells us:

We have successfully demonstrated the required calculation step-by-step, ending at exactly \(\frac{2}{3}\) as the question requested.

Final Answers:

(a) D

(b) B

(c) Shown: \(\frac{8}{12} = \frac{2}{3}\)

Total: 5 marks

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Question 8 (2 marks)

\(3 \text{ kg}\) of meat costs \(£54\)

Nina buys \(2 \text{ kg}\) of the meat.

Work out how much Nina pays.

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Question 9 (2 marks)

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

(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.

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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: Finding the Total Cost

Why we do this:

Before we can figure out what Tim alone pays, we need to know the entire cost of the holiday for the whole group. We add up all the expenses.

Working:

Total Cost = Plane Tickets + Villa + Car Hire

Total Cost = \(£1280 + £640 + £220\)

   1 2 8 0
     6 4 0
+    2 2 0
----------
   2 1 4 0
   ¹ ¹
----------

The total cost for the group is \(£2140\).

✓ (P1) for a start to the process (adding the values).

Step 2: Dividing the Cost

Why we do this:

Tim and his three friends means there are \(4\) people sharing the cost equally. We divide the total cost by \(4\) to find Tim’s share.

Working:

Tim’s share = \(2140 \div 4\)

Using the bus stop method:

     5 3 5
    -------
 4| 2²1¹4²0

\(4\) goes into \(2\) zero times, carry the \(2\).
\(4\) goes into \(21\) five times, remainder \(1\).
\(4\) goes into \(14\) three times, remainder \(2\).
\(4\) goes into \(20\) five times exactly.

✓ (P1) for a full process to find the cost per person.

✓ (A1) Correct final answer.

What this tells us:

Tim’s personal share of the holiday costs is \(£535\). We can check this by estimating: \(2000 \div 4 = 500\), so \(535\) is a very reasonable answer.

Final Answer:

\(£535\)

Total: 3 marks

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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

Step 1: Part (a) – Counter-example for Factors

Why we do this:

To prove a statement is “not correct”, we only need to find one specific example where the rule fails. This is called a counter-example. A factor is a whole number that divides exactly into another number. We need an even number that has at least one odd factor.

Working:

Let’s pick an even number, like \(12\).

The factors of \(12\) are: \(1, 2, 3, 4, 6, 12\).

Notice that \(3\) is a factor of \(12\), and \(3\) is an odd number. Also, \(1\) is a factor of every number, and \(1\) is odd.

✓ (C1) for a correct example (e.g., “\(3\) is a factor of \(12\)” or “\(1\) is a factor of \(6\)”).

Step 2: Part (b) – Counter-example for Digits in Odd Numbers

Why we do this:

An odd number is determined entirely by its last digit (it must end in \(1, 3, 5, 7,\) or \(9\)). The other digits in the number can be anything. We need to create an odd number that contains at least one even digit.

Working:

Let’s make a two-digit odd number: \(23\).

\(23\) is an odd number because it ends in \(3\).

However, the first digit is \(2\), which is an even digit. Therefore, not all digits in \(23\) are odd.

✓ (C1) for a correct example (e.g., \(23\) or \(45\)).

What this tells us:

General statements in maths can be disproven with just a single counter-example. Thinking about the fundamental definitions (like what makes a number odd) helps us find these exceptions.

Final Answers:

(a) 12 is an even number, but 3 is a factor of 12 and 3 is odd.

(b) 25 is an odd number, but the digit 2 is even.

Total: 2 marks

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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.

(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

Step 1: Part (a) – Reading the Bar Chart

Why we do this:

To find the number of desktop computers sold in 2015, we look at the 2015 column. The desktop computers are represented by the bottom section (grey box). We read the value where the top of this box aligns with the y-axis.

Working:

Looking at the 2015 column, the grey bar (desktops) goes from \(0\) up to exactly the line for \(100\).

✓ (B1) Correct answer: \(100\).

Step 2: Part (b) – Calculating Total Laptops

Why we do this:

Because this is a composite (or stacked) bar chart, the number of laptops sold each year is the difference between the top of the laptop box (orange) and the bottom of the laptop box. We must find this difference for all three years and add them together.

