Mathematics Specification A 4MA1 | Paper 4 Mock 01 | Structured / Short Answer

(a) \(OA\) is a radius and \(AT\) is tangent to the bed at \(A\), so angle \(OAT = 90^{\circ}\), a tangent always meeting its radius at right angles. [1 mark]

(b) Triangle \(OAT\) is right-angled at \(A\), so by Pythagoras, \(OT = \sqrt{OA^{2}+AT^{2}} = \sqrt{20^{2}+24^{2}} = \sqrt{976} = 31.2\) m, correct to 3 significant figures. Using the same triangle, \(\tan(\text{angle }ATO) = \dfrac{OA}{AT} = \dfrac{20}{24}\), so angle \(ATO = \tan^{-1}(0.8\overline{3}) = 39.8^{\circ}\), correct to 3 significant figures. [3 marks]

(c) The figure is symmetrical about line \(OT\), so angle \(ATB\) is twice angle \(ATO\): angle \(ATB = 2 \times 39.8 = 79.6^{\circ}\). [2 marks]

(d) Quadrilateral \(OATB\) has angles summing to \(360^{\circ}\): two right angles at \(A\) and \(B\), plus \(79.6^{\circ}\) at \(T\), leaves angle \(AOB = 360-90-90-79.6 = 100.4^{\circ}\). [1 mark]

(e) \(A\), \(B\) and \(C\) all lie on the flower bed's circle, and \(C\) is on the major arc \(AB\), so the angle at the centre, angle \(AOB\), is twice the angle at the circumference, angle \(ACB\): angle \(ACB = 100.4 \div 2 = 50.2^{\circ}\), correct to 3 significant figures. [1 mark]

(f) No. \(O\) is the centre of the circle on which \(A\), \(B\) and \(C\) lie, and the centre of a circle is never a point on the circle itself, since every point on the circle is a full radius away from the centre, not zero. [1 mark]

Part (f) checks a conceptual point rather than a calculation: a "centre" and a "point on the circumference" are different roles, so \(O\) cannot join \(A\), \(B\), \(C\) on their circle even though it governs the angles between them.

Sharing a total distance equally among a fleet of buses means dividing the total by the number of buses; carrying out this division using the standard-form value, then re-expressing the result in standard form, keeps the working exact throughout.

Evaluating the total distance: \(4.8\times10^{5}=480\,000\) km. [1 mark]

Dividing by the number of buses: \(480\,000\div480=1000\). [1 mark]

Writing \(1000\) in standard form: \(1000=1\times10^{3}\) km. [1 mark]

The fleet size, \(480\), divides exactly into \(480\,000\) because the digits \(4.8\) in the standard-form total match the fleet size's own leading digits, which is a useful mental check before dividing.

To add or subtract surds, each one must first be written in terms of the same surd, by taking out the largest perfect square factor from underneath each root.

Simplify each edging length: \(\sqrt{45} = \sqrt{9\times5} = 3\sqrt5\), and \(\sqrt{20} = \sqrt{4\times5} = 2\sqrt5\); \(\sqrt5\) is already in simplest form. [1 mark] Now every term is a multiple of \(\sqrt5\), so they can be combined like ordinary algebraic terms: \(3\sqrt5 - 2\sqrt5 + \sqrt5 = (3-2+1)\sqrt5 = 2\sqrt5\). [2 marks] Comparing with the form \(k\sqrt5\) gives \(k=2\).

Exam tip: surds can only be combined once they share the same number under the root sign - always simplify every surd in an expression before adding or subtracting.

Each bracket must be expanded separately before the subtraction is carried out, since subtracting an entire bracketed expression means changing the sign of every term inside it.

Expanding each product: \(x(x+9) = x^{2}+9x\) and \(x(x-3) = x^{2}-3x\). [1 mark]

Subtracting the second bracket (changing every one of its signs): \(x^{2}+9x-(x^{2}-3x) = x^{2}+9x-x^{2}+3x\). [1 mark]

The \(x^{2}\) terms cancel, leaving \(9x+3x = 12x\). [1 mark]

Exam tip: subtracting a whole bracket changes the sign of every term inside it, not just the first one - a common error here would be writing \(x^2+9x-x^2-3x\) instead of \(x^2+9x-x^2+3x\).

