triangles bfm and baf proving to be the most popular, but ... key...triangles bfm and baf proving to...

New test - October 20, 2015 [197 marks]

[2 marks]1a.

The diagram shows a right triangular prism, ABCDEF, in which the face ABCD is a square.

AF = 8 cm, BF = 9.5 cm, and angle BAF is 90°.

Calculate the length of AB .

Markscheme9.5 = 8 + AB (M1)

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

AB = 5.12 (cm) (5.12347…) (A1) (C2)

[2 marks]

Examiners reportThis seemed to be a good discriminatory question enabling the majority of candidates to at least score well on part (a). Challengesarose for candidates who were then required to see the problem in three dimensions for the remainder of the question. Indeed, asignificant number of candidates correctly identified the required lengths for part (b) and, provided they used Pythagoras correctly,were able to pick up the marks in this part of the question. However, in part (c), invariably the wrong triangle was chosen withtriangles BFM and BAF proving to be the most popular, but incorrect triangles, chosen.

2 2 2

1b. [2 marks]M is the midpoint of EF .

Calculate the length of BM .

Markscheme (M1)

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

= 9.84 (cm) (9.83933…) (A1)(ft) (C2)

Notes: Accept alternative methods. Follow through from their answer to part

[2 marks]

BM = +9.52 ( )5.12347...2

2− −−−−−−−−−−−−−√


1c. [2 marks]M is the midpoint of EF .

Find the size of the angle between BM and the face ADEF .

Markschemesin (M1)

Note: Award (M1) for a correctly substituted trigonometrical equation using .

= 31.4 (31.3801...) (A1)(ft) (C2)

Notes: If radians used, the answer will be 0.5476… award (M1)(A0)(ft). Degree symbol ° not required. Follow through from theiranswers to part (a) and to part (b).

[2 marks]


A B = M 5.12347...9.83933...

A BM

[2 marks]2a.

The diagram shows points A(2, 8), B(14, 4) and C(4, 2). M is the midpoint of AC.

Write down the coordinates of M.

Markscheme (M1)

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

(A1) (C2)

Note: Brackets must be present for final (A1) to be awarded.

Note: Accept , .

[2 marks]

Examiners reportOverall, there was a very good response to parts (a) and (b) with only a few candidates giving an incorrect expression for the gradientin part (b).

( , )2+42

8+22

= (3, 5)

x = 3 y = 5

[2 marks]2b. Calculate the gradient of the line AB.

Markscheme (M1)

Note: Award (M1) for correctly substituted formula.

(A1) (C2)

[2 marks]

Examiners reportOverall, there was a very good response to parts (a) and (b) with only a few candidates giving an incorrect expression for the gradientin part (b). Occasionally, the final mark in part (b) was lost because the negative sign was dropped by some candidates.

8−42−14

= − 13

( ,−0.333)−412

(−0.333333…)

[2 marks]2c. Find the equation of the line parallel to AB that passes through M.

Markscheme (M1)(A1)(ft)

OR

(M1)

(A1)(ft) (C2)

Notes: Award (M1) for substitution of their gradient into equation of line with their values from (a) correctly substituted.

Accept correct equivalent forms of the equation of the line. Follow through from their parts (a) and (b).

[2 marks]

Examiners reportMany able candidates recognized they needed to do something with the equation in part (c). Weaker candidates clearlyshowed a lack of understanding of an equation of a line and either simply gave a numerical answer for this part of the question ortried to use the coordinates of into what they believed was the required equation of the straight line. A popular incorrect answerseen was .

(y − 5) = − (x − 3)13

5 = − (3) + c13

y = − x + 613

y = mx + c

M

y = 3x + 5

3. [6 marks]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 thelines drawn above.

Markscheme

(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]

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

[2 marks]4a.

The points A (−4, 1), B (0, 9) and C (4, 2) are plotted on the diagram below. The diagram also shows the lines AB, L and L .

Find the gradient of AB.

Markscheme (M1)

= 2 (A1)(G2)

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

[2 marks]

Examiners reportThis question was in general well answered. In part (a) the gradient of the line AB was correctly found although some candidates didnot substitute well in the gradient formula and found answers as or –2. Also some students read B as (0, 8) instead of (0, 9). In part(b) many students again did not make good use of time as they found the equation of the line instead of just extending it to find the y -intercept. The equation of L in (c) was correctly found in the form y = mx + c but very few students were able to rearrange theequation in the form ax + by + d = 0 where a, b, d . In (d) many candidates found the coordinates of point D by solvingsimultaneous equations which led again to a waste of time. The last two parts of this question were well done by those students thatattempted them.

1 2

9−10−(−4)

12

2∈ Z

4b. [1 mark]L passes through C and is parallel to AB.

