Loading....
|
Press & Hold to Drag Around |
|||
|
Click Here to Close |
|||
Question 1 Report
The diagram shows a rectangle divided into \(20\) equal squares. \(8\) of the squares are shaded.
(i) Write down the fraction of the rectangle that is shaded, giving your answer in its simplest form.
(ii) Work out this fraction of \(350\).
A fraction of a shape compares the number of shaded parts to the total number of equal parts, then is simplified; that same fraction of a quantity is found by dividing by the denominator and multiplying by the numerator.
(i) 8 out of 20 squares are shaded, giving \(\frac{8}{20}\); dividing top and bottom by 4 gives \(\frac{2}{5}\) [B1].
(ii) Finding \(\frac{2}{5}\) of 350 means dividing by the denominator and multiplying by the numerator: \(350 \div 5 \times 2\) oe [M1]. Since \(350 \div 5 = 70\) and \(70 \times 2 = 140\), the answer is \(140\) [A1].
A fraction of a shape compares the number of shaded parts to the total number of equal parts, then is simplified; that same fraction of a quantity is found by dividing by the denominator and multiplying by the numerator.
(i) 8 out of 20 squares are shaded, giving \(\frac{8}{20}\); dividing top and bottom by 4 gives \(\frac{2}{5}\) [B1].
(ii) Finding \(\frac{2}{5}\) of 350 means dividing by the denominator and multiplying by the numerator: \(350 \div 5 \times 2\) oe [M1]. Since \(350 \div 5 = 70\) and \(70 \times 2 = 140\), the answer is \(140\) [A1].
Question 2 Report
Simplify \(2x^{3}y \times 4x^{2}y^{5}\).
When multiplying algebraic terms, multiply the numbers together and add the powers of matching letters: \(2 \times 4 = 8\), \(x^{3} \times x^{2} = x^{5}\), and \(y \times y^{5} = y^{6}\), so \(2x^{3}y \times 4x^{2}y^{5} = 8x^{5}y^{6}\) [B2] (1 mark is awarded for getting two of the three parts, the coefficient 8, the power \(x^{5}\), and the power \(y^{6}\), correct).
When multiplying algebraic terms, multiply the numbers together and add the powers of matching letters: \(2 \times 4 = 8\), \(x^{3} \times x^{2} = x^{5}\), and \(y \times y^{5} = y^{6}\), so \(2x^{3}y \times 4x^{2}y^{5} = 8x^{5}y^{6}\) [B2] (1 mark is awarded for getting two of the three parts, the coefficient 8, the power \(x^{5}\), and the power \(y^{6}\), correct).
Question 3 Report
Write \(45\%\) as a fraction in its lowest terms.
\(45\%=\frac{45}{100}\). Dividing numerator and denominator by \(5\) gives \(\frac{9}{20}\) [B1] cao, which has no further common factor so is already in lowest terms.
\(45\%=\frac{45}{100}\). Dividing numerator and denominator by \(5\) gives \(\frac{9}{20}\) [B1] cao, which has no further common factor so is already in lowest terms.
Question 4 Report
In triangle \(ABC\), \(AB = AC\) and angle \(BAC = 40^\circ\).
Find angle \(ABC\) and write down the mathematical name of this type of triangle.
Since \(AB = AC\), triangle \(ABC\) is isosceles, so the two base angles \(ABC\) and \(ACB\) are equal. The angles of a triangle sum to \(180^\circ\), so the two equal angles share the remaining \(180 - 40 = 140^\circ\) between them.
\((180 - 40) \div 2\) [M1]
Angle \(ABC = 70^\circ\) [A1]
Isosceles [B1]
Since \(AB = AC\), triangle \(ABC\) is isosceles, so the two base angles \(ABC\) and \(ACB\) are equal. The angles of a triangle sum to \(180^\circ\), so the two equal angles share the remaining \(180 - 40 = 140^\circ\) between them.
\((180 - 40) \div 2\) [M1]
Angle \(ABC = 70^\circ\) [A1]
Isosceles [B1]
Question 5 Report
A flight leaves at \(22\ 45\) and lands at \(06\ 20\) the next day.
Work out the length of the flight.
This tests a duration that crosses midnight. Split the journey at midnight so that both pieces are ordinary forward-counting times.
