GCSE Mathematics (1MA1/2H) – Higher Tier (Calculator)

✍️ Working:

\[ CB^2 = AC^2 – AB^2 \] \[ CB^2 = 19^2 – 10^2 \] \[ CB^2 = 361 – 100 \] \[ CB^2 = 261 \] \[ CB = \sqrt{261} \]

✓ (M1) Correct first step to find CB or \(19^2 – 10^2\) shown.

✍️ Working:

\[ CB = \sqrt{261} \approx 16.15549… \]

Rounding to 3 significant figures:

\[ CB \approx 16.2 \]

✓ (A1) For answer in range 16.1 to 16.2.

✍️ Working:

\[ 90 = 2 \times 45 \] \[ 45 = 3 \times 15 \] \[ 15 = 3 \times 5 \]

The prime factors are \(2, 3, 3, 5\).

✓ (M1) For a complete method to find prime factors, e.g., listing 2, 3, 3, 5.

✍️ Working:

\[ 90 = 2 \times 3^2 \times 5 \]

✓ (A1) For \(2 \times 3^2 \times 5\) oe.

✍️ Working:

Given \(A = 2^2 \times 3^1\) and \(B = 2^1 \times 3^2\).

Highest power of 2: \(2^2\)

Highest power of 3: \(3^2\)

\[ \text{LCM}(A, B) = 2^2 \times 3^2 \] \[ \text{LCM}(A, B) = 4 \times 9 \] \[ \text{LCM}(A, B) = 36 \]

✓ (B1) For 36.

✍️ Working:

\[ H_M = \frac{72}{6} = 12 \text{ hours} \]

✓ (M1) For method to use formula, e.g., \(72 \div 6\).

✍️ Working:

\[ H_T = \frac{72}{9} = 8 \text{ hours} \]

✓ (M1 implied by calculation) For method to use formula, e.g., \(72 \div 9\).

✍️ Working:

\[ \text{Difference} = 12 – 8 = 4 \text{ hours} \]

✓ (A1) Correct answer only (cao).

✍️ Working:

\[ \text{Total discs} = \frac{24}{0.16} = 150 \]

✓ (P1) For process to find total number of discs, e.g., \(24 \div 0.16\).

✍️ Working:

\[ \text{Red and Blue discs} = 150 – 24 = 126 \]

✓ (P1) For process to find number of Red and Blue discs, e.g., \(150 – 24\).

✍️ Working:

\[ \text{Value of one part} = 126 \div 9 = 14 \] \[ \text{Number of red discs} = 5 \times 14 = 70 \]

✓ (P1) For complete correct process to find the number of red discs, e.g., \((150 – 24) \div 9 \times 5\).

✍️ Working:

For \(x = -1\): \(y = (-1)^2 – (-1) = 1 + 1 = 2\)

For \(x = 1\): \(y = (1)^2 – (1) = 1 – 1 = 0\)

For \(x = 3\): \(y = (3)^2 – (3) = 9 – 3 = 6\)

✓ (B2) For all 3 values correct (2, 0, 6). (B1 for 1 or 2 correct values).

✍️ Working:

Points plotted: \((-2, 6), (-1, 2), (0, 0), (1, 0), (2, 2), (3, 6)\).

✓ (B2) For a fully correct graph.

✍️ Working:

Draw the line \(y = 4\).

Read the \(x\) coordinates of the two intersections:

\[ x \approx -1.6 \quad \text{and} \quad x \approx 2.6 \]

✓ (M1) For drawing the line \(y=4\) or reading off one correct solution (or pair of coordinates).

✓ (A1 ft) For answers in the range \(-1.7\) to \(-1.5\) and \(2.5\) to \(2.7\).

✍️ Working:

Initial shares (parts of 21):

\[ \text{Andy (A)} = 1 \] \[ \text{Luke (L)} = 6 \] \[ \text{Tina (T)} = 14 \]

✍️ Working:

Sweets Tina gives to Andy: \(\frac{3}{7} \times 14 = 6\) parts.

