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Edexcel GCSE Mathematics: Foundation Paper 1
Non-Calculator | May 2020 | Specification 1MA1
Mark Scheme Legend
- M1: Method mark awarded for a correct method or partial method
- P1: Process mark awarded for a correct process as part of a problem-solving question
- A1: Accuracy mark (awarded after a correct method or process)
- B1: Unconditional accuracy mark (no method needed)
- C1: Communication mark for fully correct statement(s)
Table of Contents
- Question 1 (Ordering Decimals)
- Question 2 (Multiples)
- Question 3 (Rounding)
- Question 4 (Fractions to Decimals)
- Question 5 (Place Value)
- Question 6 (Probability Scales)
- Question 7 (Pictograms)
- Question 8 (Coordinates)
- Question 9 (Ratios & Fractions)
- Question 10 (Weights Problem)
- Question 11 (Reflections)
- Question 12 (Number Machines)
- Question 13 (Bearings & Scale)
- Question 14 (Two-way Tables)
- Question 15 (Mixing Ratios)
- Question 16 (Median Analysis)
- Question 17 (Percentage Sale)
- Question 18 (Estimation)
- Question 19 (Algebraic Operations)
- Question 20 (nth term)
- Question 21 (Fractions Multiplication)
- Question 22 (Matching Graphs)
- Question 23 (Congruent Triangles)
- Question 24 (Percentage Profit)
- Question 25 (Parallel Line Angles)
- Question 26 (Distributions)
- Question 27 (Pressure & Area)
- Question 28 (Standard Form)
- Question 29 (Ratio Problems)
- Question 30 (Formulae & Indices)
Question 1 (1 mark)
Write the following numbers in order of size. Start with the smallest number.
\( 0.32 \quad 0.4 \quad 0.35 \quad 0.309 \)
Worked Solution
Step 1: Compare place values
Why we do this: To compare decimals easily, it helps to write them all with the same number of decimal places.
✏ Working:
Add placeholder zeros so all have three decimal places:
- \( 0.320 \)
- \( 0.400 \)
- \( 0.350 \)
- \( 0.309 \)
✓ (B1) for correct ordering
Final Answer:
\( 0.309, \quad 0.32, \quad 0.35, \quad 0.4 \)
Question 2 (1 mark)
Here is a list of numbers.
\( 5 \quad 11 \quad 18 \quad 22 \quad 29 \)
From the list, write down a multiple of 3.
Worked Solution
Step 1: Identify Multiples of 3
Why we do this: A multiple of 3 is a number in the 3 times table.
✏ Working:
Check each number:
- \( 5 \div 3 = 1 \text{ r } 2 \) (No)
- \( 11 \div 3 = 3 \text{ r } 2 \) (No)
- \( 18 \div 3 = 6 \) (Yes!)
- \( 22 \div 3 = 7 \text{ r } 1 \) (No)
- \( 29 \div 3 = 9 \text{ r } 2 \) (No)
✓ (B1) cao
Final Answer:
\( 18 \)
Question 3 (1 mark)
Write \( 4.666 \) correct to the nearest whole number.
Worked Solution
Step 1: Rounding Rule
Why we do this: To round to the nearest whole number, we look at the first decimal place (the tenths).
✏ Working:
The digit in the tenths place is 6.
Since 6 is “5 or more”, we round the whole number up.
\[ 4 + 1 = 5 \]
✓ (B1) cao
Final Answer:
\( 5 \)
Question 4 (1 mark)
Write \( \frac{3}{4} \) as a decimal.
Worked Solution
Step 1: Conversion Strategy
Why we do this: We can convert a fraction to a decimal by making the denominator 100, or by dividing the numerator by the denominator.
✏ Working:
Method 1: Equivalent fraction
\[ \frac{3}{4} = \frac{3 \times 25}{4 \times 25} = \frac{75}{100} = 0.75 \]
✓ (B1) cao
Final Answer:
\( 0.75 \)
Question 5 (1 mark)
Write down the value of the 7 in the number \( 8765 \)
Worked Solution
Step 1: Place Value Analysis
Why we do this: We break the number down into its constituent parts (Thousands, Hundreds, Tens, Units).
