Mathematics 0580 | Paper 3 Mock 01 | Calculator (Core)

The three angles of any triangle add up to \(180^\circ\). Here two of them are already known: the right angle at \(B\) is \(90^\circ\) and angle \(A = 34^\circ\).

\[ C = 180^\circ - 90^\circ - 34^\circ = 56^\circ \] [B1]

A useful shortcut for right-angled triangles: the two non-right angles are complementary, so they must total \(90^\circ\). That gives \(90^\circ - 34^\circ = 56^\circ\) directly. No trigonometry or measuring is needed here, which is why the question says "write down".

This question tests area and perimeter of a shape made of whole grid squares, where the shape is not a simple rectangle.

(a) Each small square has side \(1\) cm and so represents \(1\) cm\(^2\). Counting the squares that make up the cross-section gives

\[ 33 \text{ cm}^2 \] [B1]

(b) The perimeter is the total distance around the outside, so add the lengths of all the edges of the cross-section as you travel once around the boundary. [M1]

Because every edge of this shape runs along a grid line, the horizontal edges together travel the full width twice and the vertical edges together travel the full height twice. With a width of \(8\) cm and a height of \(6\) cm this gives

\[ 2 \times (8 + 6) = 28 \text{ cm} \] [A1]

Notice what this shows: the bracket has a smaller area than the \(8 \times 6 = 48\) cm\(^2\) rectangle that just contains it, but exactly the same perimeter, because each step cut into the shape removes area while simply redistributing the same total horizontal and vertical travel. This shortcut only works while every notch is a rectangular step; if the boundary had a sloping or curved edge, you would have to add the individual edge lengths one by one.

The mean height is the total height divided by the number of sunflowers.

Add the six heights:

Then divide by \(6\):

Correct to 3 significant figures the mean height is \(1.99\) m. [A1]

Round only at the end. The third significant figure here is the second decimal place, and since the next digit is \(6\) the \(8\) rounds up to \(9\). A useful check is that the mean must lie between the smallest height \(1.67\) m and the largest \(2.31\) m, which \(1.99\) m does.

(a) Order of rotational symmetry. In a regular polygon the number of sides, the number of lines of symmetry and the order of rotational symmetry are all the same number, because every vertex is interchangeable with every other. With 16 lines of symmetry the polygon has 16 sides, so its order of rotational symmetry is 16 [B1]. Each rotation of \(360\div 16 = 22.5^\circ\) moves every vertex on to the next one.

(b) Each interior angle. The exterior angles of a polygon add to \(360^\circ\), so one exterior angle of this regular polygon is \(360\div 16\). The interior angle is the supplement of the exterior angle at the same vertex:

\( 180 - 360 \div 16 \) or equivalent [M1]

Each interior angle \(= 157.5^\circ\) [A1]

The alternative method, \((16-2)\times 180 \div 16 = 2520 \div 16 = 157.5\), gives the same value and is equally acceptable. Notice the answer is not a whole number of degrees; do not round it to \(158^\circ\), since no rounding is asked for and the exact value is available.

A point lies on the curve \(y = \frac{8}{x}\) exactly when its coordinates satisfy that equation, so the value of \(y\) at \(x = 3.2\) can be calculated directly rather than read off the grid. Calculating is more accurate than reading, since a graph can usually only be read to about half a small square.

If the division is not obvious, clear the decimal by multiplying top and bottom by \(10\): \(\frac{8}{3.2} = \frac{80}{32} = \frac{5}{2} = 2.5\).

This is a reciprocal (inverse proportion) curve: as \(x\) increases, \(y\) decreases, and their product is always \(8\). Check the answer with that property: \(3.2 \times 2.5 = 8\). The common error is to multiply instead of divide, giving \(25.6\), which fails the product check immediately.

To read a gradient off an equation, it must first be arranged with \(y\) alone on the left, because only then does the number multiplying \(x\) equal the gradient.

1. Subtract \(6x\) from both sides: \[ 2y = 10 - 6x \]
2. Divide every term by \(2\): \[ y = 5 - 3x \] [M1]

Written in the standard order this is \(y = -3x + 5\), so

\[ \text{Gradient} = -3 \] [A1]

Two points are worth stressing. Every term must be divided by \(2\), not just the \(x\) term. And the gradient is negative, meaning the line falls as \(x\) increases; reading \(3\) from the original \(6x\) without rearranging is the usual mistake.