Working:

Let’s find the number of laptops sold each year:

2015: The laptop bar starts at \(100\) and ends at \(260\).
Laptops sold \(= 260 – 100 = 160\)
2016: The laptop bar starts at \(120\) and ends at \(340\).
Laptops sold \(= 340 – 120 = 220\)
2017: The laptop bar starts at \(160\) and ends at \(440\).
Laptops sold \(= 440 – 160 = 280\)

Total laptops \(= 160 + 220 + 280\)

   1 6 0
   2 2 0
+  2 8 0
--------
   6 6 0
   ¹
--------

✓ (M1) for reading at least 3 required figures from the graph correctly.

✓ (M1) for adding the three differences together.

✓ (A1) Correct final sum: \(660\).

Step 3: Part (c) – Identifying Greatest Increase

Why we do this:

We need to compare how much the sales of each item grew from 2015 to 2017. We can calculate the sales of tablets and desktops to compare with laptops.

Working:

Let’s look at the growth of each item from 2015 to 2017:

Desktops: \(100\) (in 2015) grew to \(160\) (in 2017). Increase \(= 60\).
Laptops: \(160\) (in 2015) grew to \(280\) (in 2017). Increase \(= 120\).
Tablets: \(320 – 260 = 60\) (in 2015) grew to \(780 – 440 = 340\) (in 2017). Increase \(= 340 – 60 = 280\).

Tablets clearly had the greatest increase.

✓ (B1) for identifying Tablets.

✓ (C1) for valid statement showing evidence (e.g., increased by \(280\) from \(60\) to \(340\)).

Step 4: Part (d) – Analyzing the Claim

Why we do this:

Alex is equating the “number of items sold” directly to “profit”. We need to evaluate if that logical leap is valid based *only* on the information given in the graph.

Working:

The graph only shows us the number of items sold (volume). It does not provide any information about the cost to make them or the price they are sold for.

For example, if the profit on one desktop is £500, but the profit on a tablet is only £10, selling more tablets wouldn’t necessarily mean more total profit.

Therefore, Alex is incorrect because we lack financial data.

✓ (C1) for “No” supported by a valid reason (e.g., “we do not know costs or profit margins”).

Final Answers:

(a) 100

(b) 660

(c) Tablets, because sales increased by 280 (from 60 to 340).

(d) No, because the chart only shows the amount sold, it does not tell us the profit made per item.

Total: 7 marks

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Question 13 (3 marks)

A piece of wire is \(240 \text{ cm}\) long.

Peter cuts two \(45 \text{ cm}\) lengths off the wire.

He then cuts the rest of the wire into as many \(40 \text{ cm}\) lengths as possible.

Work out how many \(40 \text{ cm}\) lengths of wire Peter cuts.

Worked Solution

Step 1: Finding the Remaining Wire Length

Why we do this:

Before we can figure out how many \(40 \text{ cm}\) lengths can be made, we need to know how much wire is left over after Peter makes his first two cuts of \(45 \text{ cm}\).

Working:

Length of the two initial cuts = \(2 \times 45 = 90 \text{ cm}\)

Remaining length = \(240 – 90\)

   ¹
   2 4 0
-    9 0
--------
   1 5 0
--------

There is \(150 \text{ cm}\) of wire remaining.

✓ (P1) for a start to the process (e.g. \(240 – 90 = 150\)).

Step 2: Calculating the Number of 40cm Cuts

Why we do this:

We need to see how many times \(40 \text{ cm}\) fits whole into the remaining \(150 \text{ cm}\). We can do this by dividing, or by repeatedly adding \(40\) until we can’t go any higher without passing \(150\).

Working:

Let’s count up in \(40\)s:

1 piece = \(40 \text{ cm}\)
2 pieces = \(80 \text{ cm}\)
3 pieces = \(120 \text{ cm}\)
4 pieces = \(160 \text{ cm}\) (Too much! We only have \(150 \text{ cm}\))

So, Peter can cut a maximum of \(3\) full \(40 \text{ cm}\) lengths. (He will have \(30 \text{ cm}\) left over as waste).

✓ (P1) for a complete process to find the number of pieces.

✓ (A1) Correct final answer.

What this tells us:

Because the question asks “as many… as possible”, we are finding the whole integer result of the division, ignoring the decimal remainder.

