Mathematics Specification A 4MA1 | Paper 2 Mock 01 | Written Paper 2 (2F/2H)

This question compares two bus intervals given in different units, so a unit conversion is needed before the intervals can be compared directly.

(a) Route X's interval is \(2^1\) minutes: \[ 2^1 = 2 \text{ minutes} \] [1 mark]

(b) Route Y's interval is \(2^{-1}\) hours; evaluate the power first, then convert to minutes: \[ 2^{-1} \text{ hours} = \frac{1}{2} \text{ hour} = 0.5 \times 60 = 30 \text{ minutes} \] [2 marks]

(c) With both intervals in the same unit, divide the longer interval by the shorter one to count how many Route X buses fit between two Route Y buses: \[ 30 \div 2 = 15 \text{ buses} \] [2 marks]

The question is a reminder that index laws only combine values with a common base and common units; converting hours to minutes first is essential before any comparison or division makes sense.

Consecutive integers increase by 1 each time, so if the smallest is \(n\), the next two are \(n+1\) and \(n+2\). Writing their sum as an equation turns the word problem into algebra that can be solved directly.

The equation is \(n + (n+1) + (n+2) = 258\).

Collecting like terms on the left gives \(3n + 3 = 258\).

Subtracting 3 from both sides gives \(3n = 255\), and dividing by 3 gives \(n = 85\). [1 mark]

Since \(n\) is the smallest integer, the three consecutive integers are 85, 86 and 87. [1 mark]

A quick check confirms this: \(85 + 86 + 87 = 258\), matching the given total exactly. This "let the unknown be the smallest term" approach works for any set of consecutive integers, since every other term can then be written using only \(n\).

This question multiplies two powers of 10 to find an area, then converts that area into hectares by a further division of powers of 10.

(a) Area is length times width, and multiplying powers with the same base adds the exponents: \[ \text{Area} = 10^3 \times 10^2 = 10^{3+2} = 10^5 \] [2 marks]

(b) Evaluating the power gives the numerical area: \[ 10^5 = 100000 \text{ m}^2 \] [1 mark]

(c) Dividing by \(10^4\) to convert to hectares subtracts the exponents: \[ 10^5 \div 10^4 = 10^{5-4} = 10^1 = 10 \text{ hectares} \] [1 mark]

Keeping the area as a power of 10 throughout makes the hectare conversion a one-step exponent subtraction, rather than a division of the large number 100000 by 10000.

This question tests identifying the highest-frequency class from a grouped table, estimating the mean using midpoints, and locating the class containing the median from cumulative frequencies.

(a) The class \( 50 < c \leq 100 \) has the highest frequency, 20, so it has the highest frequency of the five classes. [1 mark]

(b) Using the midpoint of each class as a representative capacity:

\[ \sum fx = (8 \times 25) + (20 \times 75) + (18 \times 125) + (10 \times 175) + (4 \times 225) = 200 + 1500 + 2250 + 1750 + 900 = 6600 \]

Dividing by the 60 tanks gives the estimated mean:

\[ \frac{6600}{60} = 110 \text{ litres} \] [3 marks]

(c) The running (cumulative) totals are 8, 28, 46, 56, 60. With 60 tanks, the median lies between the 30th and 31st values; since the cumulative total reaches 28 by the end of \( 50 < c \leq 100 \) but 46 by the end of \( 100 < c \leq 150 \), both the 30th and 31st values fall in \( 100 < c \leq 150 \). [1 mark]

The class with the highest frequency and the class containing the median are not automatically the same class, since the median depends on where the running total crosses the halfway point of 60 tanks, not on which single class is individually largest.

This question tests multiplying a mass per batch by the number of batches, finding the cost of that mass, and comparing the cost with a fixed budget.