✏ Working:
- 8 is in the Thousands place (\( 8000 \))
- 7 is in the Hundreds place (\( 700 \))
- 6 is in the Tens place (\( 60 \))
- 5 is in the Units place (\( 5 \))
✓ (B1) for 700 or “seven hundred”
Final Answer:
\( 700 \)
Question 6 (2 marks)
Gita spins a fair 8-sided spinner.
(a) On the probability scale, mark with a cross (X) the probability that the spinner will land on C.
(b) On the probability scale, mark with a cross (X) the probability that the spinner will land on D.
Worked Solution
Step 1: Calculate Probability of C
Why we do this: To find probability, we count successful outcomes and divide by the total number of equally likely outcomes.
✏ Working:
Total sections = 8.
Sections labelled ‘C’ = 4.
\[ P(C) = \frac{4}{8} = \frac{1}{2} \]
✓ (B1) for cross at 1/2
Step 2: Calculate Probability of D
Why we do this: We look for outcomes that match the criteria ‘D’.
✏ Working:
There are NO sections labelled ‘D’.
This is an impossible event.
\[ P(D) = 0 \]
✓ (B1) for cross at 0
Final Answer:
(a) Cross at 1/2. (b) Cross at 0.
Question 7 (4 marks)
The incomplete pictogram shows information about the number of eggs sold from a farm shop on Monday.
On Monday the shop sold 18 eggs.
On Tuesday the shop sold 24 eggs.
On Wednesday the shop sold 27 eggs.
Use this information to complete the pictogram and the key.
Worked Solution
Step 1: Determine the Key
Why we do this: We use the known value for Monday to find out what one circle represents.
✏ Working:
Monday symbols = 1 full circle + 1 half circle = 1.5 circles.
\[ 1.5 \text{ circles} = 18 \text{ eggs} \]
\[ 1 \text{ circle} = 18 \div 1.5 = 12 \text{ eggs} \]
✓ (C1) Deduces 1 ellipse = 12 eggs
Step 2: Calculate Symbols for Tuesday and Wednesday
Why we do this: We divide the number of eggs by the value of one circle.
✏ Working:
Tuesday: \( 24 \div 12 = 2 \) full circles.
Wednesday: \( 27 \div 12 = 2 \text{ remainder } 3 \). Since \( 3 = \frac{1}{4} \text{ of } 12 \), we need 2.25 circles.
✓ (C1) 2 circles for Tuesday
✓ (C1) 2.25 circles for Wednesday
✓ (C1) Key shown correctly
Final Answer:
Key: Circle = 12 eggs. Tuesday = 2 circles. Wednesday = 2 and a quarter circles.
Question 8 (3 marks)
(a) Write down the coordinates of point A.
(b) Write down the coordinates of point B.
(c) On the grid, mark with a cross (X) the point (-2, 1). Label this point C.
Worked Solution
Step 1: Reading Coordinates
Why we do this: Coordinates are always given as (x, y). First move horizontally (along the x-axis), then vertically (up or down).
✏ Working:
Point A: 2 units right, 3 units up. \((2, 3)\)
Point B: 0 units horizontally, 1 unit down. \((0, -1)\)
✓ (B1) cao for A
✓ (B1) cao for B
Step 2: Plotting Point C
Why we do this: Move to -2 on the x-axis, then up 1 on the y-axis.
✏ Working:
Start at Origin (0,0). Move left 2. Move up 1.
✓ (B1) cao for plotting C
Final Answer:
(a) (2, 3) (b) (0, -1) (c) Point marked at (-2, 1).
Question 9 (3 marks)
(a) A bag contains red counters and blue counters only.
number of red counters : number of blue counters = \( 3 : 4 \)
Write down the fraction of the counters that are red.
(b) Write the ratio \( 12 : 30 \) in the form \( 1 : n \)
Worked Solution
Step 1: Part (a) – Ratios to Fractions
Why we do this: The total number of parts in the ratio represents the denominator of the fraction.
✏ Working:
Total parts = \( 3 + 4 = 7 \)
Red parts = 3
\[ \text{Fraction red} = \frac{3}{7} \]
✓ (B1) oe
Step 2: Part (b) – Simplifying to 1:n
Why we do this: To get the first term to 1, we must divide both terms by the first term (12).
✏ Working:
\[ 12 \div 12 : 30 \div 12 \]
\[ 1 : 2.5 \]
✓ (M1) for \( 30 \div 12 \)
✓ (A1) cao
Final Answer:
(a) \( \frac{3}{7} \) (b) \( 1 : 2.5 \)
Question 10 (4 marks)
Jenny has 12 marbles.