Final Answer:

\(3\)

Total: 3 marks

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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 and spends the rest of his salary
Harry spends \(\frac{3}{4}\) of his salary and saves the rest of his salary
the amount of salary Isabel saves : the amount of salary 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: Planning a Common Format

Why we do this:

We are given information in three different formats: a percentage, a fraction, and a ratio. To compare them fairly to see who saves the most, we need to convert them all into the same format. Percentages are usually the easiest format for comparing amounts out of a whole.

Step 2: Calculating Each Person’s Savings

Why we do this:

We will calculate the percentage saved for each person one by one.

Working:

1. Gavin:

Gavin saves \(28\%\). (This is already in our desired format).

2. Harry:

Harry spends \(\frac{3}{4}\). The whole salary is represented by \(1\) (or \(\frac{4}{4}\)).

Harry saves \(= 1 – \frac{3}{4} = \frac{1}{4}\).

To convert \(\frac{1}{4}\) to a percentage, we multiply by \(100\):

\[ \frac{1}{4} \times 100 = 25\% \]

✓ (P1) for a process to work with \(\frac{3}{4}\) (e.g. finding \(25\%\)).

3. Isabel:

Isabel’s saves : spends ratio is \(3 : 7\).

Total parts in the ratio \(= 3 + 7 = 10\) parts.

This means she saves \(3\) parts out of the total \(10\) parts, which is the fraction \(\frac{3}{10}\).

To convert \(\frac{3}{10}\) to a percentage, we multiply top and bottom by \(10\) to get it out of \(100\):

\[ \frac{3 \times 10}{10 \times 10} = \frac{30}{100} = 30\% \]

✓ (P1) for a process to work with the ratio \(3:7\) (e.g. finding \(\frac{3}{10}\) or \(30\%\)).

Step 3: Comparing the Results

Why we do this:

Now we list the final saving percentages side-by-side to determine who saves the most.

Working:

Gavin = \(28\%\)
Harry = \(25\%\)
Isabel = \(30\%\)

Comparing these, \(30\%\) is the largest number, so Isabel saves the most.

✓ (A1) for all three correct comparative values (e.g. \(28\%\), \(25\%\), \(30\%\)).

✓ (C1) for clearly stating Isabel with supported evidence.

Final Answer:

Isabel

Total: 4 marks

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Question 15 (2 marks)

Work out \(15\%\) of \(160 \text{ grams}\).

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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

Step 1: Part (a) – Substituting Values into an Expression

Why we do this:

We need to find the value of \(P\). To do this, we replace the letters \(x\) and \(y\) with their given numbers, making sure to remember that \(4x\) means \(4 \times x\) and \(3y\) means \(3 \times y\). [cite: 108]

Working:

Substitute \(x = 5\) and \(y = -2\) into the formula:

\[ P = 4(5) + 3(-2) \]

Now, do the multiplications first (following order of operations):

\[ 4 \times 5 = 20 \] \[ 3 \times -2 = -6 \]

Put it back together:

\[ P = 20 + (-6) = 20 – 6 = 14 \]

✓ (M1) for showing the substitution \(4 \times 5\) and \(3 \times -2\). [cite: 108]

✓ (A1) Correct final answer. [cite: 108]

Step 2: Part (b) – Expanding Brackets

Why we do this:

“Expand” means to multiply out the brackets. We must multiply the term outside the bracket (\(4e\)) by every term inside the bracket (\(e\) and \(+2\)). [cite: 108]

Working:

Multiply \(4e\) by \(e\):

\[ 4e \times e = 4e^2 \]

Multiply \(4e\) by \(2\):

\[ 4e \times 2 = 8e \]

Combine them:

\[ 4e^2 + 8e \]

✓ (B2) Fully correct expansion. (Award B1 if only \(4e^2\) or \(8e\) is correct). [cite: 108]

Step 3: Part (c) – Solving a Linear Equation

Why we do this:

We want to find the value of \(m\) that makes the equation true. We can do this in two ways: expanding the brackets first, or dividing both sides by \(3\) immediately. Dividing by \(3\) is usually faster if the number on the other side divides evenly (which \(21\) does). [cite: 108]

Working (Method 1: Divide first):

\[ 3(m – 4) = 21 \]