1. The total mass of cement is the mass per batch multiplied by the number of batches: \(3 \times 62.4 = 187.2\) kg [2 marks].
2. The total cost is the mass multiplied by the price per kilogram: \(187.2 \times £0.20 = £37.44\) [2 marks].
3. Since \(£37.44\) is less than the £40 set aside, the budget is enough [1 mark].

Finding the total mass first, before applying the price per kilogram, avoids having to multiply three numbers together in a single step.

This question tests reading a rainfall measurement from a vertical gauge scale and then adding it to a previous day's reading.

1. The gauge is marked every \(1\) cm with minor ticks every \(0.2\) cm. The water level sits three minor ticks below the \(3\) cm mark, so the rainfall shown is \(2.9\) cm [1 mark].
2. Adding the previous day's rainfall gives the two-day total: \(2.9 + 1.65 = 4.55\) cm [2 marks].

On a gauge measured from a zero at the bottom, the reading increases upwards, so the water level is compared against the nearest labelled mark above the zero line.

This question tests multiplying fractions to find "a fraction of a fraction" of the whole channel. Since \(\frac{4}{9}\) of the opened \(\frac{3}{8}\) of the channel feeds young crops, the fraction of the whole channel is \(\frac{3}{8} \times \frac{4}{9} = \frac{3 \times 4}{8 \times 9} = \frac{12}{72}\) [1 mark].

Simplifying by dividing the numerator and denominator by their highest common factor, 12, gives \(\frac{12}{72} = \frac{1}{6}\) [1 mark]. So one sixth of the whole channel feeds the young crops nearest the gate.

Every quadrilateral has interior angles summing to \(360^\circ\), since it can be split into two triangles, each contributing \(180^\circ\). Writing this fact as an equation lets the four angle expressions be combined and solved for \(x\).

1. \(2x + 3x + (x + 40) + (4x - 10) = 360\)
   Collecting the \(x\) terms and the numbers separately: \(10x + 30 = 360\). [2 marks]
2. \(10x = 330\)
   \(x = 33\). [2 marks]

As a check, substitute \(x = 33\) back into each angle: \(2(33)=66^\circ\), \(3(33)=99^\circ\), \(33+40=73^\circ\), \(4(33)-10=122^\circ\); these total \(66+99+73+122=360^\circ\), confirming the solution. Combining like terms before solving, rather than solving with all four terms separately, is what keeps this kind of equation manageable.

(a) Substituting \(j = 12\) into \(A = 3j + 8\) gives \(A = 3(12) + 8 = 36 + 8 = 44\) pounds [2 marks].

(b) The two companies charge the same amount when \(A = B\), so \(3j + 8 = 2j + 15\). Subtracting \(2j\) and 8 from both sides gives \(j = 7\) [2 marks].

(c) Substituting \(j = 20\) into each formula gives \(A = 3(20) + 8 = 68\) and \(B = 2(20) + 15 = 55\). Since 55 is less than 68, Company B is cheaper for 20 journeys in a week [2 marks]. This makes sense because Company B has a lower rate per journey, so it becomes relatively cheaper the more journeys are made, even though its fixed charge is higher.

The diagram shows two labelled bars, one marked "8%" for Tariff A and one marked "22%" for Tariff B; these are the discount rate and surcharge rate referred to in the question, read directly from the figure.

(a) Tariff A applies an 8% discount to the £120 usage cost, then adds the £45 fixed fee:

\[120 \times 0.92 = \pounds 110.40\] [1 mark]

\[45 + 110.40 = \pounds 155.40\] [1 mark]

(b) Tariff B applies a 22% surcharge to the same £120 usage cost, with no fixed fee:

\[120 \times 1.22\] [1 mark]

\[= \pounds 146.40\] [1 mark]

(c) Tariff A's extra cost compared with Tariff B, as a percentage of Tariff B's cost, is:

\[\frac{155.40 - 146.40}{146.40} \times 100 = 6.1\%\] (1 d.p.) [1 mark]

(d) At £150 usage, Tariff A costs \((150 \times 0.92) + 45 = \pounds 183.00\) and Tariff B costs \(150 \times 1.22 = \pounds 183.00\). The two tariffs cost exactly the same at this usage level, so neither is cheaper [1 mark]. This happens because Tariff A's fixed fee makes it relatively better at low usage, while Tariff B's lack of a fixed fee makes it relatively better at high usage; £150 is the point where the two effects balance out.