Write down the y-intercept of L .

Markscheme–6 (A1)

Note: Accept (0, –6) .

[1 mark]

1

1


12

2∈ Z

4c. [3 marks]L passes through A and is perpendicular to AB.

Write down the equation of L . Give your answer in the form ax + by + d = 0 where a, b and d .

Markscheme (or equivalent) (A1)(ft)(A1)

Notes: Award (A1)(ft) for gradient, (A1) for correct y-intercept. Follow through from their gradient in (a).

x + 2y + 2 = 0 (A1)(ft)

Notes: Award (A1)(ft) from their gradient and their y-intercept. Accept any multiple of this equation with integer coefficients.

OR

(or equivalent) (A1)(ft)(A1)

Note: Award (A1)(ft) for gradient, (A1) for any point on the line correctly substituted in equation.

x + 2y + 2 = 0 (A1)(ft)

Notes: Award (A1)(ft) from their equation. Accept any multiple of this equation with integer coefficients.

[3 marks]


2

2 ∈ Z

y = − x − 112

y − 1 = − (x + 4)12

12

2∈ Z

[1 mark]4d. Write down the coordinates of the point D, the intersection of L and L .

MarkschemeD(2, –2) or x = 2, y = –2 (A1)

Note: Award (A0) if brackets not present.

[1 mark]

1 2


12

2∈ Z

4e. [2 marks]There is a point R on L such that ABRD is a rectangle.

Write down the coordinates of R.

MarkschemeR(6, 6) or x = 6, y = 6 (A1)(A1)

Note: Award at most (A0)(A1)(ft) if brackets not present and absence of brackets has not already been penalised in part (d).

[2 marks]


1

12

2∈ Z

4f. [4 marks]The distance between A and D is .

(i) Find the distance between D and R .

(ii) Find the area of the triangle BDR .

Markscheme(i) (M1)

(8.94) (A1)(ft)(G2)

Note: Award (M1) for correct substitution into the distance formula. Follow through from their D and R.

(ii) (M1)

= 30 (30.0) (A1)(ft)(G2)

Note: Award (M1) for correct substitution in the area of triangle formula. Follow through from their answer to part (f) (i).

[4 marks]

45−−√

DR = +82 42− −−−−−√

DR = 80−−√

Area = ×80√ 45√2


12

2∈ Z

5a. [4 marks]

The coordinates of point A are (−4, p) and the coordinates of point B are (2, −3) .

The mid-point of the line segment AB, has coordinates (q, 1) .

Find the value of

(i) q ;

(ii) p .

Markscheme(i) (M1)

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

q = –1 (A1)

(ii) (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]

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

= q−4+22

= 1p+(−3)2

= 1(p+3)2

2 2

[2 marks]5b. Calculate the distance AB.

Markscheme (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]

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

AB = +(2 + 4) 2 (−3 − 5) 2− −−−−−−−−−−−−−−−√

= 1(p+3)2

2 2

[1 mark]6a.

Line L is given by the equation 3y + 2x = 9 and point P has coordinates (6 , –5).

Explain why point P is not on the line L.

Markscheme3 × (–5) + 2 × 6 ≠ 9 (A1) (C1)

Note: Also accept 3 × (–5) + 2x = 9 gives x =12 ≠ 6 or 3y + 2 × (6) = 9 gives y = –1 ≠ –5.

[1 mark]

Examiners reportIn part (a), the word 'explain' required more than simply stating that 'I put the coordinates into my GDC and it did not work'. Awritten statement, showing the substitution of one or both of the coordinates leading to an inequality was required for this first mark. Itwas pleasing to see that many scripts showed correct methodology for calculating the gradient of L and the gradient of a lineperpendicular to L. The correct equation of the line perpendicular to L passing through P proved to be more elusive as poor arithmeticspoilt what could have been excellent work. In particular leading to proved to be a popular but erroneouscalculation.

−5 = × 6 + c32

c = (±)4

[2 marks]6b. Find the gradient of line L.

Markscheme3y = –2x + 9 (M1)

Note: Award (M1) for 3y = –2x + 9 or or .

(A1) (C2)

[2 marks]

y = x + 3−23

y = (−2x+9)3

gradient = − (−0.667)(−0.666666...)23


−5 = × 6 + c32

c = (±)4

6c. [3 marks](i) Write down the gradient of a line perpendicular to line L.

(ii) Find the equation of the line perpendicular to L and passing through point P.

Markscheme(i) gradient of perpendicular line (A1)(ft)

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

(ii)

(M1)

Note: Award (M1) for substitution of their perpendicular gradient and the point (6, –5) into the equation of their line.

(A1)(ft)

Note: Follow through from their perpendicular gradient. Accept equivalent forms.

