Mathematics Specification B 4MB1 | Paper 2 Mock 01 | Written Paper 2

Three standard formulae are needed here: the volume of a cone \(V = \frac{1}{3}\pi r^2 h\), the curved surface area of a cone \(A = \pi r l\) where \(l\) is the slant height, and the volume of a sphere \(V = \frac{4}{3}\pi r^3\). The final part compares two of them.

(a) Volume of the cone. [2]
Radius 3 cm, vertical height 11 cm:

\[V = \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3} \times \pi \times 3^2 \times 11 = \tfrac{1}{3} \times \pi \times 9 \times 11 = 33\pi\] \[V = 103.67\ldots = 104\ \text{cm}^3\ \text{(3 s.f.)}\]

(b) Slant height and curved surface area. [3]
The slant height is the distance from the rim to the point, and it is the hypotenuse of a right-angled triangle whose legs are the radius and the vertical height:

\[l = \sqrt{r^2 + h^2} = \sqrt{3^2 + 11^2} = \sqrt{9 + 121} = \sqrt{130} = 11.4017\ldots = 11.4\ \text{cm}\]

Then

\[A = \pi r l = \pi \times 3 \times 11.4017\ldots = 107.47\ldots = 107\ \text{cm}^2\ \text{(3 s.f.)}\]

The vertical height 11 cm and the slant height 11.4 cm are close but not the same, and the curved surface formula needs the slant one. Marks: one for the Pythagoras step, one for the slant height, one for the surface area.

(c) Volume of the scoop. [2]
A sphere of radius 3 cm:

\[V = \tfrac{4}{3}\pi r^3 = \tfrac{4}{3} \times \pi \times 3^3 = \tfrac{4}{3} \times \pi \times 27 = 36\pi\] \[V = 113.09\ldots = 113\ \text{cm}^3\ \text{(3 s.f.)}\]

Note \(r^3\), not \(r^2\): the radius is cubed for a sphere.

(d) Will the cone overflow? [2]
Compare the melted ice cream with the space inside the cone:

\[\text{scoop } 113.09\ldots\ \text{cm}^3 \ >\ \text{cone } 103.67\ldots\ \text{cm}^3\]

The scoop has the larger volume, so the cone does overflow. The excess is

\[113.09\ldots - 103.67\ldots \approx 9.4\ \text{cm}^3\]

which is roughly 9% more than the cone can hold.

The reason is worth noticing. The cone and the sphere have the same radius, but the sphere of radius 3 cm has volume \(36\pi\) while the cone needs a height of \(h\) with \(\frac{1}{3}\pi \times 9 \times h = 36\pi\), that is \(h = 12\) cm, to match it. The cone here is only 11 cm tall, so it falls just short.

Examination point: use the unrounded volumes for the comparison in part (d). Rounded to 3 significant figures the two values are 113 cm³ and 104 cm³, which still gives the right conclusion, but in a closer question rounding first can reverse the decision.

Three-dimensional problems are solved by picking out flat right-angled triangles. In a cuboid, a vertical edge is at right angles to every line drawn on the horizontal face, which is what makes each triangle right-angled.

(a) Length of AC. [2]
AC is a diagonal of the horizontal rectangular floor ABCD:

\[AC^2 = AB^2 + BC^2 = 45^2 + 28^2 = 2025 + 784 = 2809\] \[AC = \sqrt{2809} = 53\ \text{m}\]

This value is exact, which makes the later parts cleaner.

(b) Length of AG. [3]
G is directly above C, so CG is vertical and equal to the height, 21 m. Triangle ACG lies in a vertical plane and is right-angled at C:

\[AG^2 = AC^2 + CG^2 = 2809 + 21^2 = 2809 + 441 = 3250\] \[AG = \sqrt{3250} = 57.008\ldots = 57.0\ \text{m (3 s.f.)}\]

(c) Angle between AG and the floor. [3]
The angle a line makes with a plane is the angle between the line and its shadow on that plane. The point G projects straight down onto C, so the shadow of AG on the floor is AC, and the required angle is angle GAC:

\[\tan(\angle GAC) = \frac{CG}{AC} = \frac{21}{53} = 0.396226\] \[\angle GAC = \tan^{-1}(0.396226) = 21.61\ldots = 21.6^{\circ}\ \text{(3 s.f.)}\]

(d) Angle GAB. [3]
Work in triangle ABG. First find BG, which lies in the vertical face BCGF and is right-angled at C:

\[BG^2 = BC^2 + CG^2 = 28^2 + 21^2 = 784 + 441 = 1225 \quad\Rightarrow\quad BG = 35\ \text{m}\]

The edge AB is perpendicular to the whole face BCGF, so angle ABG is a right angle. In triangle ABG:

\[\tan(\angle GAB) = \frac{BG}{AB} = \frac{35}{45} = 0.77778\] \[\angle GAB = \tan^{-1}(0.77778) = 37.87\ldots = 37.9^{\circ}\ \text{(3 s.f.)}\]

(e) Show the cable is not longer than 60 m. [2]
From part (b), \(AG = 57.008\ldots\) m. Since \(57.0 < 60\), the cable is shorter than 60 m and the engineer is wrong, by about 3 m.

