The equation of the straight line \({L_1}\) is \(y = 2x - 3.\)

Write down the \(y\)-intercept of \({L_1}\) .

Write down the gradient of \({L_1}\) .

The line \({L_2}\) is parallel to \({L_1}\) and passes through the point \((0,\,\,3)\) .

Write down the equation of \({L_2}\) .

The line \({L_3}\) is perpendicular to \({L_1}\) and passes through the point \(( - 2,\,\,6).\)

Write down the gradient of \({L_3}.\)

Find the equation of \({L_3}\) . Give your answer in the form \(ax + by + d = 0\) , where \(a\) , \(b\) and \(d\) are integers.

\((0,\,\, - 3)\)       (A1)     (C1)

Note: Accept \( - 3\) or \(y =  - 3.\)

\(2\)       (A1)     (C1)

\(y = 2x + 3\)        (A1)(ft)     (C1)

Note: Award (A1)(ft) for correct equation. Follow through from part (b)
Award (A0) for \({L_2} = 2x + 3\).

\( - \frac{1}{2}\)             (A1)(ft)     (C1)

Note: Follow through from part (b).

\(6 =  - \frac{1}{2}( - 2) + c\)        (M1)

\(c = 5\)  (may be implied)

OR

\(y - 6 =  - \frac{1}{2}(x + 2)\)        (M1)

Note: Award (M1) for correct substitution of their gradient in part (d) and the point \(( - 2,\,\,6)\). Follow through from part (d).

\(x + 2y - 10 = 0\)  (or any integer multiple)        (A1)(ft)     (C2)

Note: Follow through from (d). The answer must be in the form \(ax + by + d = 0\) for the (A1)(ft) to be awarded. Accept any integer multiple.

Question 12: Linear function.

Many candidates demonstrated a good understanding of linear functions so successfully found the \(y\)-intercepts, gradient and equation in the form \(y = mx + c\). However only the very best were able to rewrite this in the form \(ax + by + d = 0\) where \(a\), \(b\) and \(d\) are integers.

Question 12: Linear function.

Many candidates demonstrated a good understanding of linear functions so successfully found the \(y\)-intercepts, gradient and equation in the form \(y = mx + c\). However only the very best were able to rewrite this in the form \(ax + by + d = 0\) where \(a\), \(b\) and \(d\) are integers.

Question 12: Linear function.

Many candidates demonstrated a good understanding of linear functions so successfully found the \(y\)-intercepts, gradient and equation in the form \(y = mx + c\). However only the very best were able to rewrite this in the form \(ax + by + d = 0\) where \(a\), \(b\) and \(d\) are integers.

Question 12: Linear function.

Many candidates demonstrated a good understanding of linear functions so successfully found the \(y\)-intercepts, gradient and equation in the form \(y = mx + c\). However only the very best were able to rewrite this in the form \(ax + by + d = 0\) where \(a\), \(b\) and \(d\) are integers.

Question 12: Linear function.

Many candidates demonstrated a good understanding of linear functions so successfully found the \(y\)-intercepts, gradient and equation in the form \(y = mx + c\). However only the very best were able to rewrite this in the form \(ax + by + d = 0\) where \(a\), \(b\) and \(d\) are integers.

Write down the gradient of the line L₁.

A second line L₂ is perpendicular to L₁.

Write down the gradient of L₂.

The point (5, 3) is on L₂.

Determine the equation of L₂.

Lines L₁ and L₂ intersect at point P.

Using your graphic display calculator or otherwise, find the coordinates of P.

\(\frac{-2}{{5}}\)     (A1)     (C1)

\(\frac{5}{{2}}\)     (A1)(ft)     (C1)

Note: Follow through from their answer to part (a).

\(3 = \frac{5}{2} \times 5 + c\)     (M1)

Notes: Award (M1) for correct substitution of their gradient into equation of line. Follow through from their answer to part (b).

\(y = \frac{5}{2}x - \frac{19}{2}\)     (A1)(ft)

OR

\(y - 3 = \frac{5}{2}(x - 5)\)     (M1)(A1)(ft)     (C2)

Notes: Award (M1) for correct substitution of their gradient into equation of line. Follow through from their answer to part (b).

(3, −2)     (A1)(ft)(A1)(ft)     (C2)

Notes: If parentheses not seen award at most (A0)(A1)(ft). Accept x = 3, y = −2. Follow through from their answer to part (c), even if no working is seen. Award (M1)(A1)(ft) for a sensible attempt to solve \(2x + 5y = −4\) and their \(y = \frac{5}{2}x - \frac{19}{2}\) or equivalent, simultaneously.

Sketch the triangle writing in the side length and angles.

Calculate the length of \({\text{AB}}\).

Find the area of triangle \({\text{ABC}}\).

     (A1)     (C1)

Note: (A1) for fully labelled sketch.

[1 mark]

Unit penalty (UP) may apply in this question.

\(\frac{{{\text{AB}}}}{{\sin 30}} = \frac{7}{{\sin 65}}\)     (M1)

(UP)     \({\text{AB}} = 3.86{\text{ cm}}\)     (A1)(ft)     (C2)

Note: (M1) for use of sine rule with correct values substituted.

[2 marks]

Unit penalty (UP) may apply in this question.

\({\text{Angle BAC}} = {85^ \circ }\)     (A1)

\({\text{Area}} = \frac{1}{2} \times 7 \times 3.86 \times \sin {85^ \circ }\)     (M1)

(UP)     \( = 13.5{\text{ }}{{\text{cm}}^2}\)     (A1)(ft)     (C3)

[3 marks]

The triangle was drawn correctly by most and a majority correctly found the length of AB - a few did not write down the units (cm) and so lost a Unit penalty mark. There was still a significant number who tried to use right-angled trigonometry to find the length.

Finding the area of the triangle was mixed with many again assuming the existence of a right angle. Some candidates had their calculators in radian mode rather than degree mode.

The triangle was drawn correctly by most and a majority correctly found the length of AB - a few did not write down the units (cm) and so lost a Unit penalty mark. There was still a significant number who tried to use right-angled trigonometry to find the length.

Finding the area of the triangle was mixed with many again assuming the existence of a right angle. Some candidates had their calculators in radian mode rather than degree mode.

The triangle was drawn correctly by most and a majority correctly found the length of AB - a few did not write down the units (cm) and so lost a Unit penalty mark. There was still a significant number who tried to use right-angled trigonometry to find the length.

Finding the area of the triangle was mixed with many again assuming the existence of a right angle. Some candidates had their calculators in radian mode rather than degree mode.

Calculate the gradient of \({R_2}\) .

The point of intersection of \({R_1}\) and \({R_2}\) is \((4{\text{, }}k)\) .
Find
(i)     the value of \(k\) ;
(ii)    the equation of \({R_2}\) .

\(y = - 2x + 8\)     (M1)

Note: Award (M1) for rearrangement of equation or for \( - 2\) seen.

\(m({\text{perp}}) = \frac{1}{2}\)     (A1)     (C2)

[2 marks]

(i)     \(2(4) + k - 8 = 0\)     (M1)

Note: Award (M1) for evidence of substituting \(x = 4\) into \({R_1}\) .

\(k = 0\)     (A1)     (C2)

(ii)   \(y = \frac{1}{2}x + c\) (can be implied)     (M1)

Note: Award (M1) for substitution of \(\frac{1}{2}\) into equation of the line.

\(0 = \frac{1}{2}(4) + c\)

\(y = \frac{1}{2}x - 2\)     (A1)(ft)     (C2)

Notes: Follow through from parts (a) and (b)(i). Accept equivalent forms for the equation of a line.

OR

\(y - {y_1} = \frac{1}{2}(x - {x_1})\)     (M1)

Note: Award (M1) for substitution of \(\frac{1}{2}\) into equation of the line.

\(y = \frac{1}{2}(x - 4)\)     (A1)(ft)     (C2)

Notes: Follow through from parts (a) and (b)(i). Accept equivalent forms for the equation of a line.

[4 marks]

Parts a and bi of Question 13 appeared to be accessible to most candidates, but part bii was not well attempted. Many candidates did not show their working and lost method marks due to their incorrect answers.

Parts a and bi of Question 13 appeared to be accessible to most candidates, but part bii was not well attempted. Many candidates did not show their working and lost method marks due to their incorrect answers.

Find the radius, \(r\), of the circular base of the cone.

Find the slant height, \(l\), of the cone.

Find the curved surface area of the cone.

\(100 = \frac{1}{3}\pi {r^2}(8)\)     (M1)

Note:     Award (M1) for correct substitution into volume of cone formula.

\(r = 3.45{\text{ (cm) }}\left( {3.45494 \ldots {\text{ (cm)}}} \right)\)     (A1)     (C2)

[2 marks]

\({l^2} = {8^2} + {(3.45494 \ldots )^2}\)     (M1)

Note:     Award (M1) for correct substitution into Pythagoras’ theorem.

\(l = 8.71{\text{ (cm) }}\left( {8.71416 \ldots {\text{ (cm)}}} \right)\)     (A1)(ft)     (C2)

Note:     Follow through from part (a).

[2 marks]

\(\pi  \times 3.45494 \ldots  \times 8.71416 \ldots \)     (M1)

Note:     Award (M1) for their correct substitutions into curved surface area of a cone formula.

\( = 94.6{\text{ c}}{{\text{m}}^2}{\text{ }}(94.5836 \ldots {\text{ c}}{{\text{m}}^2})\)     (A1)(ft)     (C2)

Note:     Follow through from parts (a) and (b). Accept \(94.4{\text{ c}}{{\text{m}}^2}\) from use of 3 sf values.

[2 marks]

Write down the gradient of L₁ .

Line L₂, is perpendicular to line L₁, and passes through the point (4, 5) .

(i) Write down the gradient of L₂ .

(ii) Find the equation of L₂ .

Line L₂, is perpendicular to line L₁, and passes through the point (4, 5) .

Write down the coordinates of the point of intersection of L₁ and L₂ .

–2     (A1)     (C1)

Note: Do not accept \(\frac{{ - 2}}{1}\)

[1 mark]

(i) \(\frac{1}{2}(0.5)\)     (A1)(ft)

Note: Follow through from their part (a).

(ii) \(5 = \frac{1}{2}(4) + c\)     (M1)

Note: Award (M1) for their gradient substituted correctly.

\(y = \frac{1}{2}x + 3\)     (A1)(ft)

Note: Follow through from their part (b)(i).

OR

\(y - 5 = \frac{1}{2}(x - 4)\)     (M1)(A1)(ft)     (C3)

Notes: Award (M1) for their gradient substituted correctly, (A1)(ft) for 5 and 4 seen in the correct places. Follow through from their part (b)(i).

[3 marks]

(0.8, 3.4) or \(\left( {\frac{4}{5},\frac{{17}}{5}} \right)\)     (A1)(ft)(A1)(ft)     (C2)

Notes: Accept x = 0.8 and y = 3.4. Award (A1)(ft) for an attempt to solve the equations analytically, (attempt to eliminate either x or y), or graphically with a sketch (two reasonably accurate straight line graphs (from their answer to part (b)) and an indication of scale). Follow through from their L₂ if it intersects L₁, OR follow through from their equation, or expression in x, from their part (b)(ii). Award at most (A1)(ft)(A0)(ft) if brackets missing. Award (A0)(ft)(A1)(ft) for an answer of (0, 5) following an equation (or expression in x) of the form y = mx + 5 (m ≠ –2) found in part (b).

[2 marks]

The majority of candidates were able to write down the gradient of the straight line in part (a) but a correct answer for the gradient of the perpendicular proved to be more elusive in part (b)(i). Many however recovered in the remainder of the question as they were able to find the equation of a line using their gradient and the coordinates of a point on the line but, in some case, did not always show clear working. A significant minority of candidates, who attempted to substitute (4, 5) into the equation y = mx + c , incorrectly identified the value of c as 5.

The majority of candidates were able to write down the gradient of the straight line in part (a) but a correct answer for the gradient of the perpendicular proved to be more elusive in part (b)(i). Many however recovered in the remainder of the question as they were able to find the equation of a line using their gradient and the coordinates of a point on the line but, in some case, did not always show clear working. A significant minority of candidates, who attempted to substitute (4, 5) into the equation y = mx + c , incorrectly identified the value of c as 5.

The majority of candidates were able to write down the gradient of the straight line in part (a) but a correct answer for the gradient of the perpendicular proved to be more elusive in part (b)(i). Many however recovered in the remainder of the question as they were able to find the equation of a line using their gradient and the coordinates of a point on the line but, in some case, did not always show clear working. A significant minority of candidates, who attempted to substitute (4, 5) into the equation y = mx + c , incorrectly identified the value of c as 5.

Find the volume of water in the container.

Find the value of \(h\).

\(\pi  \times {8^2} \times 12\)     (M1)

Note:     Award (M1) for correct substitution into the volume of a cylinder formula.

\(2410{\text{ c}}{{\text{m}}^3}{\text{ }}(2412.74 \ldots {\text{ c}}{{\text{m}}^3},{\text{ }}768\pi {\text{ c}}{{\text{m}}^3})\)     (A1)     (C2)

[2 marks]

\(\frac{4}{3}\pi  \times {2.9^3} + 768\pi  = \pi  \times {8^2}h\)     (M1)(M1)(M1)

Note:     Award (M1) for correct substitution into the volume of a sphere formula (this may be implied by seeing 102.160…), (M1) for adding their volume of the ball to their part (a), (M1) for equating a volume to the volume of a cylinder with a height of \(h\).

OR

\(\frac{4}{3}\pi  \times {2.9^3} = \pi  \times {8^2}(h - 12)\)     (M1)(M1)(M1)

Note:     Award (M1) for correct substitution into the volume of a sphere formula (this may be implied by seeing 102.160…), (M1) for equating to the volume of a cylinder, (M1) for the height of the water level increase, \(h - 12\). Accept \(h\) for \(h - 12\) if adding 12 is implied by their answer.

\((h = ){\text{ }}12.5{\text{ (cm) }}\left( {12.5081 \ldots {\text{ (cm)}}} \right)\)     (A1)(ft)     (C4)

Note:     If 3 sf answer used, answer is 12.5 (12.4944…). Follow through from part (a) if first method is used.

[4 marks]

Determine the

(i) coordinates of M, the midpoint of P and Q.

(ii) gradient of the line drawn through P and Q.

(iii) gradient of the line drawn through M, perpendicular to PQ.

The perpendicular line drawn through M meets the y-axis at R (0, k).

Find k.