Subtracting the clock readings directly, \(06\ 20 - 22\ 45\), gives a negative result and is the main pitfall. Splitting at midnight avoids it. As a check, \(22\ 45\) plus \(7\) hours is \(05\ 45\), and a further \(35\) minutes gives \(06\ 20\).
This tests a duration that crosses midnight. Split the journey at midnight so that both pieces are ordinary forward-counting times.
Subtracting the clock readings directly, \(06\ 20 - 22\ 45\), gives a negative result and is the main pitfall. Splitting at midnight avoids it. As a check, \(22\ 45\) plus \(7\) hours is \(05\ 45\), and a further \(35\) minutes gives \(06\ 20\).
Question 6 Report
There are \(30\) students in a class and \(18\) of them are girls.
Work out the percentage of the class that are girls.
The proportion of girls is written as a fraction of the whole class, then converted to a percentage by multiplying by \(100\): \(\frac{18}{30}\times 100\) [M1] oe.
\(\frac{18}{30}=\frac{3}{5}\), and \(\frac{3}{5}\times 100=60\%\) [A1].
The proportion of girls is written as a fraction of the whole class, then converted to a percentage by multiplying by \(100\): \(\frac{18}{30}\times 100\) [M1] oe.
\(\frac{18}{30}=\frac{3}{5}\), and \(\frac{3}{5}\times 100=60\%\) [A1].
Question 7 Report
By writing each number correct to 1 significant figure, estimate the value of \(\frac{39 \times 21}{4.1}\).
Rounding each value to 1 significant figure before calculating gives a quick, sensible estimate: \(39 \to 40\), \(21 \to 20\), \(4.1 \to 4\).
\(\frac{40 \times 20}{4}\) [M1]
\(= \frac{800}{4} = 200\) [A1]
Rounding each value to 1 significant figure before calculating gives a quick, sensible estimate: \(39 \to 40\), \(21 \to 20\), \(4.1 \to 4\).
\(\frac{40 \times 20}{4}\) [M1]
\(= \frac{800}{4} = 200\) [A1]
Question 8 Report
The diagram shows a right-angled triangle with shorter sides of length 9 cm and 12 cm.
(a) Calculate the length of the hypotenuse. [2]
(b) Work out the perimeter of the triangle. [2]
(c) Work out the area of the triangle. [2]
(a) Adding the squares of the two shorter sides gives the hypotenuse: \(9^2 + 12^2 = 81 + 144 = 225\) [M1]. The square root gives \(\sqrt{225} = 15\) cm [A1].
(b) The perimeter is the sum of all three sides: \(9 + 12 + 15\) [M1], giving 36 cm [A1].
(c) The two shorter sides are perpendicular, so they serve as the base and height: area \(= \frac{1}{2} \times 9 \times 12\) [M1], giving \(54 \; \text{cm}^2\) [A1].
Exam tip: 9, 12, 15 is the 3, 4, 5 triple scaled by 3, letting every part of this question be solved without a calculator once the triple is recognised.
(a) Adding the squares of the two shorter sides gives the hypotenuse: \(9^2 + 12^2 = 81 + 144 = 225\) [M1]. The square root gives \(\sqrt{225} = 15\) cm [A1].
(b) The perimeter is the sum of all three sides: \(9 + 12 + 15\) [M1], giving 36 cm [A1].
(c) The two shorter sides are perpendicular, so they serve as the base and height: area \(= \frac{1}{2} \times 9 \times 12\) [M1], giving \(54 \; \text{cm}^2\) [A1].
Exam tip: 9, 12, 15 is the 3, 4, 5 triple scaled by 3, letting every part of this question be solved without a calculator once the triple is recognised.
Question 9 Report
Each of \(50\) students chooses a favourite subject from mathematics, English, science and art.
The bar chart shows the results for mathematics, English and science only.
(a) Work out the number of students who chose art. [2]
(b) Write down the modal subject. [1]
(c) Write down whether favourite subject is qualitative or quantitative data. [1]
(d) Write down the fraction of the students who chose mathematics, giving your answer in its simplest form. [1]
This tests using a known overall total to find a missing category, then reading the chart and classifying the data.
