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Question 1 Report
The point \(P(4, 1)\) is shown on the grid. Work out the image of the point \(P(4, 1)\) after a reflection in the \(y\)-axis. Then write down the image of \(P\) after a rotation of \(180^\circ\) about the origin. Give both answers as coordinates.
Do not confuse these transformations: reflecting in the \(y\)-axis does not change the vertical coordinate.
Do not confuse these transformations: reflecting in the \(y\)-axis does not change the vertical coordinate.
Question 2 Report
Write down the coordinates of the point where the line \(y = 2x + 9\) crosses the \(y\)-axis. The diagram shows this line.
The y-axis has equation \(x=0\). Substitute \(x=0\) into \(y=2x+9\):
\[y=2(0)+9=9\]
Therefore the line crosses the y-axis at \((0,9)\). [B1]
The y-axis has equation \(x=0\). Substitute \(x=0\) into \(y=2x+9\):
\[y=2(0)+9=9\]
Therefore the line crosses the y-axis at \((0,9)\). [B1]
Question 3 Report
Use this information about a pie chart to answer the question. The pie chart shows how \(72\) students travel to school. The angle of the sector for walking is \(150^\circ\). Work out how many students walk. Then calculate the angle that stands for \(12\) students. The pie chart in the diagram shows the sector for walking.
A full pie chart is \(360^\circ\) and represents all \(72\) students.
The walking sector represents \(\frac{150}{360}\) of the students:
\[\frac{150}{360}\times72=30\]
So \(30\) students walk [M1][A1].
Each student represents \(360\div72=5^\circ\). Therefore, for \(12\) students:
\[12\times5^\circ=60^\circ\]
The required angle is \(60^\circ\) [B1].
A full pie chart is \(360^\circ\) and represents all \(72\) students.
The walking sector represents \(\frac{150}{360}\) of the students:
\[\frac{150}{360}\times72=30\]
So \(30\) students walk [M1][A1].
Each student represents \(360\div72=5^\circ\). Therefore, for \(12\) students:
\[12\times5^\circ=60^\circ\]
The required angle is \(60^\circ\) [B1].
Question 4 Report
Calculate the circumference of the circle shown in the diagram. Its diameter is \(18\) cm. Use \(\pi = 3.142\) in your working. Give your answer in cm correct to one decimal place.
The circumference of a circle is \(C=\pi d\). The diagram gives diameter \(d=18\) cm and specifies \(\pi=3.142\):
\[C=3.142\times18=56.556\text{ cm}\] [M1]
Rounded to one decimal place:
\[C=56.6\text{ cm}\] [A1]
The circumference of a circle is \(C=\pi d\). The diagram gives diameter \(d=18\) cm and specifies \(\pi=3.142\):
\[C=3.142\times18=56.556\text{ cm}\] [M1]
Rounded to one decimal place:
\[C=56.6\text{ cm}\] [A1]
Question 5 Report
Calculate the probability that a marble taken at random from a jar is green. The tree diagram shows the three colours in the jar, which holds red, green and yellow marbles only. The probability of red is \(\frac{1}{4}\) and the probability of yellow is \(\frac{2}{5}\). Give your answer as a fraction.
The probabilities of all possible colours must add to 1. First add the probabilities for red and yellow using a common denominator:
\[\frac{1}{4}+\frac{2}{5}=\frac{5}{20}+\frac{8}{20}=\frac{13}{20}.\]
Green is the remaining probability:
\[P(\text{green})=1-\frac{13}{20}=\frac{7}{20}.\]
The required probability is \(\frac{7}{20}\). [3 marks]
The probabilities of all possible colours must add to 1. First add the probabilities for red and yellow using a common denominator:
\[\frac{1}{4}+\frac{2}{5}=\frac{5}{20}+\frac{8}{20}=\frac{13}{20}.\]
Green is the remaining probability:
\[P(\text{green})=1-\frac{13}{20}=\frac{7}{20}.\]
The required probability is \(\frac{7}{20}\). [3 marks]
Question 6 Report
Calculate the column vector \(\overrightarrow{AB}\), where \(A\) is the point \((2, 9)\) and \(B\) is the point \((10, 3)\), as shown on the grid. Then work out the length of \(AB\). Give the length as a whole number.
