GCSE Maths Edexcel June 2020 Paper 3 (Foundation)

💡 What we know: There are \( 100 \) centimetres in \( 1 \) metre.

To convert from centimetres (smaller unit) to metres (larger unit), we need to divide by \( 100 \).

💡 Meaning: Finding \( \frac{1}{3} \) of a number is the same as dividing that number by \( 3 \).

🔢 Calculator Steps:

1. Press: 24
2. Press: ÷
3. Press: 3
4. Press: =

💡 Concept: “Percent” means “out of 100”. So \( 40\% \) means \( 40 \) out of \( 100 \).

We can simplify the fraction by dividing the top and bottom by \( 20 \).

💡 Meaning: \( 2.5^2 \) means multiplying \( 2.5 \) by itself.

🔢 Calculator Steps:

1. Press: 2.5
2. Press: x² (or x 2.5)
3. Press: =
4. If it shows a fraction, press S<=>D

💡 Strategy: Negative numbers are smaller than zero. The “more negative” a number is (further from zero), the smaller it is.

Let’s list the negative numbers: \( -4, -6, -3 \).

Ordering these from smallest (most negative) to largest:

• \( -6 \) is the smallest
• \( -4 \) comes next
• \( -3 \) is next

Now we add \( 0 \) and the positive numbers: \( 1, 0, 7, 2 \).

Ordering these: \( 0, 1, 2, 7 \).

💡 Strategy: Locate the line for 1.4 on the vertical axis. We need to count the bars that are strictly higher than this line.

• Year 1: 1.2 (Less)
• Year 2: 1.3 (Less)
• Year 3: 1.4 (Equal – not “more than”)
• Year 4: 1.5 (More)
• Year 5: 1.55 (More)
• Year 6: 1.6 (More)
• Year 7: 1.6 (More)
• Year 8: 1.7 (More)

💡 Strategy: Look for two bars that have exactly the same height.

Looking at the graph:

• Bar 5 is approximately 1.55
• Bar 6 is at 1.6
• Bar 7 is at 1.6

Bar 6 and Bar 7 are level with each other.

💡 Total Squares: The grid is \( 4 \times 4 \), so there are \( 16 \) total squares.

💡 Shaded Squares: Count the grey squares.

• Row 1: 2
• Row 2: 4
• Row 3: 3
• Row 4: 3

Divide the numerator and denominator by their highest common factor (4).

💡 Why: We can’t buy individual tiles, only full boxes. We need to find out how many boxes contain \( 200 \) tiles.

Multiply the number of boxes by the cost per box.

💡 Strategy: Calculate the denominator first (BIDMAS/BODMAS).

💡 Strategy: Work out the brackets first.

💡 Meaning: The reciprocal of a number \( x \) is \( \frac{1}{x} \).

💡 Strategy: Find \( 2 \) on the horizontal axis (Time). Go up to the curve, then go left to read the temperature.

💡 Strategy: Find \( 50 \) on the vertical axis (Temperature). Go right to the curve, then go down to read the time.

💡 Strategy: Calculate the temperature drop for both time periods.

💡 Fact: The angles in any quadrilateral add up to \( 360^\circ \).

We know three of the interior angles: \( 95^\circ, 130^\circ, 65^\circ \).

Let the missing interior angle at \( S \) be \( x \).

💡 Fact: Angles on a straight line add up to \( 180^\circ \).

The interior angle \( x \) and the exterior angle \( y \) are on the straight line \( PST \).

💡 Pattern: The sequence goes down by \( 5 \) each time (\( -5 \)).

\( 25 – 5 = 20 \)

\( 20 – 5 = 15 \)

💡 Strategy: Continue the sequence until we reach \( -5 \).

• Term 1: 45
• Term 2: 40
• Term 3: 35
• Term 4: 30
• Term 5: 25
• Term 6: 20
• Term 7: 15
• Term 8: 10
• Term 9: 5
• Term 10: 0
• Term 11: -5

💡 Strategy: Substitute \( n = 9 \) into the expression.

💡 Strategy: Perimeter is the total distance around the outside. We need all side lengths.

We are missing two sides:

• Right-most vertical side: Total height (left) is \( 7 \). Inner vertical is \( 5 \).
  Missing side = \( 7 – 5 = 2 \text{ cm} \).
• Bottom horizontal inner side: Wait, the diagram gives the total bottom width as \( 10 \). The top width is \( 4 \).
  Missing horizontal side = \( 10 – 4 = 6 \text{ cm} \).

💡 Strategy: Group the \( x \)’s together and the \( y \)’s together.

\( x \)’s: \( 3x + 2x \)

\( y \)’s: \( +5y – 4y \)

💡 Strategy: Isolate \( p \) by performing the inverse operations in reverse order.

1. Subtract \( 7 \) from both sides.
2. Divide by \( 5 \).

💡 Strategy: Calculate how many packs are needed for each shop to get at least 30 batteries, then calculate the total cost.