OR

(M1)(A1)(ft) (C3)

Notes: Award (M1) for substitution of their perpendicular gradient and the point (6, –5) into the equation of their line. Follow throughfrom their perpendicular gradient.

[3 marks]


= (1.5)32

y = x + c32

−5 = × 6 + c32

y = x − 1432

y + 5 = (x − 6)32

−5 = × 6 + c32

c = (±)4

[1 mark]7a. Write down the size of angle CBA.

Markscheme32° (A1) (C1)

[1 mark]

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

[1 mark]7b. Write down the size of angle CAB.


[1 mark]


[4 marks]7c. The area of triangle ABC is 360 cm . Calculate the length of side AC. Express your answer in millimetres.


Notes: Award (M1) for substitution into correct formula with 360 seen, (A1)(ft) for correct substitution, follow through from theiranswer 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 beawarded for correct conversion to mm.

[4 marks]


2

360 = × × sin12

x2 116∘

8a. [3 marks]

The Great Pyramid of Cheops in Egypt is a square based pyramid. The base of the pyramid is a square of side length 230.4 m and thevertical height is 146.5 m. The Great Pyramid is represented in the diagram below as ABCDV . The vertex V is directly above thecentre O of the base. M is the midpoint of BC.

(i) Write down the length of OM .

(ii) Find the length of VM .

Markscheme(i) 115.2 (m) (A1)

Note: Accept 115 (m)

(ii) (M1)

Note: Award (M1) for correct substitution.

186 (m) (186.368…) (A1)(ft)(G2)

Note: Follow through from part (a)(i).

[3 marks]

Examiners report(a) This part was very well done on the whole.

( + )146.52 115.22− −−−−−−−−−−−−−√

[2 marks]8b. Find the area of triangle VBC .

Markscheme (M1)

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

21500 m (21469.6…m ) (A1)(ft)(G2)

Notes: The final answer is 21500 m ; units are required. Accept 21400 m for use of 186 m and/or 115 m.

[2 marks]

× 230.4 × 186.368...12

2 2

2 2

Examiners report(b) Amazingly badly done. Many candidates used 146.4 for the height and others tried unsuccessfully to find slant heights and anglesto that they could use the area of a triangle formula .

ab sin C12

[2 marks]8c. Calculate the volume of the pyramid.

Markscheme (M1)

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

2590000 m (2592276.48 m ) (A1)(G2)

Note: The final answer is 2590000 m ; units are required but do not penalise missing or incorrect units if this has already beenpenalised in part (b).

[2 marks]

Examiners report(c) This was fairly well done.

× × 146.513

230.42

3 3

3

[2 marks]8d. Show that the angle between the line VM and the base of the pyramid is 52 ° correct to 2 significant figures.

Markscheme (M1)

Notes: Award (M1) for correct substituted trig ratio. Accept alternate correct trig ratios.

= 51.8203...= 52° (A1)(AG)

Notes: Both the unrounded answer and the final answer must be seen for the (A1) to be awarded. Accept 51.96° = 52°, 51.9° = 52°or 51.7° = 52°

Examiners report(d) Quite a few candidates managed to show this although they did not always put down the unrounded answer and so lost the lastmark. Some even tried to use 52° to verify its value.

.

( )tan−1 146.5115.2

8e. [1 mark]Ahmed is at point P , a distance x metres from M on horizontal ground, as shown in the following diagram. The size of angleVPM is 27° . Q is a point on MP .

Write down the size of angle VMP .

Markscheme128° (A1)

[1 mark]

Examiners report(e) Very well done on the whole – even if part (d) was wrong.

8f. [4 marks]Ahmed is at point P , a distance x metres from M on horizontal ground, as shown in the following diagram. The size of angleVPM is 27° . Q is a point on MP .

Using your value of VM from part (a)(ii), find the value of x.

Markscheme (A1)(M1)(A1)(ft)

Notes: Award (A1)(ft) for their angle MVP seen, follow through from their part (e). Award (M1) for substitution into sine formula,(A1) for correct substitutions. Follow through from their VM and their angle VMP.

x = 173 (m) (173.490...) (A1)(ft)(G3)

Note: Accept 174 from use of 186.4.

[4 marks]

Examiners report(f) This was well done by those who attempted it. Not all candidates used VM to find x and so lost one mark. There were quite a fewdifferent methods of finding the answer.

=186.368sin 27

x

sin 25

8g. [4 marks]Ahmed is at point P , a distance x metres from M on horizontal ground, as shown in the following diagram. The size of angleVPM is 27° . Q is a point on MP .

Ahmed walks 50 m from P to Q.

Find the length of QV, the distance from Ahmed to the vertex of the pyramid.

