Pearson Edexcel GCSE Summer 2022 Foundation Paper 3

💡 Why we do this: The symbol \( \% \) means “per cent” or “out of 100”. To convert a percentage to a fraction, we just write the number over 100.

💡 What’s next: Both numbers end in 0 or 5, so they can be divided by 5.

🖩 Calculator Method: Type 35, press the fraction key a b/c or ◻/◻, type 100, then press =.

💡 Strategy: Finding “one quarter of” a number is the same as dividing that number by 4.

The word “of” in maths usually means multiply: \( \frac{1}{4} \times 28 \).

💡 Definition: Factors are whole numbers that divide into a number exactly (without leaving a remainder).

We need to find numbers that 12 can be divided by.

💡 Strategy: When multiplying algebraic terms, we multiply the numbers (coefficients) together and keep the letters (variables).

💡 Tool: This is a square root symbol. It asks: “What number multiplied by itself gives 1.69?”

🖩 Calculator Steps:

1. Press the square root button: \( \sqrt{\square} \)
2. Type: 1.69
3. Press: =

We need a four-sided shape (quadrilateral) with two specific properties:

• No lines of symmetry: If you fold it in half, the two halves won’t match.
• Rotational symmetry of order 2: If you spin it 360°, it looks the same in 2 positions (start and halfway/180°).

Which quadrilaterals have rotational symmetry of order 2?

• Rectangle (has lines of symmetry ❌)
• Rhombus (has lines of symmetry ❌)
• Parallelogram (has NO lines of symmetry ✅, order 2 ✅)

So, we must draw a parallelogram (that isn’t a rectangle or rhombus).

💡 Strategy: Add up the number of apples for all three weeks.

💡 Strategy: Add up the number of oranges for all three weeks.

💡 Strategy: Subtract the total oranges from the total apples to find “how many more”.

💡 Finding the rule: Look at the gap between the numbers.

\( 8 – 3 = 5 \)
\( 13 – 8 = 5 \)

The rule is add 5 each time.

💡 Strategy: Look at the properties of the numbers in the sequence (3, 8, 13, 18, 23, 28, 33…).

Notice a pattern in the last digit (units digit).

💡 Definition: A face is a flat surface of the 3D shape.

• 2 triangular faces (front and back)
• 3 rectangular faces (bottom and two sides)

💡 Definition: An edge is a line where two faces meet.

• 3 edges on the front triangle
• 3 edges on the back triangle
• 3 edges connecting the front to the back

💡 Check list: Look at the list: 2, 2, 3, 5, 6, 6, 8, 9.

Is the number 7 in the list? No.

This means the probability is 0 (impossible).

💡 Count: Count numbers strictly greater than 5.

List: 2, 2, 3, 5, 6, 6, 8, 9

There are 4 numbers greater than 5.

Total numbers = 8.

Probability = \( \frac{4}{8} = \frac{1}{2} \)

💡 Identify even numbers: Numbers divisible by 2.

List: 2, 2, 3, 5, 6, 6, 8, 9

The even numbers are: 2, 2, 6, 6, 8.

💡 Goal: Find out how many nails she needs in total.

She needs 4 nails for each of the 35 frames.

💡 Goal: Find out how many nails she actually has.

She has 3 boxes, with 48 nails in each.

💡 Comparison: Is the number she has (144) greater than or equal to the number she needs (140)?

💡 Meaning: “A as a fraction of B” means \( \frac{A}{B} \).

💡 Method: Divide top and bottom by the same number until you can’t go any further.

💡 Instructions: Use a ruler to measure the lines on the original exam paper.

Note: Values depend on exact printing size, but standard exam values are usually:

• Length AB ≈ 5 cm
• Length BC ≈ 4 cm
• Length AC ≈ 7 cm

💡 Scale: 1 cm represents 150 metres.

💡 Definition: A bearing is an angle measured clockwise from North.

“Bearing of A from C” means start at C, face North, and turn clockwise until you face A.

💡 Definition: The median is the middle number when the data is ordered.

There are 12 numbers (even amount), so the median is halfway between the 6th and 7th numbers.

💡 Definition: Range = Biggest value – Smallest value.