(i) (2, – 2) parentheses not required.     (A1)

(ii) gradient of PQ \( = \left( {\frac{{ - 5 - 1}}{{0 - 4}}} \right) = \frac{6}{4} = \frac{3}{2}(1.5)\)     (M1)(A1)

(M1) for gradient formula with correct substitution

Award (A1) for \(y = \frac{3}{2}x - 5\) with no other working

(iii) gradient of perpendicular is \( - \frac{2}{3}\)     (A1)(ft)     (C4)

[4 marks]

\(\left( {\frac{{k + 2}}{{0 - 2}}} \right) =  - \frac{2}{3}\), \(k =  - \frac{2}{3}\) or \(y =  - \frac{2}{3}x + c\), \(c =  - \frac{2}{3}\therefore k =  - \frac{2}{3}\)     (M1)(A1)(ft)

Allow (\(0, - \frac{2}{3}\))

(M1) is for equating gradients or substituting gradient into \(y = mx + c\)     (C2)

[2 marks]

There were some good answers, but many candidates showed a shaky understanding of coordinate geometry and some difficulty in dealing with negative numbers. Evidently a favourite question for some centres that consistently scored high marks here.

a) This was done quite well by most candidates with the main errors being reversal of the x, y values in the formula and using the negative, rather than the negative reciprocal for the perpendicular.

There were some good answers, but many candidates showed a shaky understanding of coordinate geometry and some difficulty in dealing with negative numbers. Evidently a favourite question for some centres that consistently scored high marks here.

b) Poorly answered though many candidates did gain a mark by substituting the correct value for gradient into \(y = mx + c\).

Find the radius of the circle.

This circle is the base of a solid cylinder of height 25 cm.

Write down the volume of the solid cylinder.

This circle is the base of a solid cylinder of height 25 cm.

Find the total surface area of the solid cylinder.

πr² = 8     (M1)

Note: Award (M1) for correct area formula.

r = 1.60 (cm)     (1.59576...)     (A1)     (C2)

[2 marks]

200 cm³     (A1)(ft)     (C1)

Notes: Units are required. Follow through from their part (a). Accept 201 cm³ (201.061…) for use of r = 1.60 .

[1 mark]

Surface area = 16 + 2π(1.59576...)25     (M1)(M1)

Note: Award (M1) for correct substitution of their r into curved surface area formula, (M1) for adding 16 or 2 × π × (their answer to part (a))²

267 cm²     (266.662...cm²)     (A1)(ft)     (C3)

Note: Follow through from their part (a).

[3 marks]

There were two common errors identified on a number of scripts in this question. In part (a), the majority of candidates correctly wrote down πr² = 8 for method but a significant number of candidates then went on to write down r = 2.55 (forgetting to square root). Such candidates were able to recover in subsequent parts of the question with the allowable follow through marks, but a method mark and final accuracy mark was lost on many scripts in part (c) with incomplete methods shown for the total surface area of the cylinder. Leaving off the addition of either one or both end surfaces of the cylinder resulted in the final two marks being lost.

There were two common errors identified on a number of scripts in this question. In part (a), the majority of candidates correctly wrote down πr² = 8 for method but a significant number of candidates then went on to write down r = 2.55 (forgetting to square root). Such candidates were able to recover in subsequent parts of the question with the allowable follow through marks, but a method mark and final accuracy mark was lost on many scripts in part (c) with incomplete methods shown for the total surface area of the cylinder. Leaving off the addition of either one or both end surfaces of the cylinder resulted in the final two marks being lost.

There were two common errors identified on a number of scripts in this question. In part (a), the majority of candidates correctly wrote down πr² = 8 for method but a significant number of candidates then went on to write down r = 2.55 (forgetting to square root). Such candidates were able to recover in subsequent parts of the question with the allowable follow through marks, but a method mark and final accuracy mark was lost on many scripts in part (c) with incomplete methods shown for the total surface area of the cylinder. Leaving off the addition of either one or both end surfaces of the cylinder resulted in the final two marks being lost.

(i) Sketch a diagram to illustrate this information. Label the points \({\text{A, B, C}}\). Show the angles \({90^ \circ }\), \({60^ \circ }\) and the length \(7.3{\text{ cm}}\) on your diagram.

(ii) Find the length of \({\text{BC}}\).

Point \({\text{D}}\) is on the straight line \({\text{AC}}\) extended and is such that angle \({\text{CDB}}\) is \({20^ \circ }\).

(i) Show the point \({\text{D}}\) and the angle \({20^ \circ }\) on your diagram.

(ii) Find the size of angle \({\text{CBD}}\).

Unit penalty (UP) is applicable where indicated in the left hand column.

(i)

     (A1)

For \({\text{A}}\), \({\text{B}}\), \({\text{C}}\), \(7.3\), \({60^ \circ }\), \({90^ \circ }\), shown in correct places     (A1)

Note: The \({90^ \circ }\) should look like \({90^ \circ }\) (allow \( \pm {10^ \circ }\))

(ii) Using \(\tan 60\) or \(\tan 30\)     (M1)

(UP)     \(4.21{\text{ cm}}\)     (A1)(ft)

Note: (ft) on their diagram

Or

Using sine rule with their correct values     (M1)

(UP)     \( = 4.21{\text{ cm}}\)     (A1)(ft)

Or

Using special triangle \(\frac{{7.3}}{{\sqrt 3 }}\)     (M1)

(UP)     \(4.21{\text{ cm}}\)     (A1)(ft)

Or

Any other valid solution

Note: If A and B are swapped then \({\text{BC}} = 8.43{\text{ cm}}\)     (C3)

[3 marks]

(i) For \({\text{ACD}}\) in a straight line and all joined up to \({\text{B}}\), for \({20^ \circ }\) shown in correct place and \({\text{D}}\) labelled. \({\text{D}}\) must be on \({\text{AC}}\) extended.     (A1)

(ii) \({\text{B}}\hat {\text{C}}{\text{D}} = {120^ \circ }\)     (A1)

\({\text{C}}\hat {\text{B}}{\text{D}} = {40^ \circ }\)     (A1)     (C3)

[3 marks]

The initial diagram was well drawn by most candidates but few could extend \({\text{AC}}\) to find \({\text{D}}\). The point \({\text{D}}\) was either drawn between \({\text{A}}\) and \({\text{C}}\) or on \({\text{CA}}\) extended. When on \({\text{CA}}\) extended the candidates could be awarded A1 follow through for the angle. A surprising number of candidates could not find the correct answer for the length of \({\text{BC}}\).

The initial diagram was well drawn by most candidates but few could extend \({\text{AC}}\) to find \({\text{D}}\). The point \({\text{D}}\) was either drawn between \({\text{A}}\) and \({\text{C}}\) or on \({\text{CA}}\) extended. When on \({\text{CA}}\) extended the candidates could be awarded A1 follow through for the angle. A surprising number of candidates could not find the correct answer for the length of \({\text{BC}}\).

Write down the y-intercept of the line AB.

Calculate the gradient of the line AB.

The acute angle between the line AB and the x-axis is θ.

Show θ on the diagram.

The acute angle between the line AB and the x-axis is θ.

Calculate the size of θ.

6

OR

(0, 6)     (A1)     (C1)

[1 mark]

\(\frac{{(2 - 5)}}{{(8 - 2)}}\)     (M1)

Note: Award (M1) for substitution in gradient formula.

\( =  - \frac{1}{2}\)    (A1)     (C2)

[2 marks]

Angle clearly identified.     (A1)     (C1)

[1 mark]

\(\tan \theta  = \frac{1}{2}\) (or equivalent fraction)     (M1)

\(\theta = 26.6^\circ\)     (A1)(ft)     (C2)

Note: (ft) from (b).

Accept alternative correct trigonometrical methods.

[2 marks]

This was generally well answered.

This was generally well answered, the errors coming from incorrect substitution into the gradient formula rather than using the two intercepts.

There was often a lack of accuracy in the answers. Also, the use of the sine rule overly complicated matters for many.

The following diagrams show six lines with equations of the form y =mx +c.

In the table below there are four possible conditions for the pair of values m and c. Match each of the given conditions with one of the lines drawn above.

     (A6)     (C6)

Notes: Award (A6) for all correct, (A5) for 3 correct, (A3) for 2 correct, (A1) for 1 correct.

Deduct (A1) for any repetition.

[6 marks]

Many candidates received full marks and a number received 3 marks for giving two correct answers. Very few candidates were awarded zero marks. As most candidates did not show working for this question it is difficult to comment on the errors that might have been made.

A ladder is standing on horizontal ground and leaning against a vertical wall. The length of the ladder is \(4.5\) metres. The distance between the bottom of the ladder and the base of the wall is \(2.2\) metres.

Use the above information to sketch a labelled diagram showing the ground, the ladder and the wall.

Calculate the distance between the top of the ladder and the base of the wall.

Calculate the obtuse angle made by the ladder with the ground.

(A1)   (C1)

Notes: Award (A1) for drawing an approximately right angled triangle, with correct labelling of the distances \(4.5\,({\text{m}})\) and \(2.2\,({\text{m}})\).

\(\sqrt {\,{{4.5}^2} - {{2.2}^2}} \,\,\,({\text{accept eqivalent }}eg\,\,{d^2} + {2.2^2} = {4.5^2})\)         (M1)

\( = 3.93\,({\text{m}})\,\,\,\left( {\sqrt {\,15.41} \,({\text{m}}),\,\,3.92555...\,({\text{m}})} \right)\)          (A1)    (C2)

Note: Award (M1) for a correct substitution in the Pythagoras formula.

\(180^\circ  - {\cos ^{ - 1}}\left( {\frac{{2.2}}{{4.5}}} \right)\)        (M1)(M1)

OR

\(180^\circ  - {\tan ^{ - 1}}\left( {\frac{{3.92555...}}{{2.2}}} \right)\)        (M1)(M1)

OR

\(180^\circ  - {\sin ^{ - 1}}\left( {\frac{{3.92555...}}{{4.5}}} \right)\)        (M1)(M1)

Note: Award (M1) for a correct substitution in the correct trigonometric ratio.
Award (M1) for subtraction from \(180^\circ \) (this may be implied if the sum of their inverse of the trigonometric ratio and their final answer equals \(180\)).

\( = 119^\circ \,\,\,(119.267...^\circ )\)        (A1)(ft)    (C3)

Note: Follow through from their part (b) if cosine is not used. Accept \(119.239...\) or \(119.151...\) from use of \(3\) sf values.

Question 3: Right angle trigonometry.
Candidates sketched the ladder leaning against the wall and recognized that Pythagoras’ theorem was needed to find the distance between the top of the ladder and the base of the wall (but not always correctly). Although it was a right triangle a number of the candidates used the law of sines (instead of Pythagoras’ theorem) and law of cosines (instead of a trigonometry ratio). Many candidates failed to find the obtuse angle made by the ladder with the ground even though the word obtuse was in bold type in the question.

Question 3: Right angle trigonometry.
Candidates sketched the ladder leaning against the wall and recognized that Pythagoras’ theorem was needed to find the distance between the top of the ladder and the base of the wall (but not always correctly). Although it was a right triangle a number of the candidates used the law of sines (instead of Pythagoras’ theorem) and law of cosines (instead of a trigonometry ratio). Many candidates failed to find the obtuse angle made by the ladder with the ground even though the word obtuse was in bold type in the question.

Question 3: Right angle trigonometry.
Candidates sketched the ladder leaning against the wall and recognized that Pythagoras’ theorem was needed to find the distance between the top of the ladder and the base of the wall (but not always correctly). Although it was a right triangle a number of the candidates used the law of sines (instead of Pythagoras’ theorem) and law of cosines (instead of a trigonometry ratio). Many candidates failed to find the obtuse angle made by the ladder with the ground even though the word obtuse was in bold type in the question.

Find the length of AC in centimetres.

Find the size of angle ADC.

\(\frac{{{\text{AC}}}}{{\sin 100^\circ }} = \frac{{10}}{{\sin 50^\circ }}\)     (M1)(A1)

Note: Award (M1) for substitution in the sine rule formula, (A1) for correct substitutions.

\(=12.9 (12.8557...)\)     (A1)     (C3)

Note: Radian answer is 19.3, award (M1)(A1)(A0).

\(\frac{{{{12}^2} + {7^2} - {{12.8557...}^2}}}{{2 \times 12 \times 7}}\)     (M1)(A1)(ft)

Note: Award (M1) for substitution in the cosine rule formula, (A1)(ft) for correct substitutions.

= 80.5° (80.4994...°)     (A1)(ft)     (C3)

Notes: Follow through from their answer to part (a). Accept 80.9° for using 12.9. Using the radian answer from part (a) leads to an impossible triangle, award (M1)(A1)(ft)(A0).

Give a reason why L₁ does not pass through A.

Find the gradient of L₁.

L₂ is a line perpendicular to L₁. The equation of L₂ is \(y = mx + c\).

Write down the value of m.

L₂ does pass through A.

Find the value of c.

\(2 \times 0 - 3 \times 6 \ne 11\)     (R1)

Note: Stating \(2 \times 0 - 3 \times 6 =  - 18\) without a conclusion is not sufficient.

OR

Clear sketch of L₁ and A.

     (R1)

OR

Point A is (6, 0) and \(2y - 3x = 11\) has x-intercept at \(- \frac{11}{3}\) or the line has only one x-intercept which occurs when x is negative.     (R1)     (C1)

\(2y = 3x + 11\) or \(y - \frac{3}{2}x = \frac{{11}}{2}\)     (M1)

Note: Award (M1) for a correct first step in making y the subject of the equation.

\(({\text{gradient equals}}) = \frac{3}{2}(1.5)\)     (A1)     (C2)

Note: Do not accept 1.5x.

\((m = ) - \frac{2}{3}\)     (A1)(ft)     (C1)

Notes: Follow through from their part (b).

\(0 =  - \frac{2}{3}(6) + c\)     (M1)

Note: Award (M1) for correct substitution of their gradient and (6, 0) into any form of the equation.

(c =) 4     (A1)(ft)     (C2)

Note: Follow through from part (c).

There were multiple acceptable reasons why the line did not pass through a given point (including numerically substituting values in the equation; drawing a graph or algebraically finding the x-intercept of the line). This was one of two reasoning marks in the paper.

There were multiple acceptable reasons why the line did not pass through a given point (including numerically substituting values in the equation; drawing a graph or algebraically finding the x-intercept of the line). This was one of two reasoning marks in the paper.

There were multiple acceptable reasons why the line did not pass through a given point (including numerically substituting values in the equation; drawing a graph or algebraically finding the x-intercept of the line). This was one of two reasoning marks in the paper.

There were multiple acceptable reasons why the line did not pass through a given point (including numerically substituting values in the equation; drawing a graph or algebraically finding the x-intercept of the line). This was one of two reasoning marks in the paper.