(a) The three bars shown are mathematics \(15\), English \(12\) and science \(13\). Since all \(50\) students chose one of the four subjects, the number choosing art is \(50 - (15 + 12 + 13)\) [M1]. The bracket gives \(15 + 12 + 13 = 40\), so \(50 - 40 = 10\) students chose art [A1].
(b) The modal subject is the one with the highest frequency. Comparing \(15\), \(12\), \(13\) and \(10\), the largest is \(15\), so the mode is mathematics [B1].
(c) Favourite subject is a category described in words, not a measurement or count of something, so it is qualitative data [B1].
(d) The fraction choosing mathematics is \(\frac{15}{50}\). Dividing numerator and denominator by \(5\) gives \(\frac{3}{10}\) [B1].
Part (b) must be answered after part (a), because art could in principle have been the largest group. Here it is not, since \(10 \lt 15\). The mode is the name of the subject, not the frequency \(15\).
This tests using a known overall total to find a missing category, then reading the chart and classifying the data.
(a) The three bars shown are mathematics \(15\), English \(12\) and science \(13\). Since all \(50\) students chose one of the four subjects, the number choosing art is \(50 - (15 + 12 + 13)\) [M1]. The bracket gives \(15 + 12 + 13 = 40\), so \(50 - 40 = 10\) students chose art [A1].
(b) The modal subject is the one with the highest frequency. Comparing \(15\), \(12\), \(13\) and \(10\), the largest is \(15\), so the mode is mathematics [B1].
(c) Favourite subject is a category described in words, not a measurement or count of something, so it is qualitative data [B1].
(d) The fraction choosing mathematics is \(\frac{15}{50}\). Dividing numerator and denominator by \(5\) gives \(\frac{3}{10}\) [B1].
Part (b) must be answered after part (a), because art could in principle have been the largest group. Here it is not, since \(10 \lt 15\). The mode is the name of the subject, not the frequency \(15\).
Question 10 Report
The diagram shows the net of a cuboid measuring \(10\) cm by \(4\) cm by \(3\) cm.
(a) Work out the total area of the net. [3]
(b) Work out the volume of the cuboid made from this net. [2]
(a) The net of a cuboid is its six rectangular faces laid flat, and opposite faces are identical, so they form three matching pairs.
The three different rectangles from edges \(10\) cm, \(4\) cm and \(3\) cm are \(10 \times 4 = 40\ \mathrm{cm}^2\), \(10 \times 3 = 30\ \mathrm{cm}^2\) and \(4 \times 3 = 12\ \mathrm{cm}^2\) [M1].
Doubling for the pairs: \(2 \times (40 + 30 + 12)\) [M1], which is \(2 \times 82 = 164\ \mathrm{cm}^2\) [A1]. This is also the surface area of the cuboid, since folding does not change area.
(b) Volume of a cuboid is the product of its three perpendicular edges: \(10 \times 4 \times 3\) [M1], giving \(120\ \mathrm{cm}^3\) [A1].
Watch the units, since they separate the two parts: multiplying two lengths gives an area in \(\mathrm{cm}^2\), while multiplying three gives a volume in \(\mathrm{cm}^3\). The commonest error in the first part is stopping at \(82\ \mathrm{cm}^2\), which counts only three of the six faces.
(a) The net of a cuboid is its six rectangular faces laid flat, and opposite faces are identical, so they form three matching pairs.
The three different rectangles from edges \(10\) cm, \(4\) cm and \(3\) cm are \(10 \times 4 = 40\ \mathrm{cm}^2\), \(10 \times 3 = 30\ \mathrm{cm}^2\) and \(4 \times 3 = 12\ \mathrm{cm}^2\) [M1].
Doubling for the pairs: \(2 \times (40 + 30 + 12)\) [M1], which is \(2 \times 82 = 164\ \mathrm{cm}^2\) [A1]. This is also the surface area of the cuboid, since folding does not change area.
(b) Volume of a cuboid is the product of its three perpendicular edges: \(10 \times 4 \times 3\) [M1], giving \(120\ \mathrm{cm}^3\) [A1].
Watch the units, since they separate the two parts: multiplying two lengths gives an area in \(\mathrm{cm}^2\), while multiplying three gives a volume in \(\mathrm{cm}^3\). The commonest error in the first part is stopping at \(82\ \mathrm{cm}^2\), which counts only three of the six faces.
Question 11 Report
Solve the equation \(\frac{m}{4} = 9\) to find the value of \(m\).