MarkschemeVQ = (186.368...) + (123.490...) − 2 × (186.368...) × (123.490...) × cos128 (A1)(ft)(M1)(A1)(ft)

Notes: Award (A1)(ft) for 123.490...(123) seen, follow through from their x (PM) in part (f), (M1) for substitution into cosineformula, (A1)(ft) for correct substitutions. Follow through from their VM and their angle VMP.

OR

173.490... − 50 = 123.490... (123) (A1)(ft)

115.2 + 123.490... = 238.690... (A1)(ft)

(M1)

VQ = 280 (m) (280.062...) (A1)(ft)(G3)

Note: Accept 279 (m) from use of 3 significant figure answers.

[4 marks]

2 2 2

VQ = ( + )146.52 238.690...2− −−−−−−−−−−−−−−−−√

Examiners report(g) Again this was well done by those who attempted it. Again there were many different ways to reach the correct answer.

[3 marks]9a.

The diagram shows triangle ABC in which , , and .

Calculate the size of angle ADB.

Markscheme (M1)(A1)

Notes: Award (M1) for substituted cosine rule formula, (A1) for correct substitutions.

\(\angle {\text{ADB}} = 120\) (A1)(G2)

[3 marks]

Examiners reportThe vast majority of candidates scored very well on this question. Those who did not attempted it using the trigonometry associatedwith right angled triangles. There were few problems with the use of radians and part (d), which was expected to prove challenging,was successfully overcome by more than half of the candidature. Problems arose mainly because of a lack of clarity in identifying thecorrect triangle.

AB = 28 cm BC = 13 cm BD = 12 cm AD = 20 cm

cosADB = + −122 202 282

2(12)(20)

[3 marks]9b. Find the area of triangle ADB.


Notes: Award (M1) for substituted area formula, (A1)(ft) for their correct substitutions.

( ) (A1)(ft)(G2)

Note: The final answer is , the units are required. Accept .

[3 marks]


Area = (12)(20) sin 120∘

2

= 104 cm2 103.923… cm2

104 cm2 100 cm2

[4 marks]9c. Calculate the size of angle BCD.

Markscheme (A1)(ft)(M1)(A1)

Note: Award (A1)(ft) for their 60 seen, (M1) for substituted sine rule formula, (A1) for correct substitutions.

( ) (A1)(G3)

Note: Accept , do not accept or .

[4 marks]


=sin BCD12

sin 60∘

13

BCD = 53.1∘ 53.0736…

53 50 53.0

[4 marks]9d. Show that the triangle ABC is not right angled.

=sin BAC13

sin 53.1∘

28

=sin BAD12

sin 120∘

28

BAC = BAD = 21.8∘ 21.7867…

22 20 21.7

− ( + ) ≠180∘ 53.1∘ 21.8∘ 90∘

=CDsin 66.9∘

13sin 60∘

CD = 13.8 (13.8075…)

+ ≠133 282 33.82


[2 marks]10a.

A room is in the shape of a cuboid. Its floor measures m by m and its height is m.

Calculate the length of AC.

Markscheme (M1)

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

(A1) (C2)

[2 marks]

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

7.2 9.6 3.5

A = +C2 7.22 9.62

AC = 12 m

[2 marks]10b. Calculate the length of AG.

Markscheme (M1)

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

(A1)(ft) (C2)

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

[2 marks]


A = +G2 122 3.52

AG = 12.5 m

[2 marks]10c. Calculate the angle that AG makes with the floor.

Markscheme or or (M1)

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

(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]


tanθ = 3.512

sin θ = 3.512.5

cosθ = 1212.5

θ = 16.3∘

11a. [3 marks]

The diagram shows the straight lines and . The equation of is .

Find(i) the gradient of ;

(ii) the equation of .

Markscheme(i) (M1)

(A1) (C2)

(ii) (A1)(ft) (C1)

Notes: Follow through from their gradient in part (a)(i). Accept equivalent forms for the equation of a line.

[3 marks]

Examiners reportIn this question, many candidates did not use the and intercepts to find the slope and attempted to read ordered pairs from thegraph.

L1 L2 L2 y = x

L1

L1

0−26−0

= − (− , − 0.333)13

26

y = − x + 213

x y

[2 marks]11b. Find the area of the shaded triangle.

Markscheme (A1)(M1)

Note: Award (A1) for seen, (M1) for use of triangle formula with seen.

(A1) (C3)

[2 marks]

Examiners reportPart b proved difficult for many candidates, often using trigonometry rather than the more straight forward area of the triangle.

area = 6×1.52

1.5 6

= 4.5

[2 marks]12a.

The equation of the line is . The line is perpendicular to .

Calculate the gradient of .