The shape is a rectangle with a semicircle sitting on the top edge. That top edge of the rectangle is inside the shape, so it counts for the area but not for the perimeter.

(a) Going round the outline: the bottom of the rectangle is \(20\) cm, the two vertical sides are \(14\) cm each, and the curved edge is half the circumference of a circle of diameter \(20\) cm.

\[ P=20+2\times 14+\frac{\pi\times 20}{2} \] [M1]

\[ P=20+28+31.415...=79.415... \]

Correct to 3 significant figures, the perimeter is \(79.4\) cm [A1] (\(79.415...\) is accepted).

(b) For the area, add the rectangle and the semicircle. The semicircle has diameter \(20\) cm, so its radius is \(10\) cm:

\[ A=20\times 14+\frac{\pi\times 10^2}{2} \] [M1]

\[ A=280+157.07...=437.07... \]

Correct to 3 significant figures, the area is \(437\) cm\(^2\) [A1] (\(437.07...\) is accepted).

Notice how the \(20\) cm is used differently in the two parts: as a diameter in the arc length \(\frac{\pi d}{2}\), and halved to a radius of \(10\) cm in the area \(\frac{\pi r^2}{2}\). Mixing these up is the usual source of error in composite circle questions.

The right angle is at \( C \), so \( AB = 15.4 \) cm is the hypotenuse. Relative to the \( 52^\circ \) angle at \( A \), the side \( BC \) is opposite, so the ratio of opposite to hypotenuse is required, which is the sine.

To \( 3 \) significant figures, \( BC = 12.1 \) cm. [A1]

Identifying the hypotenuse first is the key step: it is always the side facing the right angle, here \( AB \), never a side of the right angle itself. Since a sine is always less than \( 1 \), \( BC \) must be shorter than the hypotenuse, and \( 12.1 \lt 15.4 \) confirms this.

Finding \(D\). The given vertices are \(A(2,1)\), \(B(6,1)\) and \(C(6,5)\). Going \(A\) to \(B\) moves 4 units right, and \(B\) to \(C\) moves 4 units up, confirming a square of side 4. In the square \(ABCD\) the vertices are taken in order, so \(D\) must complete the circuit: from \(C\) move 4 units left (the reverse of \(A\) to \(B\)) to reach \(D\), or equivalently from \(A\) move 4 units up.

The horizontal line of symmetry. A line parallel to the \(x\)-axis is horizontal, so its equation has the form \(y=\) a number. The square has its horizontal sides at \(y=1\) (through \(A\) and \(B\)) and \(y=5\) (through \(C\) and \(D\)), so the mirror line lies midway between them:

Equation: \( y = 3 \) [B1] (or any equivalent form)

The check that the square really is a square matters: only then does the midway horizontal line pass through the centre and act as a mirror. A quick sketch on axes prevents the common slip of writing \(x=3\) for a line parallel to the \(x\)-axis.

The number line gives the midnight temperature as \(-7^\circ\)C, a point \(7\) units to the left of zero. A rise in temperature moves to the right along the line, so add the rise:

\[ -7+19=12 \]

The temperature at noon is \(12^\circ\)C. [B1]

Counting it out makes the arithmetic clear. Moving \(7\) units to the right from \(-7\) reaches \(0\), which uses up \(7\) of the \(19\) degrees. The remaining \(19-7=12\) degrees carry the temperature to \(12^\circ\)C above zero.

The common error is to add the digits and ignore the sign, giving \(26^\circ\)C, or to subtract and give \(-26^\circ\)C. A rise always moves the value in the positive direction, and here the rise is larger than the size of the starting temperature, so the result must end up above zero.

