GCSE Mathematics Foundation Paper 2 (Calculator) – June 2017

Working:

\[ 5p – 3p + 1p \] \[ = 2p + 1p \] \[ = 3p \]

✓ (B1) Correct answer only [cite: 86]

Working:

\[ 1m^3 + 1m^3 = 2m^3 \]

✓ (B1) Correct answer only [cite: 86]

Working:

Let’s rearrange the terms to group them:

\[ 10 + (3c – 7c) + (5d + 1d) \]

Now simplify each group:

• Numbers: \(10\)
• The \(c\) terms: \(3c – 7c = -4c\)
• The \(d\) terms: \(5d + 1d = +6d\)

\[ 10 – 4c + 6d \]

✓ (M1) For finding either \(-4c\) or \(+6d\) correctly [cite: 86]

✓ (A1) For the fully correct expression [cite: 86]

Working:

Number: \(56.78\)

• The 1st significant figure is the 5 (in the tens column).
• The next digit (the decider) is the 6 (in the units column).

Because the decider is 5 or more (it’s 6), we must round the 5 up to a 6. The position of the number must be maintained, so the tens column becomes 60.

✓ (B1) Correct answer only [cite: 86]

Working:

Reading the heights of the dark bars:

• Car = 5
• Bus = 6
• Walk = 9 (This is the tallest)
• Cycle = 4
• Other = 2

✓ (B1) Correct answer: Walk [cite: 86]

Working:

On your exam paper, you would draw a light grey bar in the “Walk” category that reaches a height of 7 on the frequency axis.

✓ (B1) For drawing a bar of height 7 for girls walking [cite: 86]

Working:

Total for Car = 5 (boys) + 9 (girls) = 14

Total for Bus = 6 (boys) + 4 (girls) = 10

Difference = \(14 – 10 = 4\)

✓ (B1) Correct answer: 4 [cite: 86]

Working:

Step 1: Count Year 6 Total.

• Car: \(5 + 9 = 14\)
• Bus: \(6 + 4 = 10\)
• Walk: \(9 + 7 = 16\) (using the 7 from part b)
• Cycle: \(4 + 1 = 5\)
• Other: \(2 + 1 = 3\)

Year 6 Total = \(14 + 10 + 16 + 5 + 3 = 48\)

✓ (M1) For a method to find number of Year 6 students [cite: 86]

Step 2: Total for Years 5 and 6.

Year 5 + Year 6 = \(48 + 48 = 96\)

✓ (A1) Correct answer: 96 [cite: 86]

Working (Fraction Method):

Change each fraction to have a denominator of 30:

• \(\frac{2}{5} = \frac{2 \times 6}{5 \times 6} = \frac{12}{30}\)
• \(\frac{11}{30}\) stays as \(\frac{11}{30}\)
• \(\frac{1}{2} = \frac{1 \times 15}{2 \times 15} = \frac{15}{30}\)
• \(\frac{7}{15} = \frac{7 \times 2}{15 \times 2} = \frac{14}{30}\)

✓ (M1) Converts fractions to a common form [cite: 86]

Working:

Compare the numerators out of 30:

\(11\) is the smallest, then \(12\), then \(14\), then \(15\).

Order out of 30: \(\frac{11}{30}, \frac{12}{30}, \frac{14}{30}, \frac{15}{30}\)

Translate back to original forms:

\[ \frac{11}{30}, \quad \frac{2}{5}, \quad \frac{7}{15}, \quad \frac{1}{2} \]

✓ (A1) Correct order [cite: 86]

Working:

Monday’s tally marks show two groups of 5, and one group of 4.

Total tallies = \(5 + 5 + 4 = 14\)

However, the frequency column says “12”. These do not match.

✓ (C1) For seeing the difference in tally marks and frequency for Monday [cite: 87]

Working:

Let’s check Tuesday: The frequency is 18. If 1 circle = 3 CDs, we need \(18 \div 3 = 6\) circles. But there are only 5 circles drawn (which would equal 15 CDs). So Tuesday is missing a picture.