Markscheme (M1)

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

(A1) (C2)

[2 marks]

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

R1 2x + y − 8 = 0 R2 R1

R2

y = −2x + 8

−2

m(perp) = 12

12b. [4 marks]The point of intersection of and is .Find

(i) the value of ;(ii) the equation of .

R1 R2 (4, k)

k

R2

Markscheme(i) (M1)

Note: Award (M1) for evidence of substituting into .

(A1) (C2)

(ii) (can be implied) (M1)

Note: Award (M1) for substitution of into equation of the line.

(A1)(ft) (C2)

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

OR

(M1)

Note: Award (M1) for substitution of into equation of the line.

(A1)(ft) (C2)

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

[4 marks]

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

2(4) + k − 8 = 0

x = 4 R1

k = 0

y = x + c12

12

0 = (4) + c12

y = x − 212

y − = (x − )y112

x1

12

y = (x − 4)12

[2 marks]13a.

The straight line passes through the points and .

Calculate the gradient of .

Markscheme (M1)

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

(A1) (C2)

[2 marks]

Examiners reportGenerally, a well answered question with many candidates achieving full marks. Indeed, marks which tended to be lost were as aresult of premature rounding rather than method. On a number of scripts, part (a) produced a rather curious wrong answer of following a correct gradient expression. It would seem that this was as a result of typing into the calculator .

L A(−1, 4) B(5, 8)

L

8−45−(−1)

( , 0.667)23

46

8.28 − 4 ÷ 5 + 1

[2 marks]13b. Find the equation of .L

Markscheme (A1)(ft)

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

(A1)(ft) (C2)

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

OR

(A1)(ft)

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

OR (A1)(ft) (C2)

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

[2 marks]

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

y = x + c23

y = x +23

143

y − = (x − )y123

x1

y − 4 = (x + 1)23

y − 8 = (x − 5)23

[2 marks]13c. The line also passes through the point . Find the value of .

Markscheme OR OR (M1)

Note: Award (M1) for substitution of into their equation.

( ) (A1)(ft) (C2)


[2 marks]

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

L P(8, y) y

y = × 8 +23

143

y − 4 = (8 + 1)23

y − 8 = (8 − 5)23

x = 8

y = 10 10.0

14a. [1 mark]

The diagram shows a triangle ABC in which AC = 17 cm. M is the midpoint of AC.Triangle ABM is equilateral.

Write down the size of angle MCB.


[1 mark]

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

[1 mark]14b. Write down the length of BM in cm.

Markscheme8.5 (cm) (A1)

[1 mark]


14c. [1 mark]Write down the size of angle BMC.

Markscheme120° (A1)

[1 mark]


[3 marks]14d. Calculate the length of BC in cm.


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

(A1)(ft)

[3 marks]


=BCsin 120

8.5sin 30

BC = 14.7( )17 3√2

[1 mark]15a.

The straight line, L , has equation .

Write down the y intercept of L .

Markscheme–2 (A1) (C1)

Note: Accept (0, –2).

[1 mark]

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

1 y = − x − 212

1

[1 mark]15b. Write down the gradient of L .

Markscheme (A1) (C1)

[1 mark]


1

− 12

15c. [1 mark]The line L is perpendicular to L and passes through the point (3, 7).

Write down the gradient of the line L .

2 1

2

Markscheme2 (A1)(ft) (C1)


[1 mark]


15d. [3 marks]The line L is perpendicular to L and passes through the point (3, 7).

Find the equation of L . Give your answer in the form ax + by + d = 0 where .

Markschemey = 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]


2 1

2 a, b, d ∈ Z

[2 marks]16a.

A rectangular cuboid has the following dimensions.

Length 0.80 metres (AD)

Width 0.50 metres (DG)

Height 1.80 metres (DC)

Calculate the length of AG.

Markscheme (M1)

AG = 0.943 m (A1) (C2)

[2 marks]

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

AG = +0.82 0.52− −−−−−−−−√

[2 marks]16b. Calculate the length of AF.

Markscheme (M1)

= 2.03 m (A1)(ft) (C2)


[2 marks]


AF = A +G2 1.802− −−−−−−−−−√

[2 marks]16c. Find the size of the angle between AF and AG.

Markscheme (M1)

(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]


cosG F =A 0.943(39…)

2.03(22…)

G F =A 62.3∘

[1 mark]17a.

The diagram shows triangle ABC. Point C has coordinates (4, 7) and the equation of the line AB is x + 2y = 8.

Find the coordinates of A.

MarkschemeA(0, 4) Accept x = 0, y = 4 (A1)

[1 mark]

Examiners reportThis question had many correct solutions, but a large number of candidates were unable to follow the logical flow of the question tothe end and many gave up. It should be pointed out to future candidates that parts (e) and (f) could be attempted independently fromthe rest and that care must be taken not to abandon hope too early in the longer questions of paper 2.