Shares after Step 2:

\[ A = 1 + 6 = 7 \] \[ L = 6 \] \[ T = 14 – 6 = 8 \]

✓ (P1) For process to find sweets Tina gives to Andy (6 parts or equivalent fraction).

✍️ Working:

Sweets Tina gives to Luke: \(12.5\% \text{ of } 8 = \frac{1}{8} \times 8 = 1\) part.

Final shares:

\[ A = 7 \] \[ L = 6 + 1 = 7 \] \[ T = 8 – 1 = 7 \]

✓ (P1) For process to find sweets Tina gives to Luke (1 part or equivalent fraction).

✓ (P1) For finding the final amounts (7, 7, 7) for at least two people.

✍️ Working:

Since \(A=7\), \(L=7\), and \(T=7\), Tina’s statement is correct.

✓ (C1) Yes, supported by full working and accurate figures.

✍️ Working:

\[ (2x + 15) + (4x + 15) + (4x + 8) + (3x – 3) = 360 \]

Group \(x\) terms and constants:

\[ (2x + 4x + 4x + 3x) + (15 + 15 + 8 – 3) = 360 \] \[ 13x + 35 = 360 \]

✓ (M1) For deriving a suitable equation, e.g., \(13x + 35 = 360\).

✍️ Working:

\[ 13x = 360 – 35 \] \[ 13x = 325 \] \[ x = \frac{325}{13} \] \[ x = 25 \]

✓ (M1 dep) For a method to isolate terms in \(x\) and solve for \(x\).

✓ (A1) For solving equation to \(x = 25\).

✍️ Working:

Substitute \(x=25\):

\[ \angle DAB \text{ (A)}: 2(25) + 15 = 65^\circ \] \[ \angle ABC \text{ (B)}: 4(25) + 15 = 115^\circ \] \[ \angle BCD \text{ (C)}: 4(25) + 8 = 108^\circ \] \[ \angle CDA \text{ (D)}: 3(25) – 3 = 72^\circ \]

Check consecutive interior angles between sides AD and BC:

\[ \angle DAB + \angle ABC = 65^\circ + 115^\circ = 180^\circ \]

Since \(\angle DAB + \angle ABC = 180^\circ\), the lines \(AD\) and \(BC\) are parallel.

Check consecutive interior angles between sides AB and DC:

\[ \angle ABC + \angle BCD = 115^\circ + 108^\circ = 223^\circ \] \[ \angle BCD + \angle CDA = 108^\circ + 72^\circ = 180^\circ \]

Since both pairs of adjacent angles between the sides AD and BC AND CD and AD sum to 180, this means \(AD \parallel BC\) AND \(AB \parallel DC\). This shape is a parallelogram. Since a parallelogram is a special type of trapezium (it has at least one pair of parallel sides), the proof holds.

Conclusion: Since \(AD \parallel BC\) (co-interior angles sum to \(180^\circ\)), \(ABCD\) is a trapezium.

✓ (C1) For substituting \(x=25\) into a pair of adjacent angles, showing the sum is \(180^\circ\), and giving a suitable statement (co-interior/allied angles sum to 180, or since \(A+B=180\) the lines are parallel).

✍️ Working:

\[ \text{Real Area} = 28 \text{ cm}^2 \times (5)^2 \] \[ \text{Real Area} = 28 \times 25 \] \[ \text{Real Area} = 700 \text{ m}^2 \]

✓ (P1) For one correct use of the given scale, e.g., \(28 \times 5^2\).

✍️ Working:

\[ 700 = \frac{1}{2} \times PQ \times 40 \] \[ 700 = 20 \times PQ \] \[ PQ = \frac{700}{20} \] \[ PQ = 35 \text{ m} \]

✓ (P1) For process to use area of triangle, e.g., \(700 = \frac{1}{2} \times PQ \times 40\).