Divide both sides by \(3\):

\[ m – 4 = 7 \]

Add \(4\) to both sides to isolate \(m\):

\[ m = 7 + 4 = 11 \]

✓ (M1) for a correct first step (e.g., \(m – 4 = 7\) or expanding to \(3m – 12 = 21\)). [cite: 108]

✓ (A1) Correct final answer of \(11\). [cite: 108]

Final Answers:

(a) \(14\)

(b) \(4e^2 + 8e\)

(c) \(m = 11\)

Total: 6 marks

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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 \(1 : n\)

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Question 18 (3 marks)

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

\(B =\) {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: Listing the Sets

Why we do this:

Before working out the union (\(\cup\)) or intersection (\(\cap\)), we need to write out exactly which numbers belong in set \(A\) and set \(B\). [cite: 109]

Working:

Multiples of \(5\) between \(14\) and \(26\):
\(A = \{15, 20, 25\}\)

Odd numbers between \(14\) and \(26\):
\(B = \{15, 17, 19, 21, 23, 25\}\)

✓ (M1) for correctly listing either set with no incorrect numbers. [cite: 109]

Step 2: Part (a) – Finding the Union (\(A \cup B\))

Why we do this:

The symbol \(\cup\) means “Union”, which is the collection of all elements that are in set \(A\), OR set \(B\), OR both. We simply combine both lists together into one big list, but we do not write duplicates twice. [cite: 109]

Working:

Combining \(A\) and \(B\):
\(A \cup B = \{15, 17, 19, 20, 21, 23, 25\}\)

✓ (A1) Correct list of numbers with no repeats. [cite: 109]

Step 3: Part (b) – Finding the Intersection (\(A \cap B\))

Why we do this:

The symbol \(\cap\) means “Intersection”, which means the elements that exist in BOTH set \(A\) AND set \(B\) at the same time. The question asks us to “describe” them, but stating the actual numbers is also an acceptable description. [cite: 109]

Working:

Looking at our two lists:
\(A = \{ \mathbf{15}, 20, \mathbf{25} \}\)
\(B = \{ \mathbf{15}, 17, 19, 21, 23, \mathbf{25} \}\)

The numbers that appear in both are \(15\) and \(25\). These are the odd multiples of \(5\) in that range.

✓ (C1) for either listing \(15, 25\) or describing them as “odd multiples of \(5\)”. [cite: 109]

Final Answers:

(a) \(15, 17, 19, 20, 21, 23, 25\)

(b) \(15\) and \(25\)

Total: 3 marks

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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

Step 1: Part (a) – Adding Mixed Numbers

Why we do this:

To add fractions, we must have a common denominator. We can add the whole numbers together separately, and then add the fraction parts by finding a common multiple of \(7\) and \(4\) (which is \(28\)). [cite: 109]

Working:

First, add the whole numbers:

\[ 2 + 1 = 3 \]

Now, make the fractions have a common denominator of \(28\):

\[ \frac{1 \times 4}{7 \times 4} = \frac{4}{28} \] \[ \frac{1 \times 7}{4 \times 7} = \frac{7}{28} \]

Add the fractions:

\[ \frac{4}{28} + \frac{7}{28} = \frac{11}{28} \]

Combine the whole number and fraction:

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

✓ (M1) for method to find common denominator with at least one fraction correct. [cite: 109]

✓ (A1) Correct final mixed number or improper fraction (\(\frac{95}{28}\)). [cite: 109]

Step 2: Part (b) – Dividing Mixed Numbers

Why we do this:

When multiplying or dividing mixed numbers, we MUST convert them to improper fractions (top-heavy fractions) first. Then, to divide by a fraction, we multiply by its reciprocal (flip the second fraction upside down). [cite: 109]

Working:

Convert \(1\frac{1}{5}\) to an improper fraction:

\[ 1\frac{1}{5} = \frac{(1 \times 5) + 1}{5} = \frac{6}{5} \]

The problem is now:

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

Keep, Change, Flip (Multiply by the reciprocal):

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

✓ (M1) for converting to improper and multiplying by reciprocal. [cite: 109]

Now, simplify by dividing top and bottom by their greatest common factor, which is \(3\):

\[ \frac{24 \div 3}{15 \div 3} = \frac{8}{5} \]

The question requires a mixed number answer. How many \(5\)s fit into \(8\)?