Let’s check Wednesday: The frequency is 8. Two circles = 6 CDs. We need to represent 2 more CDs. However, the drawing shows half a circle. Half of 3 CDs is \(1.5\) CDs. It’s impossible to sell half a CD, so drawing half a symbol doesn’t make logical sense here, and it would sum to 7.5 CDs, not 8.

You can state either of these errors.

✓ (C1) For suitable comment, eg extra picture for Tuesday needed or explains that 0.5 of a CD is not possible [cite: 87]

Working:

Number of £1 coins = \(495 \div 3 = 165\)

Value of £1 coins = \(165 \times \text{£}1 = \text{£}165\)

✓ (P1) For a process to work out the value of the £1 coins [cite: 87]

Working:

Value of 50p coins = \(124 \times \text{£}0.50 = \text{£}62\)

Working:

Coins already accounted for = \(165\) (£1 coins) \(+\) \(124\) (50p coins) \(= 289\) coins

Number of 20p coins = \(495 – 289 = 206\) coins

✓ (P1) For a process to find the number of 20p coins [cite: 87]

Value of 20p coins = \(206 \times \text{£}0.20 = \text{£}41.20\)

Working:

Total Value = \(\text{£}165 + \text{£}62 + \text{£}41.20\)

Total Value = \(\text{£}268.20\)

✓ (P1) For a complete process to find total value using consistent units [cite: 87]

✓ (A1) Correct answer only [cite: 87]

Working:

\[ P(\text{no fault}) = 1 – P(\text{has a fault}) \] \[ P(\text{no fault}) = 1 – 0.015 \]

You can use your calculator or perform column subtraction: \(1.000 – 0.015 = 0.985\)

✓ (B1) Correct answer only [cite: 87]

Working:

• \(3 \times 3 = 9\)
• \(4 \times 4 = 16\)
• \(5 \times 5 = 25\) (This is in our list!)
• \(6 \times 6 = 36\)

✓ (B1) Correct answer: 25 [cite: 87]

Working:

Multiples of 4 in our range: \(24, 28\)

Multiples of 6 in our range: \(24\)

The only number that appears in both lists is \(24\).

✓ (B1) Correct answer: 24 [cite: 87]

Working:

Let’s eliminate non-primes:

• Even numbers (except 2) are not prime: eliminate \(22, 24, 26, 28\).
• Numbers ending in 5 are divisible by 5: eliminate \(25\).
• Check if divisible by 3 (sum of digits is multiple of 3): \(21\) (\(2+1=3\), divisible by 3) and \(27\) (\(2+7=9\), divisible by 3) are not prime.

We are left with \(23\) and \(29\). Neither of these can be divided by \(2, 3, 4, 5, 6, 7\) etc.

✓ (B1) For 23 and 29 and no extras [cite: 87]

Working:

Angles around a point = \(360^\circ\)

\[ 3x + 2x + 90 = 360 \]

✓ (M1) For a method to form equation [cite: 87]

Working:

Combine the \(x\) terms:

\[ 5x + 90 = 360 \]

Subtract \(90\) from both sides:

\[ 5x = 360 – 90 \] \[ 5x = 270 \]

✓ (M1) For \(5x = 270\) (or equivalent step) [cite: 87]

Divide by \(5\):

\[ x = \frac{270}{5} \]

Calculator Step: Press 270 ÷ 5 =

\[ x = 54 \]

✓ (A1) Correct answer only [cite: 87]

Working:

Number of packs needed = \(150 \div 25 = 6\) packs

✓ (P1) For the start of a process to find comparable costs (e.g. \(150 \div 25 = 6\)) [cite: 89]

Total cost at Letters2send = \(6 \times \text{£}3.49\)

Calculator Step: 6 x 3.49 = 20.94

Total cost = £\(20.94\)

✓ (P1) For a process to find the cost from Letters2send [cite: 89]

