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

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 reading a pie chart: since the full circle (360°) represents the whole event (180 minutes), any sector's time is found by taking that sector's fraction of 360° and applying it to 180 minutes.

1. The races sector is 200° out of 360°, so the time for races is \(\frac{200}{360} \times 180 = \) 100 minutes [2].
2. The opening ceremony is \(\frac{40}{360} \times 180 = 20\) minutes, and the prize-giving is \(\frac{120}{360} \times 180 = 60\) minutes. The ratio of ceremony time to prize-giving time is \(20:60\), which simplifies to 1:3 [2].

As a check, the three sector times should add back to the total: \(20+100+60=180\) minutes, matching the 3-hour sports day exactly, which confirms the working in both parts.

This question tests using the tangent ratio to find a height from an angle of elevation, then the cosine ratio to find the direct distance, and comparing the result with a stated rule.

1. (a) [2] The height is opposite the \(27^\circ\) angle and the ground distance (\(12\) m) is adjacent to it: \[\text{height} = 12 \times \tan 27^\circ = 6.114\text{ m (3 s.f.)}\]
2. (b) [2] The direct distance is the hypotenuse, with the ground distance adjacent to the angle: \[\text{distance} = \frac{12}{\cos 27^\circ} = 13.468\text{ m (3 s.f.)}\]
3. (c) [1] Since the height, \(6.114\) m, is greater than the required \(5.5\) m, the flagpole does meet the rule.

Once the height is found using tangent, the direct distance can be found either with cosine (ground distance over direct distance) or with Pythagoras' theorem using the height and ground distance; both give the same result.

The firewood store is a triangular prism: a right-angled triangular cross-section that runs a fixed length along the wall. Its volume is \( V = \text{cross-sectional area} \times \text{length} \).

1. The triangular end has a right angle, so its base and height are the two shorter sides: \[ \text{Area} = \frac{1}{2} \times 40 \times 25 = 500 \text{ cm}^2 \] [2] (1 for \(\frac{1}{2} \times 40 \times 25\), 1 for 500 cm\(^2\))
2. Multiplying the cross-sectional area by the length the store runs along the wall:\[ V = 500 \times 150 = 75000 \text{ cm}^3 \]Since \(1\) litre \(= 1000\) cm\(^3\), this is \( 75000 \div 1000 = 75 \) litres. [2] (1 for \(500 \times 150\), 1 for converting to 75 litres)

Treating any solid with a constant cross-section (prism) as area times length is the general method that also applies to cylinders, and here it avoids having to find any missing hypotenuse.

This question tests converting two fractions with different totals into percentages so that progress can be compared fairly.

(a) Each site's percentage progress is the number of units completed divided by the number planned, multiplied by 100:

(b) Comparing the percentages, Site A has made the greater percentage progress. [1]

Even though Site B has completed fewer housing units in total, 133 against 156, converting to a common scale of 100 shows that Site A is proportionally further ahead of its own target, which is what "percentage progress" measures.

This question chains index laws, an exponential equation, a negative index, and surd subtraction, finishing by testing why roots cannot be subtracted "inside" a single root.