17b. [1 mark]Find the coordinates of B.

MarkschemeB(8, 0) Accept x = 8, y = 0 (A1)(ft)

Note: Award (A0) if coordinates are reversed in (i) and (A1)(ft) in (ii).

[1 mark]


[2 marks]17c. Show that the distance between A and B is 8.94 correct to 3 significant figures.

Markscheme (M1)

AB = 8.944 (A1)

= 8.94 (AG)

[2 marks]


AB = =+82 42− −−−−−√ 80−−√

17d. [3 marks]N lies on the line AB. The line CN is perpendicular to the line AB.

Find the gradient of CN.

Markschemey = –0.5x + 4 (M1)Gradient AB = –0.5 (A1)

Note: Award (A2) if –0.5 seen.

OR

Gradient (M1)

(A1)

Note: Award (M1) for correct substitution in the gradient formula. Follow through from their answers to part (a).

Gradient CN = 2 (A1)(ft)(G2)

Note: Special case: Follow through for gradient CN from their gradient AB.

[3 marks]

AB = (0−4)

(8−0)

= − 12


17e. [2 marks]N lies on the line AB. The line CN is perpendicular to the line AB.

Find the equation of CN.

MarkschemeCN: y = 2x + c

7 = 2(4) + c (M1)

Note: Award (M1)for correct substitution in equation of a line.

y = 2x – 1 (A1)(ft)(G2)

Note: Accept alternative forms for the equation of a line including y – 7 = 2(x – 4) . Follow through from their gradient in (i).

Note: If c = –1 seen but final answer is not given, award (A1)(d).

[2 marks]


17f. [3 marks]N lies on the line AB. The line CN is perpendicular to the line AB.

Calculate the coordinates of N.

Markschemex + 2(2x – 1) = 8 or equivalent (M1)

N(2, 3) (x = 2, y = 3) (A1)(A1)(ft)(G3)

Note: Award (M1) for attempt to solve simultaneous equations or a sketch of the two lines with an indication of the point ofintersection.

[3 marks]


17g. [3 marks]It is known that AC = 5 and BC = 8.06.

Calculate the size of angle ACB.

MarkschemeCosine rule: (M1)(A1)

Note: Award (M1) for use of cosine rule with numbers from the problem substituted, (A1) for correct substitution.

(A1)(G2)

Note: If alternative right-angled trigonometry method used award (M1) for use of trig ratio in both triangles, (A1) for correctsubstitution of their values in each ratio, (A1) for answer.

Note: Accept 82.8° with use of 8.94.

[3 marks]


cos(A B) =C + −52 8.062 8.9442

2×5×8.06

A B =C 82.9∘

17h. [3 marks]It is known that AC = 5 and BC = 8.06.

Calculate the area of triangle ACB.

MarkschemeArea (M1)(A1)(ft)

Note: Award (M1) for substituted area formula, (A1) for correct substitution. Follow through from their angle in part (e).

OR

Area (M1)(M1)(ft)

Note: Award (M1) substituted area formula with their values, (M1) for substituted distance formula. Follow through fromcoordinates of N.

Area ACB = 20.0 (A1)(ft)(G2)

Note: Accept 20

[3 marks]


ACB = 5×8.06 sin(82.9)2

ACB = =AB×CN2

8.94× +(4−2) 2 (7−3) 2√2

[3 marks]18a.

In the diagram below A, B and C represent three villages and the line segments AB, BC and CA represent the roads joining them.The lengths of AC and CB are 10 km and 8 km respectively and the size of the angle between them is 150°.

Find the length of the road AB.

MarkschemeAB = 10 + 8 – 2 × 10 × 8 × cos150° (M1)(A1)

AB = 17.4 km (A1)(G2)

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

[3 marks]

Examiners reportThe weak students answered parts (a) and (b) using right-angled trigonometry. Different types of mistakes were seen in (a) whenapplying the cosine rule: some forgot to square root their answer and others calculated each part separately and then missed the 2minuses. Part (b) was better done than (a). Follow through was applied from (a) to (c). Part (d) was not well done. Most of thestudents lost one mark in this part question as they did not show the unrounded answer (2.0550...). Part (e) was fairly well done bythose who attempted it. In (f) there were very few correct answers. Students found it difficult to find the time when the average speedand distance were given.

2 2 2

[3 marks]18b. Find the size of the angle CAB.

Markscheme (M1)(A1)

(A1)(ft)(G2)

Notes: Award (M1) for substitution into correct formula, (A1) for correct substitution, (A1) for correct answer. Follow through fromtheir answer to part (a).