(a) The \(46\) French speakers include the \(15\) who also speak Spanish. Removing them leaves the French-only region.

\[ 46 - 15 = 31 \text{ people} \] [B1]

(b) Adding the two language totals counts the \(15\) bilingual people twice, so subtract the overlap once to get the number who speak at least one language.

\[ 46 + 38 - 15 = 69 \text{ people speak French or Spanish or both} \] [M1]

\[ 80 - 69 = 11 \text{ people speak neither} \] [A1]

(c) All \(38\) Spanish speakers count here, whether or not they also speak French, out of the \(80\) people:

\[ P(\text{Spanish}) = \frac{38}{80} = \frac{19}{40} \] [B1]

Equivalent forms such as \(\frac{38}{80}\) or \(0.475\) are accepted. The completed diagram has \(31\) French only, \(15\) both, \(23\) Spanish only and \(11\) neither, and these four regions add to \(80\), which is the check worth doing before writing any answers.

This question tests reading points off a scatter diagram and then adding the values read.

Mass is on the horizontal axis, so the three heaviest parcels are the three points furthest to the right. Their costs are read off the vertical axis, giving \(\$26\), \(\$30\) and \(\$32\).

\[ 26 + 30 + 32 \] [M1]

\[ = 88 \] [A1]

The total cost of the three heaviest parcels is \(\$88\).

The trap is to pick the three most expensive parcels, that is the three highest points, rather than the three heaviest. On a scatter diagram the two variables usually correlate but are not identical, so always check which axis the question is selecting on: "heaviest" means read along the mass axis first, then go up to the point and across to the cost axis.

A regular polygon has equal sides, so perimeter \(=\) number of sides \(\times\) length of one side. This question uses that single relationship four times.

(a) With \(9\) sides each of length \((x - 2)\) cm:

\[ P = 9(x - 2) = 9x - 18 \ \text{cm} \] [B1]

The \(9\) multiplies both terms, so the constant becomes \(-18\), not \(-2\).

(b) Set the expression equal to the perimeter given:

\[ 9x - 18 = 72 \] [M1]

\[ 9x = 90 \quad \Rightarrow \quad x = 10 \] [A1]

Dividing first is an equally valid method: one side is \(72 \div 9 = 8\) cm, so \(x - 2 = 8\) and \(x = 10\).

(c) The side length is \((x - 2)\) cm, so

\[ 10 - 2 = 8 \ \text{cm} \] [B1]

(d) The new polygon is regular with the same side length of \(8\) cm and a perimeter of \(96\) cm. Dividing the perimeter by the side length gives the number of sides:

\[ \frac{96}{8} = 12 \] [B1]

So it has \(12\) sides. The reasoning in the last part is the same relationship read the other way round: knowing any two of perimeter, side length and number of sides determines the third. Note that \(x\) itself, \(10\), is not a side length and must not be used in that division.

A net is the flat shape that folds up to make a solid, so the faces on the net are exactly the faces of the finished solid. Every face here is a rectangle, and a closed solid whose faces are all rectangles has six of them, meeting three at each corner.

The solid is a cuboid. [B1]

A cuboid has \( 6 \) rectangular faces, \( 12 \) edges and \( 8 \) vertices, with opposite faces identical and parallel. A cube is the special case in which all six rectangles happen to be squares, so "cube" is only correct if the net shows squares. "Rectangle" on its own is not accepted here, because that names a flat two-dimensional shape rather than the solid; questions asking for the name of a solid want a three-dimensional name such as cuboid, prism, pyramid, cylinder or cone.

This question tests reading the type of correlation from a scatter diagram.

Correlation describes how two quantities change together. If the points fall as you move to the right, so that a larger value of one quantity goes with a smaller value of the other, the correlation is negative. If the points rise to the right it is positive, and if there is no clear pattern there is no correlation.

Here the horizontal axis is the age of the car and the vertical axis is its value. Older cars have had more use and wear, so they sell for less, and the points slope downwards from left to right. The correlation is therefore negative [B1]

Notice that "negative" describes the direction of the trend, not the values themselves; the car values are all positive numbers. Also, correlation here shows a strong association between age and value, but you should describe the trend rather than claim that age is the only cause of the fall in price.

Two unknown prices need two equations, one for each purchase described. Here \(n\) is the cost of a notebook and \(f\) the cost of a folder, both in dollars.