Working:

Total packs needed = \(150 \div 10 = 15\) packs

Because the offer is “Buy 2 get 1 free”, they come in groups of \(3\). Let’s see how many groups of \(3\) there are in \(15\) packs:

\(15 \div 3 = 5\) groups

For each group of \(3\), she only pays for \(2\) packs. Therefore, the number of packs she actually pays for is:

\(5 \times 2 = 10\) packs

Total cost at Stationery World = \(10 \times \text{£}2.10\)

Total cost = £\(21.00\)

✓ (P1) For a process to find the cost at Stationery World [cite: 89]

Working:

Letters2send cost: £\(20.94\)

Stationery World cost: £\(21.00\)

£\(20.94\) is less than £\(21.00\), so Letters2send is cheaper.

✓ (C1) For correct conclusion with correct values from each shop (20.94 and 21) [cite: 89]

Working:

• Find \(74\) on the ‘centimetres’ axis. Each small square represents \(2 \text{ cm}\), so \(74\) is two small squares below \(80\) (or seven small squares above \(60\)).
• Move right to the black line.
• Move straight down to the ‘inches’ axis.
• Read the value: It lands just before the \(30\) mark, at approximately \(29\).

✓ (B1) Correct answer in the range 29 to 30 [cite: 90]

Working:

Step 1: Convert feet to inches.

\(6 \text{ feet} = 6 \times 12 \text{ inches} = 72 \text{ inches}\)

Total inches = \(72 + 3 = 75 \text{ inches}\)

✓ (M1) For changing 6ft 3 inches to inches (e.g., \(6 \times 12 + 3 = 75\)) [cite: 90]

Step 2: Convert to centimetres.

The graph only goes up to \(50\) inches, so we can’t look up \(75\) directly. Instead, we can look up \(25\) inches on the graph.

• Find \(25\) on the ‘inches’ axis.
• Go up to the line and across to the ‘centimetres’ axis.
• It reads approximately \(63.5 \text{ cm}\).

Since \(75 \text{ inches}\) is exactly \(3 \times 25 \text{ inches}\), we multiply the cm value by \(3\):

\[ 63.5 \times 3 = 190.5 \text{ cm} \]

(Alternatively, you could look up \(10\) inches (\(\approx 25.4 \text{ cm}\)) and multiply by \(7.5\).)

✓ (M1) For a method to convert to cm using the graph [cite: 90]

✓ (A1) For an answer in the range 186 to 195 [cite: 90]

Working:

Numerator (Top):

Calculator Steps: press \(\sqrt{}\), type 13.4 - 1.5, press =

\[ \sqrt{13.4 – 1.5} = \sqrt{11.9} = 3.449637662… \]

Denominator (Bottom):

Calculator Steps: type ( 6.8 + 0.06 ), press x², press =

\[ (6.8 + 0.06)^2 = (6.86)^2 = 47.0596 \]

✓ (M1) For correct numerator (\(3.4496…\)) OR correct denominator (\(47.0596\)) [cite: 90]

Working:

Calculator Steps: type 3.449637662 ÷ 47.0596 =

\[ \frac{3.449637662…}{47.0596} = 0.07330359081… \]

✓ (A1) For exactly \(0.073303…\) (accept the full string of digits your calculator produces) [cite: 90]

Working:

Let’s find the original vertices of shape A and apply the transformation:

• Bottom-left corner: \((1, -4) \rightarrow (-4, -1)\)
• Top-left corner: \((1, -3) \rightarrow (-3, -1)\)
• Top-right corner: \((2, -1) \rightarrow (-1, -2)\)
• Bottom-right corner: \((4, -4) \rightarrow (-4, -4)\)

On the grid, you would plot these new points and join them to draw the rotated shape.

✓ (B2) For a fully correct rotation at (-4,-1), (-3,-1), (-4,-4), (-1,-2) [cite: 90]

Working:

The transformation is a “Reflection”.