Write down the length of \({\rm{MB}}\).

Find the size of angle \({\rm{C\hat MB}}\).

Find the length of \({\rm{CB}}\).

\(10\) m     (A1)(C1)

\({\rm{A\hat MC}} = 70^\circ \;\;\;\)OR\(\;\;\;{\rm{A\hat CM}} = 55^\circ \)     (A1)

\({\rm{C\hat MB}} = 110^\circ \)     (A1)     (C2)

\({\text{C}}{{\text{B}}^2} = {10^2} + {10^2} - 2 \times 10 \times 10 \times \cos 110^\circ \)     (M1)(A1)(ft)

Notes: Award (M1) for substitution into the cosine rule formula, (A1)(ft) for correct substitution. Follow through from their answer to part (b).

OR

\(\frac{{{\text{CB}}}}{{\sin 110^\circ }} = \frac{{10}}{{\sin 35^\circ }}\)     (M1)(A1)(ft)

Notes: Award (M1) for substitution into the sine rule formula, (A1)(ft) for correct substitution. Follow through from their answer to part (b).

OR

\({\rm{A\hat CB}} = 90^\circ \)     (A1)

\(\sin 55^\circ  = \frac{{{\text{CB}}}}{{55}}\;\;\;\)OR\(\;\;\;\cos 35^\circ  = \frac{{{\text{CB}}}}{{20}}\)     (M1)

Note: Award (A1) for some indication that \({\rm{A\hat CB}} = 90^\circ \), (M1) for correct trigonometric equation.

OR

Perpendicular \({\rm{MN}}\) is drawn from \({\rm{M}}\) to \({\rm{CB}}\).     (A1)

\(\frac{{\frac{1}{2}{\text{CB}}}}{{10}} = \cos 35^\circ \)     (M1)

Note: Award (A1) for some indication of the perpendicular bisector of \({\rm{BC}}\), (M1) for correct trigonometric equation.

\({\text{CB}} = 16.4{\text{ (m)}}\;\;\;\left( {16.3830 \ldots {\text{ (m)}}} \right)\)     (A1)(ft)(C3)

Notes: Where a candidate uses \({\rm{C\hat MB}} = 90^\circ \) and finds \({\text{CB}} = 14.1{\text{ (m)}}\) award, at most, (M1)(A1)(A0).

Where a candidate uses \({\rm{C\hat MB}} = 60^\circ \) and finds \({\text{CB}} = 10{\text{ (m)}}\) award, at most, (M1)(A1)(A0).

Find the value of

(i) q ;

(ii) p .

Calculate the distance AB.

(i) \(\frac{{ - 4 + 2}}{2} = q\)     (M1)

Note: Award (M1) for correct substitution in the correct formula.

q = –1     (A1)

(ii) \(\frac{{p + ( - 3)}}{2} = 1\)     (M1)

Note: Award (M1) for correct substitution into the correct formula or consistent with their equation in (i).

p = 5     (A1)     (C4)

Notes: Award A marks for integer values. Penalise if answers left as a fraction the first time a fraction is seen.

[4 marks]

\({\text{AB}} = \sqrt {{{(2 + 4)}^2} + {{( - 3 - 5)}^2}} \)     (M1)

Note: Award (M1) for the correct substitution of their coordinates for A and B in the correct formula.

AB = 10     (A1)(ft)     (C2)

Note: Follow through from their answer to part (a)(ii).

[2 marks]

(a) Despite some good answers in this part of the question, sign errors in setting up one or both of the equations meant that marks were lost by some candidates. This error was particularly prevalent in finding the value of p and the equation \(\frac{{(p + 3)}}{2} = 1\) was seen quite often. In part (b), there was a requirement for a correct substitution of their coordinates for A and B into the correct formula for Pythagoras and, while (2 + 4)² was often seen, sign errors in the second component of (–3 – 5)² proved to be the downfall of a significant number of candidates. As a consequence, the final two marks were lost. Given these errors, there were still a significant number of full mark responses to this question.

(a) Despite some good answers in this part of the question, sign errors in setting up one or both of the equations meant that marks were lost by some candidates. This error was particularly prevalent in finding the value of p and the equation \(\frac{{(p + 3)}}{2} = 1\) was seen quite often. In part (b), there was a requirement for a correct substitution of their coordinates for A and B into the correct formula for Pythagoras and, while (2 + 4)² was often seen, sign errors in the second component of (–3 – 5)² proved to be the downfall of a significant number of candidates. As a consequence, the final two marks were lost. Given these errors, there were still a significant number of full mark responses to this question.

Use the diagram above to calculate the gradient of the ramp.

The gradient for a safe descending wheelchair ramp is \( - \frac{1}{{12}}\).

Using your answer to part (a), comment on why wheelchair ramp \({\text{A}}\) is not safe.

The equation of a second wheelchair ramp, B, is \(2x + 24y - 1920 = 0\).

(i)     Determine whether wheelchair ramp \({\text{B}}\) is safe or not. Justify your answer.

(ii)     Find the horizontal distance of wheelchair ramp \({\text{B}}\).

\( - \frac{{80}}{{940}}{\text{ }}\left( {-0.0851, -0.085106 \ldots , -\frac{4}{{47}}} \right)\)     (A1)     (C1)

[1 mark]

\(-0.0851{\text{ }} (-0.085106 \ldots ) <  - \frac{1}{{12}}(-0.083333 \ldots )\)     (A1)(ft)     (C1)

Notes: Accept “less than” in place of inequality.

     Award (A0) if incorrect inequality seen.

     Follow through from part (a).

[1 mark]

(i)     ramp \({\text{B}}\) is safe     (A1)

the gradient of ramp \({\text{B}}\) is \( - \frac{1}{{12}}\)     (R1)

Notes: Award (R1) for “the gradient of ramp \({\text{B}}\) is \( - \frac{1}{{12}}\)” seen.

     Do not award (A1)(R0).

(ii)     \(2x = 1920\)     (M1)

Note: Accept alternative methods.

\(960 {\text{ (cm)}}\)     (A1)     (C4)

[4 marks]

Write down and simplify a quadratic equation in \(x\) which links the three sides of the triangle.

Solve the quadratic equation found in part (a).

Write down the value of the perimeter of the triangle.

\({(x + 8)^2} = {(x + 7)^2} + {x^2}\)     (A1)

Note: Award (A1) for a correct equation.

\({x^2} + 16x + 64 = {x^2} + 14x + 49 + {x^2}\)     (A1)

Note: Award (A1) for correctly removed parentheses.

\({x^2} - 2x - 15 = 0\)     (A1)     (C3)

Note: Accept any equivalent form.

[3 marks]

\(x = 5\), \(x = - 3\)     (A1)(ft)(A1)(ft)     (C2)

Notes: Accept (A1)(ft) only from the candidate’s quadratic equation.

[2 marks]

\(30{\text{ cm}}\)     (A1)(ft)     (C1)

Note: Follow through from a positive answer found in part (b).

[1 mark]

This question proved to be difficult for the majority of candidates. Many simply were unable to see that, to relate the three given lengths, a Pythagorean equation needed to be produced. Indeed, many did not seem to appreciate the concept of a quadratic equation and, as a consequence, either wrote down a linear equation linking one length to the sum of the other two lengths or multiplied all three lengths together. For the minority who stated a correct Pythagorean equation, many could not remove brackets successfully and arrived at \({x^2} = 15\) . Consequently, very few candidates earned more than one mark for part (a). Where the correct quadratic equation was seen in part (a), many were able to solve this quadratic correctly in part (b) and arrive at the required value of \(x = 5\) for the answer for part (c).

This question proved to be difficult for the majority of candidates. Many simply were unable to see that, to relate the three given lengths, a Pythagorean equation needed to be produced. Indeed, many did not seem to appreciate the concept of a quadratic equation and, as a consequence, either wrote down a linear equation linking one length to the sum of the other two lengths or multiplied all three lengths together. For the minority who stated a correct Pythagorean equation, many could not remove brackets successfully and arrived at \({x^2} = 15\) . Consequently, very few candidates earned more than one mark for part (a). Where the correct quadratic equation was seen in part (a), many were able to solve this quadratic correctly in part (b) and arrive at the required value of \(x = 5\) for the answer for part (c).

This question proved to be difficult for the majority of candidates. Many simply were unable to see that, to relate the three given lengths, a Pythagorean equation needed to be produced. Indeed, many did not seem to appreciate the concept of a quadratic equation and, as a consequence, either wrote down a linear equation linking one length to the sum of the other two lengths or multiplied all three lengths together. For the minority who stated a correct Pythagorean equation, many could not remove brackets successfully and arrived at \({x^2} = 15\) . Consequently, very few candidates earned more than one mark for part (a). Where the correct quadratic equation was seen in part (a), many were able to solve this quadratic correctly in part (b) and arrive at the required value of \(x = 5\) for the answer for part (c).

Write down the coordinates of M, the midpoint of line segment AB.

Calculate the gradient of \({L_1}\).

The line \({L_2}\) is parallel to \({L_1}\) and passes through the point \((3,{\text{ }}2)\).

Find the equation of \({L_2}\). Give your answer in the form \(y = mx + c\).

\((3,{\text{ }}1)\)     (A1)(A1)     (C2)

Note: Accept \(x = 3\), \(y = 1\). Award (A0)(A1) if parentheses are missing.

\(\frac{{2 - 0}}{{0 - 6}}\)     (M1)

Note: Award (M1) for correct substitution into gradient formula.

\( =  - \frac{1}{3}( - 0.333333 \ldots )\)     (A1)     (C2)

Note: Accept \( - \frac{2}{6}\).

\((y - 2) =  - \frac{1}{3}(x - 3)\)     (M1)

OR

\(2 =  - \frac{1}{3}(3) + c\)     (M1)

Note: Award (M1) for substitution of their gradient from part (b).

\(y =  - \frac{1}{3}x + 3\)     (A1)(ft)     (C2)

Note: Follow through from part (b).

The answer must be an equation in the form \(y = mx + c\) for the (A1)(ft) to be awarded.

Parts (a), finding the midpoint, and (b) finding the gradient, of this question were done well by the majority of candidates.

Parts (a), finding the midpoint, and (b) finding the gradient, of this question were done well by the majority of candidates. Some candidates substituted incorrectly into the gradient formula or reversed the numerator and denominator.

There was a significant number of candidates who calculated the equation of the normal to the given line and not the equation of a parallel line. It seemed those candidates answered the question they expected and not the question asked.

Find the gradient of \({L_1}\) .

The equation of line \({L_2}\) is \(y = mx\) . \({L_2}\) is perpendicular to \({L_1}\) .

Find the value of \(m\).

The equation of line \({L_2}\) is \(y = mx\) . \({L_2}\) is perpendicular to \({L_1}\) .

Find the coordinates of the point of intersection of the lines \({L_1}\) and \({L_2}\) .

\(4y = - x - 34\) or similar rearrangement     (M1)

\({\text{Gradient}} = - \frac{1}{4}\)     (A1)     (C2)

\(m = 4\)     (A1)(ft)(A1)(ft)     (C2)

Note: (A1) Change of sign
(A1) Use of reciprocal

[2 marks]

Reasonable attempt to solve equations simultaneously     (M1)

\(( - 2{\text{, }} - 8)\)     (A1)(ft)     (C2)

Note: Accept \(x = - 2\)  \(y = - 8\). Award (A0) if brackets not included.

[2 marks]

The omission of the negative sign was a common fault.

Most candidates managed to answer this correctly from their (a).

This part proved challenging for the majority. Once again, the use of the GDC was expected.

The equation of line \({L_1}\) is \(y = 2.5x + k\). Point \({\text{A}}\) \(\,(3,\, - 2)\) lies on \({L_1}\).

Find the value of \(k\).

The line \({L_2}\) is perpendicular to \({L_1}\) and intersects \({L_1}\) at point \({\text{A}}\).

Write down the gradient of \({L_2}\).

Find the equation of \({L_2}\). Give your answer in the form \(y = mx + c\) .

Write your answer to part (c) in the form \(ax + by + d = 0\)  where \(a\), \(b\) and \(d \in \mathbb{Z}\).

\( - 2 = 2.5\, \times 3 + k\)       (M1)

Note: Award (M1) for correct substitution of \((3,\, - 2)\) into equation of \({L_1}\).

\((k = ) - 9.5\)       (A1) (C2)

\( - 0.4\,\left( { - \frac{2}{5}} \right)\)       (A1)  (C1)

\(y - ( - 2) =  - 0.4\,(x - 3)\)       (M1)

OR

\( - 2 =  - 0.4\,(3) + c\)       (M1)

Note: Award (M1) for their gradient and given point substituted into equation of a straight line. Follow through from part (b).

\(y =  - 0.4x - 0.8\)       \(\left( {y =  - \frac{2}{5}x - \frac{4}{5}} \right)\)       (A1)(ft)    (C2)

\(2x + 5y + 4 = 0\) (or any integer multiple)      (A1)(ft) (C1)

Note: Follow through from part (c).

Question 7: Perpendicular Line
The response to this question was mixed.
Part (a) was well attempted by the majority.

In part (b), the gradient was not fully calculated (being left as a reciprocal) by a large number of candidates.

In part (c), the common error was the use of c from part (a) in the line.

In part (d), the notation for integer was not understood by a large number of candidates.

Write down the y intercept of L₁.

Write down the gradient of L₁.

The line L₂ is perpendicular to L₁ and passes through the point (3, 7).

Write down the gradient of the line L₂.

The line L₂ is perpendicular to L₁ and passes through the point (3, 7).

Find the equation of L₂. Give your answer in the form ax + by + d = 0 where \(a,{\text{ }}b,{\text{ }}d \in \mathbb{Z}\).

–2     (A1)     (C1)

Note: Accept (0, –2).

[1 mark]

\( - \frac{1}{2}\)     (A1)     (C1)

[1 mark]

2     (A1)(ft)     (C1)

Note: Follow through from their answer to part (b).

[1 mark]

y = 2x + c (can be implied)
7 = 2 × 3 + c     (M1)
c = 1     (A1)(ft)
y = 2x + 1

Notes: Award (M1) for substitution of (3, 7), (A1)(ft) for c.

Follow through from their answer to part (c).

OR

y – 7 = 2(x – 3)     (M1)(M1)

Note: Award (M1) for substitution of their answer to part (c), (M1) for substitution of (3, 7).

2x – y + 1 = 0  or  –2x + y – 1 = 0     (A1)(ft)     (C3)

Note: Award (A1)(ft) for their equation in the stated form.