\( \frac{1}{4} \) of these 12 marbles are large. The rest of these 12 marbles are small.
Each large marble has a weight of 70 grams. Each small marble has a weight of 50 grams.
Work out the total weight of the 12 marbles.
Worked Solution
Step 1: Find the number of each type of marble
Why we do this: We need to know exactly how many large and small marbles Jenny has before calculating total weight.
✏ Working:
Large marbles = \( \frac{1}{4} \times 12 = 3 \)
Small marbles = \( 12 – 3 = 9 \)
✓ (P1) for finding marble counts
Step 2: Calculate weights
Why we do this: Multiply the quantity by the unit weight for each group.
✏ Working:
Weight of large = \( 3 \times 70 = 210 \text{ g} \)
Weight of small = \( 9 \times 50 = 450 \text{ g} \)
✓ (P1) for partial weight calculations
Step 3: Total Weight
Why we do this: Sum the weights of the two groups.
✏ Working:
Total = \( 210 + 450 = 660 \text{ g} \)
✓ (P1) for complete process
✓ (A1) cao
Final Answer:
\( 660 \text{ grams} \)
Question 11 (2 marks)
Reflect the shaded shape in the mirror line.
Worked Solution
Step 1: Point-by-point reflection
Why we do this: Each vertex of the reflected shape must be the same distance from the mirror line as the original, but on the opposite side.
✏ Working:
- Point at (240, 120) is 1 unit right of line -> Reflected to 1 unit left.
- Point at (280, 40) is 2 units right of line -> Reflected to 2 units left.
- Point at (280, 160) is 2 units right of line -> Reflected to 2 units left.
- Point at (260, 160) is 1.5 units right of line -> Reflected to 1.5 units left.
✓ (M1) for correct reflection of at least 3 vertices
✓ (A1) cao
Final Answer:
Shape drawn mirroring the original across the dashed line.
Question 12 (3 marks)
The diagram shows a number machine.
(a) Find the output when the input is 7.
(b) Find the input when the output is 41.
Worked Solution
Step 1: Part (a) – Forward direction
Why we do this: Simply follow the operations in the order they appear.
✏ Working:
\[ 7 \times 2 = 14 \]
\[ 14 – 3 = 11 \]
✓ (B1) cao
Step 2: Part (b) – Reverse direction (Inverse operations)
Why we do this: To find the input, we work backward from the output using the opposite (inverse) of each operation.
✏ Working:
Output is 41.
Inverse of -3 is +3: \[ 41 + 3 = 44 \]
Inverse of ×2 is ÷2: \[ 44 \div 2 = 22 \]
✓ (M1) for 41+3 or sight of inverse ops
✓ (A1) cao
Final Answer:
(a) 11 (b) 22
Question 13 (4 marks)
The diagram shows two points, A and B, on a map. Scale: 1 to 25 000
(a) Find the bearing of B from A.
(b) Work out the real distance between A and B. Give your answer in kilometres.
Worked Solution
Step 1: Part (a) – Measuring the Bearing
Why we do this: Bearings are measured from North, in a clockwise direction, using three digits.
✏ Working:
Using a protractor on the original diagram at A, measure the angle from North clockwise to line AB.
The angle is 25°.
As a 3-digit bearing: 025°
✓ (B1) for angle in range 23 to 27
Step 2: Part (b) – Distance and Scale
Why we do this: Real distance = map distance × scale factor.
✏ Working:
1. Measure distance AB on the paper. Distance ≈ 5 cm.
2. Multiply by scale: \( 5 \times 25,000 = 125,000 \text{ cm} \).
3. Convert to kilometres:
\[ 125,000 \div 100 = 1,250 \text{ metres} \]
\[ 1,250 \div 1,000 = 1.25 \text{ km} \]
✓ (M1) for measurement 4.8 to 5.2 cm
✓ (M1) for “5” × 25,000
✓ (A1) for answer 1.2 to 1.3 km
Final Answer:
(a) 025° (b) 1.25 km
Question 14 (3 marks)
Ishmael asked 30 students at college to tell him the sport they each like the best from cricket or tennis or swimming.
- 11 of the 20 female students said swimming.
- 2 of the male students said tennis.
- 5 students said cricket.
- The number of male students who said cricket was the same as the number of male students who said swimming.
