Mathematics Specification A 4MA1 | Paper 1 Mock 01 | Written Paper 1 (1F/1H)

This question tests forming a quadratic equation from the area of a rectangle shown in a diagram, solving it, and using the result to state a missing dimension.

(a) The area is length times width; expanding the product and setting it equal to 312:

(b) Look for two numbers that multiply to \(-330\) and add to \(7\): these are \(22\) and \(-15\), giving the factorisation:

Since \(x\) must be positive for the dimensions to make sense, \(x=15\). [3]

(c) Substituting \(x=15\) into the width expression:

Both dimensions, \((x+9)\) and \((x-2)\), must give positive lengths once \(x\) is found; checking \(x-2>0\) confirms that \(x=15\) is the physically sensible root, since \(x=-22\) would make both sides of the rectangle negative.

Expressing one measurement as a percentage of another uses \(\frac{\text{part}}{\text{whole}} \times 100\), here comparing the height of water in the butt with its full height.

(a) Water height as a percentage of the full height:

[3]

The water butt is treated as having a uniform cross-section from top to bottom, which is why the height reading alone is enough to give the percentage full; without that assumption, height and volume would not be proportional.

A Venn diagram represents sets as regions inside a rectangle (the universal set \(\mathscr{E}\)). Shading identifies a specific combination of set membership, which can be written using set notation such as intersection (\(\cap\)), union (\(\cup\)), or complement (\(\,'\)).

1. The shading covers the part of \(P\) that does not overlap with \(Q\): every element in \(P\) that is not also in \(Q\). This region is written as:\[ P \cap Q' \][2] (1 for identifying the region as "in \(P\) but not \(Q\)", 1 for the correct notation \(P \cap Q'\))
2. The number of elements in \(P\) only is the total in \(P\) minus those shared with \(Q\):\[ n(P \cap Q') = n(P) - n(P \cap Q) = 12 - 4 = 8 \][2] (1 for \(n(P) - n(P \cap Q)\), 1 for the value 8)

Whenever two circles overlap on a Venn diagram, the region belonging to one circle only must have the overlapping part subtracted out first; simply reading off \(n(P) = 12\) for the shaded region would wrongly include the 4 elements shared with \(Q\).

This question tests forming and solving a linear equation from a description of two related charges, then using the result to find one of the original quantities.

(a) The fixed charge is \(x\), and the usage charge is \(\pounds8\) less than double the fixed charge, so it is \(2x-8\). The two charges total \(\pounds61\):

Combine like terms:

The fixed charge is \(\pounds23\). [4] (1 for the correct equation, 1 for combining to \(3x-8=61\), 1 for isolating \(3x=69\), 1 for \(x=23\))

(b) Substitute \(x=23\) into the usage-charge expression:

The usage charge for that month is \(\pounds38\). [1]

Checking the answer by adding the two charges, \(23+38=61\), confirms both the fixed charge and the usage charge are consistent with the total bill given, which is good practice whenever a problem asks for a second value derived from the first.

The table links a cube's side length to its volume by cubing; both parts use this relationship in opposite directions.

1. (a) [2]. If the side length cubed is \(125\), the side length is \(\sqrt[3]{125} = 5\) m, since \(5\times5\times5 = 125\). One mark for identifying the cube root as the correct operation, one for the value \(5\) m.
2. (b) [2]. For a side length of \(6\) m, the volume is \(6^3 = 6\times6\times6 = 216\ \text{m}^3\). One mark for using the correct formula (side length cubed), one for the value \(216\ \text{m}^3\).

Cubing a side length gives volume; taking a cube root of a volume reverses this to give the side length. Keep clear which direction the table is being read in each part.

This question tests taking two separate fractions of the same total and finding what remains.

(a) Bars sold individually are \(\frac{2}{3}\) of the total:

[2]

(b) Bars donated are \(\frac{1}{9}\) of the total:

The bars remaining after both are subtracted from the total:

[2]

Both fractions here are taken from the same original 90 bars, so the remaining amount is found by direct subtraction rather than by working through a reduced remainder at each stage.

This question tests solving one equation with the unknown on both sides and one equation with the unknown inside a fraction.

(a) For \(6x+11=2x+39\), subtract \(2x\) from both sides and subtract 11 from both sides:

(b) For \(\dfrac{2x-3}{5}=3\), multiply both sides by 5 first:

In part (b), multiplying by 5 clears the denominator because the entire expression \(2x-3\) is divided by 5, not just the \(2x\) term, so the whole numerator must be multiplied by 5 together.

This question tests repeated percentage decrease (compound depreciation) and finding when a quantity first drops below a threshold.

1. After one year, the car retains \(100\% - 18\% = 82\%\) of its value: \(12500 \times 0.82\) [1] \(= \) £10,250 [1].
2. Because the depreciation is 18% of the value at the start of each year, the value after 3 years is found by multiplying by 0.82 three times: \(12500 \times 0.82^3\) [1] \(= 12500 \times 0.551368 = \) £6892 to the nearest pound [1].
3. Continuing the pattern: after 4 years, \(12500 \times 0.82^4 \approx £5651.52\), still above £5000; after 5 years, \(12500 \times 0.82^5 \approx £4634.25\), which is below £5000 [1]. The value first falls below £5000 after 5 complete years [1].