[3 marks]

Although the first three parts of this question were well answered, with most candidates knowing how to find the y intercept, gradient of a given line and gradient of the perpendicular line, very few candidates could find the equation of the perpendicular line and correctly state it in the required form.

Although the first three parts of this question were well answered, with most candidates knowing how to find the y intercept, gradient of a given line and gradient of the perpendicular line, very few candidates could find the equation of the perpendicular line and correctly state it in the required form.

Although the first three parts of this question were well answered, with most candidates knowing how to find the y intercept, gradient of a given line and gradient of the perpendicular line, very few candidates could find the equation of the perpendicular line and correctly state it in the required form.

Although the first three parts of this question were well answered, with most candidates knowing how to find the y intercept, gradient of a given line and gradient of the perpendicular line, very few candidates could find the equation of the perpendicular line and correctly state it in the required form.

Write down an equation for \(x\) and \(y\).

The area of the swimming pool is \({\text{112}}{{\text{m}}^2}\).

Write down a second equation for \(x\) and \(y\).

Use your graphic display calculator to find the value of \(x\) and the value of \(y\).

An Olympic sized swimming pool is \(50\) m long and \(25\) m wide.

Determine the area of the hotel swimming pool as a percentage of the area of an Olympic sized swimming pool.

\(2x + 2y = 44\)     (A1)     (C1)

Note: Accept equivalent forms.

\(xy = 112\)     (A1) (C1)

\(8\), \(14\)     (A1)(ft)(A1)(ft)     (C2)

Notes: Accept \(x = 8\), \(y = 14\)   OR   \(x = 14\), \(y = 8\)

Follow through from their answers to parts (a) and (b) only if both values are positive.

\(\frac{{112}}{{1250}} \times 100\)     (M1)

Note: Award (M1) for \(112\) divided by \(1250\).

\( = 8.96\)     (A1)     (C2)

Note: Do not penalize if percentage sign seen.

Calculate the exact value of the volume of the cuboid, in cm³.

Write your answer to part (a) correct to

(i) one decimal place;

(ii) three significant figures.

Write your answer to part (b)(ii) in the form \(a \times 10^k\), where \(1 \leqslant a < 10 , k \in \mathbb{Z}\).

\({\text{V}} = 8.7 \times 5.6 \times 3.4\)     (M1)

Note: Award (M1) for multiplication of the 3 given values.

\(=165.648\)     (A1)     (C2)

(i) 165.6     (A1)(ft)

Note: Follow through from their answer to part (a).

(ii) 166     (A1)(ft)     (C2)

Note: Follow through from their answer to part (a).

\(1.66 \times 10^2\)     (A1)(ft)(A1)(ft)     (C2)

Notes: Award (A1)(ft) for 1.66, (A1)(ft) for \(10^2\). Follow through from their answer to part (b)(ii) only. The follow through for the index should be dependent on the value of the mantissa in part (c) and their answer to part (b)(ii).

Calculate the gradient of the line AB.

Given that the line AC is perpendicular to the line AB

write down the gradient of the line AC.

Given that the line AC is perpendicular to the line AB

find the value of p.

\(m({\rm{AB}}) = \frac{{ - 3 - 3}}{{7 - 4}} = - 2\)     (M1)(A1)     (C2)

Note: Award (M1) for attempt to substitute into correct gradient formula.

[2 marks]

\(m({\rm{AC}}) = \frac{1}{2}\)     (A1)(ft)

[1 mark]

\(\frac{{p - 3}}{{0.5 - 4}} = \frac{1}{2}\)  (or equivalent method)     (M1)(A1)(ft)

Note: Award (M1) for equating gradient to \(\frac{1}{2}\). (A1) for correct substitution.

\(p = 1.25\)     (A1)(ft)     (C4)

[3 marks]

While parts (a) and (b)(i) were attempted with some success, few candidates made progress in (b)(ii). Some candidates used the coordinates of point B rather than C and others could not find the unknown value p as they did not realise they had to equate their substituted formula for the gradient to the answer to part (b)(i). A large number of candidates did not attempt this part of the question.

While parts (a) and (b)(i) were attempted with some success, few candidates made progress in (b)(ii). Some candidates used the coordinates of point B rather than C and others could not find the unknown value p as they did not realise they had to equate their substituted formula for the gradient to the answer to part (b)(i). A large number of candidates did not attempt this part of the question.

While parts (a) and (b)(i) were attempted with some success, few candidates made progressin (b)(ii). Some candidates used the coordinates of point B rather than C and others could not find the unknown value p as they did not realise they had to equate their substituted formula for the gradient to the answer to part (b)(i). A large number of candidates did notattempt this part of the question.

Find the coordinates of M.

Find the gradient of \({L_1}\).

Write down the gradient of \({L_2}\).

Write down, in the form \(y = mx + c\), the equation of \({L_2}\).

\((0,{\text{ }}2.5)\)\(\,\,\,\)OR\(\,\,\,\)\(\left( {0,{\text{ }} - \frac{5}{2}} \right)\)     (A1)(A1)     (C2)

Note:     Award (A1) for 0 and (A1) for –2.5 written as a coordinate pair. Award at most (A1)(A0) if brackets are missing. Accept “\(x = 0\) and \(y =  - 2.5\)”.

[2 marks]

\(\frac{{2 - ( - 7)}}{{ - 6 - 6}}\)     (M1)

Note:     Award (M1) for correct substitution into gradient formula.

\( =  - \frac{3}{4}{\text{ }}( - 0.75)\)     (A1)     (C2)

[2 marks]

\(\frac{4}{3}{\text{ }}(1.33333 \ldots )\)     (A1)(ft)     (C1)

Note:     Award (A0) for \(\frac{1}{{0.75}}\). Follow through from part (b).

[1 mark]

\(y = \frac{4}{3}x - \frac{5}{2}{\text{ }}(y = 1.33 \ldots x - 2.5)\)     (A1)(ft)     (C1)

Note:     Follow through from parts (c)(i) and (a). Award (A0) if final answer is not written in the form \(y = mx + c\).

[1 mark]

Write down the radius of the satellite’s orbit.

Calculate the distance travelled by the satellite in one orbit of the Earth. Give your answer correct to the nearest km.

Write down your answer to (b) in the form \(a \times {10^k}\) , where \(1 \leqslant a < 10{\text{, }}k \in \mathbb{Z}\) .

\(6900\) km     (A1)     (C1)

[1 mark]

\(2\pi (6900)\)     (M1)(A1)(ft)

Notes: Award (M1) for substitution into circumference formula, (A1)(ft) for correct substitution. Follow through from part (a).

\( = 43354\)     (A1)(ft)     (C3)

Notes: Follow through from part (a). The final (A1) is awarded for rounding their answer correct to the nearest km. Award (A2) for \(43 400\) shown with no working.

[3 marks]

\(4.3354 \times {10^4}\)     (A1)(ft)(A1)(ft)     (C2)

Notes: Award (A1)(ft) for \(4.3354\), (A1)(ft) for \( \times {10^4}\) . Follow through from part (b). Accept \(4.34 \times {10^4}\) .

[2 marks]

Candidates appeared to be confused by the context in this question. They had difficulty identifying the radius and many used the formula for the area of a circle, rather than the circumference.

Candidates appeared to be confused by the context in this question. They had difficulty identifying the radius and many used the formula for the area of a circle, rather than the circumference. A large number of candidates misread the final sentence in part b and did not write their answer to the nearest kilometre.

Candidates appeared to be confused by the context in this question. They had difficulty identifying the radius and many used the formula for the area of a circle, rather than the circumference.

Find the volume of 1 chocolate.

Write down the volume of 20 chocolates.

The diagram shows the chocolate box from above. The 20 chocolates fit perfectly in the box with each chocolate touching the ones around it or the sides of the box.

Calculate the volume of the box.

The diagram shows the chocolate box from above. The 20 chocolates fit perfectly in the box with each chocolate touching the ones around it or the sides of the box.

Calculate the volume of empty space in the box.

The first time a correct answer has incorrect or missing units, the final (A1) is not awarded.

\(\frac{4}{3}\pi {(1)^3}\)     (M1)

Notes: Award (M1) for correct substitution into correct formula.

\( = 4.19{\text{ }}\left( {{\text{4.18879}} \ldots ,{\text{ }}\frac{4}{3}\pi } \right){\text{ c}}{{\text{m}}^3}\)     (A1)     (C2)

[2 marks]

The first time a correct answer has incorrect or missing units, the final (A1) is not awarded.

\(83.8{\text{ }}\left( {{\text{83.7758}} \ldots ,{\text{ }}\frac{{80}}{3}\pi } \right){\text{ c}}{{\text{m}}^3}\)     (A1)(ft)     (C1)

Note: Follow through from their answer to part (a).

[1 mark]

The first time a correct answer has incorrect or missing units, the final (A1) is not awarded.

\(10 \times 8 \times 2\)     (M1)

Note: Award (M1) for correct substitution into correct formula.

\( = 160{\text{ c}}{{\text{m}}^3}\)     (A1)     (C2)

[2 marks]

The first time a correct answer has incorrect or missing units, the final (A1) is not awarded.

\(76.2{\text{ }}\left( {{\text{76.2241}} \ldots ,{\text{ }}\left( {160 - \frac{{80}}{3}\pi } \right)} \right){\text{ c}}{{\text{m}}^3}\)     (A1)(ft)     (C1)

Note: Follow through from their part (b) and their part (c).

[1 mark] 

Write down the size of angle CBA.

Write down the size of angle CAB.

The area of triangle ABC is 360 cm². Calculate the length of side AC. Express your answer in millimetres.

32°     (A1)     (C1)

[1 mark]

116°     (A1)     (C1)

[1 mark]

\(360 = \frac{1}{2} \times {x^2} \times \sin 116^\circ \)     (M1)(A1)(ft)

Notes: Award (M1) for substitution into correct formula with 360 seen, (A1)(ft) for correct substitution, follow through from their answer to part (b).

x = 28.3 (cm)     (A1)(ft)

x = 283 (mm)     (A1)(ft)     (C4)

Notes: The final (A1)(ft) is for their cm answer converted to mm. If their incorrect cm answer is seen the final (A1)(ft) can be awarded for correct conversion to mm.

[4 marks]

Candidates had difficulties finding the length of the side of the isosceles triangle and chose an incorrect angle in their substitution into the area formula. Many candidates thought this question related to right angle triangle trigonometry.

Candidates had difficulties finding the length of the side of the isosceles triangle and chose an incorrect angle in their substitution into the area formula. Many candidates thought this question related to right angle triangle trigonometry.

Candidates had difficulties finding the length of the side of the isosceles triangle and chose an incorrect angle in their substitution into the area formula. Many candidates thought this question related to right angle triangle trigonometry.

Find the length of BC.

Find the area of the postage stamp.

\(\frac{{{\text{BC}}}}{{\sin 34^\circ }} = \frac{5}{{\sin 120^\circ }}\)     (M1)(A1)

Note:     Award (M1) for substituted sine rule formula, (A1) for correct substitutions.

\({\text{BC}} = 3.23{\text{ (cm) }}\left( {3.22850 \ldots {\text{ (cm)}}} \right)\)     (A1)     (C3)

[3 marks]

\(\frac{1}{2}(5)(3.22850)\sin 26^\circ \)     (M1)(A1)(ft)

Note:     Award (M1) for substituted area of a triangle formula, (A1) for correct substitutions.

\( = 3.54{\text{ }}({\text{c}}{{\text{m}}^2}){\text{ }}\left( {3.53820 \ldots {\text{ }}({\text{c}}{{\text{m}}^2})} \right)\)     (A1)(ft)     (C3)

Note:     Follow through from part (a).

[3 marks]

Write down

(i)     the gradient of \({L_1}\);

(ii)     the \(y\)-intercept of \({L_1}\).

The line \({L_2}\) is parallel to \({L_1}\) and passes through the point \({\text{P}}(0,{\text{ }}3)\).

Write down the equation of \({L_2}\).

The line \({L_2}\) is parallel to \({L_1}\) and passes through the point \({\text{P}}(0,{\text{ }}3)\).

Find the \(x\)-coordinate of the point where \({L_2}\) crosses the \(x\)-axis.

(i)     \( - 2\)     (A1)     (C1)

(ii)     \(10\)     (A1)     (C1)

\(2x + y - 3 = 0\)     (A1)(ft)(A1)     (C2)

Notes: Award (A1)(ft) for gradient, (A1) for correct \(y\)-intercept.

The answer must be an equation.

\( - 2x + 3 = 0\;\;\;\)or equivalent     (M1)

\((x = ){\text{ }}1.5\)     (A1)(ft)     (C2)

Notes: Follow through from their equation in part (b). If answer given as coordinates \((1.5,{\text{ }}0)\) award at most (M1)(A0) if working seen or (A1)(A0) if no working seen.

Calculate \({\text{VM}}\).

Calculate the volume of the pyramid.

\({\text{A}}{{\text{C}}^2} = {8^2} + {6^2}\)     (M1)

Note: Award (M1) for correct substitution into Pythagoras, or recognition of Pythagorean triple.

\({\text{AC}} = 10\)     (A1)

Note: Award (A2) for \({\text{AC}} = 10\;\;\;\)OR\(\;\;\;{\text{AM}} = 5\) with no working seen.

\({\text{V}}{{\text{M}}^2} = {13^2} - {5^2}\)     (M1)

Note: Award (M1) for correct second use of Pythagoras, using the result from the first use of Pythagoras.

\({\text{VM}} = 12{\text{ (cm)}}\)     (A1)     (C4)

Notes: Accept alternative methods and apply the markscheme as follows: Award (M1)(A1) for first correct use of Pythagoras with lengths from the question, (M1) for a correct second use of Pythagoras, consistent with the method chosen, (A1) for correct height.

\(\frac{1}{3} \times 8 \times 6{\kern 1pt}  \times 12\)     (M1)

Note: Award (M1) for their correct substitutions into volume formula.

\( = 192{\text{ c}}{{\text{m}}^3}\)     (A1)(ft)     (C2)

Notes: Follow through from part (a), only if working seen.

In part (a) many candidates struggled to identify right angled triangles correctly. A regular mistake was to calculate the slant height and not the vertical height. Often values were used which did not correspond to a right angled triangle in the diagram, such as 13 and 8. Another common mistake was incorrect use of Pythagoras, where the hypotenuse was not correctly identified or was incorrectly substituted into the formula.

Despite the problems to obtain a correct answer for part (a), in part (b) many candidates wrote down a correctly substituted formula for the volume of a pyramid (with their height substituted) and received follow through marks. Very few, having calculated their volume correctly, failed to give the correct units. Some candidates used the perimeter (28) of the base and not the area.