💡 Strategy: Re-calculate the cost assuming we can buy exactly 30 batteries. Or observe the unit prices.

💡 Formula: \( \text{Probability} = \frac{\text{Number of successful outcomes}}{\text{Total number of possible outcomes}} \)

First, find the total number of cards.

💡 Fact: The probability of something happening plus the probability of it not happening equals \( 1 \).

💡 Method: Since it is an isosceles triangle with two sides of length \( 6\text{ cm} \), we use a compass.

1. Open your compass to exactly \( 6\text{ cm} \) using a ruler.
2. Place the point of the compass on the left end of the \( 8\text{ cm} \) line and draw an arc above the line.
3. Keep the compass width the same (\( 6\text{ cm} \)).
4. Place the point on the right end of the line and draw another arc to cross the first one.
5. The intersection point is the top vertex of the triangle.
6. Use a ruler to join this point to both ends of the base line.

💡 Strategy: The graph does not go up to \( 345 \) dollars. We need to find a value we can read from the graph and multiply it.

Let’s pick a convenient point on the line. For example, check \( 50 \) pounds or \( 60 \) dollars.

From the graph: \( £50 \approx \$65 \).

Or look for \( \$60 \). \( \$60 \approx £46 \)? Let’s find a clear point.

Let’s find the exchange rate: \( \text{Rate} = \frac{\$65}{£50} = 1.3 \)

We have dollars, we want pounds. We divide by the exchange rate (or use a multiplier).

Alternatively, find a factor. \( 345 \) dollars.

Can we break \( 345 \) down? \( 345 = 69 \times 5 \)? No obvious simple multiple from graph reading.

Let’s use the rate method.

💡 Strategy: We know that \( 40\% \) represents \( 56 \) sandwiches. We need to find \( 100\% \) (the total).

Method 1: Find \( 1\% \) first.

\( 1\% = 56 \div 40 \)

\( 100\% = (56 \div 40) \times 100 \)

💡 Formula: \( \frac{\text{Part}}{\text{Total}} \times 100 \)

We want \( 18 \) out of \( 56 \) as a percentage.

💡 Strategy: We know the mean for 3 people is £80.

\( \text{Total} = \text{Mean} \times \text{Count} \)

Subtract Akhtar and Ben’s money from the total.

Carl has: 3 x £5 notes + 1 x £20 note + some £10 notes.

Calculate value of known notes:

\( 3 \times 5 = 15 \)

\( 1 \times 20 = 20 \)

Total known = \( 15 + 20 = 35 \).

Remaining amount must be in £10 notes.

💡 Concept: A fair coin has a probability of \( 0.5 \) (or \( \frac{1}{2} \)) for heads.

\( \text{Expected Number} = \text{Probability} \times \text{Total Trials} \)

💡 Principle: In probability experiments, performing more trials usually leads to more reliable (accurate) results.

Simon performed \( 100 \) trials, while Paula only performed \( 10 \).

💡 Concept: The “plan” is the view from directly above.

• Looking down, we see the square base of the pyramid (which matches the top of the cube).
• This is a square of side \( 6\text{ cm} \).
• We also see the edges of the pyramid meeting at the peak in the centre.
• This looks like a square with diagonals crossing in the middle (an ‘X’).

💡 Rule: When multiplying terms with the same base, add the powers.

\( a^x \times a^y = a^{x+y} \)

💡 Rule: When dividing terms with the same base, subtract the powers.

Note: \( d \) on its own is \( d^1 \).

💡 Strategy: Solve exactly like an equation, keeping the \( > \) sign.

💡 Formula: \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \)

💡 Strategy: The number is rounded to 1 decimal place (0.1).

The bounds are half a unit above and below.

Unit = \( 0.1 \). Half unit = \( 0.05 \).

💡 Strategy: Calculate the area of both lawns. Find the seed needed per square metre.

💡 Strategy: Recalculate the total seed she actually has.

💡 Strategy: Determine individual probabilities.

Spinner A (1, 2, 3):

• P(lands on 2) = \( \frac{1}{3} \)
• P(does not land on 2) = \( \frac{2}{3} \)

Spinner B (1, 2, 3): Same probabilities.

• P(lands on 2) = \( \frac{1}{3} \)
• P(does not land on 2) = \( \frac{2}{3} \)

💡 Rule: Multiply along the branches for “AND”.

Path: A lands on 2 AND B does not land on 2.

💡 Strategy: Find the point where the two lines intersect (cross).

Read the coordinates \( (x, y) \).

💡 Strategy: We are given the graph of \( y = x^2 – 4x + 2 \).

We need to solve \( x^2 – 4x + 2 = 0 \).

This means we find where \( y = 0 \), which is where the curve crosses the x-axis.

💡 Formula: \( \text{Volume of prism} = \text{Area of cross-section} \times \text{Length} \)

Cross-section is a triangle: \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \)

💡 Formula: \( \text{Mass} = \text{Density} \times \text{Volume} \)