(a) A translation by the vector \(\binom{x}{3}\) adds \(x\) to the starting \(x\)-coordinate and adds 3 to the starting \(y\)-coordinate. Point \(A\) starts at \((2, 1)\) and reaches \((6, 4)\), so looking only at the \(x\)-coordinates, \(2 + x = 6\), which gives \(x = 4\) [2 marks]. This can be checked against the \(y\)-coordinates too: \(1 + 3 = 4\), which matches the given image point.

(b) The image point after the first translation is \((6, 4)\). Applying the second translation \(\binom{-1}{y}\) takes this to \((5, 10)\), so looking at the \(y\)-coordinates, \(4 + y = 10\), which gives \(y = 6\) [2 marks]. Checking the \(x\)-coordinates: \(6 + (-1) = 5\), which matches the final point given.

This question links a specific evaluation of the revenue model, a quadratic equation for a target revenue, and a real-world restriction shown on a diagram, all built from the same formula \(R(x) = -x^2 + 30x\).

(a) Substituting \(x = 5\): \(R(5) = -5^2 + 30(5) = -25 + 150 = £125\). [1 mark]

(b) Setting revenue to £200: \(-x^2 + 30x = 200\), which rearranges to \(x^2 - 30x + 200 = 0\). [1 mark]

(c) Factorising: two numbers that multiply to \(200\) and add to \(-30\) are \(-10\) and \(-20\), so \((x-10)(x-20) = 0\), giving \(x = 10\) or \(x = 20\). [3 marks]

(d) The diagram shows the owner will only accept a price increase up to \(15\) pence, so \(x = 20\) is outside this allowed range and must be rejected, leaving \(x = 10\) pence as the owner's choice. [1 mark]

As in the previous question, both roots give the same £200 revenue mathematically, but the diagram's shaded range (\(0\) to \(15\) pence) is what rules out the larger solution.

(a) This part tests adding fractions with different denominators. Writing both fractions over the common denominator 20 gives \(\frac{2}{5} = \frac{8}{20}\) and \(\frac{1}{4} = \frac{5}{20}\), so the silo is \(\frac{8}{20} + \frac{5}{20} = \frac{13}{20}\) full after both harvests [2 marks].

(b) The empty fraction is \(1 - \frac{13}{20} = \frac{7}{20}\), and this equals the 49 tonnes of empty space [2 marks]. Dividing 49 by \(\frac{7}{20}\) (multiplying by its reciprocal, \(\frac{20}{7}\)) gives the total capacity: \(49 \div \frac{7}{20} = 49 \times \frac{20}{7} = 140\) tonnes [1 mark].

(c) The mass added during the second harvest is \(\frac{1}{4} \times 140 = 35\) tonnes [1 mark].

(d) Since the silo needs at least 150 tonnes of empty space for safe ventilation and only 49 tonnes are empty, and \(49 < 150\), the silo does not currently meet this requirement [1 mark].

The mean relates the total length of all four shelves to the number of shelves, so the missing shelf length is found once the total is known.

Total length = \(4\times80=320\) cm. Sum of the three known shelves = \(75+85+78=238\) cm. Fourth shelf = \(320-238=82\) cm. [2 marks]

To identify the single transformation mapping triangle \(Y\) onto triangle \(Z\), corresponding vertices are compared on the grid. Each vertex of \(Z\) lies the same distance from the point \((0, 1)\) as the matching vertex of \(Y\), but on the exact opposite side of it, which is the signature of a half-turn rotation [1 mark], of \(180^\circ\) [1 mark], about the centre \((0, 1)\), the fixed point of the turn [2 marks]. For a \(180^\circ\) rotation, describing the turn as clockwise or anticlockwise gives the same final position either way, since a half turn in either direction lands on exactly the same point.