Depreciation and other repeated percentage changes are modelled by a single multiplier raised to a power (\(0.82^n\)), never by multiplying \(n \times 18\%\) and subtracting it all at once, which would overstate the fall in value.

This question tests reversing a percentage increase and then stating the size of the increase in pounds.

1. Letting \(x\) be the cost before the rise, a 12% increase means \(1.12x = 78.40\), so \(x = 78.40 \div 1.12 = \) £70 [3].
2. The increase in cost is \(78.40 - 70 = \) £8.40 [1].

Once the original £70 cost is found, the increase follows directly by subtraction from the new £78.40 cost; there is no need to recompute 12% of £70 separately, though it gives the same £8.40.

The car park bay is a parallelogram, so its area is \( \text{base} \times \text{perpendicular height} \), not base times a slanted side.

The area of the bay is \(11.52\) m\(^2\). [3] (1 for identifying base and perpendicular height, 1 for \(4.8 \times 2.4\), 1 for 11.52 m\(^2\))

The dashed line in the diagram marks the perpendicular height, drawn at right angles to the base; the slanted side of the parallelogram is longer than this and must never be substituted into the area formula in its place.

This question tests whether individual ticket prices or a combined family price gives the lower total cost, so both totals must be calculated and then compared.

(a) Buying 2 adult and 3 child tickets separately:

[2]

(b) Buying 1 family ticket (covering 2 adults and 2 children) plus 1 extra child ticket:

The saving compared with part (a) is:

[2]

With combined ticket offers, always check exactly what the offer covers, here 2 adults and 2 children, so that the extra tickets needed to cover the whole group are added on correctly.

This question tests using column vectors between named points to find a displacement and its magnitude, then comparing that direct distance with a given two-stage route.

(a) The vector from \(H\) to \(S\) is found by subtracting the position vector of \(H\) from that of \(S\):

(b) The same method gives the vector from \(H\) to \(B\):

Its magnitude, the straight-line distance from home to the bank, is found by Pythagoras' theorem:

(c) The saving is the difference between the two-stage route and the direct route, using the more precise unrounded value for \(|\vec{HB}|\) to avoid compounding rounding error:

Subtracting the two vectors, as in parts (a) and (b), gives the actual displacement between two points regardless of the path taken to get there; comparing its magnitude with a longer, indirect route is exactly how "distance saved by a shortcut" problems are solved.

This question builds a full angle-sum proof for a triangle in algebraic form, then verifies the result using an independent parallel-line argument.

(a) Using the angle sum of triangle \(ABC\):

[3]

(b) Substituting \(x=30\): angle \(BAC = 2(30)+10 = 70^\circ\), angle \(ABC = 30+5 = 35^\circ\), angle \(ACB = 3(30)-15 = 75^\circ\). [2]

Check: \(70+35+75=180^\circ\).

(c) Angle \(ACD\) is the exterior angle at \(C\), on the straight line \(BD\) extended beyond \(C\), so it is supplementary to angle \(ACB\): \(180^\circ - 75^\circ = 105^\circ\). This agrees with the exterior angle theorem, which gives it directly as the sum of the two opposite interior angles: \(70^\circ + 35^\circ = 105^\circ\). [2]

(d) A line through \(A\) parallel to \(BC\) creates alternate angles with the transversal \(AB\); the angle alternate to angle \(ABC\) is equal to it, since alternate angles between parallel lines are equal, so it is \(35^\circ\). [2]

(e) At vertex \(A\), three angles now lie along the straight line formed by the parallel through \(A\): the angle alternate to \(B\) (\(35^\circ\)), angle \(BAC\) itself (\(70^\circ\)), and the angle alternate to \(C\) (\(75^\circ\)). Since these three angles lie on a straight line, they must sum to \(180^\circ\). But the alternate angles equal angle \(B\) and angle \(C\) exactly, so this sum is the same as angle \(A\) + angle \(B\) + angle \(C\), which proves the triangle's three angles must add to \(180^\circ\). [2]

Part (e) is the classic parallel-line proof of the triangle angle sum rule; it is worth understanding this construction (a line through one vertex, parallel to the opposite side) as the reason the \(180^\circ\) rule is always true, not just something to accept.

This question tests inverse proportion between the number of mechanics and the time taken to service a fixed fleet of vans.

1. With 4 mechanics the constant is \(4 \times 9 = 36\) mechanic-hours. With 6 mechanics, the time is \(36 \div 6 = \) 6 hours [2].
2. To finish within 4 hours, mechanics \(\geq 36 \div 4 = 9\). The least number of mechanics needed is 9 [2].

The 36 mechanic-hours is the fixed total workload of servicing the fleet; more mechanics working together share that same fixed workload in less time, which is the defining feature of inverse proportion.