A vector from \(A\) to \(B\) is found by subtracting the coordinates of \(A\) from the corresponding coordinates of \(B\):
\[\overrightarrow{AB}=\begin{pmatrix}10-2\\3-9\end{pmatrix}=\begin{pmatrix}8\\-6\end{pmatrix}.\]
Its length is found using Pythagoras’ theorem:
\[AB=\sqrt{8^2+(-6)^2}=\sqrt{64+36}=\sqrt{100}=10.\]
Therefore \(\overrightarrow{AB}=\begin{pmatrix}8\\-6\end{pmatrix}\) and \(AB=\mathbf{10}\). [4 marks]
A vector from \(A\) to \(B\) is found by subtracting the coordinates of \(A\) from the corresponding coordinates of \(B\):
\[\overrightarrow{AB}=\begin{pmatrix}10-2\\3-9\end{pmatrix}=\begin{pmatrix}8\\-6\end{pmatrix}.\]
Its length is found using Pythagoras’ theorem:
\[AB=\sqrt{8^2+(-6)^2}=\sqrt{64+36}=\sqrt{100}=10.\]
Therefore \(\overrightarrow{AB}=\begin{pmatrix}8\\-6\end{pmatrix}\) and \(AB=\mathbf{10}\). [4 marks]
Question 7 Report
Write the formula \(V = \dfrac{4}{3}\pi r^{3}\) with \(r\) as the subject.
This formula gives the volume of the sphere of radius \(r\) shown in the diagram. Show each step of your rearrangement.
The task is to isolate \(r\) in the sphere-volume formula. First remove the denominator by multiplying both sides by \(3\):
\[3V=4\pi r^3\quad\text{[M1]}\]
Then divide by \(4\pi\):
\[r^3=\frac{3V}{4\pi}\quad\text{[M1]}\]
Finally, take the cube root of both sides because \(r\) is cubed:
\[\boxed{r=\sqrt[3]{\frac{3V}{4\pi}}}\quad\text{[A1]}\]
A square root would not undo \(r^3\); the inverse operation required is a cube root.
The task is to isolate \(r\) in the sphere-volume formula. First remove the denominator by multiplying both sides by \(3\):
\[3V=4\pi r^3\quad\text{[M1]}\]
Then divide by \(4\pi\):
\[r^3=\frac{3V}{4\pi}\quad\text{[M1]}\]
Finally, take the cube root of both sides because \(r\) is cubed:
\[\boxed{r=\sqrt[3]{\frac{3V}{4\pi}}}\quad\text{[A1]}\]
A square root would not undo \(r^3\); the inverse operation required is a cube root.
Question 8 Report
Calculate the capacity of the larger of two mathematically similar bottles. The diagram shows the two bottles.
The smaller bottle is \(15\) cm high and holds \(540\) ml, and the larger bottle is \(20\) cm high. Give your answer in millilitres.
For similar three-dimensional objects, capacities and volumes scale by the cube of the length scale factor.
\[\text{length scale factor}=\frac{20}{15}=\frac{4}{3}\]
[M1]
\[\text{volume scale factor}=\left(\frac{4}{3}\right)^3=\frac{64}{27}\]
[M1 A1]
\[\text{larger capacity}=540\times\frac{64}{27}=20\times64=1280\text{ ml}\]
[A1] The larger bottle holds \(1280\text{ ml}\).
Use a cube for capacity because capacity measures three-dimensional space.
For similar three-dimensional objects, capacities and volumes scale by the cube of the length scale factor.
\[\text{length scale factor}=\frac{20}{15}=\frac{4}{3}\]
[M1]
\[\text{volume scale factor}=\left(\frac{4}{3}\right)^3=\frac{64}{27}\]
[M1 A1]
\[\text{larger capacity}=540\times\frac{64}{27}=20\times64=1280\text{ ml}\]
[A1] The larger bottle holds \(1280\text{ ml}\).
Use a cube for capacity because capacity measures three-dimensional space.
Question 9 Report
Use the bar chart in the diagram to answer this question about reading. The chart shows how many books each of six students read in one month. Work out the mean number of books. Then state the modal number and the range.