Find the gradient of L.

Sarah wishes to draw the tangent to \(f (x) = x^4\) parallel to L.

Write down \(f ′(x)\).

Find the x coordinate of the point at which the tangent must be drawn.

Write down the value of \(f (x)\) at this point.

y = 13.5x + 4.5     (M1)

Note: Award (M1) for 13.5x seen.

gradient = 13.5     (A1)     (C2)

[2 marks]

4x³     (A1)     (C1)

[1 mark]

4x³ = 13.5     (M1)

Note: Award (M1) for equating their answers to (a) and (b).

x = 1.5     (A1)(ft)

[2 marks]

\(\frac{{81}}{{16}}\)   (5.0625, 5.06)     (A1)(ft)     (C3)

Note: Award (A1)(ft) for substitution of their (c)(i) into x4 with working seen.

[1 mark]

The structure of this question was not well understood by the majority; the links between parts not being made. Again, this question was included to discriminate at the grade 6/7 level.

Most were successful in this part.

The structure of this question was not well understood by the majority; the links between parts not being made. Again, this question was included to discriminate at the grade 6/7 level.

This part was usually well attempted.

The structure of this question was not well understood by the majority; the links between parts not being made. Again, this question was included to discriminate at the grade 6/7 level.

Only the best candidates succeeded in this part.

The structure of this question was not well understood by the majority; the links between parts not being made. Again, this question was included to discriminate at the grade 6/7 level.

Only the best candidates succeeded in this part.

The length of one side of a rectangle is 2 cm longer than its width.

If the smaller side is x cm, find the perimeter of the rectangle in terms of x.

The length of one side of a rectangle is 2 cm longer than its width.

The perimeter of a square is equal to the perimeter of the rectangle in part (a).

Determine the length of each side of the square in terms of x.

The length of one side of a rectangle is 2 cm longer than its width.

The perimeter of a square is equal to the perimeter of the rectangle in part (a).

The sum of the areas of the rectangle and the square is \(2x^2 + 4x +1\) (cm²).

(i) Given that this sum is 49 cm², find x.

(ii) Find the area of the square.

Unit penalty (UP) is applicable where indicated in the left hand column.

(UP) \({\text{P (rectangle)}} = 2x + 2(x + 2) = 4x + 4{\text{ cm}}\)     (A1)     (C1)

(UP) Simplification not required

[1 mark]

Unit penalty (UP) is applicable where indicated in the left hand column.

(UP) Side of square = (4x + 4)/4 = x + 1 cm     (A1)(ft)     (C1)

[1 mark]

(i) \(2x^2 + 4x + 1 = 49\) or equivalent     (M1)

\((x + 6)(x – 4) = 0\)

\(x = - 6\) and \(4\)     (A1)

Note: award (A1) for the values or for correct factors    

Choose \(x = 4\)     (A1)(ft)

Award (A1)(ft) for choosing positive value.     (C3)

(ii) \({\text{Area of square}} = 5 \times 5 = 25{\text{ c}}{{\text{m}}^2}\)     (A1)(ft)

Note: Follow through from both (b) and (c)(i).     (C1)

[4 marks]

a) and b) Two thirds of the candidates found the perimeter of the rectangle and the side of the square correctly, though most of them did not include units (thereby incurring a unit penalty).

a) and b) Two thirds of the candidates found the perimeter of the rectangle and the side of the square correctly, though most of them did not include units (thereby incurring a unit penalty).

c) Although a majority of candidates produced the quadratic equation many were unable to solve it correctly. This could easily be done using the GDC so it was disappointing.

On the diagram, draw and label with an x the angle of depression of B from P.

Find the size of angle APB.

Find the size of the angle of depression of B from P.

(A1) (C1)

[1 mark]

\(\frac{{40}}{{{\text{sin APB}}}} = \frac{{30}}{{{\text{sin 48}}^\circ }}\)     (M1)(A1)

Note: Award (M1) for substitution into sine rule, (A1) for correct substitution.

(angle APB =) 82.2°  (82.2473…°)     (A1)  (C3)

[3 marks]

180 − 48 − 82.2473…     (M1)

49.8°  (49.7526…°)     (A1)(ft)  (C2)

Note: Follow through from parts (a) and (b).

[2 marks]

Write down the size of angle AOB.

Find the area of the triangle AOB.

The height of the prism is 20 cm.

Find the volume of the prism.

60°     (A1)     (C1)

[1 mark]

\(\frac{{15 \times \sqrt {{{15}^2} - {{7.5}^2}} }}{2} = 97.4{\text{ c}}{{\text{m}}^2}\)     (97.5 cm²)     (A1)(M1)(A1)

Notes: Award (A1) for correct height, (M1) for substitution in the area formula, (A1) for correct answer.

Accept 97.5 cm² from taking the height to be 13 cm.

OR

\(\frac{1}{2} \times {15^2} \times \sin 60^\circ  = 97.4{\text{ c}}{{\text{m}}^2}\)     (M1)(A1)(A1)(ft)     (C3)

Notes: Award (M1) for substituted formula of the area of a triangle, (A1) for correct substitution, (A1)(ft) for answer.

Follow through from their answer to part (a).

If radians used award at most (M1)(A1)(A0).

[3 marks]

97.4 × 120 = 11700 cm³     (M1)(A1)(ft)     (C2)

Notes: Award (M1) for multiplying their part (b) by 120.

[2 marks]

This question proved to be difficult for a number of candidates. Most were able find the size of the angle in part a), but many had problems finding the area of the triangle in part b). A significant number of candidates were unable to use the Pythagoras Theorem correctly to find the height of the triangle AOB. Those who used the formula for the area of a triangle \({\text{A}} = \frac{1}{2}{\text{ab}}\sin {\text{C}}\)  were more successful in this part of the question. It was surprising that a great number of candidates were unable to find the volume of the prism – many incorrectly used the formula for calculating volume of a pyramid rather than a hexagonal prism.

This question proved to be difficult for a number of candidates. Most were able find the size of the angle in part a), but many had problems finding the area of the triangle in part b). A significant number of candidates were unable to use the Pythagoras Theorem correctly to find the height of the triangle AOB. Those who used the formula for the area of a triangle \({\text{A}} = \frac{1}{2}{\text{ab}}\sin {\text{C}}\)  were more successful in this part of the question. It was surprising that a great number of candidates were unable to find the volume of the prism – many incorrectly used the formula for calculating volume of a pyramid rather than a hexagonal prism.

This question proved to be difficult for a number of candidates. Most were able find the size of the angle in part a), but many had problems finding the area of the triangle in part b). A significant number of candidates were unable to use the Pythagoras Theorem correctly to find the height of the triangle AOB. Those who used the formula for the area of a triangle \({\text{A}} = \frac{1}{2}{\text{ab}}\sin {\text{C}}\)  were more successful in this part of the question. It was surprising that a great number of candidates were unable to find the volume of the prism – many incorrectly used the formula for calculating volume of a pyramid rather than a hexagonal prism.

Find the gradient of the line AB.

Let M be the midpoint of the line segment AB.

Write down the coordinates of M.

Let M be the midpoint of the line segment AB.

Find the equation of the line perpendicular to AB and passing through M.

\({\text{Gradient}} = \frac{{(5 - 1)}}{{(4 - 2)}}\)     (M1)

Note: Award (M1) for correct substitution in the gradient formula.

\(= 2\)     (A1)     (C2)

[2 marks]

Midpoint = (3, 3) (accept x = 3, y = 3 )     (A1)     (C1)

[1 mark]

\({\text{Gradient of perpendicular}} = -\frac{{1}}{{2}}\)     (A1)(ft)

\(y = - \frac{{1}}{{2}} x + c\)     (M1)

\(3 = - \frac{{1}}{{2}} \times 3 + c\)

\(c = 4.5\)

\(y = -0.5x + 4.5\)     (A1)(ft)

OR

\(y - 3 = -0.5(x - 3)\)     (A1)(A1)(ft)

Note: Award (A1) for –0.5, (A1) for both threes.

OR

\(2y + x = 9\)     (A1)(A1)(ft)     (C3)

Note: Award (A1) for 2, (A1) for 9.

[3 marks]

While parts (a) and (b) were answered or at least attempted with various success, few candidates made progress in part (c). Some candidates used the coordinates of point A or B rather than M and others could not find the gradient of the perpendicular line.

While parts (a) and (b) were answered or at least attempted with various success, few candidates made progress in part (c). Some candidates used the coordinates of point A or B rather than M and others could not find the gradient of the perpendicular line.

While parts (a) and (b) were answered or at least attempted with various success, few candidates made progress in part (c). Some candidates used the coordinates of point A or B rather than M and others could not find the gradient of the perpendicular line.

Calculate the amount of money he has in the account after 5 years.

Write down the amount of interest he earned after 5 years.

Karl decides to donate this interest to a charity in France. The charity receives 170 euros (EUR). The exchange rate is 1 USD = t EUR.

Calculate the value of t.

\(1000{\left( {1 + \frac{{3.5}}{{4 \times 100}}} \right)^{4 \times 5}}\)     (M1)(A1)

Note: Award (M1) for substitution in compound interest formula, (A1) for correct substitution.

OR

N = 5

I = 3.5

PV = 1000

P/Y = 1

C/Y = 4

Note: Award (A1) for C/Y = 4 seen, (M1) for other correct entries.

OR

N = 5 × 4

I = 3.5

PV = 1000

P/Y = 1

C/Y = 4

Note: Award (A1) for C/Y = 4 seen, (M1) for other correct entries.

= 1190.34 (USD)     (A1)

Note: Award (M1) for substitution in compound interest formula, (A1) for correct substitution.

[3 marks]

190.34 (USD)      (A1)(ft) (C4)

Note: Award (A1)(ft) for subtraction of 1000 from their part (a)(i). Follow through from (a)(i).

[1 mark]

\(\frac{{170}}{{190.34}}\)     (M1)

Note: Award (M1) for division of 170 by their part (a)(ii).

= 0.89     (A1)(ft) (C2)

Note: Follow through from their part (a)(ii).

[2 marks]

Find the perimeter of the rectangle, giving your answer in the form \(a \times {10^k}\), where \(1 \leqslant a \leqslant 10\) and \(k \in \mathbb{Z}\).

Find the area of the rectangle, giving your answer correct to the nearest thousand square centimetres.

Note: Unit penalty (UP) applies in this part

(2680 + 1970) × 2     (M1)

(UP)     = 9.30 × 10³ cm     (A1)(A1)     (C3)

Notes: Award (M1) for correct formula.

(A1) for 9.30 (Accept 9.3).

(A1) for 10³.

[3 marks]

2680 × 1970     (M1)

= 5279600     (A1)

= 5,280,000 (5280 thousand)     (A1)(ft)     (C3)

Note: Award (M1) for correctly substituted formula.

Accept 5.280 × 10⁶.

Note: The last (A1) is for specified accuracy, (ft) from their answer.

The (AP) for the paper is not applied here.

[3 marks]

This question was well answered by many candidates although the majority lost a mark as a unit penalty in part (a). Some candidates used the wrong formula for the perimeter. Most could give their answer in standard form.

This question was well answered by many candidates although the majority lost a mark as a unit penalty in part (a). Some candidates used the wrong formula for the perimeter. Most could give their answer in standard form.

Using simplest terms, write an equation in b and m for Thursday’s sales.

Find b and m.

Draw a sketch, in the space provided, to show how the prices can be found graphically.

Thursday's sales, \(6b + 9m = 23.40\)     (A1)

\(2b + 3m = 7.80\)     (A1)     (C2)

[2 marks]

\(m = 1.40\)     (accept 1.4)     (A1)(ft)

\(b = 1.80\)     (accept 1.8)     (A1)(ft)

Award (A1)(d) for a reasonable attempt to solve by hand and answer incorrect.     (C2)

[2 marks]

     (A1)(A1)(ft)

(A1) each for two reasonable straight lines. The intersection point must be approximately correct to earn both marks, otherwise penalise at least one line.

Note: The follow through mark is for candidate’s line from (a).     (C2)

[2 marks]

a) Nearly all the candidates were able to write the equation but very few simplified it.

b) A majority of candidates were able to find the values of b and m. Some used the right method but made arithmetical errors, many of which were due to them using the method of substitution which involved fractions. GDC use was expected.

c) A majority of candidates did not attempt this part. For those who did, very few were able to sketch the graph correctly. Common errors were to plot the point (1.4, 1.8) or draw a straight line through that point and the origin.

Calculate the gradient of \(L\) .

Find the equation of \(L\) .

The line \(L\) also passes through the point \({\text{P}}(8{\text{, }}y)\) . Find the value of \(y\) .

\(\frac{{8 - 4}}{{5 - ( - 1)}}\)     (M1)

Note: Award (M1) for correct substitution into the gradient formula.

\(\frac{2}{3}{\text{ }}\left( {\frac{4}{6}{\text{, }}0.667} \right)\)     (A1)     (C2)

[2 marks]

\(y = \frac{2}{3}x + c\)     (A1)(ft)

Note: Award (A1)(ft) for their gradient substituted in their equation.

\(y = \frac{2}{3}x + \frac{{14}}{3}\)     (A1)(ft)     (C2)

Notes: Award (A1)(ft) for their correct equation. Accept any equivalent form. Accept decimal equivalents for coefficients to 3 sf.

OR

\(y - {y_1} = \frac{2}{3}(x - {x_1})\)     (A1)(ft)

Note: Award (A1)(ft) for their gradient substituted in the equation.

\(y - 4 = \frac{2}{3}(x + 1)\)     OR     \(y - 8 = \frac{2}{3}(x - 5)\)     (A1)(ft)     (C2)

Note: Award (A1)(ft) for correct equation.

[2 marks]

\(y = \frac{2}{3} \times 8 + \frac{{14}}{3}\)     OR     \(y - 4 = \frac{2}{3}(8 + 1)\)     OR     \(y - 8 = \frac{2}{3}(8 - 5)\)     (M1)

Note: Award (M1) for substitution of \(x = 8\) into their equation.

\(y = 10\) (\(10.0\))     (A1)(ft)     (C2)

Note: Follow through from their answer to part (b).

[2 marks]

Generally, a well answered question with many candidates achieving full marks. Indeed, marks which tended to be lost were as a result of premature rounding rather than method. On a number of scripts, part (a) produced a rather curious wrong answer of \(8.2\) following a correct gradient expression. It would seem that this was as a result of typing into the calculator \(8 - 4 ÷ 5 + 1\).

Generally, a well answered question with many candidates achieving full marks. Indeed, marks which tended to be lost were as a result of premature rounding rather than method.