With edge \(XY = 5\) cm already marked, the third vertex \(Z\) is fixed by two arcs, one measuring its distance from \(X\) and one measuring its distance from \(Y\).

Open the compasses to \(12\) cm (the length \(ZX\)) and draw an arc centred on \(X\); then open the compasses to \(9\) cm (the length \(YZ\)) and draw an arc centred on \(Y\), so the arcs intersect. [1 mark]

Each radius must be set carefully against the ruler and the compasses kept undisturbed while drawing, since a shifted setting moves the crossing point and makes \(Z\) inaccurate. [1 mark]

The arcs cross at the point that is \(12\) cm from \(X\) and \(9\) cm from \(Y\) at once: that is \(Z\). Triangle \(XYZ\) is completed with straight lines from \(X\) to \(Z\) and \(Y\) to \(Z\), leaving all construction arcs visible. [1 mark]

Exam tip: check that \(5 + 9 > 12\) (the triangle inequality) before starting; here \(14 > 12\), so the triangle is possible and the arcs will meet as expected.

Profit is the amount received from selling stock minus what it cost to buy, and profit as a percentage compares that profit to the original cost price using \(\dfrac{\text{profit}}{\text{cost price}} \times 100\).

1. \(60 \times 1.20 = 72\), so \(\$72\) is received from selling all 60 candles. [1 mark]
2. Profit \( = 72 - 54 = \$18\). [2 marks]
3. \(\dfrac{18}{54} \times 100 = 33.33\ldots\%\), which rounds to \(33.3\%\) (3 s.f.). [2 marks]

Percentage profit is always calculated against the cost price, not the selling price; using \(\$72\) as the base instead of \(\$54\) would give a different, incorrect percentage (\(25\%\)), because the two prices are not interchangeable in this formula.

Standard-form numbers with different powers of \(10\) cannot be added directly; only their equivalent ordinary numbers can. [1 mark]

Converting first: \(3 \times 10^{4} = 30\,000\) and \(5 \times 10^{3} = 5\,000\). Adding these gives \(30\,000 + 5\,000 = 35\,000\), which in standard form is \(3.5 \times 10^{4}\). [1 mark]

A common error is to add the leading values directly (\(3 + 5 = 8\)) without first matching the powers of \(10\); this only works when the powers are already equal.

This question tests reading the median, the two quartiles and a specific cumulative frequency value directly from a cumulative frequency graph.

(a) With 40 competitors, the median lies at a cumulative frequency of \( 40 \div 2 = 20 \). Reading across from 20 on the vertical axis to the curve, then down to the horizontal axis, gives a median distance of 4.8 m (accept 4.6 to 5.0). [1 mark]

(b) The lower quartile is read at a cumulative frequency of \( 40 \div 4 = 10 \), giving 4.0 m, and the upper quartile is read at \( 3 \times 40 \div 4 = 30 \), giving 5.7 m. The interquartile range is

\[ 5.7 - 4.0 = 1.7 \text{ m (accept 1.5 to 1.9)} \] [2 marks]

(c) Reading up from 5.5 m on the horizontal axis to the curve gives a cumulative frequency of about 28, meaning 28 competitors jumped 5.5 m or less. The number who jumped further than 5.5 m is therefore

\[ 40 - 28 = 12 \text{ competitors (accept 10 to 14)} \] [1 mark]

A cumulative frequency graph always reads "up to and including" a distance, so finding "more than" a value, as in part (c), requires subtracting the reading from the total.

The fraction in the first equation is cleared by multiplying through by 2 before the equations are combined.