(a) The total area is the rectangle plus the semicircular herb bed. The rectangle has area \(8x\), and the semicircle, of diameter \(8\) m so radius \(4\) m, has area \(\dfrac{1}{2}\pi(4)^{2} = 8\pi\). Setting the total equal to \(100\) m\(^{2}\) gives \(8x + 8\pi = 100\), so \(8x = 100-8\pi\), giving \(x = \dfrac{100-8\pi}{8} = 9.36\) m, correct to 3 significant figures. [3 marks]

(b) The perimeter is made up of the two long sides, each of length \(x\), the one remaining short side of length \(8\) m, and the curved semicircular arc that replaces the other short side, of length \(\pi \times 4\). Using the unrounded value \(x = 9.3584...\): perimeter \(= 2(9.3584...) + 8 + 4\pi = 39.3\) m, correct to 3 significant figures. [3 marks]

(c) The fencing cost, using the unrounded perimeter \(39.283...\) m, is \(39.283... \times \pounds6 = \pounds235.70\), which is more than the \(\pounds150\) budget, so the allotment holder does not have enough. [2 marks]

Using the full, unrounded value of \(x\) in part (b), rather than the value already rounded to 3 significant figures, keeps the perimeter accurate; only the final answer in each part should be rounded.

Vertically opposite angles, formed where two straight lines cross, are always equal; this fact turns the two expressions for the marked angles directly into an equation in \(x\).

Setting the two vertically opposite angles equal: \(6x-10=4x+30\). [1 mark]

Subtracting \(4x\) from both sides and adding \(10\) to both sides: \(2x=40\), so \(x=20\). [1 mark]

Checking both expressions with \(x=20\): \(6(20)-10=110\) and \(4(20)+30=110\), which agree, confirming the vertically opposite angles are indeed equal.

The diagram shows the long diagonal of a cuboid, running from one bottom corner to the opposite top corner. This diagonal is found in one step using the three-dimensional version of Pythagoras' theorem, which extends the two-dimensional case to three mutually perpendicular edges meeting at a corner.

\[ d = \sqrt{4^2 + 3^2 + 2^2} \] [1 mark]

\[ d = \sqrt{16 + 9 + 4} = \sqrt{29} \] [1 mark]

\[ d = 5.39 \text{ m (3 s.f.)} \] [1 mark]

An equivalent method finds the diagonal of the floor first, \(\sqrt{4^2+3^2}=5\) m, and then treats that floor diagonal and the \(2\) m height as the two legs of a second right-angled triangle: \(\sqrt{5^2+2^2}=\sqrt{29}\), the same result.

Reading the height of each bar against the labelled gridlines gives the mass sold each week; totals and ratios are then found directly from those readings.

The bar heights read from the chart's axis (in thousands of kg) are Week 1 \(=3.2\), Week 2 \(=4.5\), Week 3 \(=2.8\), Week 4 \(=5.1\).

(a) Adding all four weeks: \(3.2+4.5+2.8+5.1 = 15.6\) thousand kg \(= 15\,600\) kg. In standard form this is \(1.56\times10^{4}\) kg. [2 marks]

(b) Dividing Week 4's mass by Week 3's: \(5.1\div2.8 = 1.8214\ldots\), which rounds to \(1.82\) (3 s.f.). [2 marks]

Exam tip: "how many times greater" means divide the larger reading by the smaller one, not subtract them.

The support post's cap is a cylinder topped with a cone of the same radius; Pythagoras' theorem finds the cone's slant height from its radius and height, and the total volume equation (cylinder plus cone) can then be solved for the cylinder's unknown height \(H\).