Complete the two-way table.
Worked Solution
Step 1: Fill in direct values
Why we do this: We start with the facts explicitly given in the text.
✏ Working:
- Total Males = \( 30 – 20 = 10 \).
- Female Swimming = 11.
- Male Tennis = 2.
- Total Cricket = 5.
Step 2: Solve the logic constraints
Why we do this: We use the rule about male cricket and swimming being equal.
✏ Working:
Let male cricket = \( x \). Then male swimming = \( x \).
Male total is 10: \[ x + x + 2 = 10 \]
\[ 2x = 8 \Rightarrow x = 4 \]
So Male Cricket = 4 and Male Swimming = 4.
✓ (M1) for correctly entering two of 11, 2, 5, 10
Step 3: Complete the table
✏ Working:
- Female Cricket = Total Cricket (5) – Male Cricket (4) = 1.
- Female Tennis = Female Total (20) – Cricket (1) – Swimming (11) = 8.
- Total Swimming = 4 + 11 = 15.
- Total Tennis = 2 + 8 = 10.
✓ (M1) for using rules correctly
✓ (A1) for complete correct table
Final Answer:
Males: 4, 2, 4 (Total 10). Females: 1, 8, 11 (Total 20). Totals: 5, 10, 15 (Total 30).
Question 15 (3 marks)
Jamil makes a drink by mixing 1 part of orange squash with 9 parts of water.
He uses 750 millilitres of orange squash.
Jamil is going to put the drink he has mixed into 1 litre bottles.
Work out the greatest number of 1 litre bottles that Jamil can completely fill.
Worked Solution
Step 1: Find total volume of drink
Why we do this: We need the total mixture amount. 1 part is 750ml, and there are 10 parts in total (1 squash + 9 water).
✏ Working:
Total parts = \( 1 + 9 = 10 \)
\[ 10 \times 750 = 7500 \text{ ml} \]
✓ (P1) for \( 750 \times 9 = 6750 \) or finding total parts
Step 2: Convert to litres
Why we do this: The bottles are measured in litres, so we need consistent units. (1 litre = 1000 ml).
✏ Working:
\[ 7500 \div 1000 = 7.5 \text{ litres} \]
✓ (P1) for dividing by 1000 or converting units
Step 3: Determine number of full bottles
Why we do this: We only want “completely filled” bottles, so we round down to the nearest whole number.
✏ Working:
He has 7.5 litres. He can fill 7 full bottles and will have half a bottle left over.
✓ (A1) cao
Final Answer:
\( 7 \)
Question 16 (2 marks)
The table gives information about the number of points scored by each of 16 students in a game.
Points | Frequency
0 | 1
1 | 3
2 | 5
3 | 4
4 | 3
Tina worked out the median of the number of points scored to be 5.
(a) Explain why it is not possible for the median to be 5.
Tina also worked out the total number of points scored by the 16 students. Here is her working: \[ (0 \times 1) + (1 \times 3) + (2 \times 5) + (3 \times 4) + (4 \times 3) = 1 + 3 + 10 + 12 + 12 = 38 \]
(b) Describe the mistake that Tina made.
Worked Solution
Step 1: Part (a) – Median definition
Why we do this: The median must be a value from the data set. The highest number of points anyone scored was 4.
✏ Working:
The maximum score in the list is 4. The median must fall within the range of scores given (0 to 4).
✓ (C1) for explanation e.g. “max score is 4” or “median is 2”
Step 2: Part (b) – Identifying arithmetic error
Why we do this: Look closely at the first term of her sum.
✏ Working:
Tina calculated \( 0 \times 1 = 1 \).
In reality, \( 0 \times 1 = 0 \).
✓ (C1) for identifying \( 0 \times 1 = 0 \) or similar
Final Answer:
(a) The scores only go up to 4. (b) She thought \( 0 \times 1 = 1 \), but it is 0.
Question 17 (5 marks)
In a shop, a TV has a normal price of £500. The shop has a sale.
On Monday, the normal price of the TV is reduced by \( \frac{1}{10} \) to give the sale price.
On Tuesday, the sale price of the TV is reduced by 20%.
Chris wants to buy the TV. He has £400 to spend. Does Chris have enough money? You must show your working.
Worked Solution
Step 1: Calculate Monday’s price
Why we do this: First, find the first reduction.