To solve an equation of the form \(\frac{m}{a} = b\), multiply both sides by \(a\) to undo the division.
Multiplying both sides of \(\frac{m}{4} = 9\) by 4 gives \(m = 9 \times 4 = 36\) [B1].
Checking by substitution: \(\frac{36}{4} = 9\), which confirms the solution.
To solve an equation of the form \(\frac{m}{a} = b\), multiply both sides by \(a\) to undo the division.
Multiplying both sides of \(\frac{m}{4} = 9\) by 4 gives \(m = 9 \times 4 = 36\) [B1].
Checking by substitution: \(\frac{36}{4} = 9\), which confirms the solution.
Question 12 Report
By writing each number correct to 1 significant figure, estimate the value of \(\dfrac{412 + 189}{29}\).
Round each number to 1 significant figure before working out the simplified calculation.
\(412 \to 400\), \(189 \to 200\), \(29 \to 30\), giving \(\dfrac{400 + 200}{30}\) [M1].
\(400 + 200 = 600\), and \(600 \div 30 = 20\), so the estimate is \(20\) [A1].
Round each number to 1 significant figure before working out the simplified calculation.
\(412 \to 400\), \(189 \to 200\), \(29 \to 30\), giving \(\dfrac{400 + 200}{30}\) [M1].
\(400 + 200 = 600\), and \(600 \div 30 = 20\), so the estimate is \(20\) [A1].
Question 13 Report
A map has a scale of 1 : 100 000.
A straight road measures 8.5 cm on the map. Work out the real length of the road, in kilometres.
A scale of \(1:100\,000\) means \(1\) cm on the map represents \(100\,000\) cm in real life, so the map length is multiplied by \(100\,000\) to find the real length in centimetres.
\(8.5 \times 100\,000 = 850\,000\) cm [M1].
Converting to kilometres (\(1\) km \(=100\,000\) cm), the road is \(8.5\) km long [A1].
Because \(100\,000\) cm equals exactly \(1\) km, the value in centimetres and the value in kilometres share the same digits here, a useful check.
A scale of \(1:100\,000\) means \(1\) cm on the map represents \(100\,000\) cm in real life, so the map length is multiplied by \(100\,000\) to find the real length in centimetres.
\(8.5 \times 100\,000 = 850\,000\) cm [M1].
Converting to kilometres (\(1\) km \(=100\,000\) cm), the road is \(8.5\) km long [A1].
Because \(100\,000\) cm equals exactly \(1\) km, the value in centimetres and the value in kilometres share the same digits here, a useful check.
Question 14 Report
The cross-section of a solid prism is a right-angled triangle with sides 3 cm, 4 cm and 5 cm. The prism is 10 cm long.
Calculate the total surface area of the prism.
The total surface area of a prism is the two triangular end faces plus the rectangular side faces, whose combined area equals the triangle's perimeter multiplied by the prism's length.
Area of one triangle \(=\frac{1}{2}\times3\times4=6\) cm\(^2\) [M1].
Perimeter of triangle \(=3+4+5=12\) cm [M1].
Total surface area \(=2\times6+12\times10\) [M1]
\(=132\) cm\(^2\) [A1].
The total surface area of a prism is the two triangular end faces plus the rectangular side faces, whose combined area equals the triangle's perimeter multiplied by the prism's length.
Area of one triangle \(=\frac{1}{2}\times3\times4=6\) cm\(^2\) [M1].
Perimeter of triangle \(=3+4+5=12\) cm [M1].
Total surface area \(=2\times6+12\times10\) [M1]
\(=132\) cm\(^2\) [A1].
Question 15 Report
The bar chart shows how the students in a class travel to school.
(a) Work out the total number of students in the class. [2]
(b) Work out the fraction of the class who walk to school. Give your answer in its simplest form. [2]
(a) The total number of students in the class is the sum of the frequencies for every travel method shown on the bar chart: \(18 + 12 + 24 + 6\) [M1], which gives 60 [A1].
(b) The fraction who walk is the walking frequency over the total: \(\frac{24}{60}\) [M1], which simplifies (dividing numerator and denominator by 12) to \(\frac{2}{5}\) [A1].
Finding the total first in part (a) is essential before part (b) can be answered, since the fraction in part (b) depends on the total number of students found in part (a).