Examination point: the angle between a line and a plane is never measured to an edge of the solid but to the projection of the line onto the plane. Here A to C is that projection, so angle GAC is the answer and angle GAB, which is between two lines rather than a line and a plane, is a different and larger angle.

Triangle \(Q\) has vertices (1, 1), (4, 1) and (1, 3). Both transformations here have a centre or mirror line away from the origin, so the rules must be built from the geometry rather than quoted: a half turn about \((a, b)\) sends \((x, y)\) to \((2a - x, \ 2b - y)\), and a reflection in \(y = b\) sends \((x, y)\) to \((x, \ 2b - y)\).

1. Rotation of 180 degrees about (3, 2) [3 marks]
   The centre is (3, 2), so the rule is \[(x, y) \rightarrow (6 - x, \ 4 - y) \tag{1 mark}\] \[(1, 1) \rightarrow (5, 3), \qquad (4, 1) \rightarrow (2, 3) \tag{1 mark}\] \[(1, 3) \rightarrow (5, 1) \tag{1 mark}\] The image has vertices (5, 3), (2, 3) and (5, 1). Each image point is the same distance from (3, 2) as its original but on the opposite side, which is a useful check: (1, 1) is 2 left and 1 below the centre, and (5, 3) is 2 right and 1 above it.
2. Reflection in the line \(y = 2\) [2 marks]
   The mirror is horizontal, so \(x\) is unchanged and \(y\) maps to \(4 - y\): \[(1, 1) \rightarrow (1, 3), \qquad (4, 1) \rightarrow (4, 3) \tag{1 mark}\] \[(1, 3) \rightarrow (1, 1) \tag{1 mark}\] The second image has vertices (1, 3), (4, 3) and (1, 1).
3. Transformation from the second image to the first [2 marks]
   Match the vertices: (1, 3) goes to (5, 3), (4, 3) goes to (2, 3) and (1, 1) goes to (5, 1). The \(y\) coordinates are unchanged and each \(x\) becomes \(6 - x\), so the mirror line is \(x = 3\). The single transformation is a reflection in the line \(x = 3\). This is consistent with the earlier parts, since a reflection in \(y = 2\) followed by a reflection in \(x = 3\) gives the half turn about their crossing point (3, 2).
4. Why all three triangles have equal areas [2 marks]
   A rotation and a reflection are both isometries: they preserve every length and every angle. Each image is therefore congruent to \(Q\), and congruent shapes have equal areas. Only an enlargement with scale factor other than \(\pm 1\) would change the area. Here that area is \(\tfrac{1}{2} \times 3 \times 2 = 3\) square units for all three triangles.

Examination reminder: two reflections in perpendicular lines always combine to give a half turn about the point where the lines cross, which is a quick way to check answers like these against each other.

Two journeys over the same distance at different speeds, with a stated difference in time, always lead to a quadratic. The relation used throughout is \(\text{time} = \dfrac{\text{distance}}{\text{speed}}\).

(a) Time on Monday. [1]

\[t_{\text{Mon}} = \frac{84}{v}\ \text{hours}\]

(b) Show that \(v^2 + 7v - 588 = 0\). [4]
Tuesday's speed is \((v + 7)\) km/h, so Tuesday's time is \(\dfrac{84}{v + 7}\) hours. Tuesday was the faster crossing, so it was the shorter one, and the difference is 1 hour:

\[\frac{84}{v} - \frac{84}{v + 7} = 1\]

Multiply every term by \(v(v + 7)\):

\[84(v + 7) - 84v = v(v + 7)\] \[84v + 588 - 84v = v^2 + 7v\] \[588 = v^2 + 7v \quad\Rightarrow\quad v^2 + 7v - 588 = 0\]

Marks: one for Tuesday's time, one for the equation with the subtraction the right way round, one for clearing fractions, one for the final form. Subtracting in the wrong order gives \(-1\) on the right and leads to \(v^2 + 7v + 588 = 0\), which has no real roots at all.

