GCSE 2019 Edexcel Foundation Paper 3

💡 What are we asked to do?

We need to round \( 478 \) to the nearest hundred. This means we are checking if \( 478 \) is closer to \( 400 \) or \( 500 \).

💡 What is a multiple?

A multiple of \( 8 \) is a number in the \( 8 \) times table.

💡 What range do we need?

We need a number between \( 41 \) and \( 60 \).

💡 How many metres in a kilometre?

There are \( 1000 \) metres in \( 1 \) kilometre.

To change from km to m, we multiply by \( 1000 \).

💡 What is a power of 3?

A power of \( 3 \) is a number obtained by multiplying \( 3 \) by itself a certain number of times.

💡 Which numbers from the list match?

We check our list of powers against the numbers given in the question.

💡 What does “percent” mean?

“Percent” means “out of \( 100 \)”.

💡 Strategy:

A common way to find percentages without a calculator (though you can use one here) is to find \( 10\% \) first. To find \( 10\% \), we divide by \( 10 \).

💡 How do we get from 10% to 20%?

\( 20\% \) is double \( 10\% \). So we multiply our previous answer by \( 2 \).

Alternative Calculator Method:

Convert \( 20\% \) to a decimal (\( 0.2 \)) and multiply.

💡 Which columns represent green counters?

We need to look for “green circle” and “green square”.

💡 Which columns represent red counters?

We need “red circle” and “red square”.

💡 What does “how many more” mean?

It means we need to subtract the smaller number from the larger number.

💡 What do we know?

The total capacity is \( 52 \) litres. The tank currently has \( \frac{1}{4} \) of this amount.

💡 How much more is needed?

We need to fill it up to \( 52 \) litres. We subtract what we have from the total.

💡 Thinking about fractions

If the tank is \( \frac{1}{4} \) full, then it is \( \frac{3}{4} \) empty.

We need to find \( \frac{3}{4} \) of \( 52 \).

💡 What are like terms?

Like terms are terms that have the same variable (letter). We can collect the ‘e’ terms and the ‘f’ terms separately.

Remember that the sign before the term belongs to that term.

💡 Combine the coefficients

Add the numbers in front of the letters.

💡 How many did he give in total?

Add up the amounts given to Sameena, Henry, and Lucas.

💡 How many remain?

Subtract the total given away from the starting amount.

💡 Format: Part over Whole

The question asks for the fraction left out of the total.

💡 Reduce to simplest form

Divide top and bottom by common factors until you can’t go further.

💡 Identify the parts

The ratio is \( 6 : 9 : 10 \).

• Ali gets \( 6 \) parts.
• Ben gets \( 9 \) parts.
• Cathy gets \( 10 \) parts.

💡 The “Whole”

To make a fraction, the denominator (bottom number) is the total number of parts.

💡 Part over Whole

Ben gets \( 9 \) parts out of \( 25 \) total parts.

💡 Order of Operations (BIDMAS)

The fraction bar acts like brackets. We must calculate the top and bottom before dividing.

💡 Use Calculator

Divide the top result by the bottom result.

💡 Analyze the shape

The shape is made of two identical rectangles. Each rectangle is \( 6 \text{ cm} \) by \( 4 \text{ cm} \).

We need to find the length of every outer side to calculate the perimeter.

💡 Calculating the hidden lengths

Let’s trace the perimeter clockwise from the top-left corner.

💡 Measure the diagram

Since we are on a screen, we must rely on relative sizes.

Visually compare the width of the box to the height of the mast.

The width of the box represents \( 1.5 \) m.

💡 Multiply

Multiply the real width of the box by the scale factor we found.

💡 Move the constant

We want \( x \) on its own. First, move the \( +4 \) to the other side by subtracting \( 4 \).

💡 Divide by the coefficient

Currently \( x \) is multiplied by \( 2 \). To undo this, divide the whole equation by \( 2 \).

💡 Properties of a Square

\( ABDE \) is a square. The diagonal \( EB \) cuts the corner angle (\( 90^\circ \)) in half.