(a) Four notebooks and three folders cost \(\$23.40\), so \(4n+3f=23.40\) [B1]. Six notebooks and five folders cost \(\$37.00\), so \(6n+5f=37.00\) [B1].

(b) To eliminate \(f\), match its coefficients using the lowest common multiple of \(3\) and \(5\), which is \(15\). Multiply the first equation by \(5\) and the second by \(3\):

\[20n+15f=117\qquad\text{and}\qquad 18n+15f=111\]

which is the method mark [M1]. The \(f\) terms are identical and both positive, so subtract: \(2n=6\), giving \(n=3.00\), so a notebook costs \(\$3.00\) [A1].

Substituting into \(4n+3f=23.40\) gives \(12+3f=23.40\), so \(3f=11.40\) and a folder costs \(\$3.80\) [A1].

Check the second purchase: \(6(3.00)+5(3.80)=18.00+19.00=37.00\), as stated.

Multiply every term, including the money value on the right hand side, when scaling an equation. Leaving \(23.40\) unchanged while multiplying the left by \(5\) is the error that most often derails this question.

This tests careful calculator use with a fraction. The horizontal bar acts as a bracket around the whole numerator and the whole denominator, so both must be completed before the division is carried out.

Numerator: \(\sqrt{18.6}=4.3127\ldots\) and \(4.2^{3}=74.088\), so the numerator is \(78.400\ldots\)

Denominator: \(1.9^{2}=3.61\), so the denominator is \(3.61-0.75=2.86\) [M1] for either of these intermediate values seen

(a) Dividing, \(78.400\ldots\div 2.86=27.41285\ldots\) [A1]. The question asks for all the figures on the display, so the digits must be copied out rather than rounded.

(b) To \(3\) significant figures, look at the fourth significant figure. In \(27.41285\ldots\) the first three significant figures are \(2\), \(7\), \(4\) and the next digit is \(1\), which is below \(5\), so the \(4\) stays:

\(27.4\) [B1] (following through from your part (a))

The error that costs both marks is typing the expression without brackets, so the calculator divides only \(4.2^{3}\) by \(3.61\) and then handles the \(-0.75\) separately. Enter it as \((\sqrt{18.6}+4.2^{3})\div(1.9^{2}-0.75)\), or evaluate the top and bottom separately and divide at the end.

The cube root asks which number, multiplied by itself three times, gives \(2744\).

So \(\sqrt[3]{2744}=14\). [B1]

You can narrow the answer down before reaching for the calculator: \(10^{3}=1000\) and \(20^{3}=8000\), so the root lies between \(10\) and \(20\), and since \(2744\) ends in \(4\) the root must end in \(4\), because \(4^{3}=64\) is the only single digit cube ending in \(4\). Do not confuse the cube root with dividing by \(3\), which would give a far larger number.

The table is read as: \(1\) dollar buys \(0.79\) pounds, \(148\) yen, \(0.93\) euros or \(18.6\) rand. That single fact decides both parts. Going from dollars into a foreign currency you multiply by the rate; coming back from a foreign currency into dollars you divide by the rate.

(a) Dollars to pounds, so multiply by \(0.79\):

\[ 340\times 0.79 \] [M1]

\[ =268.6 \]

\(\$340\) is \(268.60\) pounds. [A1] Money is written to two decimal places, so \(268.60\) rather than \(268.6\).

(b) Yen to dollars, so divide by \(148\):

\[ 5550\div 148 \] [M1]

\[ =37.5 \]

\(5550\) yen is \(\$37.50\). [A1]

If you are unsure which way round to go, sanity-check the size of the answer. Since one dollar is worth \(148\) yen, a few thousand yen must be only a few tens of dollars, so the answer has to be much smaller than \(5550\). Multiplying would have given \(821\,400\), which is clearly wrong.

First confirm the shape. From \(A(1,2)\) to \(B(7,2)\) is 6 units across, and from \(B(7,2)\) to \(C(7,8)\) is 6 units up, so \(ABCD\) is a square of side 6 with centre at \((4,5)\).

(a) Lines of symmetry parallel to the axes. Each passes through the centre and through the midpoints of a pair of opposite sides.