The bar heights give the values \(3,5,4,5,8,5\).
\[3+5+4+5+8+5=30\text{ books}\] [M1]
\[\text{mean}=30\div6=5\text{ books}\]
The mean is \(5\) books [A1]. The value occurring most often is \(5\), so the mode is \(5\) books [B1]. The largest value is \(8\) and the smallest is \(3\), hence \(8-3=5\): the range is \(5\) books [B1].
The bar heights give the values \(3,5,4,5,8,5\).
\[3+5+4+5+8+5=30\text{ books}\] [M1]
\[\text{mean}=30\div6=5\text{ books}\]
The mean is \(5\) books [A1]. The value occurring most often is \(5\), so the mode is \(5\) books [B1]. The largest value is \(8\) and the smallest is \(3\), hence \(8-3=5\): the range is \(5\) books [B1].
Question 10 Report
Work out the number of red counters in a bag. The bag holds \(n\) red counters and \(6\) blue counters. The probability of taking a red counter at random is \(\dfrac{2}{5}\).
(a) Work out the value of \(n\).
(b) Two counters are then taken at random without replacement, as shown on the tree diagram. Work out the probability that both are blue.
Without replacement is important: the second denominator decreases from 10 to 9.
Without replacement is important: the second denominator decreases from 10 to 9.
Question 11 Report
The graph of \(y = 2x + 5\) is shown. Solve the inequality \(-3 \leqslant 2x + 5 < 13\).
Then write down all of the integers that satisfy your inequality.
Subtract \(5\) from every part of the compound inequality:
\[-8\leq2x<8\quad\text{[M1]}\]
Divide throughout by positive \(2\):
\[\boxed{-4\leq x<4}\quad\text{[A1]}\]
The lower boundary is included but the upper boundary is not. Therefore the integers are
\[\boxed{-4,-3,-2,-1,0,1,2,3}\quad\text{[A1]}\]
Subtract \(5\) from every part of the compound inequality:
\[-8\leq2x<8\quad\text{[M1]}\]
Divide throughout by positive \(2\):
\[\boxed{-4\leq x<4}\quad\text{[A1]}\]
The lower boundary is included but the upper boundary is not. Therefore the integers are
\[\boxed{-4,-3,-2,-1,0,1,2,3}\quad\text{[A1]}\]
Question 12 Report
Work out the equation of a straight line. The diagram shows the two points and the line through them.
The line passes through the points \((2, -1)\) and \((6, 11)\). Give your answer in the form \(y = mx + c\).
First find the gradient from the two given points:
\[m=\frac{11-(-1)}{6-2}=\frac{12}{4}=3\] [M1][A1]
Use \(y=mx+c\) and substitute \((2,-1)\):
\[-1=3(2)+c\]
\[c=-7\]
Therefore the equation is:
\[y=3x-7\] [A1]
A check using \(x=6\) gives \(3(6)-7=11\), which matches the other point.
First find the gradient from the two given points:
\[m=\frac{11-(-1)}{6-2}=\frac{12}{4}=3\] [M1][A1]
Use \(y=mx+c\) and substitute \((2,-1)\):
\[-1=3(2)+c\]
\[c=-7\]
Therefore the equation is:
\[y=3x-7\] [A1]
A check using \(x=6\) gives \(3(6)-7=11\), which matches the other point.
Question 13 Report
Work out the values of \(t\) at which a particle is at rest. The diagram shows \(s\) plotted against \(t\).
The particle moves along a straight line, and its displacement from a fixed point after \(t\) seconds is \(s = t^{3} - 6t^{2} + 9t\) metres.
A particle is at rest when its velocity is zero. Velocity is the rate of change of displacement:
\[v=\frac{ds}{dt}=3t^2-12t+9\] [M1][A1]
Set \(v=0\) and factorise:
\[3t^2-12t+9=0\]
\[3(t-1)(t-3)=0\] [M1]
Therefore:
\[t=1\text{ s}\quad\text{or}\quad t=3\text{ s}\] [A1]
A particle is at rest when its velocity is zero. Velocity is the rate of change of displacement:
\[v=\frac{ds}{dt}=3t^2-12t+9\] [M1][A1]
Set \(v=0\) and factorise:
\[3t^2-12t+9=0\]
\[3(t-1)(t-3)=0\] [M1]
Therefore:
\[t=1\text{ s}\quad\text{or}\quad t=3\text{ s}\] [A1]
Question 14 Report
Use the information below and the diagram to answer parts (a) and (b).