The shapes are an equal distance away from the vertical y-axis. For example, point \((-1, 0)\) on shape B maps to \((1, 0)\) on shape C. The mirror line is directly between them at \(x = 0\).

Therefore, it is a reflection in the y-axis.

✓ (B1) For stating “reflection” [cite: 90]

✓ (B1) For stating “in the y-axis” (or \(x=0\)) [cite: 90]

Working:

The terms are \(5\) and \(10m\).

The highest number that divides into both \(5\) and \(10\) is \(5\).

Pulling \(5\) outside the bracket:

\[ 5( \dots – \dots ) \]

Inside the bracket: \(5 \div 5 = 1\), and \(10m \div 5 = 2m\).

\[ 5(1 – 2m) \]

✓ (B1) Correct answer only [cite: 91]

Working:

The terms are \(2a^2b\) and \(6ab^2\).

• Numbers: The HCF of \(2\) and \(6\) is \(2\).
• Letters (\(a\)): The terms are \(a^2\) and \(a\). The highest common factor is \(a\).
• Letters (\(b\)): The terms are \(b\) and \(b^2\). The highest common factor is \(b\).

So, the term outside the bracket is \(2ab\).

✓ (M1) For identifying a common factor like \(2ab(\dots)\), \(2a(\dots)\), or \(ab(\dots)\) [cite: 91]

Now, divide the original terms by \(2ab\):

• \(\frac{2a^2b}{2ab} = a\)
• \(\frac{6ab^2}{2ab} = 3b\)

Put these inside the bracket:

\[ 2ab(a + 3b) \]

✓ (A1) For the fully correct expression [cite: 91]

Working:

\[ 4.7 \div 10 = 0.47 \]

✓ (B1) Correct answer only [cite: 91]

Working:

Rearrange the calculation:

\[ (2.4 \times 9.5) \times (10^3 \times 10^5) \]

1. Multiply the numbers (using calculator):

\[ 2.4 \times 9.5 = 22.8 \]

2. Multiply the powers of \(10\) (add the indices):

\[ 10^3 \times 10^5 = 10^{3+5} = 10^8 \]

Combine them:

\[ 22.8 \times 10^8 \]

✓ (M1) For correct value but not in standard form (e.g. \(22.8 \times 10^8\)) [cite: 91]

Final Step: Adjusting to Standard Form

\(22.8\) is bigger than \(10\), so it’s not proper standard form. We need to turn \(22.8\) into \(2.28\) by dividing it by \(10\). To keep the overall value the same, we must multiply the power part by \(10\) (which increases the index by \(1\)).

\[ 2.28 \times 10^{8+1} \] \[ 2.28 \times 10^9 \]

✓ (A1) Correct answer only [cite: 91]

Working:

Map distance = \(250 \div 100 = 2.5\text{ cm}\)

Using a pair of compasses, set the radius to exactly \(2.5\text{ cm}\). Place the point of the compass on A and draw a circle (or an arc).

✓ (C1) For drawing a circle/arc of radius 2.5 cm centered at A

Working:

1. Open your compass to a width greater than half the distance between B and C.

2. Put the point on B and draw an arc above and below the line connecting B and C.

3. Keep the compass at the SAME width, put the point on C, and draw intersecting arcs.

4. Draw a straight ruler line through the two points where your arcs cross. This is the perpendicular bisector.

✓ (C1) For showing a perpendicular bisector between B and C

Working:

Mark with an ‘X’ or a dot and label it ‘T’ where your \(2.5\text{ cm}\) circle crosses the bisector line. (There may be two points where they cross; the question says “show one of the possible positions”, so picking either intersection is fine).