(c) Solve for \(v\). [3]
Using the formula with \(a = 1\), \(b = 7\), \(c = -588\):

\[b^2 - 4ac = 49 + 2352 = 2401, \qquad \sqrt{2401} = 49\] \[v = \frac{-7 \pm 49}{2}\] \[v = \frac{42}{2} = 21 \qquad\text{or}\qquad v = \frac{-56}{2} = -28\]

A speed cannot be negative, so \(v = 21\) km/h.

(d) Time on Tuesday. [2]

\[\text{speed} = 21 + 7 = 28\ \text{km/h}, \qquad t = \frac{84}{28} = 3\ \text{hours}\]

Check the story: Monday took \(\frac{84}{21} = 4\) hours, and \(4 - 3 = 1\) hour, exactly the stated difference.

(e) Average speed of the Islander. [2]
On the distance-time graph the line for the Islander runs from the origin up to 84 km at 3.5 hours. Its gradient is the average speed:

\[\text{speed} = \frac{84}{3.5} = 24\ \text{km/h}\]

(f) Does the Islander meet the rule? [3]
The rule is that a crossing must take less than 3 hours 30 minutes, that is less than 3.5 hours. The Islander takes exactly 3.5 hours, which is not less than 3.5, so it does not meet the rule.

For a crossing under 3.5 hours the speed must satisfy

\[\frac{84}{v} < 3.5 \quad\Rightarrow\quad 84 < 3.5v \quad\Rightarrow\quad v > 24\]

The least whole number speed that works is 25 km/h, which gives a crossing time of \(\frac{84}{25} = 3.36\) hours, that is 3 hours 21.6 minutes.

Examination point: "less than" excludes the boundary value. An exact 3.5 hour crossing fails a "less than 3.5 hours" rule, and \(v > 24\) means 24 km/h itself is not allowed, so the least whole number is 25 and not 24.

A walkway of the same width all the way round reduces each dimension of the pitch by \(x\) at both ends, so each one loses \(2x\), not \(x\). Everything else follows from that.

(a) Length and width of the display. [2]
The pitch is 18 m by 12 m and the walkway is \(x\) m wide on all four sides:

\[\text{length} = (18 - 2x)\ \text{m}, \qquad \text{width} = (12 - 2x)\ \text{m}\]

(b) Show that \(x^2 - 15x + 26 = 0\). [3]
The display has area 112 m²:

\[(18 - 2x)(12 - 2x) = 112\]

Expand carefully, watching the signs:

\[216 - 36x - 24x + 4x^2 = 112\] \[4x^2 - 60x + 216 = 112 \quad\Rightarrow\quad 4x^2 - 60x + 104 = 0\]

Every coefficient is divisible by 4:

\[x^2 - 15x + 26 = 0\]

Marks: one for forming the product, one for the expansion, one for dividing through to the printed form. Both middle terms are negative, so they combine to \(-60x\); making one of them positive is the commonest error.

(c) Solve \(x^2 - 15x + 26 = 0\). [2]
Two numbers multiply to \(+26\) and add to \(-15\): those are \(-2\) and \(-13\).

\[(x - 2)(x - 13) = 0 \quad\Rightarrow\quad x = 2 \ \text{ or }\ x = 13\]

(d) Rejecting a root. [1]
Both roots are positive, so "widths cannot be negative" is not enough on its own here. Substitute into the display dimensions instead. If \(x = 13\) then the display width is \(12 - 2(13) = -14\) m, which is impossible, and the walkway would in any case be wider than the pitch itself. So \(x = 2\).

(e) Area of the walkway. [2]
The walkway is what is left of the pitch after the display is removed:

\[\text{pitch area} = 18 \times 12 = 216\ \text{m}^2\] \[\text{walkway area} = 216 - 112 = 104\ \text{m}^2\]

Check with \(x = 2\): the display measures \(18 - 4 = 14\) m by \(12 - 4 = 8\) m, and \(14 \times 8 = 112\) m², as stated.

Examination point: when both roots of a border problem are positive, test them in the reduced dimensions rather than quoting the usual "a length cannot be negative". The impossible root is the one that makes the inner rectangle vanish or turn negative.

A balance is a picture of an equation: whatever is on the left pan weighs the same as whatever is on the right pan. This question uses that idea to form a linear equation, solve it, use the result in a new balancing situation, and finish with an inequality.