(a) The cone's radius (7 cm), height (24 cm) and slant height form a right-angled triangle, so \(\text{slant}^{2}=7^{2}+24^{2}=49+576=625\). Taking the square root: slant \(=\sqrt{625}=25\) cm. [2 marks]

(b) The cylinder's volume is \(\pi(7)^{2}H=49\pi H\); the cone's volume is \(\dfrac13\pi(7)^{2}(24)=\dfrac13\pi(49)(24)=392\pi\). Adding and setting equal to the total: \(49\pi H+392\pi=882\pi\). Dividing every term by \(\pi\): \(49H+392=882\), so \(49H=490\) and \(H=10\) cm. [3 marks]

(c) The curved surface area is the cylinder's curved surface plus the cone's curved surface: \(2\pi(7)(10)+\pi(7)(25)=140\pi+175\pi=315\pi\) cm\(^{2}\). [1 mark]

(d) Evaluating \(315\pi\approx989.6\) cm\(^{2}\), which is less than \(1000\) cm\(^{2}\), so one roll of canopy material is enough to cover one post cap. [1 mark]

The cone's curved surface area formula, \(\pi r l\), uses the slant height \(l\) found in part (a), not the vertical height of 24 cm; mixing these two lengths up is a common error when a cone's surface area follows a "show that" height question.

Direct proportion means cost divides by mass to give a constant price per kilogram; once that constant is found from the known pair of values, it can be used to price any other mass.

\[ k = 6.25\div2.5 = \pounds2.50 \text{ per kg} \] [1 mark]

Cost of \(4\) kg: \[ 2.50\times4 = \pounds10.00 \] [2 marks]

Because cost is directly proportional to mass, doubling or scaling the mass by any factor scales the cost by exactly the same factor, which is the property that makes the price-per-kilogram method valid here.

To compare value for money, find the cost per single egg in each box. For the box of \(6\): \(2.16\div6=\pounds0.36\) per egg. [1 mark]

For the box of \(15\): \(5.10\div15=\pounds0.34\) per egg. [1 mark]

Since \(\pounds0.34\) is less than \(\pounds0.36\), each egg costs less in the larger box, so the box of \(15\) eggs gives better value for money. [1 mark]

Comparing unit prices, rather than the total prices of the two boxes, is the only fair way to judge value when the boxes contain different numbers of eggs.

Substituting a table value into the cost formula finds the missing entry \(k\); setting the formula equal to a target cost and rearranging (multiplying through by \(n\) to clear the fraction) then solves for \(n\).

(a) Substituting \(n=10\): \(C = \dfrac{2(10)+30}{10} = \dfrac{50}{10} = 5\), so \(k=5\). [1 mark]

(b) Setting \(C=2.5\): \(\dfrac{2n+30}{n}=2.5\). Multiplying both sides by \(n\): \(2n+30=2.5n\). Subtracting \(2n\) from both sides: \(30=0.5n\). Dividing by 0.5: \(n=60\) items. [3 marks]

An enlargement centred at the origin multiplies every coordinate of every vertex by the scale factor; comparing the resulting vertices with the boundary of the fenced plot at \((8,8)\) then decides whether the enlarged shape stays inside the fence.

(a) Applying scale factor \(2\) to shape A's vertices \((1,1)\), \((3,1)\), \((1,3)\): \((2,2)\), \((6,2)\), \((2,6)\). [2 marks]

(b) Applying scale factor \(3\) instead gives \((3,3)\), \((9,3)\), \((3,9)\). Since the vertex \((9,3)\) has an x-coordinate of \(9\), which is greater than the fence's limit of \(8\), this enlarged image does not fit entirely inside the fenced plot. [2 marks]

Only one vertex needs to fall outside the boundary for the whole enlarged shape to fail to fit; here it is enough to spot that \(9\gt8\) for the vertex \((9,3)\), without needing to check every vertex in detail.

The total bill is made up of a fixed standing charge plus a rate for every unit of electricity used, so setting up an equation with \(u\) as the number of units used and solving it finds how many units were used that month.