✓ (A1) Correct answer only (cao).

✍️ Working:

\[ \text{Lower Bound} = 7.3 – 0.05 = 7.25 \]

✓ (B1) For 7.25 oe.

✍️ Working:

Leila’s statement implies that \(N \le 7.349\). However, numbers between 7.349 and 7.35 (e.g., 7.3499) still round down to 7.3 (to 2 s.f.).

Therefore, Leila is incorrect because \(N\) can be greater than 7.349, provided \(N < 7.35\).

✓ (C1) No, with supporting reason. Acceptable reasons: No, 7.3491 is greater; No, there are numbers between 7.349 and 7.350.

✍️ Working:

Triangle 1 (angle \(a\)): Opposite is \(7-2x\), Hypotenuse is \(8\).

\[ \sin a = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{7 – 2x}{8} \]

Triangle 2 (angle \(b\)): Opposite is \(1+x\), Adjacent is \(4\).

\[ \tan b = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{1 + x}{4} \]

✓ (B1) For a correct expression for \(\sin a\) or \(\tan b\).

✍️ Working:

\[ \frac{7 – 2x}{8} = \frac{1 + x}{4} \]

Multiply both sides by 8:

\[ 7 – 2x = 2(1 + x) \] \[ 7 – 2x = 2 + 2x \]

Rearrange terms:

\[ 7 – 2 = 2x + 2x \] \[ 5 = 4x \] \[ x = \frac{5}{4} = 1.25 \]

✓ (P1) For start of a process to solve the equation, e.g., \(7 – 2x = 2(1 + x)\).

✓ (A1) For 1.25 oe.

✍️ Working:

\[ 0 < w \le 2: \text{CF} = 7 \] \[ 0 < w \le 4: \text{CF} = 7 + 21 = 28 \] \[ 0 < w \le 6: \text{CF} = 28 + 15 = 43 \] \[ 0 < w \le 8: \text{CF} = 43 + 11 = 54 \] \[ 0 < w \le 10: \text{CF} = 54 + 6 = 60 \]

✓ (B1) For all values correct (7, 28, 43, 54, 60).

✍️ Working:

Plot the points: \((0, 0), (2, 7), (4, 28), (6, 43), (8, 54), (10, 60)\).

Draw a smooth curve connecting these points (as shown in the SVG diagram).

✓ (M1) For 4 or 5 points plotted correctly.

✓ (A1) For a fully correct graph.

✍️ Working:

LQ (CF=15): Reading from the graph, LQ \(\approx 2.7 \text{ kg}\) (MS range 2.6 to 2.8).

UQ (CF=45): Reading from the graph, UQ \(\approx 6.3 \text{ kg}\) (MS range 6.2 to 6.4).

\[ \text{IQR} = \text{UQ} – \text{LQ} \] \[ \text{IQR} = 6.3 – 2.7 = 3.6 \text{ kg} \]

✓ (M1) For at least one correct quartile reading or ft their graph.

✓ (A1 ft) For answer in the range 3.4 to 4.0, or ft their graph.

✍️ Working:

Read CF at \(w = 7.4\) kg:

From the graph, CF \(\approx 52\) (MS range 50 to 53).

Number of parcels greater than 7.4 kg:

\[ \text{Count} = 60 – 52 = 8 \]

✓ (M1) For CF reading at weight \(7.4\) in the range 50 to 53 or ft their graph.

✓ (A1 ft) For answer in the range 7 to 10, or ft their graph.

✍️ Working:

\[ f = \frac{k}{d^2} \] \[ 3.5 = \frac{k}{8^2} \] \[ k = 3.5 \times 64 = 224 \]

✓ (M1) For \(3.5 = \frac{k}{8^2}\) oe (k=224).

✍️ Working:

\[ f = \frac{224}{d^2} \]

✓ (A1) For the correct equation.