Multiplying the first equation by 2 clears the fraction: \(x + 2y = 20\). [1 mark]

From the second equation, \(x = y + 2\); substituting gives \((y + 2) + 2y = 20\). [1 mark]

\(3y = 18\), so \(y = 6\), the amount put into the treat fund. [1 mark]

Then \(x = y + 2 = 8\), the amount saved. [1 mark]

This question needs a fixed-distance locus (a circle) and an angle bisector (equidistance from two walls).

(a) The locus of points within \(3.5\) cm of the alarm is a circle of radius \(3.5\) cm centred on the alarm's marked position; draw it with compasses, keeping the radius undisturbed throughout, so every point on the circle is exactly \(3.5\) cm from the centre. [2 marks]

(b) For the wall angle, draw an arc from its vertex crossing both walls, then draw equal-radius arcs from those two crossing points so they intersect inside the angle; a line from the vertex through that intersection bisects the angle, since it is equally spaced from both walls. [2 marks]

Both constructions completed on the floor plan:

Exam tip: "covers rooms up to a distance" is the language of a circular locus around a single point, while "meet at a corner" and being asked to bisect it is the language of an angle bisector; matching the wording to the construction avoids confusing the two.

Each lorry load adds a fixed volume of concrete on top of the reserve already on site, so the total volume increases linearly with the number of loads, matching the pattern in the table.

(a) Using 1 load and 2 loads from the table:

\[\frac{15 - 9}{1} = 6\] [1 mark]

(b) The reserve already on site before any load arrives is the \(y\)-intercept, 3 cubic metres, so with gradient 6:

\[V = 6L + 3\] [2 marks]

(c) Substituting \(V = 51\):

\[51 = 6L + 3 \implies 6L = 48 \implies L = 8\] [1 mark]

Exam tip: the "reserve already on site" described in the question is exactly the \(y\)-intercept of the equation, the amount present before any loads (when \(L=0\)); look for this kind of wording to identify the intercept in context.

Reading frequencies from a bar chart makes it possible to find the mode (the tallest bar), the median (the middle value once the data is placed in order) and the mean (the total divided by the number of students), just as with a frequency table.

(a) Adding every bar's frequency: \(3 + 5 + 8 + 6 + 2 + 1 = 25\) students took part in the survey. [1 mark]

(b) The tallest bar, with frequency 8, is at 2 books, so the mode is 2 books. [1 mark]

(c) With 25 students, the median is the 13th value once ordered. The cumulative frequencies are 3, 8, 16, 22, 24, 25; the 13th value falls in the "2 books" group, since the cumulative total passes 8 (end of "1 book") and reaches 16 within "2 books," so the median is 2 books. [1 mark]

(d) The total number of books read is \(0(3) + 1(5) + 2(8) + 3(6) + 4(2) + 5(1) = 0 + 5 + 16 + 18 + 8 + 5 = 52\); dividing by 25 gives a mean of \(52 \div 25 = 2.08\) books. [2 marks]

Here the mode and median agree (both 2 books), but the mean (2.08) is slightly higher, pulled up by the smaller number of students who read 3, 4 or 5 books; all three measures are valid summaries, but they answer slightly different questions about the data.

This question combines the multiplication law of indices with converting a power to a number, then converting units.

(a) The supersized box is \(5^2\) times heavier than the standard \(5^3\) g box, so multiply the powers by adding exponents: \[ 5^3 \times 5^2 = 5^{3+2} = 5^5 \] [1 mark]

(b) Evaluating gives the mass in grams: \[ 5^5 = 3125 \text{ grams} \] [1 mark]

(c) Converting grams to kilograms means dividing by 1000: \[ 3125 \div 1000 = 3.125 \text{ kg} \] which rounds to \(3.13\) kg (2 d.p.) [2 marks]

Rounding to 2 decimal places means looking at the third decimal digit; here it is 5, which rounds the second digit up from 2 to 3.