(a) The total number of students in the class is the sum of the frequencies for every travel method shown on the bar chart: \(18 + 12 + 24 + 6\) [M1], which gives 60 [A1].
(b) The fraction who walk is the walking frequency over the total: \(\frac{24}{60}\) [M1], which simplifies (dividing numerator and denominator by 12) to \(\frac{2}{5}\) [A1].
Finding the total first in part (a) is essential before part (b) can be answered, since the fraction in part (b) depends on the total number of students found in part (a).
Question 16 Report
Work out \(\sqrt[3]{216} + \sqrt{169} - 2^4\).
This tests recognising a cube root, a square root and a power, then combining them in the correct order. Roots and indices are evaluated before the addition and subtraction.
The values \(216\), \(169\) and \(16\) are all worth knowing by sight for a non-calculator paper. Note that \(\sqrt[3]{216}\) is \(6\), not \(72\); dividing by \(3\) instead of taking the cube root is a frequent error.
This tests recognising a cube root, a square root and a power, then combining them in the correct order. Roots and indices are evaluated before the addition and subtraction.
The values \(216\), \(169\) and \(16\) are all worth knowing by sight for a non-calculator paper. Note that \(\sqrt[3]{216}\) is \(6\), not \(72\); dividing by \(3\) instead of taking the cube root is a frequent error.
Question 17 Report
Simplify \(4a + 7b - a + 2b\).
Simplifying collects the \(a\)-terms and the \(b\)-terms separately. For the \(a\)-terms: \(4a - a = 3a\). For the \(b\)-terms: \(7b + 2b = 9b\). So \(4a + 7b - a + 2b = 3a + 9b\) [B2] (1 mark is awarded for correctly simplifying just the \(a\)-terms to \(3a\) or just the \(b\)-terms to \(9b\) if the full simplification is not reached).
Simplifying collects the \(a\)-terms and the \(b\)-terms separately. For the \(a\)-terms: \(4a - a = 3a\). For the \(b\)-terms: \(7b + 2b = 9b\). So \(4a + 7b - a + 2b = 3a + 9b\) [B2] (1 mark is awarded for correctly simplifying just the \(a\)-terms to \(3a\) or just the \(b\)-terms to \(9b\) if the full simplification is not reached).
Question 18 Report
Find the equation of the straight line that is parallel to \(y = 4x - 1\) and passes through the point \((0,6)\).
Two facts fix a straight line in the form \(y = mx + c\): its gradient \(m\) and its \(y\)-intercept \(c\).
Parallel lines have the same gradient, because they slope at the same angle and never meet. The line \(y = 4x - 1\) has gradient \(4\), so the new line also has gradient \(=4\) [B1].
The point \((0,6)\) has \(x=0\), so it lies on the \(y\)-axis and is therefore the \(y\)-intercept itself, giving \(c = 6\). Putting the two together, the equation is \(y = 4x + 6\) [B1].
The value of \(c\) can also be found by substituting: \(6 = 4 \times 0 + c\) gives \(c = 6\). Note that only the constant changes between parallel lines. Keeping the \(-1\) from the original equation, or changing the \(4\), would produce a line that is either not parallel or does not pass through the given point.
Two facts fix a straight line in the form \(y = mx + c\): its gradient \(m\) and its \(y\)-intercept \(c\).
Parallel lines have the same gradient, because they slope at the same angle and never meet. The line \(y = 4x - 1\) has gradient \(4\), so the new line also has gradient \(=4\) [B1].
The point \((0,6)\) has \(x=0\), so it lies on the \(y\)-axis and is therefore the \(y\)-intercept itself, giving \(c = 6\). Putting the two together, the equation is \(y = 4x + 6\) [B1].
The value of \(c\) can also be found by substituting: \(6 = 4 \times 0 + c\) gives \(c = 6\). Note that only the constant changes between parallel lines. Keeping the \(-1\) from the original equation, or changing the \(4\), would produce a line that is either not parallel or does not pass through the given point.
Question 19 Report
In triangle \(ABC\), angle \(ABC = 68^\circ\) and angle \(BCA = 44^\circ\).