💡 Strategy: To compare distributions, you must mention:

1. An average (median or mean) to compare the “typical” value.
2. A measure of spread (range) to compare consistency/variety.

🖩 Calculator: Type 2.75 × 14.6 =

🖩 Calculator: Type 10 - 1.97 =

💡 Formula: \( \text{Area} = \frac{\text{base} \times \text{height}}{2} \)

We need the area to be 12.

\[ 12 = \frac{\text{base} \times \text{height}}{2} \]

Possible pairs for 24:

• \( 4 \times 6 \)
• \( 6 \times 4 \)
• \( 8 \times 3 \)

Since it must be an isosceles triangle (two sides equal), it’s easiest to pick an even number for the base, so we can put the peak exactly in the middle.

Let’s use Base = 6 cm and Height = 4 cm.

💡 Method: Multiply the number outside the bracket by each term inside.

💡 Method: Rearrange to get \( y \) on its own.

Equation: \( \frac{3y}{4} = 12 \)

💡 Method: Put the expression back into brackets by finding the highest common factor (HCF).

Expression: \( 4p + 6 \)

💡 Method: Count the first two non-zero digits from the left.

Number: 2 5 3 0

The 2nd sig fig is 5. The next digit is 3.

Since 3 is less than 5, we round down (keep the 5 as it is).

Replace remaining digits with zeros to keep the value correct.

💡 Method: Zeros at the start don’t count. Find the first non-zero digit.

Number: 0.0 8 7 4

The 1st sig fig is 8. The next digit is 7.

Since 7 is 5 or more, we round up (change 8 to 9).

💡 Method: Find \( \frac{3}{8} \) of 400.

\( (400 \div 8) \times 3 \)

💡 Method: Subtract red and yellow from the total.

Red = 150

Yellow = 82

Total = 400

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

💡 Reason: Vertically opposite angles are equal.

Angle \( APB \) and Angle \( QPR \) are vertically opposite (where two straight lines cross).

💡 Reason: Base angles of an isosceles triangle are equal.

Triangle \( PQR \) is isosceles (\( PQ = PR \)), so \( \angle PQR = \angle PRQ \).

Angles in a triangle add to 180°.

💡 Reason: Alternate angles are equal (Z-angles) OR Allied angles (C-angles).

Method 1 (Alternate): Angle \( CQP \) and Angle \( QPR \) are alternate angles because lines \( APR \) and \( CQD \) are parallel. But wait, \( APR \) is a line, not just \( AP \). Let’s check the lines again.

Actually, let’s use Allied (Co-interior) angles or Corresponding angles depending on the parallel lines.

Lines \( APR \) and \( CQD \) are parallel. Line \( PQR \) is a transversal? No, \( PQ \) is part of the line \( BPQ \).

Let’s look at the parallel lines \( APR \) and \( CQD \).

Line \( PR \) acts as a transversal connecting parallel lines \( AR \) and \( CD \)? No.

Let’s use Alternate angles with parallel lines \( AR \) and \( CD \) and transversal \( PR \). This gives \( \angle PRQ = \angle DQR \)? No.

Let’s look at the diagram simpler:

• \( \angle CQP \) and \( \angle APQ \) are Allied (add to 180)? No, \( P \) and \( Q \) are on the transversal.
• Actually, \( \angle CQP \) corresponds to \( \angle APB \)? No.
• \( \angle CQP \) is alternate to \( \angle QPA \)? Yes!

\( \angle APQ \) is on a straight line with \( 56^\circ \)? No, \( BPQ \) is straight. \( A \) is on a parallel line.

Wait, let’s look at the parallel line relation:

Parallel lines: \( APR \) and \( CQD \).

Transversal: \( BPQ \).

Therefore, \( \angle CQP \) = \( \angle APB \) (Corresponding angles)? No, \( A \) and \( C \) are on the same side?

Let’s use Corresponding angles: \( \angle DQP = \angle RPV \)? No.

Correct Parallel Line Rule:

\( \angle CQP \) and \( \angle APQ \) are Alternate? No.

Let’s look at \( \angle CQP \) and \( \angle QPR \).