(a) Equation from the balance. [2]
The left pan carries 3 bags and a 2 kg weight, so its total mass is \(3x + 2\) kg. The right pan carries 1 bag and an 8 kg weight, so its total mass is \(x + 8\) kg. Because the pans balance, the two totals are equal:

\[3x + 2 = x + 8\]

One mark for each side written correctly in terms of \(x\).

(b) Mass of one bag. [2]
Remove one bag from each pan (subtract \(x\)) and remove 2 kg from each pan:

\[3x - x = 8 - 2 \quad\Rightarrow\quad 2x = 6 \quad\Rightarrow\quad x = 3\]

Each bag has a mass of 3 kg. Check on the balance: the left pan holds \(3 \times 3 + 2 = 11\) kg and the right pan holds \(3 + 8 = 11\) kg, so it does balance. Doing exactly the same thing to both pans is what keeps the balance level, and that is why the same operation must be applied to both sides of an equation.

(c) Value of \(w\). [2]
Now the left pan holds 5 bags, mass \(5 \times 3 = 15\) kg. The right pan holds 1 bag plus weights totalling \(w\) kg, mass \(3 + w\) kg. Balancing:

\[15 = 3 + w \quad\Rightarrow\quad w = 12\]

So 12 kg of weights are needed. In effect the four extra bags on the left must be matched, and \(4 \times 3 = 12\) kg confirms it.

(d) Least number of bags with total mass at least 20 kg. [2]
If there are \(n\) bags, the total mass is \(3n\) kg, and "at least 20 kg" means

\[3n \ge 20 \quad\Rightarrow\quad n \ge 6.66\ldots\]

Bags come whole, and the number must be at least 6.66..., so \(n = 7\). Checking: 6 bags weigh 18 kg, which is under 20 kg, while 7 bags weigh 21 kg, which meets the condition. Marks: one for the inequality, one for rounding up to 7.

Examination point: the rounding direction depends on the inequality, not on the decimal. Here 6.66... rounds up to 7 because the total must reach 20 kg, even though ordinary rounding of a "just over 6" value might tempt you to write 6.

The profit model \(P = -2x^2 + bx + c\) is a downward parabola in the ticket price \(x\) pounds. Two facts are supplied: the vertex (stationary point) is at \(x = 15\), and the profit there is 250 pounds. Each fact yields one equation, which is enough to pin down the two unknown constants.

1. Show that b = 60 [2 marks]
   Differentiate the model with respect to \(x\): \[\frac{dP}{dx} = -4x + b\] At a stationary point the gradient is zero, and this happens at \(x = 15\): \[-4(15) + b = 0 \quad \Rightarrow \quad -60 + b = 0 \quad \Rightarrow \quad b = 60\] as required. (The same value follows from the symmetry formula \(x = -\dfrac{b}{2a}\) with \(a = -2\).)
2. Find the value of c [2 marks]
   Substitute \(x = 15\) and \(b = 60\) into the model and set the profit to 250: \[P(15) = -2(15)^2 + 60(15) + c = -450 + 900 + c = 450 + c\] \[450 + c = 250 \quad \Rightarrow \quad c = -200\] So the model is \(P = -2x^2 + 60x - 200\). The negative constant makes sense: at a price of zero the club would make a loss of 200 pounds, its fixed costs for staging the match.
3. Solve P = 0 [3 marks]
   \[-2x^2 + 60x - 200 = 0\] Dividing through by \(-2\) simplifies the arithmetic: \[x^2 - 30x + 100 = 0 \tag{1 mark}\] This does not factorise, so use the quadratic formula with \(a = 1\), \(b = -30\), \(c = 100\): \[x = \frac{30 \pm \sqrt{(-30)^2 - 4(1)(100)}}{2} = \frac{30 \pm \sqrt{900 - 400}}{2} = \frac{30 \pm \sqrt{500}}{2} \tag{1 mark}\] Since \(\sqrt{500} = 22.3606\ldots\), \[x = \frac{30 + 22.3606}{2} = 26.18\ldots \quad \text{or} \quad x = \frac{30 - 22.3606}{2} = 3.8196\ldots\] To 3 significant figures the ticket prices are £3.82 and £26.2.
4. Interpretation [2 marks]
   These two prices are the break-even prices: at either of them the model predicts a profit of exactly zero. Because the parabola opens downwards, \(P\) is positive only between the roots, so the club makes a profit only for prices strictly between about £3.82 and £26.20. Charge less and the ticket income fails to cover the fixed costs; charge more and (in the model) too few supporters come. The most profitable price is the vertex, £15, giving £250.

Examination reminder: for a negative quadratic, the region of positive values lies between the roots, so break-even points define the whole permitted range of prices.