Generally, a well answered question with many candidates achieving full marks. Indeed, marks which tended to be lost were as a result of premature rounding rather than method.

Write down an expression for A in terms of x.

Find the value of x given that A = 109.25 m².

The owner of the garden puts a fence around the shaded region. Find the length of this fence.

(x + 1)² – 1  or  x² + 2x     (A1)     (C1)

[1 mark]

(x + 1)² – 1 = 109.25     (M1)

x² + 2x – 109.25 = 0     (M1)

Notes: Award (M1) for writing an equation consistent with their expression in (a) (accept equivalent forms), (M1) for correctly removing the brackets.

OR

(x + 1)² – 1 = 109.25     (M1)

x + 1 = \(\sqrt {110.25} \)     (M1)

Note: Award (M1) for writing an equation consistent with their expression in (a) (accept equivalent forms), (M1) for taking the square root of both sides.

OR

(x + 1)² – 10.5² = 0     (M1)

(x – 9.5) (x + 11.5) = 0     (M1)

Note: Award (M1) for writing an equation consistent with their expression in (a) (accept equivalent forms), (M1) for factorised left side of the equation.

x = 9.5     (A1)(ft)     (C3)

Note: Follow through from their expression in part (a).

The last mark is lost if x is non positive.

If the follow through equation is not quadratic award at most (M1)(M0)(A1)(ft).

[3 marks]

4 × (9.5 + 1) = 42 m     (M1)(A1)(ft)     (C2)

Notes: Award (M1) for correct method for finding the length of the fence. Accept equivalent methods.

[2 marks]

Some candidates were able to answer this question correctly, but the majority experienced difficulty in finding the correct expression for the area of the shaded region. Those who showed working could then be awarded follow through marks for correctly equating their expressions to the given area and for their found value of x. Many candidates also could not find the perimeter of the shaded region in part c) even though they had found the value of x correctly.

Some candidates were able to answer this question correctly, but the majority experienced difficulty in finding the correct expression for the area of the shaded region. Those who showed working could then be awarded follow through marks for correctly equating their expressions to the given area and for their found value of x. Many candidates also could not find the perimeter of the shaded region in part c) even though they had found the value of x correctly.

Some candidates were able to answer this question correctly, but the majority experienced difficulty in finding the correct expression for the area of the shaded region. Those who showed working could then be awarded follow through marks for correctly equating their expressions to the given area and for their found value of x. Many candidates also could not find the perimeter of the shaded region in part c) even though they had found the value of x correctly.

Calculate the length of AG.

Calculate the length of AF.

Find the size of the angle between AF and AG.

\({\text{AG}} = \sqrt {{{0.8}^2} + {{0.5}^2}} \)     (M1)

AG = 0.943 m     (A1)     (C2)

[2 marks]

\({\text{AF}} = \sqrt {{\text{A}}{{\text{G}}^2} + {{1.80}^2}} \)     (M1)

= 2.03 m     (A1)(ft)     (C2)

Note: Follow through from their answer to part (a).

[2 marks]

\(\cos {\rm{G\hat AF}} = \frac{{0.943(39 \ldots )}}{{2.03(22 \ldots )}}\)     (M1)

\(\operatorname{G\hat AF} = 62.3^\circ \)     (A1)(ft)     (C2)

Notes: Award (M1) for substitution into correct trig ratio.

Accept alternative ratios which give 62.4° or 62.5°.

Follow through from their answers to parts (a) and (b).

[2 marks]

This question was well answered. Surprisingly few candidates used the basic trigonometric ratios (for right angle triangles), opting instead to use the sine or cosine laws.

This question was well answered. Surprisingly few candidates used the basic trigonometric ratios (for right angle triangles), opting instead to use the sine or cosine laws.

This question was well answered. Surprisingly few candidates used the basic trigonometric ratios (for right angle triangles), opting instead to use the sine or cosine laws.

Write down the size of angle MCB.

Write down the length of BM in cm.

Write down the size of angle BMC.

Calculate the length of BC in cm.

30°     (A1)     (C3)

[1 mark]

8.5 (cm)     (A1)

[1 mark]

120°     (A1)

[1 mark]

\(\frac{{{\text{BC}}}}{{\sin 120}} = \frac{{8.5}}{{\sin 30}}\)     (M1)(A1)(ft)

Note: Award (M1) for correct substituted formula, (A1) for correct substitutions.

\({\text{BC}} = {\text{14}}{\text{.7}}\left( {\frac{{17\sqrt 3 }}{2}} \right)\)     (A1)(ft)

[3 marks]

Part (a) was generally well answered with many candidates gaining full marks. Some candidates went on to make incorrect assumptions about triangle BMC being right angled and used Pythagorus theorem incorrectly. Those who used either the Sine rule or the Cosine rule correctly were generally able to substitute correctly and gain at least two marks.

Part (a) was generally well answered with many candidates gaining full marks. Some candidates went on to make incorrect assumptions about triangle BMC being right angled and used Pythagorus theorem incorrectly. Those who used either the Sine rule or the Cosine rule correctly were generally able to substitute correctly and gain at least two marks.

Part (a) was generally well answered with many candidates gaining full marks. Some candidates went on to make incorrect assumptions about triangle BMC being right angled and used Pythagorus theorem incorrectly. Those who used either the Sine rule or the Cosine rule correctly were generally able to substitute correctly and gain at least two marks.

Part (a) was generally well answered with many candidates gaining full marks. Some candidates went on to make incorrect assumptions about triangle BMC being right angled and used Pythagorus theorem incorrectly. Those who used either the Sine rule or the Cosine rule correctly were generally able to substitute correctly and gain at least two marks.

Calculate, SB, the distance between the ship and the bottom of the cliff.

Calculate, SF, the distance between the ship and the fishing boat. Give your answer correct to the nearest metre.

\(150\tan 50\)     (M1)

OR

\(\frac{{150}}{{\tan 40}}\)     (M1)

\( = 179{\text{ (m)}}\) (\(178.763 \ldots \))     (A1)     (C2)

\(150\tan 50 - 150\tan 35\)     (M1)(M1)

Note: Award (M1) for \(150\tan 35\), (M1) for subtraction from their part (a).

\( = 73.7{\text{ (m)}}\) (\(73.7319 \ldots \))     (A1)(ft)
\( = 74{\text{ (m)}}\)     (A1)(ft)     (C4)

Note: The final (A1) is awarded for the correct rounding of their answer to (b).
Note: There will always be one answer with a specified degree of accuracy on each paper.

Find the gradient of \(L\).

Find the equation of \(L\) in the form \(y = mx + c\).

Find the \(x\)-coordinate of point A.

\( - 3\)     (A1)(A1)     (C2)

Notes:     Award (A1) for 3 and (A1) for a negative value.

Award (A1)(A0) for either \(3x\) or \( - 3x\).

[2 marks]

\(6 =  - 3(2) + c\)\(\,\,\,\)OR\(\,\,\,\)\((y - 6) =  - 3(x - 2)\)     (M1)

Note:     Award (M1) for substitution of their gradient from part (a) into a correct equation with the coordinates \((2,{\text{ }}6)\) correctly substituted.

\(y =  - 3x + 12\)     (A1)(ft)     (C2)

Notes:     Award (A1)(ft) for their correct equation. Follow through from part (a).

If no method seen, award (A1)(A0) for \(y =  - 3x\).

Award (A1)(A0) for \( - 3x + 12\).

[2 marks]

\(0 =  - 3x + 12\)     (M1)

Note:     Award (M1) for substitution of \(y = 0\) in their equation from part (b).

\((x = ){\text{ }}4\)     (A1)(ft)     (C2)

Notes:     Follow through from their equation from part (b). Do not follow through if no method seen. Do not award the final (A1) if the value of \(x\) is negative or zero.

[2 marks]

Find the length of AC.

Find the area of triangle ABC.

\(\frac{{{\text{AC}}}}{{\sin 30^\circ }} = \frac{8}{{\sin 40^\circ }}\)     (M1)(A1)

Note: Award (M1) for substitution in the sine rule formula, (A1) for correct substitutions.

AC = 6.22 (m) (6.22289…)     (A1)     (C3)

[3 marks]

Area of triangle \({\text{ABC}} = \frac{1}{2} \times 8 \times 6.22289... \times \sin 110^\circ \)     (M1)(A1)(ft)

Note: Award (M1) for substitution in the correct formula, (A1)(ft) for their correct substitutions. Follow through from their part (a).

Area triangle ABC = 23.4 m² (23.3904...m²)     (A1)(ft)     (C3)

Note: Follow through from a positive answer to their part (a). The answer is 23.4 m², units are required.

[3 marks]

Whilst using the sine rule to find an angle was tested in question 10, here the sine rule was required to be used to find a length. Many scripts showed the correct value of 6.22 m, but a significant number of candidates calculated AB instead of AC and the answer of 11.7 m proved to be a popular, but erroneous, answer. Some candidates turned the problem into two right angled triangles by dropping the perpendicular from C to the line AB. Much working was then required to find AC and again, some of these candidates simply determined the length of AB. Despite a significant number of candidates identifying the incorrect length in part (a), many of these recovered in part (b) to use their value correctly in the formula for the area of a triangle and many correct calculations were seen. Unfortunately, this good work was spoilt as some candidates either missed the units or gave the incorrect units in their final answer and, as a consequence, lost the last mark.

Whilst using the sine rule to find an angle was tested in question 10, here the sine rule was required to be used to find a length. Many scripts showed the correct value of 6.22 m, but a significant number of candidates calculated AB instead of AC and the answer of 11.7 m proved to be a popular, but erroneous, answer. Some candidates turned the problem into two right angled triangles by dropping the perpendicular from C to the line AB. Much working was then required to find AC and again, some of these candidates simply determined the length of AB. Despite a significant number of candidates identifying the incorrect length in part (a), many of these recovered in part (b) to use their value correctly in the formula for the area of a triangle and many correct calculations were seen. Unfortunately, this good work was spoilt as some candidates either missed the units or gave the incorrect units in their final answer and, as a consequence, lost the last mark.

Calculate the volume of the Earth in \({\text{k}}{{\text{m}}^3}\). Give your answer in the form \(a \times {10^k}\), where \(1 \leqslant a < 10\) and \(k \in \mathbb{Z}\).

The volume of the Moon is \(2.1958 \times {10^{10}}\;{\text{k}}{{\text{m}}^3}\).

Calculate how many times greater in volume the Earth is compared to the Moon.

Give your answer correct to the nearest integer.

\(\frac{4}{3}\pi {(6371)^3}\)     (M1)

Note: Award (M1) for correct substitution into volume formula.

\( = 1.08 \times {10^{12}}\;\;\;(1.08320 \ldots  \times {10^{12}})\)     (A2)     (C3)

Notes: Award (A1)(A0) for correct mantissa between 1 and 10, with incorrect index.

Award (A1)(A0) for \(1.08\rm{E}12\)

Award (A0)(A0) for answers of the type: \(108 \times {10^{10}}\).

\(\frac{{1.08320 \ldots  \times {{10}^{12}}}}{{2.1958 \times {{10}^{10}}}}\)     (M1)

Note: Award (M1) for dividing their answer to part (a) by \(2.1958 \times {10^{10}}\).

\( = 49.3308 \ldots \)     (A1)(ft)

Note: Accept \(49.1848...\) from use of 3 sf answer to part (a).

\( = 49\)     (A1)     (C3)

Notes: Follow through from part (a).

The final (A1) is awarded for their unrounded non-integer answer seen and given correct to the nearest integer.

Do not award the final (A1) for a rounded answer of 0 or if it is incorrect by a large order of magnitude.

In part (a) many candidates correctly substituted the volume formula and wrote correctly their answer using scientific notation. The calculator notation E12 was very rarely used. A minority converted to metres, resulting in an incorrect exponent. Some candidates used an incorrect equation or used their calculator incorrectly.

In part (b) many candidates subtracted the values, where they should be divided, resulting in an answer of an unrealistic magnitude. Some reversed the numerator and denominator, leading to an answer of 0.02, which would have rounded to the unrealistic answer of 0. When a reasonable answer was found, the final mark for rounding was lost by some candidates when there was no rounding or when rounding was incorrect. There seemed to be little understanding of whether or not an answer was reasonable.

Find the length of \({\text{AB}}\).

\({\text{D}}\) is the point on \({\text{AB}}\) such that \(\tan ({\rm{D\hat CB}}) = 0.6\).

Find the length of \({\text{DB}}\).

\({\text{D}}\) is the point on \({\text{AB}}\) such that \(\tan ({\rm{D\hat CB}}) = 0.6\).

Find the area of triangle \({\text{ADC}}\).

\(({\text{A}}{{\text{B}}^2}) = {20^2} - {12^2}\)     (M1)

Note: Award (M1) for correctly substituted Pythagoras formula.

\({\text{AB}} = 16{\text{ cm}}\)     (A1)     (C2)

[2 marks]

\(\frac{{{\text{DB}}}}{{12}} = 0.6\)     (M1)

Note: Award (M1) for correct substitution in tangent ratio or equivalent ie seeing \(12 \times 0.6\).

\({\text{DB}} = 7.2{\text{ cm}}\)     (A1)     (C2)

Note: Award (M1)(A0) for using \(\tan 31\) to get an answer of \(7.21\).

     Award (M1)(A0) for \(\frac{{12}}{{\sin 59}} = \frac{{{\text{DB}}}}{{\sin 31}}\) to get an answer of \(7.2103…\) or any other incorrect answer.

[2 marks]

\(\frac{1}{2} \times 12 \times (16 - 7.2)\)     (M1)

Note: Award (M1) for their correct substitution in triangle area formula.

OR

\(\frac{1}{2} \times 12 \times 16 - \frac{1}{2} \times 12 \times 7.2\)     (M1)

Note: Award (M1) for subtraction of their two correct area formulas.

\( = 52.8{\text{ c}}{{\text{m}}^2}\)     (A1)(ft)     (C2)

Notes: Follow through from parts (a) and (b).

     Accept alternative methods.

[2 marks]

This question was not well answered. Many candidates could not use Pythagoras’ theorem correctly and many failed to appreciate the significance of \(\tan ( < {\text{DCB}}) = 0.6\) and calculated the size of the angle DCB, rounding it to 31°. Unfortunately, this method led to an inaccurate value for DB. Finding the area of triangle ADC was also difficult for many who did not realize that they needed to do a subtraction of triangle areas. Candidates who tried to find side lengths and angles for triangle ADC were generally unsuccessful in calculating its area.