One \(5\) fits in, with \(3\) left over, so it’s \(1\frac{3}{5}\).

✓ (A1) Correct mixed number in simplest form. [cite: 109]

What this tells us:

Always double-check the final format requested by the question. Stopping at \(\frac{8}{5}\) or \(\frac{24}{15}\) would lose the final accuracy mark because it specifically asks for a “mixed number in its simplest form”.

Final Answers:

(a) \(3\frac{11}{28}\)

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

Total: 4 marks

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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: Finding the Number of Flats

Why we do this:

We are given the exact number of bungalows (\(50\)). We know the ratio linking flats to bungalows is \(8 : 5\). We can use this to figure out how many flats there are. Once we know the flats, we can link that back to houses. [cite: 112]

Working:

Ratio of Flats : Bungalows = \(8 : 5\)

We know actual Bungalows = \(50\)

To get from the ratio part (\(5\)) to the actual amount (\(50\)), we multiply by \(10\).

\[ 50 \div 5 = 10 \]

Since we multiplied the bungalows by \(10\), we must multiply the flats ratio by \(10\) to maintain the proportion:

\[ \text{Actual Flats} = 8 \times 10 = 80 \]

✓ (P1) process started, finding \(80\) flats. [cite: 112]

Step 2: Finding the Number of Houses

Why we do this:

Now we know there are \(80\) flats. We can use the first ratio (Houses : Flats = \(7 : 4\)) to find the number of houses. [cite: 112]

Working:

Ratio of Houses : Flats = \(7 : 4\)

We know actual Flats = \(80\)

To get from the ratio part (\(4\)) to the actual amount (\(80\)), we must multiply by \(20\).

\[ 80 \div 4 = 20 \]

Multiply the houses ratio by \(20\) to find the actual number of houses:

\[ \text{Actual Houses} = 7 \times 20 = 140 \]

✓ (P1) full process to solve problem. [cite: 112]

✓ (A1) Correct final answer of \(140\). [cite: 112]

What this tells us:

This is a classic “ratio linking” problem. The flats acted as a bridge between the houses and the bungalows. An alternative method would be creating a combined \(3\)-part ratio (Houses : Flats : Bungalows \(\rightarrow\) \(14 : 8 : 5\)) which yields the same result.

Final Answer:

\(140\)

Total: 3 marks

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Question 21 (4 marks)

Renee buys \(5 \text{ kg}\) of sweets to sell.
She pays \(£10\) for the sweets.

Renee puts all the sweets into bags.
She puts \(250 \text{ g}\) of sweets into each bag.

She sells each bag of sweets for \(65\text{p}\).

Renee sells all the bags of sweets.

Work out her percentage profit.

Worked Solution

Step 1: Finding the Number of Bags

Why we do this:

We need to figure out how much money Renee makes (revenue). To do this, we first need to know how many bags she sells. Since the total weight is in kilograms and the bag weight is in grams, we must convert units to match.

Working:

Convert kg to g (\(1 \text{ kg} = 1000 \text{ g}\)):

\[ 5 \text{ kg} = 5 \times 1000 = 5000 \text{ g} \]

Divide the total weight by the weight of one bag:

\[ 5000 \div 250 \]

Simplify by dividing both by \(10\) (crossing off a zero):

\[ 500 \div 25 = 20 \text{ bags} \]

✓ (P1) for full process to find the number of bags (\(20\)).

Step 2: Calculating Total Revenue

Why we do this:

Now we multiply the number of bags by the selling price per bag to find the total money she receives.

Working:

Revenue = \(20 \text{ bags} \times 65\text{p}\)

\[ 2 \times 65\text{p} = 130\text{p} \] \[ 20 \times 65\text{p} = 1300\text{p} \]

Convert pence back to pounds:

\[ 1300\text{p} = £13.00 \]

✓ (P1) for process to find total revenue (\(20 \times 0.65 = 13\)).

Step 3: Calculating Percentage Profit

Why we do this:

Percentage profit is calculated as the actual profit made divided by the original cost, multiplied by \(100\). Profit is what’s left after she pays back the original \(£10\).