[3 marks]


=8

sin C BA

17.4sin 150∘

C B =A 13.3∘

18c. [1 mark]Village D is halfway between A and B. A new road perpendicular to AB and passing through D is built. Let T be the pointwhere this road cuts AC. This information is shown in the diagram below.

Write down the distance from A to D.

MarkschemeAD = 8.70 km (8.7 km) (A1)(ft)


[1 mark]


[2 marks]18d. Show that the distance from D to T is 2.06 km correct to three significant figures.

MarkschemeDT = tan (13.29...°) × 8.697... = 2.0550... (M1)(A1)

= 2.06 (AG)

Notes: Award (M1) for correct substitution in the correct formula, award (A1) for the unrounded answer seen. If 2.06 not seen awardat most (M1)(AO).

[2 marks]


18e. [3 marks]A bus starts and ends its journey at A taking the route AD to DT to TA.

Find the total distance for this journey.

Markscheme (A1)(M1)

= 19.7 km (A1)(ft)(G2)

Note: Award (A1) for AT, (M1) for adding the three sides of the triangle ADT, (A1)(ft) for answer. Follow through from theiranswer to part (c).

[3 marks]


+ 8.70 + 2.06+8.702 2.062− −−−−−−−−−√

18f. [4 marks]The average speed of the bus while it is moving on the road is 70 km h . The bus stops for 5 minutes at each of D and T .

Estimate the time taken by the bus to complete its journey. Give your answer correct to the nearest minute.

Markscheme (M1)(M1)

= 26.9 (A1)(ft)

Note: Award (M1) for time on road in minutes, (M1) for adding 10, (A1)(ft) for unrounded answer. Follow through from theiranswer to (e).

= 27 (nearest minute) (A1)(ft)(G3)

Note: Award (A1)(ft) for their unrounded answer given to the nearest minute.

[4 marks]


–1

× 60 + 1019.770

[2 marks]19a.

A line joins the points A(2, 1) and B(4, 5).

Find the gradient of the line AB.

Markscheme (M1)

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

(A1) (C2)

[2 marks]

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

Gradient = (5−1)

(4−2)

= 2

19b. [1 mark]Let M be the midpoint of the line segment AB.

Write down the coordinates of M.

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

[1 mark]


19c. [3 marks]Let M be the midpoint of the line segment AB.

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

Markscheme (A1)(ft)

(M1)

(A1)(ft)

OR

(A1)(A1)(ft)

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

OR

(A1)(A1)(ft) (C3)

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

[3 marks]

Gradient of perpendicular = − 12

y = − x + c12

3 = − × 3 + c12

c = 4.5

y = −0.5x + 4.5

y − 3 = −0.5(x − 3)

2y + x = 9


[1 mark]20a.

The diagram below shows a square based right pyramid. ABCD is a square of side 10 cm. VX is the perpendicular height of 8 cm. Mis the midpoint of BC.

Write down the length of XM.

MarkschemeUP applies in this question

(UP) XM = 5 cm (A1)

[1 mark]

Examiners reportThis part proved accessible to the great majority of candidates. The common errors were (1) the inversion of the tangent ratio (2) theomission of the units and (3) the incorrect rounding of the answer; with 58° being all too commonly seen.

20b. [1 mark]

In a mountain region there appears to be a relationship between the number of trees growing in the region and the depth of snow inwinter. A set of 10 areas was chosen, and in each area the number of trees was counted and the depth of snow measured. The resultsare given in the table below.

Use your graphic display calculator to find the standard deviation of the number of trees.

Markscheme16.8 (G1)

[1 mark]

Examiners reportA straightforward question that saw many fine attempts. Given its nature – where much of the work was done on the GDC – it mustbe emphasised to candidates that incorrect entry of data into the calculator will result in considerable penalties; they must check theirdata entry most carefully.

The use of the inappropriate standard deviation was seen, but infrequently.

[2 marks]20c. Calculate the length of VM.


VM = 5 + 8 (M1)

Note: Award (M1) for correct use of Pythagoras Theorem.

(UP) VM = = 9.43 cm (A1)(ft)(G2)

[2 marks]


2 2 2

89−−√

[2 marks]20d. Calculate the angle between VM and ABCD.

Markscheme (M1)

Note: Other trigonometric ratios may be used.

(A1)(ft)(G2)

[2 marks]


tanVMX = 85

V X =M 58.0∘

[4 marks]20e.

A path goes around a forest so that it forms the three sides of a triangle. The lengths of two sides are 550 m and 290 m. These twosides meet at an angle of 115°. A diagram is shown below.