This question was not well answered. Many candidates could not use Pythagoras’ theorem correctly and many failed to appreciate the significance of \(\tan ( < {\text{DCB}}) = 0.6\) and calculated the size of the angle DCB, rounding it to 31°. Unfortunately, this method led to an inaccurate value for DB. Finding the area of triangle ADC was also difficult for many who did not realize that they needed to do a subtraction of triangle areas. Candidates who tried to find side lengths and angles for triangle ADC were generally unsuccessful in calculating its area.

This question was not well answered. Many candidates could not use Pythagoras’ theorem correctly and many failed to appreciate the significance of \(\tan ( < {\text{DCB}}) = 0.6\) and calculated the size of the angle DCB, rounding it to 31°. Unfortunately, this method led to an inaccurate value for DB. Finding the area of triangle ADC was also difficult for many who did not realize that they needed to do a subtraction of triangle areas. Candidates who tried to find side lengths and angles for triangle ADC were generally unsuccessful in calculating its area.

Calculate CD, the height of the observation deck above the ground.

Calculate the angle of depression from A to B.

\(\tan 48^\circ  = \frac{{{\text{CD}}}}{{250}}\)    (M1)

Note:     Award (M1) for correct substitution into the tangent ratio.

\(({\text{CD}} = ){\text{ }}278{\text{ }}({\text{m}}){\text{ }}(277.653 \ldots )\)    (A1)     (C2)

[2 marks]

\(\tan {\text{ABC (or equivalent)}} = \frac{{\frac{4}{3} \times 277.653 \ldots }}{{250}}\)     (M1)(M1)(M1)

Note:     Award (M1) for \(\frac{4}{3}\) multiplying their part (a), (M1) for substitution into the tangent ratio, (M1) for correct substitution.

OR

\(90 - {\tan ^{ - 1}}\left( {\frac{{250}}{{\frac{4}{3} \times 277.653 \ldots }}} \right)\)    (M1)(M1)(M1)

Note:     Award (M1) for \(\frac{4}{3}\) multiplying their part (a), (M1) for substitution into the tangent ratio, (M1) for subtracting from 90 and for correct substitution.

\({\text{(angle of depression}} = {\text{) }}56.0^\circ {\text{ }}(55.9687 \ldots )\)    (A1)(ft)     (C4)

Note:     Follow through from part (a).

[4 marks]

Complete the Venn diagram by placing the letter corresponding to each polygon in the appropriate region. For example, R has already been placed, and represents the rectangle.

State which polygons from Tuti’s list are elements of

(i)     \(A \cap B\);

(ii)     \((A \cup B)'\).

     (A3)     (C3)

Note: Award (A3) if all four letters placed correctly,

(A2) if three letters are placed correctly,

(A1) if two letters are placed correctly.

(i)     Rhombus and rectangle OR H and R     (A1)(ft)

(ii)     Scalene triangle OR T     (A2)(ft)     (C3)

Notes: Award (A1) for a list R, H, I, P seen (identifying the union).

Follow through from their part (a).

Find the distance between A and C.

Find the size of angle BAC.

Unit penalty (UP) is applicable in question part (a) only.

\({\text{AC}}^2 = 625^2 + 986^2 - 2 \times 625 \times 986 \times \cos102^\circ\)     (M1)(A1)

( = 1619072.159)

AC = 1272.43

(UP) =1270 m     (A1)     (C3)

[3 marks]

\(\frac{{986}}{{\operatorname{sinA} }} = \frac{{1270}}{{\sin 102^\circ }}\)     (M1)(A1)(ft)

\({\text{A}} = 49.4^\circ\)     (A1)(ft)

OR

\(\frac{{986}}{{\operatorname{sinA} }} = \frac{{1272.43}}{{\sin 102^\circ }}\)     (M1)(A1)(ft)

\({\text{A}} = 49.3^\circ\)     (A1)(ft)

OR

\(\cos {\text{A}} = \left( {\frac{{{{625}^2} + {{1270}^2} - {{986}^2}}}{{2 \times 625 \times 1270}}} \right)\)     (M1)(A1)(ft)

\({\text{A}} = 49.5^\circ\)     (A1)(ft)     (C3)

[3 marks]

The candidates who used the cosine and sine rules for this question were successful on the whole. Some had their calculators in radian mode (and hence the second answer for the angle was unrealistic) but this was less frequent than in previous sessions. Those candidates who used right-angled trigonometry scored no marks. Many candidates lost an accuracy penalty in this question.

The candidates who used the cosine and sine rules for this question were successful on the whole. Some had their calculators in radian mode (and hence the second answer for the angle was unrealistic) but this was less frequent than in previous sessions. Those candidates who used right-angled trigonometry scored no marks. Many candidates lost an accuracy penalty in this question.

Calculate the volume of the observatory.

The hemispherical roof is to be painted.

Calculate the area that is to be painted.

\(V = \pi {(15)^2}(12) + 0.5 \times \frac{{4\pi {{(15)}^3}}}{3}\)     (M1)(M1)(M1)

Note: Award (M1) for correctly substituted cylinder formula, (M1) for correctly substituted sphere formula, (M1) for dividing the sphere formula by 2.

\( = 15550.8 \ldots \)

\( = 15600{\text{ }}{{\text{m}}^3}{\text{ }}(4950\pi {\text{ }}{{\text{m}}^3})\)     (A1)     (C4)

Notes: The final answer is \(15600{\text{ }}{{\text{m}}^3}\); the units are required. The use of \(\pi  = 3.14\) which gives a final answer of \(15 500\) (\(15 543\)) is premature rounding; the final (A1) is not awarded.

[4 marks]

\(SA = 0.5 \times 4\pi {\left( {15} \right)^2}\)     (M1)

\( = 1413.71 \ldots \)

\( = 1410{\text{ }}{{\text{m}}^2}{\text{ }}(450\pi {\text{ }}{{\text{m}}^2})\)     (A1)     (C2)

Notes: The final answer is \(1410{\text{ }}{{\text{m}}^2}\) ; do not penalize lack of units if this has been penalized in part (a).

[2 marks]

This question was only done well by the more able candidates. For the lower quartile of candidates, about a half of these scored no more than one mark in total for this question. In many cases, formulae were either misquoted or misused. Indeed, in part (a) many ignored the hemisphere, choosing to use the formula for the volume of a sphere instead. Some candidates who correctly used a hemisphere in part (a) then treated the surface area as a sphere in part (b). A minority of candidates thought the area to be painted in the last part of the question was a circle rather than a hemisphere.

This question was only done well by the more able candidates. For the lower quartile of candidates, about a half of these scored no more than one mark in total for this question. In many cases, formulae were either misquoted or misused. Indeed, in part (a) many ignored the hemisphere, choosing to use the formula for the volume of a sphere instead. Some candidates who correctly used a hemisphere in part (a) then treated the surface area as a sphere in part (b). A minority of candidates thought the area to be painted in the last part of the question was a circle rather than a hemisphere.

Write down the gradient of the line \(y = 3x + 4\).

Find the gradient of the line which is perpendicular to the line \(y = 3x + 4\).

Find the equation of the line which is perpendicular to \(y = 3x + 4\) and which passes through the point \((6{\text{, }}7)\).

Find the coordinates of the point of intersection of these two lines.

\(3\)     (A1)     (C1)

[1 mark]

\( - 1/3\)     (ft) from (a)     (A1)(ft)     (C1)

[1 mark]

Substituting \((6{\text{, }}7)\) in \(y ={\text{their }}mx + c\) or equivalent to find \(c\).     (M1)

\(y = \frac{{ - 1}}{3}x + 9\) or equivalent     (A1)(ft)      (C2)

[2 marks]

\((1.5{\text{, }}8.5)\)      (A1)(A1)(ft)     (C2)

Note: Award (A1) for \(1.5\), (A1) for \(8.5\). (ft) from (c), brackets not required.

[2 marks]

This question was well answered by some candidates and poorly answered by others. It seemed to be part of the syllabus that might have been fully taught by some schools and not by others. It was surprising to see how many candidates could not find the gradient of a perpendicular line when this has been tested for many years.

This question was well answered by some candidates and poorly answered by others. It seemed to be part of the syllabus that might have been fully taught by some schools and not by others. It was surprising to see how many candidates could not find the gradient of a perpendicular line when this has been tested for many years.

This question was well answered by some candidates and poorly answered by others. It seemed to be part of the syllabus that might have been fully taught by some schools and not by others. It was surprising to see how many candidates could not find the gradient of a perpendicular line when this has been tested for many years.

This question was well answered by some candidates and poorly answered by others. It seemed to be part of the syllabus that might have been fully taught by some schools and not by others. It was surprising to see how many candidates could not find the gradient of a perpendicular line when this has been tested for many years.

Calculate the length of AC.

Calculate the length of AG.

Calculate the angle that AG makes with the floor.

\({\text{A}}{{\text{C}}^2} = {7.2^2} + {9.6^2}\)     (M1) 

Note: Award (M1) for correct substitution in Pythagoras Theorem.

\({\text{AC}} = 12{\text{ m}}\)     (A1)     (C2)

[2 marks]

\({\text{A}}{{\text{G}}^2} = {12^2} + {3.5^2}\)     (M1) 

Note: Award (M1) for correct substitution in Pythagoras Theorem.

\({\text{AG}} = 12.5{\text{ m}}\)     (A1)(ft)     (C2)

Note: Follow through from their answer to part (a).

[2 marks]

\(\tan \theta  = \frac{{3.5}}{{12}}\) or \(\sin \theta  = \frac{{3.5}}{{12.5}}\) or \(\cos \theta  = \frac{{12}}{{12.5}}\)     (M1)

Note: Award (M1) for correct substitutions in trig ratio.

\(\theta  = {16.3^ \circ }\)     (A1)(ft)     (C2)

Notes: Follow through from parts (a) and/or part (b) where appropriate. Award (M1)(A0) for use of radians (0.284).

[2 marks]

Question 7 was surprisingly difficult for many candidates, especially part b. Many candidates did not recognize that ACG was a right angled triangle and tried to use the law of cosines to find angle A. Although correct substitution and manipulation provided the correct answer, many candidates attempting this method made arithmetical errors.

Question 7 was surprisingly difficult for many candidates, especially part b. Many candidates did not recognize that ACG was a right angled triangle and tried to use the law of cosines to find angle A. Although correct substitution and manipulation provided the correct answer, many candidates attempting this method made arithmetical errors.

Question 7 was surprisingly difficult for many candidates, especially part b. Many candidates did not recognize that ACG was a right angled triangle and tried to use the law of cosines to find angle A. Although correct substitution and manipulation provided the correct answer, many candidates attempting this method made arithmetical errors.

Write down \(f'(x)\).

Point \({\text{P}}(2,6)\) lies on the graph of \(f\).

Find the gradient of the tangent to the graph of \(y = f(x)\) at \({\text{P}}\).

Point \({\text{P}}(2,16)\) lies on the graph of \(f\).

Find the equation of the normal to the graph at \({\text{P}}\). Give your answer in the form \(ax + by + d = 0\), where \(a\), \(b\) and \(d\) are integers.

\(\left( {f'(x) = } \right)\)   \(4{x^3}\)     (A1)     (C1)

[1 mark]

\(4 \times {2^3}\)     (M1)

Note: Award (M1) for substituting 2 into their derivative.

\( = 32\)     (A1)(ft)     (C2)

Note: Follow through from their part (a).

[2 marks]

\(y - 16 =  - \frac{1}{{32}}(x - 2)\)   or   \(y =  - \frac{1}{{32}}x + \frac{{257}}{{16}}\)     (M1)(M1)

Note: Award (M1) for their gradient of the normal seen, (M1) for point substituted into equation of a straight line in only \(x\) and \(y\) (with any constant ‘\(c\)’ eliminated).

\(x + 32y - 514 = 0\) or any integer multiple     (A1)(ft)     (C3)

Note: Follow through from their part (b).

[3 marks]

Find the length of EB.

Write down the angle of elevation of B from E.

Find the vertical height of B above the ground.

Units are required in parts (a) and (c).

\(\frac{{{\text{EB}}}}{{\sin 53{\rm{^\circ }}}} = \frac{{1.2}}{{\sin 7{\rm{^\circ }}}}\)     (M1)(A1)

Note:     Award (M1) for substitution into sine formula, (A1) for correct substitution.

\(({\text{EB}} = ){\text{ }}7.86{\text{ m}}\)\(\,\,\,\)OR\(\,\,\,\)\(786{\text{ cm }}(7.86385 \ldots {\text{ m}}\)\(\,\,\,\)OR\(\,\,\,\)\(786.385 \ldots {\text{ cm}})\)     (A1)     (C3)

[3 marks]

34°     (A1)     (C1)

[1 mark]

Units are required in parts (a) and (c).

\(\sin 34^\circ  = \frac{{{\text{height}}}}{{7.86385 \ldots }}\)     (M1)

Note:     Award (M1) for correct substitution into a trigonometric ratio.

\(({\text{height}} = ){\text{ }}4.40{\text{ m}}\)\(\,\,\,\)OR\(\,\,\,\)\(440{\text{ cm }}(4.39741 \ldots {\text{ m}}\)\(\,\,\,\)OR\(\,\,\,\)\(439.741 \ldots {\text{ cm}})\)     (A1)(ft)     (C2)

Note:     Accept “BT” used for height. Follow through from parts (a) and (b). Use of 7.86 gives an answer of 4.39525….

[2 marks]

Find the value of \(a\).

Find the value of \(x\) that makes the volume a maximum.

\(72 = 12x + 4h\;\;\;\)(or equivalent)     (M1)

Note: Award (M1) for a correct equation obtained from the total length of the edges.

\(V = 2{x^2}(18 - 3x)\)     (A1)

\((a = ){\text{ }}36\)     (A1)     (C3)

\(\frac{{{\text{d}}V}}{{{\text{d}}x}} = 72x - 18{x^2}\)     (A1)

\(72x - 18{x^2} = 0\;\;\;\)OR\(\;\;\;\frac{{{\text{d}}V}}{{{\text{d}}x}} = 0\)     (M1)

Notes: Award (A1) for  \( - 18{x^2}\)  seen. Award (M1) for equating derivative to zero.

\((x = ){\text{ 4}}\)     (A1)(ft)     (C3)

Note: Follow through from part (a).

OR

Sketch of \(V\) with visible maximum     (M1)

Sketch with \(x \geqslant 0,{\text{ }}V \geqslant 0\) and indication of maximum (e.g. coordinates)     (A1)(ft)

\((x = ){\text{ 4}}\)     (A1)(ft)     (C3)

Notes: Follow through from part (a).

Award (M1)(A1)(A0) for \((4,{\text{ }}192)\).

Award (C3) for \(x = 4,{\text{ }}y = 192\).