• Vertical sides lie at \(x=1\) and \(x=7\), so the vertical mirror is \( x = \frac{1+7}{2} = 4 \), giving \( x=4 \) [B1].
• Horizontal sides lie at \(y=2\) and \(y=8\), so the horizontal mirror is \( y = \frac{2+8}{2} = 5 \), giving \( y=5 \) [B1].

Equivalent forms are accepted.

(b) Order of rotational symmetry. All four sides are equal and all four angles are \(90^\circ\), so a quarter turn about \((4,5)\) sends each side onto the next. Matches occur at \(90^\circ\), \(180^\circ\), \(270^\circ\) and \(360^\circ\), so the order is 4 [B1].

(c) The diagonal line of symmetry with positive gradient. A square also has both diagonals as mirror lines. The diagonal with positive gradient runs from \(A(1,2)\) up to \(C(7,8)\).

Gradient 1 seen, or the line through \((1,2)\) and \((7,8)\) identified [M1]

Substituting \((1,2)\) into \(y = x + c\) gives \(2 = 1 + c\), so \(c = 1\).

Equation: \( y = x + 1 \) [A1] (or any equivalent form)

Check it against the centre: \((4,5)\) satisfies \(y=x+1\), as every line of symmetry of the square must pass through the centre. The other diagonal, through \((1,8)\) and \((7,2)\), has gradient \(-1\) and is the one excluded by the words "positive gradient".

The scale \( 1 : 12 \) means that \( 1 \) cm on the model represents \( 12 \) cm on the real statue, so the real object is found by multiplying the model measurement by \( 12 \).

1. Multiply by the scale factor: \[ 27.5 \times 12 = 330 \text{ cm} \] [M1]
2. Convert to metres, as the question requires: \( 330 \div 100 = 3.3 \) m.

The real statue is \( 3.3 \) m tall. [A1]

Because the model was measured in centimetres, the product is also in centimetres; the ratio itself has no unit and never changes the unit you are working in. Converting only at the very end keeps the arithmetic simple and avoids dividing by \( 100 \) at the wrong point.

Simplify the number part and each letter separately, using \( \frac{x^{m}}{x^{n}} = x^{m-n} \) throughout.

1. Numbers: \( 20 \div 4 = 5 \).
2. The \( p \) terms: \( \dfrac{p^{5}}{p^{-2}} = p^{5-(-2)} = p^{7} \). Subtracting a negative index adds, so the index grows.
3. The \( q \) terms: \( \dfrac{q^{8}}{q^{3}} = q^{8-3} = q^{5} \).

The answer is \( 5p^{7}q^{5} \) [B3], awarded as one mark each for the \( 5 \), the \( p^{7} \) and the \( q^{5} \).

The \( p \) term is the discriminating part. A helpful way to see it is that \( p^{-2} \) in the denominator equals \( \dfrac{1}{p^{2}} \), and dividing by \( \dfrac{1}{p^{2}} \) multiplies by \( p^{2} \), turning \( p^{5} \) into \( p^{7} \). Answering \( p^{3} \) means the negative sign was dropped.

On a travel graph the vertical axis is distance from the depot and the horizontal axis is time. While the bus is moving, the distance changes, so the graph slopes. While the bus is stopped, the distance from the depot stays the same, so the graph is a horizontal line.

The stop at the village is therefore the flat section at the top of the graph, and its length along the time axis gives how long the bus waited.

\( 10 \) minutes [B1]

Read the two time values at the ends of the horizontal section and subtract; do not read the time at which the flat part ends and quote that as the answer. A horizontal section on a distance-time graph always means zero speed, never a constant speed.

Test each letter by asking whether a mirror placed on the letter, vertically or horizontally, reproduces the letter exactly. Use the standard printed capital forms.

• E: a horizontal mirror through the middle bar matches the top arm with the bottom arm. One line of symmetry.
• F: the arms are only at the top, so neither a horizontal nor a vertical mirror works. No lines of symmetry.
• T: a vertical mirror down the stem matches the left half of the crossbar with the right half. One line of symmetry.
• X: a vertical mirror works, a horizontal mirror also works, so X has more than one line (two), not exactly one.
• Y: a vertical mirror down the stem matches the two upper arms. One line of symmetry.