Calculate the length of the third side of the triangle. Give your answer correct to the nearest 10 m.


l = 290 + 550 − 2 × 290 × 550 × cos115° (M1)(A1)

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

l = 722 (A1)(G2)

(UP) = 720 m (A1)

Note: If 720 m seen without working award (G3).

The final (A1) is awarded for the correct rounding of their answer.

[4 marks]

Examiners reportAgain, this part proved accessible to the majority with a large number of candidates attaining full marks. However, there were also anumber of candidates who seemed not to have been prepared in the use of trigonometry in non right-angled triangles. Also, failing toround the answer in (a) to the nearest 10m was a common omission.

2 2 2

[3 marks]20f. Calculate the area enclosed by the path that goes around the forest.


(M1)(A1)

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

(UP) = (A1)(G2)

[3 marks]

Area = × 290 × 550 × sin 11512

72 300 m2

Examiners reportAgain, this part proved accessible to the majority with a large number of candidates attaining full marks. However, there were also anumber of candidates who seemed not to have been prepared in the use of trigonometry in non right-angled triangles. Also, failing toround the answer in (a) to the nearest 10 m was a common omission.

20g. [4 marks]Inside the forest a second path forms the three sides of another triangle named ABC. Angle BAC is 53°, AC is 180 m and BCis 230 m.

Calculate the size of angle ACB.

Markscheme (M1)(A1)

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

B = 38.7° (A1)(G2)

(A1)(ft)

[4 marks]

Examiners reportAgain, this part proved accessible to the majority with a large number of candidates attaining full marks. However, there were also anumber of candidates who seemed not to have been prepared in the use of trigonometry in non right-angled triangles. Also, failing toround the answer in (a) to the nearest 10 m was a common omission.

=180sin B

230sin 53

A B = 180 − ( + )C 53∘ 38.7∘

= 88.3∘

[1 mark]21a.

A and B are points on a straight line as shown on the graph below.

Write down the y-intercept of the line AB.

Markscheme6

OR

(0, 6) (A1) (C1)

[1 mark]

Examiners reportThis was generally well answered.

[2 marks]21b. Calculate the gradient of the line AB.

Markscheme (M1)

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

(A1) (C2)

[2 marks]

Examiners reportThis was generally well answered, the errors coming from incorrect substitution into the gradient formula rather than using the twointercepts.

(2−5)

(8−2)

= − 12

21c. [1 mark]The acute angle between the line AB and the x-axis is θ.

Show θ on the diagram.

MarkschemeAngle clearly identified. (A1) (C1)

[1 mark]

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

21d. [2 marks]The acute angle between the line AB and the x-axis is θ.

Calculate the size of θ.

Markscheme (or equivalent fraction) (M1)

(A1)(ft) (C2)

Note: (ft) from (b).

Accept alternative correct trigonometrical methods.

[2 marks]

Examiners report[N/A]

tanθ = 12

θ = 26.6∘

[3 marks]22a.

The diagram below shows the line PQ, whose equation is x + 2y = 12. The line intercepts the axes at P and Q respectively.

Find the coordinates of P and of Q.

Markscheme or (M1)

P(0, 6) (accept , ) (A1)

Q(12, 0) (accept , ) (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]

Examiners reportMost candidates could find the x and y intercepts but many wrote the coordinates the wrong way around. A number of candidates didnot label their coordinates as P and Q or did not include parentheses.

0 + 2y = 12 x + 2(0) = 12

x = 0 y = 6

x = 12 y = 0

[3 marks]22b. A second line with equation x − y = 3 intersects the line PQ at the point A. Find the coordinates of A.

Markscheme (M1)

(6, 3) (accept , ) (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]

Examiners reportIn this part many had trouble recognising the need to solve the two simultaneous equations.

x + 2(x − 3) = 12

x = 6 y = 3

23a. [4 marks]

The diagram shows a rectangular based right pyramid VABCD in which , and the height of the pyramid, .

Calculate(i) the length of AC;

(ii) the length of VC.

AD = 20 cm DC = 15 cmVN = 30 cm

Printed for North hills Preparatory

© International Baccalaureate Organization 2015 International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®

Markscheme(i) (M1)

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

(A1) (C2)

(ii) (M1)

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

(A1)(ft) (C2)

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


+152 202− −−−−−−−√

AC = 25 (cm)

+12.52 302− −−−−−−−−√

VC = 32.5 (cm)

[2 marks]23b. Calculate the angle between VC and the base ABCD.

Markscheme OR OR (M1)

( ) (A1)(ft) (C2)

Note: Accept alternative methods. Follow through from part (a) and/or part (b).


sin VCN = 3032.5

tanVCN = 3012.5

cosVCN = 12.532.5

= 67.4∘ 67.3801…

triangles bfm and baf proving to be the most popular, but ... key...triangles bfm and baf proving to...

Documents