(a) Work out the size of angle \(BAC\). [2]
(b) The side \(AB\) is produced to \(D\). Work out the size of angle \(CBD\). [2]
(c) Write down the sum of the three exterior angles of triangle \(ABC\). [1]
This question tests the angle sum of a triangle and the exterior angle formed when a side is extended.
(a) The three angles of triangle \(ABC\) sum to \(180^\circ\), so angle \(BAC = 180 - (68 + 44)\) [M1], giving \(68^\circ\) [A1].
(b) Producing \(AB\) to \(D\) creates a straight line through \(B\), so angle \(ABC\) and angle \(CBD\) lie on a straight line and sum to \(180^\circ\). Hence angle \(CBD = 180 - 68\) [M1], giving \(112^\circ\) [A1]. (Equivalently, this exterior angle equals the sum of the two opposite interior angles, \(44 + 68 = 112^\circ\).)
(c) The exterior angles of any polygon, including a triangle, always sum to \(360^\circ\) [B1], regardless of the triangle's individual angles.
This question tests the angle sum of a triangle and the exterior angle formed when a side is extended.
(a) The three angles of triangle \(ABC\) sum to \(180^\circ\), so angle \(BAC = 180 - (68 + 44)\) [M1], giving \(68^\circ\) [A1].
(b) Producing \(AB\) to \(D\) creates a straight line through \(B\), so angle \(ABC\) and angle \(CBD\) lie on a straight line and sum to \(180^\circ\). Hence angle \(CBD = 180 - 68\) [M1], giving \(112^\circ\) [A1]. (Equivalently, this exterior angle equals the sum of the two opposite interior angles, \(44 + 68 = 112^\circ\).)
(c) The exterior angles of any polygon, including a triangle, always sum to \(360^\circ\) [B1], regardless of the triangle's individual angles.
Question 20 Report
Four of the five interior angles of a pentagon are \(100^\circ\), \(120^\circ\), \(90^\circ\) and \(130^\circ\). Work out the size of the fifth angle.
The sum of the interior angles of a pentagon is \((5-2) \times 180^{\circ} = 540^{\circ}\) [M1].
The four known angles add to \(100^{\circ}+120^{\circ}+90^{\circ}+130^{\circ} = 440^{\circ}\), so the fifth angle is \(540^{\circ} - 440^{\circ}\) [M1], giving \(100^{\circ}\) [A1].
Exam tip: always find the total sum for the polygon first using \((n-2)\times180^{\circ}\), then subtract the known angles - never assume a pentagon's angles sum to \(360^{\circ}\), which is only true for angles at a point or exterior angles.
The sum of the interior angles of a pentagon is \((5-2) \times 180^{\circ} = 540^{\circ}\) [M1].
The four known angles add to \(100^{\circ}+120^{\circ}+90^{\circ}+130^{\circ} = 440^{\circ}\), so the fifth angle is \(540^{\circ} - 440^{\circ}\) [M1], giving \(100^{\circ}\) [A1].
Exam tip: always find the total sum for the polygon first using \((n-2)\times180^{\circ}\), then subtract the known angles - never assume a pentagon's angles sum to \(360^{\circ}\), which is only true for angles at a point or exterior angles.
Question 21 Report
Write \(\frac{1}{81}\) as a power of \(3\).
Writing a number as a power of \(3\) means expressing it in the form \(3^{k}\). Deal with the \(81\) first, then with the fact that the number is a fraction.
Building up the powers of \(3\) by repeated multiplication: \(3^1=3\), \(3^2=9\), \(3^3=27\), \(3^4=81\). So \(81=3^4\) [M1].
That makes the number \(\frac{1}{3^4}\). A reciprocal of a power is written with a negative index, using \(\frac{1}{a^{n}} = a^{-n}\). Therefore \(\frac{1}{81}=3^{-4}\) [A1].
The negative sign records that the power is on the bottom of the fraction, not that the value is negative. Since \(\frac{1}{81}\) is a small positive number, an answer such as \(-3^4\) would be wrong in sign as well as in form.
Writing a number as a power of \(3\) means expressing it in the form \(3^{k}\). Deal with the \(81\) first, then with the fact that the number is a fraction.
Building up the powers of \(3\) by repeated multiplication: \(3^1=3\), \(3^2=9\), \(3^3=27\), \(3^4=81\). So \(81=3^4\) [M1].