1. (a) [2]. \(\dfrac{x^7 \times x^{-2}}{x^3} = x^{7 + (-2) - 3} = x^{2}\). One mark for combining the indices in the numerator (\(7-2=5\)), one for the final subtraction giving \(x^2\).
2. (b) [3]. \(32 = 2^5\), so \(2^{3x-1} = 2^5\) gives \(3x - 1 = 5\), so \(3x = 6\) and \(x = 2\). One mark for \(32=2^5\), one for the equation \(3x-1=5\), one for \(x=2\).
3. (c) [2]. Using \(x=2\) from part (b), \(x^{-3} = 2^{-3} = \dfrac{1}{2^3} = \dfrac{1}{8}\). One mark for rewriting as a reciprocal, one for the fraction \(\dfrac{1}{8}\).
4. (d) [2]. \(\sqrt{75} = \sqrt{25\times3} = 5\sqrt3\) and \(\sqrt{12} = \sqrt{4\times3} = 2\sqrt3\), so \(\sqrt{75} - \sqrt{12} = 5\sqrt3 - 2\sqrt3 = 3\sqrt3\). One mark for correctly simplifying both surds, one for the final difference \(3\sqrt3\).
5. (e) [2]. \(\sqrt{75-12} = \sqrt{63} = \sqrt{9\times7} = 3\sqrt7 \approx 7.94\), which is not equal to \(3\sqrt3 \approx 5.20\) from part (d). One mark for evaluating \(\sqrt{63}\) and comparing it numerically with \(3\sqrt3\), one for the correct explanation: subtracting the numbers under separate roots is not the same operation as subtracting the roots themselves, so \(\sqrt{a}-\sqrt{b} \ne \sqrt{a-b}\) in general.

Square roots do not distribute over subtraction (or addition); always simplify each surd separately and combine like terms in \(\sqrt{k}\), rather than combining the numbers underneath a single root sign.

This question tests algebraic manipulation to verify a Pythagorean identity, and substituting a value into the general expressions.

1. (a) [3] Expand each square on the left-hand side: \[(2k)^2 = 4k^2\] \[(k^2-1)^2 = k^4 - 2k^2 + 1\] Adding these: \[4k^2 + (k^4 - 2k^2 + 1) = k^4 + 2k^2 + 1\] Expanding the right-hand side: \[(k^2+1)^2 = k^4 + 2k^2 + 1\] Since both sides simplify to \(k^4 + 2k^2 + 1\), the two expressions are equal, so \((2k)^2 + (k^2-1)^2 = (k^2+1)^2\) for every value of \(k\), meaning these three lengths always satisfy Pythagoras' theorem.
2. (b) [1] Substituting \(k=3\): \[2k = 6, \qquad k^2 - 1 = 8, \qquad k^2 + 1 = 10\] The sides are \(6\), \(8\) and \(10\) cm.

This expression is a standard way of generating Pythagorean triples for any \(k>1\); the case \(k=3\) reproduces the familiar \(6\text{-}8\text{-}10\) triple.

As with any unknown-total relative-frequency problem, the "Poor" frequency can be expressed two ways, once as whatever is left after the known categories, and once as a proportion of the total, and the two expressions are set equal to find the total.

1. (a) [2]. The known categories sum to \(15+9=24\), so the "Poor" frequency is \(T-24\), where \(T\) is the total. Since the relative frequency of "Poor" is \(0.2\), the "Poor" frequency is also \(0.2T\). So: \[T - 24 = 0.2T \implies 0.8T = 24 \implies T = 30 \text{ fields}\]
2. (b) [1]. Fields classed "Poor" number \(30 - 24 = 6\).
3. (c) [2]. "Not Poor" fields number \(24\), so \(P(\text{not Poor}) = \dfrac{24}{30} = \dfrac{4}{5}\).

This question tests rearranging a formula containing a fraction and two variables added together to make one of them the subject.

Starting from \(A=\frac{h(x+y)}{2}\), multiply both sides by \(2\) to clear the fraction:

Divide both sides by \(h\) to isolate the bracket:

Finally, subtract \(x\) from both sides to isolate \(y\):

[3] (1 for clearing the fraction, 1 for dividing by \(h\), 1 for the fully correct final formula)

The bracket \((x+y)\) must be treated as a single quantity until the very last step; dividing by \(h\) before clearing the \(2\) or before removing the bracket would introduce fractions inside fractions and is a common source of error.

Reading values from a number line and expressing them in standard form combines a data-handling skill with the standard form notation used throughout this topic.

1. [1] The fastest (shortest) time is the one furthest left on the scale, \(R_1 = 10.5\) seconds, which in standard form is \[1.05\times10^1.\]
2. [2] The slowest time is \(R_4 = 14.2\) seconds. The difference between the fastest and slowest times is \[14.2 - 10.5 = 3.7,\] which in standard form is \[3.7\times10^0.\]

Because \(3.7\) already lies between 1 and 10, it is left as \(3.7\times10^0\) rather than needing any further adjustment.