✏ Working:
Reduction = \( \frac{1}{10} \text{ of } 500 = 50 \).
Monday price = \( 500 – 50 = 450 \).
✓ (P1) for process to find 1/10 of 500
✓ (P1) for Monday price £450
Step 2: Calculate Tuesday’s price
Why we do this: Apply the second reduction (20%) to the new price, not the original.
✏ Working:
10% of 450 = 45.
20% of 450 = \( 45 \times 2 = 90 \).
Tuesday price = \( 450 – 90 = 360 \).
✓ (P1) for process to find 20% of 450
✓ (P1) for Tuesday price £360
Step 3: Conclusion
✏ Working:
Chris has £400. The TV costs £360.
\[ 400 > 360 \]
✓ (C1) for Yes + supported figures
Final Answer:
Yes, Chris has enough money (TV costs £360).
Question 18 (3 marks)
Work out an estimate for \[ \frac{790 \times 289}{49} \]
Worked Solution
Step 1: Round to 1 significant figure
Why we do this: Estimation requires simplifying the numbers into values that are easy to calculate mentally.
✏ Working:
- \( 790 \approx 800 \)
- \( 289 \approx 300 \)
- \( 49 \approx 50 \)
✓ (M1) for rounding at least two figures correctly
Step 2: Perform the simplified calculation
✏ Working:
800 × 300 240000
───────── = ──────
50 50
\[ 24000 \div 5 = 4800 \]
✓ (M1) for correct calculation with rounded values
✓ (A1) for answer in range 4550 to 4800
Final Answer:
\( 4800 \)
Question 19 (4 marks)
(a) Expand \( x(x – 4) \)
(b) Factorise \( 15y – 10 \)
(c) Solve \( 7(f – 5) = 28 \)
Worked Solution
Step 1: Part (a) – Expansion
✏ Working:
\[ x \times x – x \times 4 \]
\[ x^2 – 4x \]
✓ (B1) cao
Step 2: Part (b) – Factorising
Why we do this: Find the highest common factor (HCF) of 15y and 10, which is 5.
✏ Working:
\[ 5(3y – 2) \]
✓ (B1) cao
Step 3: Part (c) – Solving Equation
✏ Working:
Method 1: Divide by 7 first
\[ f – 5 = 28 \div 7 \]
\[ f – 5 = 4 \]
\[ f = 9 \]
✓ (M1) for correct first stage (expand or divide)
✓ (A1) cao
Final Answer:
(a) \( x^2 – 4x \) (b) \( 5(3y – 2) \) (c) \( f = 9 \)
Question 20 (2 marks)
The first five terms of an arithmetic sequence are: \[ 1, \quad 4, \quad 7, \quad 10, \quad 13 \]
Write down an expression, in terms of \( n \), for the \( n \text{th} \) term of this sequence.
Worked Solution
Step 1: Find the common difference
Why we do this: The common difference determines the “\( n \)” part of the formula.
✏ Working:
Sequence: 1 (+3) 4 (+3) 7 (+3) 10 (+3) 13
Common difference = 3. This gives us 3n.
Step 2: Find the adjustment (Zero term)
Why we do this: We subtract the difference from the first term to find the starting point.
✏ Working:
If \( n=1 \), \( 3(1) = 3 \). We need 1, so subtract 2.
Formula = \( 3n – 2 \)
✓ (B2) for 3n – 2 (B1 for 3n + k)
Final Answer:
\[ 3n – 2 \]
Question 21 (3 marks)
Show that \[ 2 \frac{1}{3} \times 3 \frac{3}{4} = 8 \frac{3}{4} \]
Worked Solution
Step 1: Convert to improper fractions
Why we do this: It is much easier to multiply fractions if they are written in improper form (numerator > denominator).
✏ Working:
\[ 2 \frac{1}{3} = \frac{(2 \times 3) + 1}{3} = \frac{7}{3} \]
\[ 3 \frac{3}{4} = \frac{(3 \times 4) + 3}{4} = \frac{15}{4} \]
✓ (M1) for correct conversion to improper
Step 2: Multiply the fractions
✏ Working:
\[ \frac{7}{3} \times \frac{15}{4} = \frac{7 \times 15}{3 \times 4} = \frac{105}{12} \]
Alternatively, cancel common factors: \[ \frac{7}{\cancel{3}} \times \frac{\cancel{15}^5}{4} = \frac{7 \times 5}{4} = \frac{35}{4} \]
✓ (M1) for correct method to multiply
Step 3: Convert back to mixed number
✏ Working:
\[ 35 \div 4 = 8 \text{ remainder } 3 \]
\[ \frac{35}{4} = 8 \frac{3}{4} \]
✓ (C1) for complete working shown
Final Answer:
Shown: \( \frac{35}{4} = 8 \frac{3}{4} \)
Question 22 (2 marks)
The diagram shows four graphs.