✓ (C1) For marking T in the correct position (intersection of arc and bisector)

Working:

Sum of known probabilities:

\[ 0.17 + 0.18 + 0.09 + 0.15 + 0.1 = 0.69 \]

Probability of 1, \(P(1)\):

\[ P(1) = 1 – 0.69 = 0.31 \]

✓ (P1) For a process to find P(1)

Working:

\[ P(1 \text{ or } 3) = P(1) + P(3) \] \[ P(1 \text{ or } 3) = 0.31 + 0.18 = 0.49 \]

Working:

Expected frequency = Probability \(\times\) Total Rolls

\[ \text{Expected} = 0.49 \times 200 \] \[ \text{Expected} = 98 \]

✓ (P1) For a process to find expected frequency (e.g. \(0.49 \times 200\))

✓ (A1) Correct answer only

Working:

\(\frac{1}{4}\) of children = \(117\)

Total children = \(117 \times 4\)

\[ 117 \times 4 = 468 \text{ children} \]

✓ (P1) For a process to work out the total number of children

Working:

Adults : Children = \(5 : 2\)

\(2 \text{ parts} = 468\)

\(1 \text{ part} = 468 \div 2 = 234\)

Total number of parts = \(5\) (Adults) \(+\) \(2\) (Children) \(= 7 \text{ parts total}\)

Total people = \(7 \times 234\)

\[ 7 \times 234 = 1638 \text{ people} \]

(Alternatively, you could find Adults: \(5 \times 234 = 1170\), then add \(1170 + 468 = 1638\))

✓ (P1) For a process to work out total number of people (or adults)

✓ (A1) For calculating 1638 (or 1170 adults)

Working:

Total seats = \(2600\)

\(60\% \text{ of } 2600 = 0.6 \times 2600\)

\[ 0.6 \times 2600 = 1560 \text{ seats} \]

✓ (P1) For a process to work out 60% of 2600

Working:

• Front base: \(2\text{m} \times 2 = 4\text{ cm}\)
• Tall left vertical: \(2\text{m} \times 2 = 4\text{ cm}\)
• Short right vertical: \(0.5\text{m} \times 2 = 1\text{ cm}\)
• Depth (Side base): \(1\text{m} \times 2 = 2\text{ cm}\)

Working:

On the grid, you would draw:

• A horizontal base measuring \(4\) squares.
• A vertical line up from the left edge measuring \(4\) squares.
• A vertical line up from the right edge measuring \(1\) square.
• A straight line connecting the top of the two vertical lines.

✓ (C2) For the front elevation as a trapezium in correct orientation with base 4 cm, parallel sides 1 cm and 4 cm

Working:

On the grid, you would draw:

• A rectangle with a width of \(2\) squares (the depth) and a total height of \(4\) squares (the tallest point).
• A horizontal line drawn across the rectangle, \(1\) square up from the bottom, to show where the slope ends.

✓ (C2) For the side elevation (4 cm by 2 cm rectangle with a solid line drawn 1 cm from the 2 cm edge)

Working:

Time for first leg = \(56 \div 70\)

\[ 56 \div 70 = 0.8 \text{ hours} \]

✓ (P1) For a process to find time from Liverpool to Manchester (e.g. 0.8 hrs)

Working:

Total Distance = \(56 + 61 = 117\text{ km}\)

✓ (P1) For a process to find the total distance (117)

Convert 75 mins to hours: \(75 \div 60 = 1.25\text{ hours}\)

Total Time = \(0.8 + 1.25 = 2.05\text{ hours}\)

✓ (P1) For a process to find total time with consistent units

Working:

Average Speed = \(117 \div 2.05\)

Calculator Steps: 117 ÷ 2.05 =

\[ = 57.07317… \]

✓ (P1) For a correct process to find average speed (\(117 \div 2.05\))

✓ (A1) For an answer in the range 57 to 57.1

Working:

If her mean speed method worked, it tells us she spent the exact same amount of time traveling from Barnsley to Leeds as she did traveling from Leeds to York.

✓ (C1) For explaining that the time taken for the two parts of the journey must be the same

Working:

The base of the large triangle is \(EC = 8.1 \text{ cm}\).

The base of the small triangle is \(DC = 5.4 \text{ cm}\).