This question tests similar triangles: the scale factor found from one pair of corresponding sides converts every other length, and the area is then found from the two scaled dimensions.

1. The bases 18 cm and 27 cm are corresponding sides, giving a scale factor of \(27 \div 18 = 1.5\). The height of flag B is \(24 \times 1.5 = \) 36 cm [2].
2. The area of flag B is \(\frac{1}{2} \times 27 \times 36 = \) 486 cm² [2].

Since the base and height of flag B are now both known exactly, its area is calculated directly rather than by squaring the 1.5 scale factor and applying it to flag A's area, though both routes agree: flag A's area is \(\frac{1}{2} \times 18 \times 24 = 216\) cm\(^2\), and \(216 \times 1.5^2 = 486\) cm\(^2\).

This question tests rearranging formulae to change the subject, including one involving a fraction and one involving a square, and substituting into a formula to find an unknown letter.

1. Multiplying both sides of \(y = \dfrac{3x+5}{2}\) by 2 gives \(2y = 3x + 5\); subtracting 5 and dividing by 3 gives \(x = \dfrac{2y-5}{3}\). [2]
2. Dividing both sides of \(A = \pi r^2\) by \(\pi\) gives \(r^2 = \dfrac{A}{\pi}\); taking the square root of both sides gives \(r = \sqrt{\dfrac{A}{\pi}}\). [2]
3. Substituting \(v = 25\), \(u = 7\), \(t = 3\) into \(v = u + at\): \(25 = 7 + 3a\), so \(3a = 18\), giving \(a = 6\). [2]

In each part, the operations applied to the original letter (multiply, add, square) are undone in reverse order to isolate the new subject; a square root must always come last, once every other term has already been cleared, which is why part (b) squares before it roots.

This question tests finding the space diagonal of a cuboid by applying Pythagoras' theorem twice: once across the base, then again up through the solid.

First find the base diagonal \(AC\), using the length \(AB = 8\) cm and width \(BC = 6\) cm: [2]

The space diagonal \(AG\) is the hypotenuse of a second right-angled triangle formed by the base diagonal \(AC\) and the vertical edge \(CG = 4\) cm: [3]

Finding the space diagonal of a cuboid always works this way: find a face (or base) diagonal first with Pythagoras, then use that diagonal as one side of a second right-angled triangle with the remaining edge.

On a single-circle Venn diagram, the two regions shown make up the whole universal set; a fraction comparing \(n(F)\) to \(n(\mathscr{E})\) should be reduced to its simplest form by dividing by the highest common factor.

1. The number of fresh-produce items inside the circle is 26, so \(n(F) = 26\). [1]
2. The number outside the circle is 44, so \(n(F') = 44\); checking, \(26 + 44 = 70\), matching the total items sold. [1]
3. Writing \(n(F)\) over \(n(\mathscr{E})\) and simplifying: \[\frac{26}{70} = \frac{13}{35}\] (dividing numerator and denominator by their highest common factor, 2). [1]

Check a simplified fraction by trying to divide again: since 13 and 35 share no common factor (13 is prime and does not divide 35), \(\frac{13}{35}\) cannot be reduced further.

This question tests writing an expression for a cost per person, solving a resulting equation with the unknown in a denominator, rearranging the formula to make the number of students the subject, and proving algebraically that a stated value is impossible.