That makes the number \(\frac{1}{3^4}\). A reciprocal of a power is written with a negative index, using \(\frac{1}{a^{n}} = a^{-n}\). Therefore \(\frac{1}{81}=3^{-4}\) [A1].
The negative sign records that the power is on the bottom of the fraction, not that the value is negative. Since \(\frac{1}{81}\) is a small positive number, an answer such as \(-3^4\) would be wrong in sign as well as in form.
Question 22 Report
A shirt costs \(\$40\). In a sale the price is reduced by \(30\%\).
Work out the sale price of the shirt.
A reduction of \(30\%\) leaves \(100\%-30\%=70\%\) of the original price, so the sale price can be found in a single multiplication by the decimal multiplier \(0.7\).
The calculation is \(40\times 0.7\) oe [M1], which gives a sale price of \(\$28\) [A1].
The two-step route gives the same result and is often easier without a calculator: \(10\%\) of \(\$40\) is \(\$4\), so \(30\%\) is \(3\times \$4=\$12\), and \(\$40-\$12=\$28\).
A common error is to stop after finding the \(\$12\) reduction and offer that as the sale price. The question asks for the price paid, so the discount must be subtracted from the original amount, or the multiplier \(0.7\) used directly.
A reduction of \(30\%\) leaves \(100\%-30\%=70\%\) of the original price, so the sale price can be found in a single multiplication by the decimal multiplier \(0.7\).
The calculation is \(40\times 0.7\) oe [M1], which gives a sale price of \(\$28\) [A1].
The two-step route gives the same result and is often easier without a calculator: \(10\%\) of \(\$40\) is \(\$4\), so \(30\%\) is \(3\times \$4=\$12\), and \(\$40-\$12=\$28\).
A common error is to stop after finding the \(\$12\) reduction and offer that as the sale price. The question asks for the price paid, so the discount must be subtracted from the original amount, or the multiplier \(0.7\) used directly.
Question 23 Report
Work out \(\sqrt[3]{64} \times \sqrt{49}\).
\(\sqrt[3]{64} = 4\) since \(4^3 = 64\), and \(\sqrt{49} = 7\) since \(7^2 = 49\).
\(4 \times 7\) [M1]
\(= 28\) [A1]
\(\sqrt[3]{64} = 4\) since \(4^3 = 64\), and \(\sqrt{49} = 7\) since \(7^2 = 49\).
\(4 \times 7\) [M1]
\(= 28\) [A1]
Question 24 Report
Solve the equation \(3(2x + 5) - 4 = 23\).
To solve an equation with a bracket and an extra constant term, expand the bracket first, then simplify and solve as usual.
Expanding \(3(2x + 5) - 4 = 23\) gives \(6x + 15 - 4 = 23\) [M1].
Simplifying the left side gives \(6x + 11 = 23\), so subtracting 11 from both sides gives \(6x = 12\) [M1].
Dividing both sides by 6 gives \(x = 2\) [A1].
Checking by substitution: \(3(2(2) + 5) - 4 = 3(9) - 4 = 23\), which confirms the solution.
To solve an equation with a bracket and an extra constant term, expand the bracket first, then simplify and solve as usual.
Expanding \(3(2x + 5) - 4 = 23\) gives \(6x + 15 - 4 = 23\) [M1].
Simplifying the left side gives \(6x + 11 = 23\), so subtracting 11 from both sides gives \(6x = 12\) [M1].
Dividing both sides by 6 gives \(x = 2\) [A1].
Checking by substitution: \(3(2(2) + 5) - 4 = 3(9) - 4 = 23\), which confirms the solution.
Question 25 Report
The diagram shows a pattern of shaded squares on a \(6\) by \(6\) grid. The bottom left corner of the grid is at the origin and each square has side \(1\) unit.
(a) Write down the number of shaded squares. [1]
(b) Write down the number of lines of symmetry of the pattern. [1]
(c) Write down the order of rotational symmetry of the pattern. [1]
(d) Write down the equations of the two lines of symmetry that are parallel to the axes. [2]
(e) Write down the coordinates of the centre of rotation. [1]
(f) Write down the smallest angle of rotation, in degrees, that maps the pattern onto itself. [1]
The grid is \(6\) by \(6\) with its bottom left corner at the origin and squares of side \(1\) unit, so it stretches from \(0\) to \(6\) in each direction and its centre is at \((3,\,3)\). All the later parts follow from that centre.