Scale Factor = \(\frac{\text{Large Base}}{\text{Small Base}} = \frac{8.1}{5.4}\)

Calculator Step: 8.1 ÷ 5.4 = 1.5

✓ (M1) For calculating the ratio \(\frac{8.1}{5.4} = 1.5\)

Now apply this scale factor to the side \(DB\) to find \(AE\).

\[ AE = 1.5 \times DB \] \[ AE = 1.5 \times 2.6 = 3.9 \text{ cm} \]

✓ (A1) Correct answer only

Working:

Find the length of the small triangle’s right side (\(BC\)):

\[ BC = \frac{AC}{\text{Scale Factor}} \] \[ BC = \frac{6.15}{1.5} \]

Calculator Step: 6.15 ÷ 1.5 = 4.1

✓ (M1) For a correct method to find a relevant length (e.g. \(BC = 4.1\))

Now find \(AB\):

\[ AB = AC – BC \] \[ AB = 6.15 – 4.1 = 2.05 \text{ cm} \]

✓ (A1) Correct answer only

Working:

Multiplier = \(1.02\)

\[ \text{Amount} = 25000 \times 1.02^3 \]

Calculator Steps: 25000 x 1.02 xʸ (or ^) 3 =

\[ \text{Amount} = \text{£}26530.20 \]

✓ (P1) For a process to work out the interest (or total value) after one year for either bank (e.g. \(25000 \times 1.02 = 25500\))

Working:

Year 1 Multiplier = \(1.043\)

Years 2 & 3 Multiplier = \(1.009\)

\[ \text{Amount} = 25000 \times 1.043 \times 1.009 \times 1.009 \]

or

\[ \text{Amount} = 25000 \times 1.043 \times 1.009^2 \]

Calculator Steps: 25000 x 1.043 x 1.009 x² =

\[ \text{Amount} = \text{£}26546.46… \]

✓ (P1) For a process to find the value of the investment after 3 years for at least one bank

Working:

Personal Bank Total = £\(26530.20\)

Secure Bank Total = £\(26546.46\)

£\(26546.46\) is greater than £\(26530.20\), so Secure Bank provides more money.

✓ (C1) For Secure Bank with correct figures stated to justify the choice

Working:

Lower bound = \(4.76 – 0.005 = 4.755\)

Upper bound = \(4.76 + 0.005 = 4.765\)

Any number from \(4.755\) up to (but NOT including) \(4.765\) will round to \(4.76\).

Working:

List factors of \(-24\):

• \(-1\) and \(24\) (adds to \(23\))
• \(-2\) and \(12\) (adds to \(10\))
• \(-3\) and \(8\) (adds to \(5\)) ← These are our numbers!

Put these into double brackets:

\[ (x – 3)(x + 8) = 0 \]

✓ (M1) For factorisation (getting the brackets right)

Working:

Either \(x – 3 = 0\), which means \(x = 3\)

Or \(x + 8 = 0\), which means \(x = -8\)

✓ (M1) For identifying the solutions from the brackets

✓ (A1) Correct answer only

Working:

The sequence goes: \(3 \rightarrow 8 \rightarrow 13 \rightarrow 18\)

It goes up by \(+5\) each time. This means the formula starts with \(5n\).

To find the number at the end, subtract \(5\) from the first term to step backwards:

\[ 3 – 5 = -2 \]

So the full expression is \(5n – 2\).

✓ (B2) For \(5n – 2\) (B1 for \(5n \pm k\))

Working:

Substitute \(n = 4\):

\[ \text{4th term} = 3 \times (4)^2 \]

Square the \(4\) first:

\[ = 3 \times 16 \]

Now multiply by \(3\):

\[ = 48 \]

Since \(48\) is not equal to \(144\), Nathan is incorrect.

(Nathan probably did \((3 \times 4)^2 = 12^2 = 144\), which is mathematically wrong).

✓ (C1) For ‘No’ with evidence, e.g., \(3 \times 4^2 = 48\)