Complete the table by matching equations to graphs.
Equation | Graph Letter
y = -x³ | ?
y = x³ | ?
y = x² | ?
y = 1/x | ?
Worked Solution
Step 1: Recognise Standard Shapes
Why we do this: Each type of function has a unique visual signature.
✏ Working:
- \( y = x^2 \) is a parabola (U-shape). -> Graph D
- \( y = 1/x \) is a hyperbola (split curve). -> Graph A
- \( y = x^3 \) goes from bottom-left to top-right. -> Graph C
- \( y = -x^3 \) goes from top-left to bottom-right. -> Graph B
✓ (B2) for all correct (B1 for 2-3 correct)
Final Answer:
B, C, D, A
Question 23 (1 mark)
The diagram shows four triangles.
Two of these triangles are congruent. Write down the letters of these two triangles.
Worked Solution
Step 1: Check Congruency Criteria
Why we do this: Triangles are congruent if they satisfy ASA, SSS, SAS, or RHS.
✏ Working:
- Triangle A: Angle 55°, side 10, angle 45°. (ASA)
- Triangle D: Two angles are 80° and 45°. The third angle is \( 180 – 80 – 45 = 55° \).
- Triangle D has angles 55° and 45° with a 10cm side between them.
Both A and D have Angle 55°, Side 10cm, Angle 45° in the same configuration.
✓ (B1) cao
Final Answer:
A and D
Question 24 (3 marks)
Sean pays £10 for 24 chocolate bars. He sells all 24 chocolate bars for 50p each.
Work out Sean’s percentage profit.
Worked Solution
Step 1: Calculate total income
✏ Working:
Income = \( 24 \times 0.50 \text{ (or 50p)} \)
Income = £12.00
✓ (P1) for finding selling price
Step 2: Calculate Profit
✏ Working:
Profit = \( \text{Selling Price} – \text{Cost Price} \)
Profit = \( 12 – 10 = £2 \)
Step 3: Calculate Percentage Profit
Why we do this: Percentage profit is always calculated relative to the original cost price.
✏ Working:
\[ \frac{\text{Profit}}{\text{Cost}} \times 100 = \frac{2}{10} \times 100 \]
\[ 0.2 \times 100 = 20\% \]
✓ (P1) for profit/cost process
✓ (A1) cao
Final Answer:
\( 20\% \)
Question 25 (5 marks)
ADC is a triangle. AED and ABC are straight lines. EB is parallel to DC.
Angle EBC = 148°, Angle ADC = 63°. Work out the size of angle EAB. Give reasons for each stage.
Worked Solution
Step 1: Use Corresponding Angles
Why we do this: Parallel lines create equal angles in corresponding positions.
✏ Working:
Angle AEB = Angle ADC = 63°
Reason: Corresponding angles are equal.
✓ (M1) for AEB = 63
Step 2: Find Angle ABE
Why we do this: Angles on a straight line add up to 180°.
✏ Working:
Angle ABE = \( 180 – 148 = 32° \)
Reason: Angles on a straight line add up to 180°.
Step 3: Solve for Angle EAB
Why we do this: The sum of angles in any triangle is always 180°.
✏ Working:
Angle EAB = \( 180 – 63 – 32 \)
Angle EAB = \( 180 – 95 = 85° \)
Reason: Angles in a triangle add up to 180°.
✓ (M1) for complete method
✓ (A1) for 85 identified
✓ (C2) for all reasons correct
Final Answer:
\( 85° \)
Question 26 (3 marks)
The table shows heights, in cm, of a group of Year 9 girls:
least height : 150 cm
median : 165 cm
greatest height: 170 cm
The stem and leaf diagram shows heights of 15 Year 9 boys:
15 | 8 9 9
16 | 4 5 7 7 8
17 | 0 3 4 4 7
18 | 0 2
Key: 15 | 8 represents 158 cm
Compare the distribution of the heights of the girls with the boys.