Setting up the equation for the total bill: \(18+0.72u=86.40\). [1 mark]

Subtracting the standing charge from both sides: \(0.72u=86.40-18=68.40\). [1 mark]

Dividing both sides by \(0.72\) isolates \(u\): \(u=68.40\div0.72=95\) units. [2 marks]

Checking the answer: \(18+0.72\times95=18+68.40=\pounds86.40\), which matches the given total bill exactly.

Each route forms a triangle with two known side lengths and the angle turned through at the middle point; the cosine rule, \(a^{2}=b^{2}+c^{2}-2bc\cos A\), converts each turning angle and pair of leg lengths into the direct "as the crow flies" distance, and comparing the two direct distances at a fixed speed decides which route is quicker.

(a) Applying the cosine rule to Route A's triangle: \(HS_2^{2}=6^{2}+9^{2}-2(6)(9)\cos78^{\circ}=36+81-108\cos78^{\circ}\approx94.5\), so \(HS_2\approx9.72\) km (3 s.f.). [2 marks]

(b) Applying the cosine rule to Route B's triangle: \(HS_4^{2}=7^{2}+8^{2}-2(7)(8)\cos95^{\circ}=49+64-112\cos95^{\circ}\approx122.8\), so \(HS_4\approx11.1\) km (3 s.f.). [2 marks]

(c) At the same flying speed, time is proportional to distance, so the route with the shorter direct distance is quicker: since \(9.72\lt11.1\), Route A is quicker. [2 marks]

The turning angle used in each cosine rule calculation is the angle actually turned through at the intermediate stop, not its supplement; using the wrong one of these two related angles is a common source of error whenever a journey's direct distance is found from two legs and a turn.

(a) The numerator, \(t^{2}-9\), is a difference of two squares, since \(9=3^{2}\), so it factorises as \((t-3)(t+3)\). Cancelling the common factor \((t-3)\) with the denominator (valid since \(t>3\) means \(t-3\) is never zero) gives \(S=t+3\). [2 marks]

(b) Substituting \(t=9\) into the simplified form: \(S=9+3=12\). [1 mark]

(c) Adding the fixed \(\pounds15\) joining bonus to \(S\): \(T(t)=(t+3)+15=t+18\). [1 mark]

(d) Setting \(T(t)=50\): \(t+18=50\), so \(t=32\). [2 marks]

(e) Since \(t=32\) months is longer than the \(24\)-month savings plan being considered, this value of \(t\) is not realistic within that plan. [1 mark]

The restriction \(t>3\) given in the question is exactly what makes the cancellation in part (a) valid; without it, \(t=3\) would need special treatment, since the original fraction would be \(\dfrac{0}{0}\) there.

The formula \(C=12+0.25m\) gives the monthly cost once the free minutes are used up, so it can be used forwards (substitute \(m\) to find \(C\)) or backwards (substitute a known \(C\) to find \(m\)).

(a) Substituting \(m=40\): \(C=12+0.25(40)=12+10=22\), so \(p=22\). [1 mark]

(b) Setting \(C=33\): \(12+0.25m=33\). Subtracting 12: \(0.25m=21\). Dividing by \(0.25\): \(m=84\) minutes beyond the allowance. [2 marks]

Because the cost rises by exactly \(\pounds0.25\) for every extra minute, each entry in the table rises by \(0.25\times20=5\) as \(m\) increases by 20, which is a useful check on the completed table: \(12, 17, 22, 27\) does indeed rise by 5 each time.

To add numbers given in standard form with different powers of \(10\), it is simplest to convert both to ordinary decimals first, add them, then convert the result back to standard form. Here \(2.3 \times 10^{-3} = 0.0023\) and \(4.7 \times 10^{-4} = 0.00047\), so their sum is \(0.0023+0.00047 = 0.00277\). [1 mark]

Writing \(0.00277\) in standard form, \(a \times 10^{n}\) with \(1 \leqslant a < 10\), gives \(2.77 \times 10^{-3}\). [1 mark]

Adding directly in standard form only works cleanly when the powers of \(10\) already match; since they differ here, converting to decimals first avoids a common error of adding the two numbers \(2.3\) and \(4.7\) without accounting for their different place values.