✍️ Working:

\[ 10 = \frac{224}{d^2} \] \[ d^2 = \frac{224}{10} = 22.4 \] \[ d = \sqrt{22.4} \approx 4.73286… \]

✓ (M1) For \(10 = \frac{224}{d^2}\) or \(d^2 = 22.4\).

✍️ Working:

\[ d \approx 4.73286… \]

Rounded to 3 significant figures:

\[ d \approx 4.73 \]

✓ (A1) For answer in range 4.73 to 4.733.

✍️ Working:

1. \(x \le 2\): Solid vertical line \(x = 2\). (Shading to the left)
2. \(y \ge -3\): Solid horizontal line \(y = -3\). (Shading above)
3. \(y < 2x + 1\): Dashed line \(y = 2x + 1\). (Shading below)
4. \(3x + 2y \le 6\): Solid line \(3x + 2y = 6\). (Shading below)

✓ (M1) For drawing at least two correct boundary lines.

✍️ Working:

The boundary lines should be drawn (M1 for remaining two boundaries). The intersection of the shaded regions is identified and labeled R.

✓ (M1) For the remaining two correct boundaries, OR a third correct boundary with shading correct for at least two correct boundaries.

✓ (C1) For a correct region (R) clearly identified.

✍️ Working:

Draw a tangent line to the curve at \(t=5\) (M1).

Select two points on the tangent line, e.g., \((t_1, v_1) = (3, 10)\) and \((t_2, v_2) = (7, 0)\):

\[ \text{Acceleration} = \frac{v_2 – v_1}{t_2 – t_1} = \frac{0 – 10}{7 – 3} \] \[ \text{Acceleration} = \frac{-10}{4} = -2.5 \text{ m/s}^2 \]

✓ (M1 dep) For a complete method to find the gradient, e.g., \(\frac{0 – 10}{7 – 3}\).

✓ (A1) For answer in the range -2.0 to -2.8 (or 2.0 to 2.8), dependent on tangent drawn.

✍️ Working:

Reading velocities from the graph (using MS implied values for accuracy):

\[ v_0 = 25, v_2 = 16, v_4 = 9, v_6 = 4 \]

Trapezium Rule formula: Area \(\approx \frac{h}{2} [y_0 + y_n + 2(y_1 + … + y_{n-1})]\)

\[ \text{Distance} \approx \frac{2}{2} [25 + 4 + 2(16 + 9)] \] \[ \text{Distance} \approx 1 [29 + 2(25)] \] \[ \text{Distance} \approx 29 + 50 = 79 \text{ m} \]

✓ (M1) For a method to find an estimate for the area of at least 1 trapezium.

✓ (M1) For a complete method.

✓ (A1) For 79 or 78.

✍️ Working:

\[ \frac{1}{\sqrt{a} + 1} = \frac{1}{\sqrt{a} + 1} \times \frac{\sqrt{a} – 1}{\sqrt{a} – 1} \]

✓ (M1) For a correct method to rationalise the denominator, e.g., multiplying by \(\frac{\sqrt{a} – 1}{\sqrt{a} – 1}\).

✍️ Working:

\[ \frac{1 \times (\sqrt{a} – 1)}{(\sqrt{a} + 1)(\sqrt{a} – 1)} = \frac{\sqrt{a} – 1}{(\sqrt{a})^2 – 1^2} \] \[ \frac{\sqrt{a} – 1}{a – 1} \]

✓ (A1) For the correct answer in its simplest form.

✍️ Working:

Set the expression equal to zero:

\[ (4x – 3)(x + 5) = 0 \]

First factor: \(4x – 3 = 0 \implies 4x = 3 \implies x = \frac{3}{4}\)

Second factor: \(x + 5 = 0 \implies x = -5\)

The critical values are \(-5\) and \(\frac{3}{4}\).

✓ (M1) For critical values \(-5\) and \(\frac{3}{4}\) oe.