The fruit bowl is a hemisphere, exactly half of a sphere. Its volume is half of the full sphere formula: \( V = \dfrac{2}{3}\pi r^3 \).

Rounded to 3 significant figures, the volume of the bowl is \(7070\) cm\(^3\). [3] (1 for \(\frac{2}{3}\pi (15)^3\), 1 for the unrounded value, 1 for 7070 to 3 s.f.)

A hemisphere's volume formula is exactly half of the full sphere's \( \frac{4}{3}\pi r^3 \); using the full sphere formula by mistake would double the correct answer.

Simple interest is calculated only on the original amount borrowed, using \(\text{interest} = \text{principal} \times \text{rate} \times \text{time}\), and the total repayment adds this interest to the loan.

(a) Interest over 3 years:

[2]

(b) Total repaid:

[2]

Because this is simple rather than compound interest, the £225 could also be found by working out one year's interest, £75, and multiplying by 3, since the interest does not itself earn further interest.

This question tests working backwards from a known remainder to find an original total, using the fraction that is left after a portion is sold.

Since \(\frac{5}{8}\) of the crop was sold, the fraction left in the barn is \(1 - \frac{5}{8} = \frac{3}{8}\), and this fraction corresponds to the 90 tonnes stored. [1]

If \(\frac{3}{8}\) of the harvest is 90 tonnes, then one eighth of the harvest is

[1]

The whole harvest (eight eighths) is therefore

[1]

Whenever a quantity is described as a fraction of an unknown total, first identify what fraction the known amount represents, then scale from one part up to the whole; scaling from the wrong fraction (using \(\frac{5}{8}\) instead of \(\frac{3}{8}\)) is the main pitfall here.

A regular dodecagon has 12 equal edges, so the total length needed to trim every edge is 12 times the length of one edge.

1. [2] \[12 \times 0.8 = 9.6 \text{ m.}\]
2. [2] At \(£6\) per metre, \[9.6 \times £6 = £57.60.\]

These "cost of the perimeter" questions always split into two clearly separate steps: first find the total length using the shape's properties, then multiply by the rate; keeping the steps separate avoids losing marks even if one step is wrong.

A total charge made up of a fixed fee plus a daily rate is a linear relationship; if the daily part of the bill can be calculated separately, the fixed fee is found by subtracting it from the total amount billed.

1. The daily charge for \(9\) days is: \[18 \times 9 = 162\] Subtracting this from the total bill of \(\pounds272\) leaves the fixed erection fee: \[272 - 162 = 110\] The fixed erection fee is \(\pounds110\). [3] (correct daily charge, correct subtraction, correct fee)
2. Combining the fixed fee with the daily rate of \(\pounds18\) per day of hire: \[C = 110 + 18d\] [1]

Whenever a bill combines a one-off fee with a repeating daily charge, work out the repeating part first (rate \(\times\) number of days) and subtract it from the total; what remains is always the fixed part.

This tree diagram tracks two draws without replacement from a bag of 4 red and 6 blue counters (10 counters in total); the second-stage branches use a denominator of 9 because one counter has already been removed, and conditional probability (part (c)) is found by dividing the probability of the joint event by the probability of the condition.

1. (a) [2]. \(P(\text{both red}) = \dfrac{4}{10}\times\dfrac{3}{9} = \dfrac{12}{90} = \dfrac{2}{15}\).
2. (b) [2]. \(P(\text{both blue}) = \dfrac{6}{10}\times\dfrac{5}{9} = \dfrac{30}{90} = \dfrac{1}{3}\).
3. (c) [3]. \(P(\text{same colour}) = P(\text{both red}) + P(\text{both blue}) = \dfrac{12}{90}+\dfrac{30}{90} = \dfrac{42}{90}\). Conditional probability restricts attention to this "same colour" outcome and asks what share of it is "both blue": \[P(\text{both blue}\mid\text{same colour}) = \frac{P(\text{both blue})}{P(\text{same colour})} = \frac{30/90}{42/90} = \frac{30}{42} = \frac{5}{7}\]
4. (d) [2]. Adding one yellow counter increases the bag to 11 counters, but the number of red counters stays at 4, so \(P(\text{first red})\) becomes \(\dfrac{4}{11}\), which is smaller than the original \(\dfrac{4}{10}\). Since the first-draw probability of red falls and the second-draw probability of red (given a red first) also falls slightly for the same reason, \(P(\text{both red})\) would be lower than before.

This question tests finding a percentage (the complement of an on-time rate) of two different totals and comparing the resulting counts.

1. On Route A, \(100\% - 92\% = 8\%\) of buses were late: \(0.08 \times 650 = \) 52 [2].
2. On Route B, \(100\% - 88\% = 12\%\) of buses were late: \(0.12 \times 800 = 96\). Route B had \(96 - 52 = \) 44 more late buses than Route A [3].

Find the "late" percentage first by subtracting the on-time percentage from 100%, then apply it to each route's own total; the two routes cannot be compared using the same total, since Route B ran more buses than Route A.