(a) Reading from the graph, the curve returns to \(s=0\) at \(t=6\) seconds (confirmed algebraically, since \(s=6t^{2}-t^{3}=t^{2}(6-t)\) is zero at \(t=0\) and \(t=6\)). [1 mark]

(b) Differentiating \(s=6t^{2}-t^{3}\) term by term gives \(\dfrac{ds}{dt}=12t-3t^{2}\). [1 mark]

(c) The velocity is the value of \(\dfrac{ds}{dt}\) at the given time. Substituting \(t=2\): \(12(2)-3(2)^{2}=24-12=12\) m/s. [2 marks]

(d) The car is momentarily at rest when its velocity is zero: \(12t-3t^{2}=0\), which factorises as \(3t(4-t)=0\), giving \(t=0\) or \(t=4\). Other than the start, this occurs at \(t=4\) s. The distance travelled by then is \(s=6(4)^{2}-(4)^{3}=96-64=32\) m. [2 marks]

The derivative gives the car's velocity at any instant; where that derivative is zero, the distance-time graph has a turning point, here a local maximum distance from the barrier before the car starts moving back towards it.

Combining two means correctly requires weighting each mechanic's mean by how many bikes they repaired, not simply averaging the two means; the box plots then let interquartile ranges be compared directly to judge consistency, independently of the combined mean.

1. \[ \text{combined mean} = \frac{n_A\bar{x}_A+n_B\bar{x}_B}{n_A+n_B} \] [2 marks]
2. \[ = \frac{15(22)+25(30)}{40} = \frac{330+750}{40} = \frac{1080}{40} = 27 \text{ minutes} \] [2 marks]
3. Mechanic A has the smaller interquartile range, \(8\) minutes compared with Mechanic B's \(15\) minutes, so Mechanic A has more consistent repair times. [2 marks]
4. Yes, the claim is fair, since the combined mean of \(27\) minutes is less than the advertised half hour (\(30\) minutes). [1 mark]

Because Mechanic B repaired more bikes (\(25\) against \(15\)) and took longer on average, the combined mean of \(27\) minutes sits closer to Mechanic B's \(30\)-minute average than to Mechanic A's \(22\)-minute average; simply averaging \(22\) and \(30\) to get \(26\) would ignore this and understate the true combined mean slightly.

A markup and a subsequent discount both act as multipliers on the price; combining the wholesale-to-retail markup with the sale discount into a single multiplier links the wholesale cost directly to the final sale price in one equation.

(a) A \(60\%\) markup gives a multiplier of \(1.6\); a \(25\%\) reduction gives a multiplier of \(0.75\). Applying both to the wholesale cost: \(C\times1.6\times0.75=54\), which combines to \(1.2C=54\). [2 marks]

(b) Dividing both sides by \(1.2\): \(C=54\div1.2=\pounds45\). [2 marks]

(c) Profit is the sale price minus the wholesale cost: \(54-45=\pounds9\). [1 mark]

(d) Percentage profit is the profit expressed as a percentage of the wholesale cost: \(\dfrac{9}{45}\times100=20\%\). [2 marks]

Even though the shop marks the price up by \(60\%\) and then reduces it by \(25\%\), the combined multiplier of \(1.2\) shows the sale price still ends up \(20\%\) above the wholesale cost, not at a loss, because the markup was applied to the smaller wholesale figure while the discount was applied to the larger, marked-up retail price.

Rounding to the nearest whole degree means the true temperature lies within half a degree either side of the displayed reading.

\[ 26.5 \leq t \lt 27.5 \] [2 marks]

The lower bound, \(26.5\), is included because a temperature of exactly \(26.5^\circ\)C would round up to \(27^\circ\)C, while the upper bound, \(27.5\), is excluded because a temperature of exactly \(27.5^\circ\)C would round up to \(28^\circ\)C rather than down to \(27^\circ\)C.