The model in this question seemed to be too difficult for the vast majority of the candidates, and therefore was a strong discriminator between grade 6 and grade 7 candidates. An attempt to find an equation for the volume of the cube often started with V = x x 2x x h . Many struggled to translate the total length of the edges into a correct equation, and consequently were unable to substitute h. Some tried to write x in terms of h and got lost, others tried to work backwards from the expression given in the question.

As very few found a value for a, often part (b) was not attempted. When a derivative was calculated this was usually done correctly.

Calculate the size of angle ACB.

Nadia makes an accurate drawing of triangle ABC. She measures angle BAC and finds it to be 29°.

Calculate the percentage error in Nadia’s measurement of angle BAC.

\(\frac{{\sin {\text{A}}{\operatorname{\hat B}}{\text{C}}}}{{13.4}} = \frac{{\sin 30^\circ }}{{6.7}}\)     (M1)(A1)

Note: Award (M1) for correct substituted formula, (A1) for correct substitution.

\({\text{A}}{\operatorname{\hat B}}{\text{C}}\) = 90°     (A1)

\({\text{A}}{\operatorname{\hat C}}{\text{B}}\) = 60°     (A1)(ft)     (C4)

Note: Radians give no solution, award maximum (M1)(A1)(A0).

[4 marks]

\(\frac{{29 - 30}}{{30}} \times 100\)     (M1)

Note: Award (M1) for correct substitution into correct formula.

% error = −33.3 %     (A1)     (C2)

Notes: Percentage symbol not required. Accept positive answer.

[2 marks]

Use of cosine rule was common. The assumption of a right angle in the given diagram was minimal.

The incorrect denominator was often seen in the error formula.

Find \(f ′(x)\).

Let L be the line with equation y = 3x + 2.

Write down the gradient of a line parallel to L.

Let L be the line with equation y = 3x + 2.

Let P be a point on the curve of f. At P, the tangent to the curve is parallel to L. Find the coordinates of P.

\(2x\)     (A1)     (C1)

[1 mark]

3     (A1)     (C1)

[1 mark]

\(2x = 3\)     (M1)

Note: (M1) for equating their (a) to their (b).

\(x =1.5\)     (A1)(ft)

\(y = (1.5)^2 - 4\)     (M1)

Note: (M1) for substituting their x in f (x).

(1.5, −1.75)   (accept x = 1.5, y = −1.75)     (A1)(ft)     (C4)

Note: Missing coordinate brackets receive (A0) if this is the first time it occurs.

[4 marks]

This question was generally answered well in parts (a) and (b).

This question was generally answered well in parts (a) and (b).

This part proved to be difficult as candidates did not realise that to find the value of the x coordinate they needed to equate their answers to the first two parts. They did not understand that the first derivative is the gradient of the function. Some found the value of x, but did not substitute it back into the function to find the value of y.

Find the value of \(m\).

State what \(m\) represents in this context.

Find the value of \(n\).

State what \(n\) represents in this context.

\(\frac{{600 - 150}}{{6 - 1}}\)     (M1)

OR

\(600 = 150 + (6 - 1)m\)     (M1)

Note: Award (M1) for correct substitution into gradient formula or arithmetic sequence formula.

\( = 90\)     (A1)     (C2)

the annual rate of growth of the number of apartments     (A1)     (C1)

Note: Do not accept common difference.

\(150 = 90 \times (1) + n\)     (M1)

Note: Award (M1) for correct substitution of their gradient and one of the given points into the equation of a straight line.

\(n = 60\)     (A1)(ft)     (C2)

Note: Follow through from part (a).

the initial number of apartments     (A1)     (C1)

Note: Do not accept “first number in the sequence”.

Write down the length of \({\text{AO}}\).

Find the size of the angle that the sloping edge \({\text{VA}}\) makes with the base of the pyramid.

Hence, or otherwise, find the area of the triangle \({\text{CAV}}\).

\({\text{AO}} = 4{\text{ (cm)}}\)     (A1)     (C1)

\(\cos {\rm{O\hat AV}} = \frac{4}{{10}}\)     (M1)

Note: Award (M1) for their correct trigonometric ratio.

OR

\(\cos {\rm{O\hat AV}} = \frac{{{{10}^2} + {8^2} - {{10}^2}}}{{2 \times 10 \times 8}}\;\;\;\)OR\(\;\;\;\frac{{{{10}^2} + {4^2} - {{(9.16515 \ldots )}^2}}}{{2 \times 10 \times 4}}\)     (M1)

Note: Award (M1) for correct substitution into the cosine rule formula.

\({\rm{O\hat AV}} = 66.4^\circ \;\;\;(66.4218 \ldots )\)     (A1)(ft)     (C2)

Notes: Follow through from their answer to part (a).

\({\text{area}} = \frac{{8 \times 10 \times \sin (66.4218 \ldots ^\circ)}}{2}\;\;\;\)OR\(\;\;\;\frac{1}{2} \times 8 \times \sqrt {{{10}^2} - {4^{\text{2}}}} \)

OR\(\;\;\;\frac{1}{2} \times 10 \times 10 \times \sin (47.1563 \ldots ^\circ )\)     (M1)(A1)(ft)

Notes: Award (M1) for substitution into the area formula, (A1)(ft) for correct substitutions. Follow through from their answer to part (b) and/or part (a).

\({\text{area}} = 36.7{\text{ c}}{{\text{m}}^2}\;\;\;(36.6606 \ldots {\text{ c}}{{\text{m}}^2})\)     (A1)(ft)     (C3)

Notes: Accept an answer of \(8\sqrt {21} {\text{ c}}{{\text{m}}^2}\) which is the exact answer.

Write down an expression for the area, \(A\), in \({{\text{m}}^2}\), of the carpet.

The area of the carpet is \({\text{10 }}{{\text{m}}^2}\).

Calculate the value of \(x\).

The area of the carpet is \({\text{10 }}{{\text{m}}^2}\).

Hence, write down the value of the length and of the width of the carpet, in metres.

\(2x(x - 4)\)   or   \(2{x^2} - 8x\)     (A1)     (C1)

Note: Award (A0) for \(x - 4 \times 2x\).

[1 mark]

\(2x(x - 4) = 10\)     (M1)

Note: Award (M1) for equating their answer in part (a) to \(10\).

\({x^2} - 4x - 5 = 0\)     (M1)

OR

Sketch of \(y = 2{x^2} - 8x\) and \(y = 10\)     (M1)

OR

Using GDC solver \(x = 5\) and \(x =  - 1\)     (M1)

OR

\(2(x + 1)(x - 5)\)     (M1)

\(x = 5{\text{ (m)}}\)     (A1)(ft)     (C3)

Notes: Follow through from their answer to part (a).

     Award at most (M1)(M1)(A0) if both \(5\) and \(-1\) are given as final answer.

     Final (A1)(ft) is awarded for choosing only the positive solution(s).

[3 marks]

\(2 \times 5 = 10{\text{ (m)}}\)     (A1)(ft)

\(5 - 4 = 1{\text{ (m)}}\)     (A1)(ft)     (C2)

Note: Follow through from their answer to part (b).

     Do not accept negative answers.

[2 marks]

Calculate the total volume of all \(75\) metal cannon balls excavated.

The cannon balls are to be melted down to form a sculpture in the shape of a cone. The base radius of the cone is \(20{\text{ cm}}\).

Calculate the height of the cone, assuming that no metal is wasted.

\(75 \times \frac{4}{3}\pi  \times {5^3}\)     (M1)(M1)

Notes: Award (M1) for correctly substituted formula of a sphere. Award (M1) for multiplying their volume by \(75\). If \(r = 10\) is used, award (M0)(M1)(A1)(ft) for the answer 314000 cm³ .

\(39300{\text{ c}}{{\text{m}}^3}\) .     (A1)     (C3)

\(\frac{1}{3}\pi  \times {20^2} \times h = 39300\)     (M1)(M1)

Notes: Award (M1) for correctly substituted formula of a cone. Award (M1) for equating their volume to their answer to part (a).

\(h = 93.8{\text{ cm}}\)     (A1)(ft)     (C3)

Notes: Accept the exact value of \(93.75\) . Follow through from their part (a).

[3 marks]

As well as some candidates reading the diameter given as the radius, there was much confusion between the area and volume of a sphere and, although there was some recovery when multiplying by \(75\), two of the three marks were invariably lost. Recovery was possible in part (b) and many successful attempts were seen to calculate the height of the cone.

As well as some candidates reading the diameter given as the radius, there was much confusion between the area and volume of a sphere and, although there was some recovery when multiplying by \(75\), two of the three marks were invariably lost. Recovery was possible in part (b) and many successful attempts were seen to calculate the height of the cone.

Calculate the volume of the balloon.

Calculate the radius of the balloon following this increase.

Units are required in parts (a) and (b).

\(\frac{4}{3}\pi  \times {6^3}\)    (M1)

Note:     Award (M1) for correct substitution into volume of sphere formula.

\( = 905{\text{ c}}{{\text{m}}^3}{\text{ }}(288\pi {\text{ c}}{{\text{m}}^3},{\text{ }}904.778 \ldots {\text{ c}}{{\text{m}}^3})\)    (A1)     (C2)

Note:     Answers derived from the use of approximations of \(\pi \) (3.14; 22/7) are awarded (A0).

[2 marks]

Units are required in parts (a) and (b).

\(\frac{{140}}{{100}} \times 904.778 \ldots  = \frac{4}{3}\pi {r^3}\) OR \(\frac{{140}}{{100}} \times 288\pi  = \frac{4}{3}\pi {r^3}\) OR \(1266.69 \ldots  = \frac{4}{3}\pi {r^3}\)     (M1)(M1)

Note:     Award (M1) for multiplying their part (a) by 1.4 or equivalent, (M1) for equating to the volume of a sphere formula.

\({r^3} = \frac{{3 \times 1266.69 \ldots }}{{4\pi }}\) OR \(r = \sqrt[3]{{\frac{{3 \times 1266.69 \ldots }}{{4\pi }}}}\) OR \(r = \sqrt[3]{{(1.4) \times {6^3}}}\) OR \({r^3} = 302.4\)     (M1)

Note:     Award (M1) for isolating \(r\).

\((r = ){\text{ }}6.71{\text{ cm }}(6.71213 \ldots )\)     (A1)(ft)     (C4)

Note:     Follow through from part (a).

[4 marks]

Calculate the horizontal distance, \({\text{GH}}\), of the plane from Günter. Give your answer to the nearest metre.

The plane took off from a point \({\text{T}}\), which is \(250\) metres from where Günter is standing, as shown in the following diagram.

\(\frac{{350}}{{\tan 20^\circ }}\)     (M1)

\( = {\text{961.617}} \ldots \)     (A1)

\( = 962{\text{ (m)}}\)     (A1)(ft)     (C3)

Notes: Award (M1) for correct substitution into correct formula, (A1) for correct answer, (A1)(ft) for correct rounding to the nearest metre.

     Award (M0)(A0)(A0) for \(961\) without working.

[3 marks]

\({\text{961.617}} \ldots  - 250 = {\text{711.617}} \ldots \)     (A1)(ft)

\({\tan ^{ - 1}}\left( {\frac{{350}}{{{\text{711.617}} \ldots }}} \right)\)     (M1)

\( = {\text{26.2}}^\circ {\text{ (26.1896}} \ldots {\text{)}}\)     (A1)(ft)     (C3)

Notes: Accept \(26.1774…\) from use of 3 sf answer \(962\) from part (a). Follow through from their answer to part (a).

     Accept alternative methods.

[3 marks] 

Calculate the length, \(l\), of the slant height of the cone.

Calculate the area that is painted red.

\(\sqrt {{5^2} + {{12}^2}} \)     (M1)

Note: Award (M1) for correct substitution in Pythagoras Formula.

=\(13{\text{ (cm)}}\)     (A1)     (C2)

\({\text{Area}} = 2\pi {(5)^2} + \pi (5)(13)\)     (M1)(M1)(M1)

Notes: Award (M1) for surface area of hemisphere, (M1) for surface of cone, (M1) for addition of two surface areas. Follow through from their answer to part (a).

\( = 361{\text{ c}}{{\text{m}}^2}\) (\(361.283 \ldots \))     (A1)(ft)     (C4)

Note: The answer is \( 361{\text{ c}}{{\text{m}}^2}\) , the units are required.

Find the coordinates of P and of Q.

A second line with equation x − y = 3 intersects the line PQ at the point A. Find the coordinates of A.

\(0 + 2y = 12\) or \(x + 2(0) = 12\)     (M1)

P(0, 6)     (accept \(x = 0\), \(y = 6\))     (A1)

Q(12, 0)     (accept \(x = 12\), \(y = 0\))     (A1)     (C3)

Notes: Award (M1) for setting either value to zero.

Missing coordinate brackets receive (A0) the first time this occurs.

Award (A0)(A1)(ft) for P(0, 12) and Q(6, 0).

[3 marks]

\(x + 2(x - 3) = 12\)     (M1)

(6, 3)     (accept \(x = 6\), \(y = 3\))     (A1)(A1)     (C3)

Note: (A1) for each correct coordinate.

Missing coordinate brackets receive (A0)(A1) if this is the first time it occurs.

[3 marks]

Most candidates could find the x and y intercepts but many wrote the coordinates the wrong way around. A number of candidates did not label their coordinates as P and Q or did not include parentheses.

In this part many had trouble recognising the need to solve the two simultaneous equations.

Calculate the values of s and t.

Find the equation of the straight line perpendicular to AB, passing through the point M.

\(s = 6\)     (A1)

\(t = - 2\)     (A1)     (C2)

[2 marks]

\({\text{gradient of AB}} = \frac{{ - 2 - 8}}{{ - 2 - 6}} = \frac{{ - 10}}{{ - 8}} = \frac{5}{4}\)     (A1)(ft)

(A1) for gradient of AM or BM \( = \frac{5}{4}\)

\({\text{Perpendicular gradient}} = - \frac{4}{5}\)     (A1)(ft)

Equation of perpendicular bisector is

\(y = - \frac{4}{5}x + c\)

\(3 = - \frac{4}{5}(2) + c\)     (M1)

\(c = 4.6\)

\(y = -0.8x + 4.6\)

or \(5y = -4x + 23\)     (A1)(ft)     (C4)

[4 marks]

(a) It was surprising how many errors were made in finding the values for s and t

(b) The candidates had difficulty in finding the equation of a straight line. They could find the gradient of the line AB and a number could give the gradient of the perpendicular line but most did not substitute the correct midpoint to find the equation of the line.

Write down the coordinates of M.

Calculate the gradient of the line AB.