A star is \(6.3 \times 10^{14}\) km from Earth, and light travels at \(3.0 \times 10^{5}\) km per second.
(a) Calculate the time light takes to travel from the star to Earth. Give your answer in seconds, in standard form. [3]
(b) One year is \(3.15 \times 10^{7}\) seconds. Work out this time in years, to the nearest year. [2]
(a) The time is found from \(\text{time}=\text{distance}\div\text{speed}\).
\[\text{time}=\frac{6.3\times10^{14}}{3.0\times10^5}\]
Divide the coefficients and subtract the powers of ten:
\[\frac{6.3}{3.0}=2.1,\qquad\frac{10^{14}}{10^5}=10^{14-5}=10^9\]
\[\text{time}=\boxed{2.1\times10^9\text{ s}}\]
[M1 M1 A1]
(b) Convert seconds to years by dividing by the number of seconds in one year:
\[\text{years}=\frac{2.1\times10^9}{3.15\times10^7}=66.66\ldots\]
To the nearest year, the time is \(67\) years. [M1 A1]
For division in standard form, subtract exponents. The answer to part (a) is already in standard form because \(2.1\) is between \(1\) and \(10\).
(a) The time is found from \(\text{time}=\text{distance}\div\text{speed}\).
\[\text{time}=\frac{6.3\times10^{14}}{3.0\times10^5}\]
Divide the coefficients and subtract the powers of ten:
\[\frac{6.3}{3.0}=2.1,\qquad\frac{10^{14}}{10^5}=10^{14-5}=10^9\]
\[\text{time}=\boxed{2.1\times10^9\text{ s}}\]
[M1 M1 A1]
(b) Convert seconds to years by dividing by the number of seconds in one year:
\[\text{years}=\frac{2.1\times10^9}{3.15\times10^7}=66.66\ldots\]
To the nearest year, the time is \(67\) years. [M1 A1]
For division in standard form, subtract exponents. The answer to part (a) is already in standard form because \(2.1\) is between \(1\) and \(10\).
Question 15 Report
Use the kite in the diagram to find the size of angle \(y\). The kite is \(ABCD\) with \(AB = AD\) and \(CB = CD\). Angle \(DAB = 108^\circ\) and angle \(ABC = 76^\circ\). Angle \(y\) is angle \(BCD\). Give your answer in degrees.
A kite with \(AB=AD\) and \(CB=CD\) has equal opposite angles between the unequal sides. Therefore angle \(ADC\) equals angle \(ABC\), so
\[\angle ADC=76^\circ\]
[M1]
The four interior angles of the kite add to \(360^\circ\):
\[y=360^\circ-108^\circ-76^\circ-76^\circ=100^\circ\]
[M1 A1] Therefore \(y=100^\circ\).
A kite with \(AB=AD\) and \(CB=CD\) has equal opposite angles between the unequal sides. Therefore angle \(ADC\) equals angle \(ABC\), so
\[\angle ADC=76^\circ\]
[M1]
The four interior angles of the kite add to \(360^\circ\):
\[y=360^\circ-108^\circ-76^\circ-76^\circ=100^\circ\]
[M1 A1] Therefore \(y=100^\circ\).
Question 16 Report
Work out the mean and the range of set \(B\), then compare the two sets. Set \(A\) has mean \(24\) and range \(10\). Set \(B\) is \(12,\ 16,\ 20,\ 26,\ 30,\ 40\). The bar chart in the diagram shows the six numbers in set B.
(a) Work out the mean of set \(B\).
(b) Work out the range of set \(B\).
(c) Write down one comparison of the two sets.
Question 17 Report
Calculate the sum of the interior angles of the regular polygon in the diagram, which has \(9\) sides. Then calculate the size of each interior angle. Give both answers in degrees.