✍️ Working:

\[ -5 < x < \frac{3}{4} \]

✓ (A1) For the correct range (oe, e.g., \(x > -5\) and \(x < 0.75\)).

✍️ Working:

Mass ratio L : M = \(64 : 125\)

Length ratio (height L : height M):

\[ k_{L:M} = \sqrt[3]{64} : \sqrt[3]{125} \] \[ k_{L:M} = 4 : 5 \]

✓ (P1) For process to find ratio of heights of L and M, e.g., \(\sqrt[3]{64} : \sqrt[3]{125}\).

✍️ Working:

Area ratio M : P = \(144 : 16\)

Length ratio (height M : height P):

\[ k_{M:P} = \sqrt{144} : \sqrt{16} \] \[ k_{M:P} = 12 : 4 \]

Simplify M : P = \(3 : 1\)

✓ (P1) For process to find ratio of heights of M and P, e.g., \(\sqrt{144} : \sqrt{16}\).

✍️ Working:

L : M = \(4 : 5\) (\(\times 3\)) \(\implies 12 : 15\)

M : P = \(3 : 1\) (\(\times 5\)) \(\implies 15 : 5\)

Combined ratio L : M : P:

\[ 12 : 15 : 5 \]

✓ (P1 dep on P2) For process to find ratio of heights of all 3, e.g., matching the M terms.

✓ (A1) For \(12 : 15 : 5\) oe.

✍️ Working:

\[ P(\text{RRR}) = \frac{4}{8} \times \frac{3}{7} \times \frac{2}{6} = \frac{24}{336} \]

✓ (P1) For a correct product for 3 red.

✍️ Working:

\[ P(\text{RRG}) = \frac{4}{8} \times \frac{3}{7} \times \frac{1}{6} = \frac{12}{336} \] \[ P(\text{RGR}) = \frac{4}{8} \times \frac{1}{7} \times \frac{3}{6} = \frac{12}{336} \] \[ P(\text{GRR}) = \frac{1}{8} \times \frac{4}{7} \times \frac{3}{6} = \frac{12}{336} \]

Total P(2R, 1G) = \(\frac{12}{336} + \frac{12}{336} + \frac{12}{336} = \frac{36}{336}\)

✓ (P1) For a correct product for 2 red and 1 green (any order).

✓ (P1) For a correct probability for 2nd or 3rd counter (Implied by above calculation).

✍️ Working:

\[ P(\text{Success}) = P(\text{RRR}) + P(\text{2R, 1G}) \] \[ P(\text{Success}) = \frac{24}{336} + \frac{36}{336} \] \[ P(\text{Success}) = \frac{60}{336} \]

Simplify by dividing by 12:

\[ P(\text{Success}) = \frac{5}{28} \]

✓ (P1) For a complete process (summing the probabilities).

✓ (A1) For \(\frac{5}{28}\) oe.

✍️ Working:

\[ \vec{OM} = \frac{1}{2} \vec{OA} = \frac{1}{2}\mathbf{a} \] \[ \vec{OC} = \vec{OA} + \vec{AC} = \mathbf{a} + k\mathbf{b} \]

✍️ Working:

\[ \vec{BC} = \vec{OC} – \vec{OB} = (\mathbf{a} + k\mathbf{b}) – \mathbf{b} = \mathbf{a} + (k – 1)\mathbf{b} \] \[ \vec{ON} = \vec{OB} + \vec{BN} = \mathbf{b} + \frac{4}{9} \vec{BC} \] \[ \vec{ON} = \mathbf{b} + \frac{4}{9} (\mathbf{a} + (k – 1)\mathbf{b}) \] \[ \vec{ON} = \frac{4}{9}\mathbf{a} + \left(\frac{9}{9} + \frac{4k – 4}{9}\right)\mathbf{b} = \frac{4}{9}\mathbf{a} + \frac{4k + 5}{9}\mathbf{b} \]

✓ (P1) For a correct expression for \(\vec{BC}\) or \(\vec{BN}\).