Working:

Actual Profit = Revenue \(- \text{Cost}\)
Profit \(= £13 – £10 = £3\)

Percentage Profit = \( \left( \frac{\text{Profit}}{\text{Original Cost}} \right) \times 100 \)

\[ \frac{3}{10} \times 100 = 30\% \]

✓ (P1) for process to find percentage profit (\( (13 – 10) \div 10 \times 100 \)).

✓ (A1) Correct final answer of \(30\).

What this tells us:

Renee made a \(30\%\) profit on her investment, meaning for every \(£1\) she spent, she got her \(£1\) back plus an extra \(30\text{p}\).

Final Answer:

\(30\%\)

Total: 4 marks

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Question 22 (4 marks)

A cycle race across America is \(3069.25 \text{ miles}\) in length.

Juan knows his average speed for his previous races is \(15.12 \text{ miles per hour}\).

For the next race across America he will cycle for \(8 \text{ hours}\) 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 \(16.27 \text{ miles per hour}\).

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

Worked Solution

Step 1: Part (a) – Rounding for Estimation

Why we do this:

The word “Estimate” is a huge hint. We should NOT do exact calculations with the messy decimals. Instead, we round the numbers to 1 significant figure (or sensible, easy-to-use numbers) first, and then do the math.

Working:

Total Distance: \(3069.25 \approx 3000 \text{ miles}\)
Average Speed: \(15.12 \approx 15 \text{ mph}\)
Hours per day: \(8 \text{ hours}\)

✓ (P1) for using rounded values in a correct process.

Step 2: Calculating the Estimate

Why we do this:

To find the total days, we can first figure out how many miles he cycles in a single day, and then divide the total race distance by this daily distance.

Working:

Estimated distance per day = Speed \(\times\) Time

\[ 15 \text{ mph} \times 8 \text{ hours} = 120 \text{ miles per day} \]

Now find the total days:

\[ \text{Total Days} = 3000 \div 120 \]

Simplify by dividing both sides by \(10\):

\[ 300 \div 12 \]

Let’s halve both numbers: \(150 \div 6\)

Halve again: \(75 \div 3 = 25\)

So, Juan will take approximately \(25\) days.

✓ (P1) for a full process to find the number of days (e.g., \(3000 \div 15 \div 8\)).

✓ (A1) Correct estimated answer of \(25\).

Step 3: Part (b) – Analyzing the Change

Why we do this:

We need to think about how a faster speed changes the time it takes to travel a fixed distance. If you go faster, it takes less time.

Working:

Because his average speed has increased, he is covering more distance each day. Therefore, it will take him fewer days to complete the race overall.

(Alternatively, you could note that for an estimate, you might still round \(16.27\) down to \(15\), meaning the estimated answer wouldn’t change, but stating it takes “fewer days” is the most direct logical conclusion.)

✓ (C1) for a valid statement referring to the time taken.

Final Answers:

(a) 25 days

(b) It will take fewer days (less time) because he is travelling faster.

Total: 4 marks

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Question 23 (6 marks)

Here is a solid square-based pyramid, \(VABCD\).

The base of the pyramid is a square of side \(6 \text{ cm}\).

The height of the pyramid is \(4 \text{ cm}\).

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

(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

Step 1: Part (a) – Front Elevation

Why we do this:

An “elevation” is a 2D drawing of what a 3D object looks like looking straight at it from one side. If we stand directly in front of the arrow and look flat at the pyramid, we will see a triangle. The base of this triangle is the side \(AB\) (\(6 \text{ cm}\)). The peak \(V\) will be exactly in the middle above the base, at a height of \(4 \text{ cm}\).

Working (Accurate Diagram):

✓ (M1) for drawing an isosceles triangle with base \(6\) and height \(4\).

✓ (A1) for a fully correct, accurate diagram.

Step 2: Part (b) – Total Surface Area

Why we do this:

The “total surface area” means the area of the outside wrapping of the shape. A square-based pyramid is made of \(5\) flat faces:

One square base on the bottom.
Four identical triangular faces around the sides.

We need to find the area of the base, the area of one triangle, multiply the triangle by \(4\), and add it all together.