A polygon with \(n\) sides has interior angle sum \((n-2)\times180^\circ\). For 9 sides:
\[(9-2)\times180^\circ=7\times180^\circ=1260^\circ\]
[M1 A1]
Because the polygon is regular, all 9 interior angles are equal:
\[1260^\circ\div9=140^\circ\]
[A1] The interior angle sum is \(1260^\circ\), and each interior angle is \(140^\circ\).
A polygon with \(n\) sides has interior angle sum \((n-2)\times180^\circ\). For 9 sides:
\[(9-2)\times180^\circ=7\times180^\circ=1260^\circ\]
[M1 A1]
Because the polygon is regular, all 9 interior angles are equal:
\[1260^\circ\div9=140^\circ\]
[A1] The interior angle sum is \(1260^\circ\), and each interior angle is \(140^\circ\).
Question 18 Report
Work out the size of angle \(x\). The diagram shows angle \(x\) and an angle of \(118^\circ\) lying together on a straight line. Give your answer in degrees.
Angles adjacent on a straight line add to \(180^\circ\). Therefore
\[x+118^\circ=180^\circ\]
\[x=180^\circ-118^\circ=62^\circ\]
[M1 A1] Hence \(x=62^\circ\).
Angles adjacent on a straight line add to \(180^\circ\). Therefore
\[x+118^\circ=180^\circ\]
\[x=180^\circ-118^\circ=62^\circ\]
[M1 A1] Hence \(x=62^\circ\).
Question 19 Report
State the size of one interior angle of the regular hexagon in the diagram. Give your answer in degrees. You do not need to show any working for this question.
A regular hexagon has six equal exterior angles, each \(360^\circ\div6=60^\circ\). Its interior angle is therefore \(180^\circ-60^\circ=120^\circ\). The size of one interior angle is \(120^\circ\). [B1]
A regular hexagon has six equal exterior angles, each \(360^\circ\div6=60^\circ\). Its interior angle is therefore \(180^\circ-60^\circ=120^\circ\). The size of one interior angle is \(120^\circ\). [B1]
Question 20 Report
Work out the median of these ten numbers. The diagram shows the same ten numbers plotted in the order given.
\(12,\ 7,\ 15,\ 9,\ 20,\ 4,\ 11,\ 18,\ 6,\ 14\)
You must show the working that you use.
Order the data before identifying the middle values:
\[4,\ 6,\ 7,\ 9,\ 11,\ 12,\ 14,\ 15,\ 18,\ 20\] [M1]
There are ten values, so use the fifth and sixth values:
\[\text{median}=\frac{11+12}{2}=11.5\]
The median is \(11.5\) [A1].
Order the data before identifying the middle values:
\[4,\ 6,\ 7,\ 9,\ 11,\ 12,\ 14,\ 15,\ 18,\ 20\] [M1]
There are ten values, so use the fifth and sixth values:
\[\text{median}=\frac{11+12}{2}=11.5\]
The median is \(11.5\) [A1].
Question 21 Report
Use the sequence \(4,\ 11,\ 22,\ 37,\ \ldots\) to answer all three parts. The sequence is quadratic.
(a) Work out the next term.
(b) Work out an expression for the \(n\)th term.
(c) Work out the 15th term.
A quadratic sequence has constant second differences. Start by finding the differences between the given terms.
| Terms | First differences | Second differences |
|---|---|---|
| \(4,\ 11,\ 22,\ 37\) | \(7,\ 11,\ 15\) | \(4,\ 4\) |
Examination reminder: a constant second difference identifies a quadratic sequence; divide that second difference by 2 to find the coefficient of \(n^2\).
A quadratic sequence has constant second differences. Start by finding the differences between the given terms.
| Terms | First differences | Second differences |
|---|---|---|
| \(4,\ 11,\ 22,\ 37\) | \(7,\ 11,\ 15\) | \(4,\ 4\) |
Examination reminder: a constant second difference identifies a quadratic sequence; divide that second difference by 2 to find the coefficient of \(n^2\).
Question 22 Report
Work out the coordinates of the vertices of triangle \(ABC\), shown on the grid, after a translation by \(\begin{pmatrix} -2 \\ 5 \end{pmatrix}\). The vertices are \(A(1, 1)\), \(B(4, 2)\) and \(C(3, 6)\). Then state the column vector of the translation that maps the image back onto triangle \(ABC\).