(a) Counting the shaded squares in the pattern gives \(16\) [B1]. Count them row by row and add the row totals rather than trying to see them all at once.
(b) Testing the four candidate mirror lines, the vertical centre line, the horizontal centre line and the two diagonals, all four reflect shaded squares onto shaded squares, giving \(4\) lines of symmetry [B1].
(c) Turning the pattern about the centre, each quarter turn reproduces it, so the order of rotational symmetry is \(4\) [B1].
(d) The two mirror lines parallel to the axes are the centre lines of the grid: \(x=3\) [B1] and \(y=3\) [B1].
(e) The centre of rotation is where all the lines of symmetry cross, at \((3,\,3)\) [B1].
(f) The smallest angle of rotation is \(360\div 4=90\) degrees [B1].
The self-check across parts (c), (e) and (f) is that the order, the centre and the smallest angle must agree: order \(4\) forces the angle \(\frac{360^\circ}{4}=90^\circ\), and the centre of rotation must be the common point of the mirror lines found in (d).
The grid is \(6\) by \(6\) with its bottom left corner at the origin and squares of side \(1\) unit, so it stretches from \(0\) to \(6\) in each direction and its centre is at \((3,\,3)\). All the later parts follow from that centre.
(a) Counting the shaded squares in the pattern gives \(16\) [B1]. Count them row by row and add the row totals rather than trying to see them all at once.
(b) Testing the four candidate mirror lines, the vertical centre line, the horizontal centre line and the two diagonals, all four reflect shaded squares onto shaded squares, giving \(4\) lines of symmetry [B1].
(c) Turning the pattern about the centre, each quarter turn reproduces it, so the order of rotational symmetry is \(4\) [B1].
(d) The two mirror lines parallel to the axes are the centre lines of the grid: \(x=3\) [B1] and \(y=3\) [B1].
(e) The centre of rotation is where all the lines of symmetry cross, at \((3,\,3)\) [B1].
(f) The smallest angle of rotation is \(360\div 4=90\) degrees [B1].
The self-check across parts (c), (e) and (f) is that the order, the centre and the smallest angle must agree: order \(4\) forces the angle \(\frac{360^\circ}{4}=90^\circ\), and the centre of rotation must be the common point of the mirror lines found in (d).
Question 26 Report
The scatter diagram shows eight pairs of values of \(x\) and \(y\). A line of best fit has been drawn using the seven points that follow the trend.
(a) Write down the type of correlation. [1]
(b) Write down the coordinates of the point that does not fit the trend. [1]
(c) Use the line of best fit to estimate the value of \(y\) when \(x=45\). [2]
The line of best fit here was drawn using only the seven points that follow the trend, which is what makes the estimate in the last part dependable.
(a) Those points rise from bottom left to top right, so larger \(x\) goes with larger \(y\) and the correlation is positive [B1].
(b) The point that does not fit the trend lies well below the band formed by the others, at \((35,\,15)\) [B1]. Read the horizontal coordinate first, using the scale printed on the axis rather than counting squares.
(c) Go up from \(x=45\) to the line of best fit, then read across horizontally to the \(y\)-axis [M1]. This gives \(47\), with any answer from \(44\) to \(50\) accepted [A1].
The method mark is for the correct use of the line, so drawing the guide lines from the axis to the line and across is worth doing even if the final reading is slightly off. An estimate taken from a nearby plotted point instead of the line does not earn it.
The line of best fit here was drawn using only the seven points that follow the trend, which is what makes the estimate in the last part dependable.
(a) Those points rise from bottom left to top right, so larger \(x\) goes with larger \(y\) and the correlation is positive [B1].
(b) The point that does not fit the trend lies well below the band formed by the others, at \((35,\,15)\) [B1]. Read the horizontal coordinate first, using the scale printed on the axis rather than counting squares.
(c) Go up from \(x=45\) to the line of best fit, then read across horizontally to the \(y\)-axis [M1]. This gives \(47\), with any answer from \(44\) to \(50\) accepted [A1].
The method mark is for the correct use of the line, so drawing the guide lines from the axis to the line and across is worth doing even if the final reading is slightly off. An estimate taken from a nearby plotted point instead of the line does not earn it.
Would you like to proceed with this action?