Working:

Area of the square base:

The base is a square with sides of \(6 \text{ cm}\).

\[ \text{Area} = 6 \times 6 = 36 \text{ cm}^2 \]

Area of one triangular face (e.g., face VBC):

The base of the triangle is \(6 \text{ cm}\). The slanted height of the triangle down its face (line \(VM\)) is \(5 \text{ cm}\). Be careful not to use the vertical \(4 \text{ cm}\) height of the pyramid inside.

\[ \text{Area of a triangle} = \frac{1}{2} \times \text{base} \times \text{height} \] \[ \text{Area} = \frac{1}{2} \times 6 \times 5 = 3 \times 5 = 15 \text{ cm}^2 \]

✓ (M1) method to find the area of a triangular face.

Total Surface Area:

Add the base and the four triangular faces together:

\[ \text{Total Area} = 36 + (4 \times 15) \] \[ \text{Total Area} = 36 + 60 = 96 \]

✓ (M1) method to find the total surface area (\(4 \times 15 + 6 \times 6\)).

✓ (A1) Correct numerical answer.

✓ (B1) Correct units (\(\text{cm}^2\)).

What this tells us:

Breaking a 3D shape down into its flat 2D nets makes surface area problems simple calculations of rectangles and triangles. The units are \(\text{cm}^2\) because we are measuring area.

Final Answers:

(a) Diagram drawn (triangle, base \(6\), height \(4\))

(b) \(96 \text{ cm}^2\)

Total: 6 marks

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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.

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: Finding the Side Length of the Squares

Why we do this:

To find the coordinates of any point inside the pattern, we first need to know how big the squares are. We can figure this out by looking at the total horizontal width from point \(A\) to point \(B\).

Working:

The \(x\)-coordinate of \(A\) is \(6\). The \(x\)-coordinate of \(B\) is \(38\).

Total horizontal width of the diagram = \(38 – 6 = 32\)

✓ (P1) process to find width or height (\(38 – 6 = 32\)).

Horizontally, the diagram is made up of exactly \(4\) squares placed side-by-side without overlapping left-to-right.

Side length of one square = \(32 \div 4 = 8\)

✓ (P1) process to find side of square (\(32 \div 4 = 8\)).

Step 2: Finding the x-coordinate of C

Why we do this:

Point \(C\) is located horizontally at the end of the second square. We simply add the width of two squares to the starting \(x\)-coordinate at \(A\).

Working:

\(x\)-coordinate of \(C\) = (start \(x\)) + (\(2\) square widths)

\[ x = 6 + (2 \times 8) = 6 + 16 = 22 \]

✓ (P1) process to find x-coordinate (\(6 + 16\)).

Step 3: Finding the y-coordinate of C

Why we do this:

We do the same thing vertically. However, it’s easier to work backwards from \(B\) this time because \(C\) lines up with the top-right corner of the second square. From \(B\), we just go down the height of two squares to reach the height of \(C\).

Working:

\(y\)-coordinate of \(C\) = (top \(y\)) – (\(2\) square heights)

\[ y = 36 – (2 \times 8) = 36 – 16 = 20 \]

✓ (P1) process to find y-coordinate (\(36 – 16\)).

What this tells us:

The coordinates of point \(C\) are \((22, 20)\). Interestingly, working forwards from \(y=7\) yields \(7 + 2(8) = 23\), which doesn’t match \(20\). This tells us the squares actually overlap slightly in the vertical direction to make the pattern, which is why working downwards from \(B\) using the visual alignment is a safer strategy.

Final Answer:

\((22, 20)\)

Total: 5 marks

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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: Creating a Table of Values

Why we do this:

To plot a straight-line graph, we need to find some coordinates (points). We do this by choosing values for \(x\) (from \(-3\) to \(3\)), substituting them into the equation \(y = 1 – 4x\), and calculating the resulting \(y\) values.

Working:

If \(x = -3\), \(y = 1 – 4(-3) = 1 + 12 = 13\) \(\rightarrow (-3, 13)\)
If \(x = -1\), \(y = 1 – 4(-1) = 1 + 4 = 5\) \(\rightarrow (-1, 5)\)
If \(x = 0\), \(y = 1 – 4(0) = 1 – 0 = 1\) \(\rightarrow (0, 1)\)
If \(x = 2\), \(y = 1 – 4(2) = 1 – 8 = -7\) \(\rightarrow (2, -7)\)
If \(x = 3\), \(y = 1 – 4(3) = 1 – 12 = -11\) \(\rightarrow (3, -11)\)

✓ (B1) for at least 2 correct points calculated.