💡 Properties of an Equilateral Triangle

\( BCD \) is an equilateral triangle. All angles in an equilateral triangle are \( 60^\circ \).

💡 Add the parts

Angle \( EBC \) is the sum of angle \( EBD \) and angle \( DBC \).

💡 Multiply by the exchange rate

To change pounds into foreign currency, we multiply by the rate.

💡 Divide by note value

We need to see how many times \( 200 \) goes into the total rand amount.

💡 Can you have part of a note?

No, you can only get whole notes. We must round down (truncate) because she doesn’t have enough for the 79th note.

💡 Divide total miles by miles per gallon

We know how far she went (\( 560 \)) and how far one gallon takes her (\( 34.5 \)).

💡 Multiply by conversion factor

There are \( 4.55 \) litres in every gallon.

💡 Multiply litres by price per litre

Each litre costs \( £1.08 \).

💡 Two decimal places

Money is always rounded to 2 decimal places.

💡 Find the percentage of medium first

We know \( 32\% \) are small and \( 40\% \) are large. The rest must be medium.

Total percentage is \( 100\% \).

💡 Find the fraction of medium first

We know \( \frac{3}{10} \) are small and \( \frac{1}{10} \) are large.

Total fraction is \( 1 \) (or \( \frac{10}{10} \)).

💡 Use the ratio

Ratio of Small : Medium = \( 3 : 4 \)

We are told Charlie has \( 150 \) small packets. This represents the ‘3’ parts.

💡 Sum the results

Add Alan’s, Beryl’s, and Charlie’s medium packets together.

💡 Convert percentage increase to decimal

An increase of \( 1.5\% \) means we have \( 100\% + 1.5\% = 101.5\% \).

As a decimal, this is \( 1.015 \).

💡 Formula: Original × Multiplier^Years

💡 Subtract the original investment

The question asks for the interest, not the total amount.

💡 Check the scale

Look at the numbers going up: \( 6, 6.5, 7, 7.5, 8, 8.5, 9, 10 \).

The gap between \( 8.5 \) and \( 9 \) is the same visual size as \( 9 \) to \( 10 \), but represents \( 0.5 \) vs \( 1.0 \).

Also, the graph does not start at \( 0 \) (truncation), which can exaggerate changes, although it’s labelled \( 6 \).

💡 Check the labels

The labels are \( 2, 3, 4 \) for each year. What happened to quarter \( 1 \)?

The points are plotted, but connected with a curve. Time series points are usually connected with straight lines.

💡 Angles match on both sides

Since there is a line of symmetry:

• Angle \( FAB = \) Angle \( CBA = 117^\circ \)
• Angle \( CDE = \) Angle \( DEF \)
• Angle \( BCD = \) Angle \( AFE \)

💡 Formula: (n-2) × 180

A hexagon has \( 6 \) sides (\( n=6 \)).

💡 Define variables

Let Angle \( CDE = x \). Then Angle \( DEF = x \).

We are told Angle \( BCD = 2 \times \) Angle \( CDE \), so Angle \( BCD = 2x \).

By symmetry, Angle \( AFE = 2x \).

💡 Rearrange and solve

💡 Substitute back

We need Angle \( AFE \), which is \( 2x \).

💡 Formula for Cylinder Surface Area

Surface Area = \( 2 \pi r^2 \) (top and bottom circles) + \( 2 \pi r h \) (curved side).

Diameter = \( 1.6 \text{ m} \), so Radius (\( r \)) = \( 0.8 \text{ m} \).

Height (\( h \)) = \( 1.8 \text{ m} \).

💡 Multiply by 3

💡 Total coverage available

Jeremy has \( 7 \) tins, each covering \( 5 \text{ m}^2 \).

💡 Is coverage > area needed?

💡 Strategy: Match coefficients

Equation 1: \( 3x – 4y = 11 \)

Equation 2: \( 9x + 2y = 5 \)

Let’s make the \( y \) coefficients match (ignoring signs). Multiply Eq 2 by \( 2 \).

💡 Divide

💡 Substitute x back

Use Equation 2 (it looks simpler).