Worked Solution
Step 1: Calculate median and range for boys
✏ Working:
Boys Median = 8th value = 168 cm.
Boys Range = \( 182 – 158 = 24 \text{ cm} \).
Girls Range = \( 170 – 150 = 20 \text{ cm} \).
✓ (B1) for median 168 or range 24
Step 2: Compare Averages (Medians)
✏ Working:
The median height for boys (168 cm) is greater than the median for girls (165 cm).
Conclusion: On average, the boys are taller.
Step 3: Compare Spread (Ranges)
✏ Working:
The range for boys (24 cm) is greater than the range for girls (20 cm).
Conclusion: The boys’ heights are more spread out.
✓ (C2) for comparing both medians and ranges with context
Final Answer:
On average, boys are taller (median 168 vs 165) and their heights are more spread out (range 24 vs 20).
Question 27 (3 marks)
A prism is placed on a horizontal floor. \[ \text{pressure} = \frac{\text{force}}{\text{area}} \]
The prism has height 3m. The volume is 18m³. The pressure on the floor is 75 newtons/m².
Work out the force exerted by the prism on the floor.
Worked Solution
Step 1: Find the contact area
Why we do this: Volume of a prism = Area of base × height. We need the base area for the pressure formula.
✏ Working:
\[ \text{Base Area} = \frac{\text{Volume}}{\text{height}} \]
\[ \text{Area} = 18 \div 3 = 6 \text{ m}^2 \]
✓ (M1) for 18 ÷ 3
Step 2: Rearrange and Solve for Force
Why we do this: From the formula, Force = Pressure × Area.
✏ Working:
\[ \text{Force} = 75 \times 6 \]
\[ \text{Force} = 450 \text{ newtons} \]
✓ (M1) for 75 × 6
✓ (A1) cao
Final Answer:
\( 450 \text{ newtons} \)
Question 28 (2 marks)
Write these numbers in order of size. Start with the smallest number.
\( 6.72 \times 10^5 \quad 67.2 \times 10^{-4} \quad 672 \times 10^4 \quad 0.000672 \)
Worked Solution
Step 1: Convert all to normal numbers
✏ Working:
- \( 6.72 \times 10^5 = 672,000 \)
- \( 67.2 \times 10^{-4} = 0.00672 \)
- \( 672 \times 10^4 = 6,720,000 \)
- \( 0.000672 \) (Already normal)
✓ (B1) for correct conversions
Final Answer:
\( 0.000672, \quad 67.2 \times 10^{-4}, \quad 6.72 \times 10^5, \quad 672 \times 10^4 \)
Question 29 (3 marks)
Given that \( \frac{a}{b} = \frac{2}{5} \) and \( \frac{b}{c} = \frac{3}{4} \), find \( a:b:c \)
Worked Solution
Step 1: Express as ratios with a common term
Why we do this: Both ratios share the variable ‘b’. We need to make the value for ‘b’ the same in both ratios.
✏ Working:
Ratio 1: \( a:b = 2:5 \)
Ratio 2: \( b:c = 3:4 \)
LCM of 5 and 3 is 15.
✓ (P1) for equating ‘b’
Step 2: Rescale the ratios
✏ Working:
Multiply Ratio 1 by 3: \( a:b = 6:15 \)
Multiply Ratio 2 by 5: \( b:c = 15:20 \)
✓ (P1) for related terms
Final Answer:
\[ 6:15:20 \]
Question 30 (3 marks)
(a) Make \( q \) the subject of \( p = 6q + 7 \)
(b) Simplify \( (m^{-2})^{-3} \)
Worked Solution
Step 1: Part (a) – Rearrange formula
Why we do this: Isolate \( q \) using inverse operations.
✏ Working:
Subtract 7: \[ p – 7 = 6q \]
Divide by 6: \[ q = \frac{p – 7}{6} \]
✓ (M1) for correct first step
✓ (A1) for correct final expression
Step 2: Part (b) – Laws of Indices
Why we do this: When we have a power raised to another power, we multiply the indices: \( (a^m)^n = a^{m \times n} \).
✏ Working:
\[ m^{(-2) \times (-3)} = m^6 \]
✓ (B1) cao
Final Answer:
(a) \( q = \frac{p – 7}{6} \) (b) \( m^6 \)