Actually, \( \angle CQP \) corresponds to \( \angle APB \) ?? No.

Let’s use Alternate Angles (Z-angles):

The “Z” shape is formed by \( A-P-Q-D \). So \( \angle APQ = \angle PQD \).

Is \( \angle APQ \) known? Angles on a straight line at P? We know \( \angle APB = 56 \). \( \angle BPQ \) is a line. So \( \angle APQ = 180 – 56 = 124 \).

So \( \angle PQD = 124 \).

Then \( \angle CQP \) is vertically opposite \( \angle PQD \)? No.

Alternative Strategy:

\( \angle CQP \) corresponds to \( \angle APB \) IF \( AP \) and \( CQ \) are parallel.

Lines are \( APR \) and \( CQD \). So \( AP \) is parallel to \( CQ \).

Transversal is \( BPQ \).

\( \angle CQP \) and \( \angle APB \) are Corresponding Angles (F-angles). Wait, they are on the same side of the transversal and in corresponding positions.

Wait, if \( AP \) is parallel to \( CQ \), then \( \angle CQP = \angle APQ \) (Alternate)? No.

Actually, \( \angle APB \) (56) and \( \angle CQP \) are Corresponding Angles.

So \( \angle CQP = 56^\circ \).

💡 Method: Write out the times tables for both numbers until you find the first match.

💡 Pythagoras’ Theorem: \( a^2 + b^2 = c^2 \)

Where \( c \) is the longest side (hypotenuse).

Here, the hypotenuse is 8.5. The other sides are 4 and \( x \).

When finding a shorter side, we subtract the square of the other side from the square of the hypotenuse.

💡 Caution: When squaring a negative number, use brackets! \( (-3)^2 \) is positive 9.

💡 Goal: Get \( p \) on its own.

Equation: \( d = 3p + 4 \)

💡 Strategy: Use algebra. Let \( x \) be the number of counters Rick has.

The total is 54.

Ratio Rick : Tony = \( 12 : 18 \)

We need it in the form \( 1 : p \).

💡 Info: 3 rolls cost £36. We need 15 rolls.

How many sets of 3 are in 15?

💡 Info: Pack of 5 rolls costs £70. We need 15 rolls.

How many packs of 5?

💡 Rule 1: Points must be plotted at the midpoint of the interval (e.g., for \( 0 < t \leq 10 \), plot at \( x = 5 \)).

💡 Rule 2: The scale on the axes must be linear (go up in equal steps).

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

⚠️ Units: Speed is in miles per hour, so time must be in hours.

15 minutes = \( \frac{15}{60} \) hours = 0.25 hours.

Time = 40 minutes = \( \frac{40}{60} \) hours = \( \frac{2}{3} \) hours.

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

Distance = 7.5 miles.

Time = 45 minutes = \( \frac{45}{60} \) hours = 0.75 hours.

A trapezium has two parallel sides. Here, the vertical sides \( QU \) (not vertical?) No, \( STUV \) is a rectangle, so \( TU \) is parallel to \( SV \).

The parallel sides are the top and bottom edges.

Height of trapezium: The distance between parallel lines is the height of the rectangle = \( 4x \).

Parallel Side 1 (Base RV): Length = 5.

Parallel Side 2 (Top QU): We need to find this length.

💡 Formula: \( \text{Area} = \frac{a + b}{2} \times h \)

Where \( a \) and \( b \) are parallel sides, and \( h \) is height.

\( a = 5 \) (side RV)

\( b = x + 5 \) (side QU)

\( h = 4x \) (side UV)

💡 Strategy: There are 60 seconds in a minute, and 60 minutes in an hour. So there are \( 60 \times 60 = 3600 \) seconds in an hour.

If you travel 30 metres in 1 second, you travel \( 30 \times 3600 \) metres in 1 hour.

💡 Strategy: There are 1000 metres in 1 kilometre. Divide by 1000.

💡 Concept: The original price represents 100%.

If it decreased by 15%, the new price is:

\[ 100\% – 15\% = 85\% \]

So, £13,600 is equal to 85% of the original price.

💡 Method: Divide the current price by the percentage multiplier (0.85).