A translation adds the vector components to every vertex. Here, subtract 2 from every \(x\)-coordinate and add 5 to every \(y\)-coordinate.
A translation adds the vector components to every vertex. Here, subtract 2 from every \(x\)-coordinate and add 5 to every \(y\)-coordinate.
Question 23 Report
Work through all four parts of this question about the numbers \(1\) to \(15\). Set \(A\) is the multiples of \(3\) and set \(B\) is the even numbers.
(a) Complete the Venn diagram in the figure by writing each of the \(15\) numbers in the correct region. [2]
(b) Write down the numbers that are in both set \(A\) and set \(B\). [1]
(c) State how many of the numbers are in neither set. [1]
(d) Work out the fraction of the \(15\) numbers that are in set \(A\) or set \(B\) or both. Give your fraction in its simplest form. [2]
Set \(A\) contains multiples of \(3\), and set \(B\) contains even numbers. The numbers in both sets must be both multiples of \(3\) and even.
Set \(A\) contains multiples of \(3\), and set \(B\) contains even numbers. The numbers in both sets must be both multiples of \(3\) and even.
Question 24 Report
Use the diagram to answer both parts of this question.
Points \(A\), \(B\) and \(C\) lie on a circle with centre \(O\). Point \(B\) is on the major arc, and angle \(AOC = 128^{\circ}\).
(i) Work out angle \(ABC\). Give a reason for your answer.
(ii) Point \(D\) lies on the minor arc \(AC\). Work out angle \(ADC\).
For points on the same circle, the angle at the centre is twice the angle at the circumference standing on the same chord \(AC\). Point \(B\) is on the major arc, so \(\angle ABC\) stands on the minor arc \(AC\):
\[\angle ABC=\frac{128^\circ}{2}=64^\circ\] [M1 A1]
Points \(A,B,C,D\) form a cyclic quadrilateral, so opposite angles total \(180^\circ\). Therefore
\[\angle ADC=180^\circ-64^\circ=116^\circ\] [A1]
Thus \(\angle ABC=64^\circ\) and \(\angle ADC=116^\circ\).
For points on the same circle, the angle at the centre is twice the angle at the circumference standing on the same chord \(AC\). Point \(B\) is on the major arc, so \(\angle ABC\) stands on the minor arc \(AC\):
\[\angle ABC=\frac{128^\circ}{2}=64^\circ\] [M1 A1]
Points \(A,B,C,D\) form a cyclic quadrilateral, so opposite angles total \(180^\circ\). Therefore
\[\angle ADC=180^\circ-64^\circ=116^\circ\] [A1]
Thus \(\angle ABC=64^\circ\) and \(\angle ADC=116^\circ\).
Question 25 Report
The diagram shows the vectors \(\mathbf{p}\) and \(\mathbf{q}\). Calculate the column vector \(\mathbf{p} + 3\mathbf{q}\), where \(\mathbf{p} = \begin{pmatrix} 2 \\ 7 \end{pmatrix}\) and \(\mathbf{q} = \begin{pmatrix} -1 \\ 3 \end{pmatrix}\). Show your working clearly.
First multiply every component of \(\mathbf q\) by 3:
\[3\mathbf q=3\begin{pmatrix}-1\\3\end{pmatrix}=\begin{pmatrix}-3\\9\end{pmatrix}.\]
Then add corresponding components:
\[\mathbf p+3\mathbf q=\begin{pmatrix}2\\7\end{pmatrix}+\begin{pmatrix}-3\\9\end{pmatrix}=\begin{pmatrix}2-3\\7+9\end{pmatrix}=\begin{pmatrix}-1\\16\end{pmatrix}.\]
The required column vector is \(\begin{pmatrix}-1\\16\end{pmatrix}\). [2 marks]
First multiply every component of \(\mathbf q\) by 3:
\[3\mathbf q=3\begin{pmatrix}-1\\3\end{pmatrix}=\begin{pmatrix}-3\\9\end{pmatrix}.\]
Then add corresponding components:
\[\mathbf p+3\mathbf q=\begin{pmatrix}2\\7\end{pmatrix}+\begin{pmatrix}-3\\9\end{pmatrix}=\begin{pmatrix}2-3\\7+9\end{pmatrix}=\begin{pmatrix}-1\\16\end{pmatrix}.\]
The required column vector is \(\begin{pmatrix}-1\\16\end{pmatrix}\). [2 marks]
Question 26 Report
Complete the table by writing True, False or Cannot tell for each statement. Nia asks \(100\) people which drink they like best. Her results are tea \(34\), coffee \(26\), juice \(22\) and water \(18\).