Letters with exactly one line of symmetry: E, T and Y [B2] (two of these correct with no more than one extra letter earns [B1])

The trap is X. It is highly symmetrical, but the question asks for exactly one line, and X has two, so it must be left out.

Both parts need the numbers arranged in order of size, so ordering the twelve card values is the first move in either case.

(a) In order the numbers are

[M1] for the correct ordered list, or for identifying the middle pair. With twelve values, \(\frac{12+1}{2}=6.5\), so the median lies halfway between the \(6\)th and \(7\)th values, which are \(11\) and \(12\):

[A1].

(b) The range is the largest value minus the smallest:

[B1].

With an even number of values the median usually is not one of the data values, and \(11.5\) is not printed on any card. That is correct and expected. The frequent error is taking a single "middle" card from the unordered display, or averaging the \(5\)th and \(8\)th values; the two central positions must be counted from the ordered list.

The volume of a cuboid is the product of its three dimensions. Here two of the measurements are given to one decimal place, so care with the arithmetic matters more than the method.

Multiplying the three dimensions: \[ 2.5 \times 3.6 \times 8 = 9 \times 8 = 72 \] so the volume is \(72\) cm\(^3\) [B1].

Taking \(2.5 \times 3.6 = 9\) first makes the calculation straightforward, which illustrates a general point: multiplication can be done in any order, so pair the numbers that combine most neatly. The unit is cm\(^3\) because three lengths in centimetres have been multiplied together. Note that the answer is exactly \(72\), not a rounded value, so no decimal places are needed.

The diameter of a circle is the straight line through the centre joining two points on the circumference, and it is made of two radii placed end to end. So the diameter is always twice the radius.

\[ d=2r=2\times 6.5=13 \]

The diameter is \(13\) cm [B1].

No \(\pi\) is involved here. The constant \(\pi\) only appears once a curved length or an area is required, such as the circumference \(\pi d\) or the area \(\pi r^2\). The command word "write down" also signals that no calculation is expected beyond the doubling.

Finding a percentage of an amount means finding that many hundredths of it. Here you need \(35\%\) of the \(600\) students shown by the whole pie chart.

A useful check without a calculator: \(10\%\) of \(600\) is \(60\), so \(30\%\) is \(180\); \(5\%\) is half of \(10\%\), which is \(30\); and \(180+30=210\).

So \(210\) students walk to school [B1]

The pie chart itself is not needed for this part, because the percentage is given in words. Do not try to measure the sector angle; use the stated \(35\%\) and the stated total of \(600\).

The journey crosses midnight, so it is safest to split the time into the part before midnight and the part after it rather than subtracting the clock readings directly.

1. From \(21\,47\) to \(24\,00\) is \(2\) h \(13\) min.
2. From \(00\,00\) to \(06\,13\) is a further \(6\) h \(13\) min [M1].
3. Add: \(2\) h \(13\) min \(+\) \(6\) h \(13\) min \(=8\) h \(26\) min [A1].

The journey time is \(8\) hours \(26\) minutes.

Subtracting \(21\,47\) from \(06\,13\) as if they were ordinary numbers gives a negative or nonsensical result, because clock time is not decimal. If you prefer a calculator, \(06\,13\) the next day is \(30.21\overline{6}\) hours after midnight of the first day and \(21\,47\) is \(21.78\overline{3}\) hours, and the difference \(8.4\overline{3}\) hours converts to \(8\) h \(26\) min since \(0.4\overline{3}\times 60=26\). Remember that \(8.43\) hours is not \(8\) h \(43\) min.

Two rays meeting at a point create two angles: the one between them and the reflex angle going the other way round. Together they make one complete turn of \(360^\circ\).

The angle between the rays is \(78.4^\circ\), so the reflex angle is

\[360-78.4=281.6^\circ\]

giving \(281.6^\circ\) [B1].

Check that the answer really is reflex: \(281.6^\circ\) lies between \(180^\circ\) and \(360^\circ\), so it qualifies. Subtracting from \(180^\circ\) instead would give \(101.6^\circ\), which is obtuse rather than reflex and so cannot be the answer to this question.