Step 2: Plotting the Line

Why we do this:

We mark these coordinates on the grid with small crosses and then join them with a single straight line using a ruler, ensuring the line goes from \(x=-3\) all the way to \(x=3\).

Working:

✓ (B2) for a correct line segment through at least 3 points.

✓ (B3) for a fully correct line drawn continuously between \(x = -3\) and \(x = 3\).

What this tells us:

The line crosses the \(y\)-axis at \(+1\) (the \(y\)-intercept) and for every \(1\) unit we move to the right, it drops down by \(4\) units (gradient of \(-4\)). Checking these properties confirms our line is perfectly accurate.

Final Answer:

Correct line drawn passing through \((-3, 13)\) to \((3, -11)\).

Total: 3 marks

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Question 26 (2 marks)

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

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

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Edexcel GCSE MathematicsFoundation Tier – Paper 1 (Non-Calculator)June 2018

Mark Scheme Legend

Table of Contents

Question 1 (1 mark)

Worked Solution

Step 1: Understanding the Question

Step 2: Identifying the Target Column

Question 2 (2 marks)

Worked Solution

Step 1: Part (a) – Ordering Integers

Step 2: Part (b) – Ordering Decimals

Question 3 (1 mark)

Worked Solution

Step 1: Converting Percentage to a Fraction

Question 4 (1 mark)

Worked Solution

Step 1: Simplifying Fractions

Question 5 (1 mark)

Worked Solution

Step 1: Listing Multiples

Question 6 (2 marks)

Worked Solution

Step 1: Part (a) – Multiplying Algebraic Terms

Step 2: Part (b) – Collecting Like Terms

Question 7 (5 marks)

Worked Solution

Step 1: Part (a) & (b) – Understanding the Probability Scale

Step 2: Part (c) – Finding the number of Green counters

Step 3: Calculating the Probability

Question 8 (2 marks)

Worked Solution

Step 1: Find the Cost of 1 kg (Unitary Method)

Step 2: Calculate the Cost of 2 kg

Question 9 (2 marks)

Worked Solution

Step 1: Identifying Circle Parts

Question 10 (3 marks)

Worked Solution

Step 1: Finding the Total Cost

Step 2: Dividing the Cost

Question 11 (2 marks)

Worked Solution

Step 1: Part (a) – Counter-example for Factors

Step 2: Part (b) – Counter-example for Digits in Odd Numbers

Question 12 (7 marks)

Worked Solution

Step 1: Part (a) – Reading the Bar Chart

Step 2: Part (b) – Calculating Total Laptops

Step 3: Part (c) – Identifying Greatest Increase

Step 4: Part (d) – Analyzing the Claim

Question 13 (3 marks)

Worked Solution

Step 1: Finding the Remaining Wire Length

Step 2: Calculating the Number of 40cm Cuts

Question 14 (4 marks)

Worked Solution

Step 1: Planning a Common Format

Step 2: Calculating Each Person’s Savings

Step 3: Comparing the Results

Question 15 (2 marks)

Worked Solution

Step 1: Finding 10% and 5%

Step 2: Combining the Blocks

Question 16 (6 marks)

Worked Solution

Step 1: Part (a) – Substituting Values into an Expression

Step 2: Part (b) – Expanding Brackets

Step 3: Part (c) – Solving a Linear Equation

Question 17 (2 marks)

Worked Solution

Step 1: Expressing the Components as Fractions

Step 2: Forming the Ratio

Question 18 (3 marks)

Worked Solution

Step 1: Listing the Sets

Step 2: Part (a) – Finding the Union (\(A \cup B\))

Step 3: Part (b) – Finding the Intersection (\(A \cap B\))

Question 19 (4 marks)

Worked Solution

Step 1: Part (a) – Adding Mixed Numbers

Edexcel GCSE Mathematics
Foundation Tier – Paper 1 (Non-Calculator)
June 2018