✓ (P1) For a correct unsimplified expression for \(\vec{MN}\) (implied by correct final components).

✍️ Working:

\[ \vec{MN} = \vec{ON} – \vec{OM} \] \[ \vec{MN} = \left(\frac{4}{9}\mathbf{a} + \frac{4k + 5}{9}\mathbf{b}\right) – \frac{1}{2}\mathbf{a} \] \[ \vec{MN} = \left(\frac{8}{18}\mathbf{a} – \frac{9}{18}\mathbf{a}\right) + \left(\frac{8k + 10}{18}\right)\mathbf{b} \] \[ \vec{MN} = -\frac{1}{18}\mathbf{a} + \frac{8k + 10}{18}\mathbf{b} \] \[ \vec{MN} = \frac{1}{18} (-\mathbf{a} + (8k + 10)\mathbf{b}) \text{ or } \frac{1}{18} (8k\mathbf{b} + 10\mathbf{b} – \mathbf{a}) \]

✓ (A1) For the correct simplified answer.

✍️ Working:

\[ \vec{MN} = -\frac{1}{18}\mathbf{a} + \frac{4k + 5}{9}\mathbf{b} \]

Since \(\vec{MN}\) contains the non-zero term \(-\frac{1}{18}\mathbf{a}\), it cannot be written as a scalar multiple of \(\vec{OB} = \mathbf{b}\).

Conclusion: No.

✓ (C1) No, with supporting reason (e.g., since \(\vec{MN}\) is not a multiple of \(\mathbf{b}\)).

✍️ Working (Completing the Square):

\[ y = 2x^2 – 12x + 7 \]

Factor out 2 from the first two terms:

\[ y = 2(x^2 – 6x) + 7 \]

Complete the square for \(x^2 – 6x\):

\[ y = 2( (x – 3)^2 – 9 ) + 7 \]

Distribute the 2:

\[ y = 2(x – 3)^2 – 18 + 7 \] \[ y = 2(x – 3)^2 – 11 \]

The turning point occurs when \(x – 3 = 0\), so \(x=3\).

✓ (P1) For start of a process to complete the square or identifying \(x=3\).

✍️ Working:

Using the vertex form \(y = 2(x – 3)^2 – 11\), when \(x=3\):

\[ y = 2(3 – 3)^2 – 11 \] \[ y = 2(0) – 11 \] \[ y = -11 \]

✓ (P1) For substituting \(x=3\) to find \(y\).

✍️ Working:

Original key points: \((-1, 1)\), \((1, -2)\), \((3, 1)\).

New points:

\[ (-1, 1) \to (1, 3) \] \[ (1, -2) \to (-1, 0) \] \[ (3, 1) \to (-3, 3) \]

The transformed graph is drawn through these three new points.

✓ (C2) For a fully correct transformation.

✍️ Working:

Since \(AP\) and \(BP\) are tangents, we have:

\[ \angle OAP = 90^\circ \] \[ \angle OBP = 90^\circ \]

Reason: Angle between radius and tangent is \(90^\circ\).

✓ (C1) For \(\angle OAP = \angle OBP = 90^\circ\) and stating the reason.

✍️ Working:

1. \(OP\) is common to both \(\triangle OAP\) and \(\triangle OBP\).

2. \(OA = OB\)

Reason: Both are radii of the same circle.

✓ (C1) For \(OA = OB\) (both radii).

✓ (C1) For \(OP\) is common.

✍️ Working:

\(\triangle OAP\) is congruent to \(\triangle OBP\) by RHS congruence.

Since the triangles are congruent, their corresponding sides are equal:

\[ AP = BP \]

✓ (C1 dep on C3) For a complete proof with all reasons given, concluding that the triangles are congruent (RHS) implies \(AP = BP\).