(a) Sharing the total cost \(\pounds(150+8n)\) equally among \(n\) students gives a cost per student of:

[1]

(b) Set the cost per student equal to \(\pounds13\) and multiply both sides by \(n\) to clear the fraction:

Collect the \(n\)-terms:

[3] (1 for clearing the fraction correctly, 1 for collecting the \(n\)-terms, 1 for \(n=30\))

(c) Starting from \(C=150+8n\), subtract \(150\) from both sides and divide by \(8\):

[2]

(d) Substitute \(C=342\) into the rearranged formula:

[2]

(e) If the cost per student could equal \(\pounds8\), then \(\frac{150+8n}{n}=8\), so \(150+8n=8n\). Subtracting \(8n\) from both sides gives \(150=0\), which is false for every value of \(n\). Since this leads to a contradiction regardless of \(n\), the cost per student can never equal \(\pounds8\). [3] (1 for setting up the equation, 1 for correctly cancelling to \(150=0\), 1 for the concluding statement)

Part (e) is a proof by contradiction: rather than solving for \(n\), the argument shows that assuming the cost per student is \(\pounds8\) forces an equation that is never true, so that assumption itself must be impossible for any number of students.

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.

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 writing directed numbers from a word description and combining them to find a net result.

(a) Running early is treated as negative and running late as positive, so the three changes in order are

[2] (1 for the correct signs, 1 for all three values correct)

(b) Adding the three changes gives the overall lateness or earliness at the final stop:

Since the result is positive, the bus is \(2\) minutes late overall. [2] (1 for a correct running total, 1 for correctly stating the bus is 2 minutes late)

The sign of the final total tells you the direction of the overall result: here a positive total means late, but a negative total would have meant the bus was early overall.

This question tests solving a linear inequality with brackets on both sides and illustrating the solution on a number line.

1. Expand both brackets: \(5(x - 2) < 3(x + 4)\) becomes \(5x - 10 < 3x + 12\). Subtract \(3x\) from both sides: \(2x - 10 < 12\). Add 10: \(2x < 22\). Divide by 2: \(x < 11\). [3]
2. Since \(x\) is strictly less than 11, an open circle is drawn at 11, with the line shaded to the left. [2]

Exam tip: expand each bracket completely and collect all the unknown terms on one side before dividing; doing this in a clear, separate line of working (as above) avoids losing method marks even if the final answer slips.

This is a "without replacement" probability question: once the first reserve is picked, the list has one fewer name, which changes the probabilities for the second pick. The probabilities of consecutive events on one branch are multiplied together.

1. (a) [2]. \(P(\text{girl, then girl}) = \dfrac{5}{8} \times \dfrac{4}{7} = \dfrac{20}{56} = \dfrac{5}{14}\). The denominator falls from 8 to 7 because one name has already been removed from the list of 8.
2. (b) [2]. "Different genders" happens as girl-then-boy or boy-then-girl, so both mutually exclusive cases are added: \[P(\text{girl, boy}) = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}, \qquad P(\text{boy, girl}) = \frac{3}{8} \times \frac{5}{7} = \frac{15}{56}\] \[P(\text{different genders}) = \frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28}\] (Equivalently, working via "same gender": \(P(GG)+P(BB) = \frac{5}{14}+\frac{3}{28}=\frac{13}{28}\), so \(P(\text{different})=1-\frac{13}{28}=\frac{15}{28}\), the same result.)

A father rebuilding the triangular base of a garden tool shed has kept the three original measurements: base \(DE = 6\) cm, \(DF = 5\) cm and \(EF = 4\) cm. Triangle \(DEF\) is fixed uniquely once all three side lengths are known (the side-side-side case), so the compasses are used to find the one point F that is simultaneously \(5\) cm from D and \(4\) cm from E.

1. Draw the base \(DE = 6\) cm accurately with a ruler. With the compass point on D and radius \(5\) cm, draw an arc above the line: every point on it is exactly \(5\) cm from D. [1]
2. With the compass point on E and radius \(4\) cm, draw a second arc so that it crosses the first. The crossing point is exactly \(5\) cm from D and \(4\) cm from E at the same time, so it is the only possible position for F. Joining D to F and E to F with straight lines completes the triangle. [1]

Because all three sides are fixed by the construction, the two arcs must be drawn to the exact radii given, not adjusted to look right by eye.