| Statement | True, False or Cannot tell |
| More people chose tea than chose coffee. | |
| Exactly half of the people chose juice or water. | |
| More men than women chose water. |
Use only the totals given; the results contain no information about gender.
| Statement | Conclusion and reason |
|---|---|
| More people chose tea than chose coffee. | True, because \(34>26\). [B1] |
| Exactly half of the people chose juice or water. | False, because \(22+18=40\), not \(50\). [B1] |
| More men than women chose water. | Cannot tell, because no gender data were recorded. [B1] |
Use only the totals given; the results contain no information about gender.
| Statement | Conclusion and reason |
|---|---|
| More people chose tea than chose coffee. | True, because \(34>26\). [B1] |
| Exactly half of the people chose juice or water. | False, because \(22+18=40\), not \(50\). [B1] |
| More men than women chose water. | Cannot tell, because no gender data were recorded. [B1] |
Question 27 Report
Work through all three parts of this question about a recipe. A recipe for \(12\) biscuits uses \(300\) g of flour and \(180\) g of butter.
(a) Calculate the mass of flour needed for \(20\) biscuits. [2]
(b) Write the ratio of flour to butter in its simplest form. [2]
(c) State the mass of butter needed for \(4\) biscuits. [1]
(a) Find the flour for one biscuit, then scale up to 20 biscuits:
\[300\text{ g}\div12=25\text{ g}\]
\[25\text{ g}\times20=500\text{ g}\]
The required flour mass is \(500\text{ g}\). [M1 A1]
(b) Start with the quantities in the recipe, \(300:180\), and divide both terms by their highest common factor, 60:
\[300:180=5:3\]
The simplest ratio of flour to butter is \(5:3\). [M1 A1]
(c) Butter per biscuit is \(180\div12=15\) g. For 4 biscuits:
\[15\times4=60\text{ g}\]
The required butter mass is \(60\text{ g}\). [B1]
In a recipe, every ingredient must be scaled by the same factor, so the flour-to-butter ratio remains unchanged.
(a) Find the flour for one biscuit, then scale up to 20 biscuits:
\[300\text{ g}\div12=25\text{ g}\]
\[25\text{ g}\times20=500\text{ g}\]
The required flour mass is \(500\text{ g}\). [M1 A1]
(b) Start with the quantities in the recipe, \(300:180\), and divide both terms by their highest common factor, 60:
\[300:180=5:3\]
The simplest ratio of flour to butter is \(5:3\). [M1 A1]
(c) Butter per biscuit is \(180\div12=15\) g. For 4 biscuits:
\[15\times4=60\text{ g}\]
The required butter mass is \(60\text{ g}\). [B1]
In a recipe, every ingredient must be scaled by the same factor, so the flour-to-butter ratio remains unchanged.
Question 28 Report
The diagram shows a quadrilateral with interior angles \(a\), \(b\), \(c\) and \(d\). State the sum of the interior angles of a quadrilateral. Give your answer in degrees.
A quadrilateral can be divided into two triangles. Since each triangle has angle sum \(180^\circ\), the total is
\[(4-2)\times180^\circ=360^\circ\]
[B1] The sum of the interior angles is \(360^\circ\).
A quadrilateral can be divided into two triangles. Since each triangle has angle sum \(180^\circ\), the total is
\[(4-2)\times180^\circ=360^\circ\]
[B1] The sum of the interior angles is \(360^\circ\).
Question 29 Report
The diagram shows a cube. State the number of faces on a cube. A cube is a solid with square faces that are all the same size. Write your answer in the box as a whole number.
A cube has six congruent square faces: top, bottom, front, back, left and right.
Number of faces \(=6\). [B1]
A cube has six congruent square faces: top, bottom, front, back, left and right.
Number of faces \(=6\). [B1]
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