This formula combines two working rates. The important idea is that the algebra of rearranging is the same whatever the letters mean, but the context lets you check whether the rearranged form makes sense.

(a) Value of \(T\) when \(a = 30\) and \(b = 20\). [2]
Substitute into \(T = \dfrac{ab}{a + b}\), working out the top and the bottom separately before dividing:

\[T = \frac{30 \times 20}{30 + 20} = \frac{600}{50} = 12\ \text{minutes}\]

The answer is sensible: two machines together clear the trolley in 12 minutes, which is faster than either machine on its own.

(b) Make \(a\) the subject. [3]
The letter \(a\) appears twice, once on the top and once on the bottom, so the fraction must be cleared first and the \(a\) terms then collected. Multiply both sides by \((a + b)\):

\[T(a + b) = ab\] \[Ta + Tb = ab\]

Gather the terms containing \(a\) on one side and everything else on the other:

\[Tb = ab - Ta\]

Factorise the right-hand side, since \(a\) is common to both terms:

\[Tb = a(b - T)\] \[a = \frac{Tb}{b - T}\]

Marks: one for clearing the fraction, one for collecting the \(a\) terms on one side, one for factorising and dividing. Factorising is the step that is usually missed; without it \(a\) cannot be isolated at all.

(c) Value of \(a\) when \(T = 8\) and \(b = 24\). [1]

\[a = \frac{8 \times 24}{24 - 8} = \frac{192}{16} = 12\ \text{minutes}\]

Check in the original formula: \(\dfrac{12 \times 24}{12 + 24} = \dfrac{288}{36} = 8\) minutes, as required.

(d) Why \(T\) must always be smaller than \(b\). [1]
In the rearranged formula \(a = \dfrac{Tb}{b - T}\), the value \(a\) is a time taken by a real machine, so it must be positive. The numerator \(Tb\) is positive because both \(T\) and \(b\) are positive times. For the whole fraction to be positive, the denominator \(b - T\) must also be positive, which means

\[b - T > 0 \quad\Rightarrow\quad T < b\]

This matches common sense: adding a second machine can only make the job faster, so the combined time must be less than the time the second machine would take alone.

Examination point: when the subject appears more than once, the routine is always the same: clear fractions, collect that letter on one side, factorise it out, then divide. Recognise that shape early rather than trying to move terms one at a time.

A bush is a cylinder with a smaller cylinder removed, so its volume is the difference of the two. Both cylinders have the same length, which lets the length be factored out and keeps the arithmetic exact.

(a) Show the volume is \(25920\pi\) mm³. [3]
Outer cylinder of radius 30 mm minus inner cylinder of radius 18 mm, both 45 mm long:

\[V = \pi \times 30^2 \times 45 - \pi \times 18^2 \times 45\]

Take out the common factor \(45\pi\):

\[V = 45\pi(900 - 324) = 45\pi \times 576 = 25920\pi\ \text{mm}^3\]

Marks: one for each cylinder volume, one for the difference in the required form. Subtracting the radii first, as \(\pi(30 - 18)^2 \times 45\), is a common error and gives a far smaller answer; the squares must be subtracted, not the radii.

(b) Volume in cm³. [2]

\[25920\pi = 81\,430.09\ldots\ \text{mm}^3\]

Since \(1\ \text{cm} = 10\ \text{mm}\), one cubic centimetre is \(10 \times 10 \times 10 = 1000\) mm³:

\[\frac{81\,430.09}{1000} = 81.43009\ \text{cm}^3 = 81.4\ \text{cm}^3\ \text{(3 s.f.)}\]

The conversion factor for volume is 1000, not 10; that is the step this part is testing.

(c) Mass of one bush. [2]
Mass is density times volume:

\[m = 81.43009 \times 7.8 = 635.155\ldots = 635\ \text{g (nearest gram)}\]

(d) Cost of the steel in one bush. [2]
The price is per kilogram, so convert the mass:

\[635.155\ \text{g} = 0.635155\ \text{kg}\] \[\text{cost} = 0.635155 \times 4.20 = 2.6677\ldots\]

The steel costs £2.67 to the nearest penny.

(e) Steel cost as a percentage of the selling price. [1]

\[\frac{2.6677}{9.50} \times 100 = 28.08\ldots = 28.1\%\ \text{(3 s.f.)}\]

Examination point: keep the exact form \(25920\pi\) until a decimal is actually needed. Rounding to 81 400 mm³ at the end of part (a) would shift the mass by about 0.2 g and could change the rounded cost in part (d).