2014 paper 1 marking scheme - kennso - home · pdf file · 2015-01-292014 paper 1...

2014 Paper 1 marking scheme [180 marks]

[2 marks]1a.

Let

Calculate the value of when , and . Write down your full calculator display.

Markscheme (M1)

(A1) (C2) Notes: Accept and also .

[2 marks]

p = .2 cos x−tan x

−zy√

p x = 45∘ y = 8192 z = 64

2 cos −tan45∘ 45∘

−648192√

= 0.015625

164

1.5625 × 10−2

1b. [4 marks]Write down your answer to part (a)

(i) correct to two decimal places;

(ii) correct to four significant figures;

(iii) in the form , where .

Markscheme(i) 0.02 (A1)(ft)(ii) 0.01563 (A1)(ft) Notes: For parts (i) and (ii), accept equivalent standard form representations.

(iii) (A2)(ft) (C4) Notes: Award (A1)(A0) for correct mantissa, between 1 and 10, with incorrect index.

Follow through from their answer to part (a).

Where the candidate has correctly rounded their mantissa from part (a) and has the correct exponent, award (A0)(A1) Award (A0)(A0) for answers of the type: .

[4 marks]

a × 10k 1 ⩽ a < 10, k ∈ Z

1.5625 × 10−2

15.625 × 10−3

[1 mark]2a.

A class of 13 Mathematics students received the following grades in their final IB examination.

3 5 3 4 7 3 2 7 5 6 5 3 4

For these grades, find the mode;

Markscheme3 (A1) (C1)[1 mark]

[2 marks]2b. For these grades, find the median;

Markscheme4 (M1)(A1) (C2) Note: Award (M1) for ordered list of numbers seen.

[2 marks]

[1 mark]2c. For these grades, find the upper quartile;

Markscheme5.5 (A1) (C1)[1 mark]

[2 marks]2d. For these grades, find the interquartile range.

Markscheme5.5 – 3 (M1) Note: Award (M1) for 3 and their 5.5 seen.

= 2.5 (A1)(ft) (C2) Note: Follow through from their answer to part (c).

[2 marks]

3a. [3 marks]

Consider the three propositions p, q and r.

p: The food is well cooked

q: The drinks are chilled

r: Dinner is spoilt

Write the following compound proposition in words.

MarkschemeIf the food is well cooked and the drinks are chilled then dinner is not spoilt. (A1)(A1)(A1) (C3) Note: Award (A1) for “If…then” (then must be seen), (A1) for the two correct propositions connected with “and”, (A1) for “not

spoilt”.

Only award the final (A1) if correct statements are given in the correct order.

[3 marks]

(p ∧ q) ⇒ ¬r

3b. [3 marks]Complete the following truth table.

Markscheme

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

Notes: Award (A1) for each correct column.

The final column must follow through from the previous two columns.

[3 marks]

[1 mark]4a.

A study was carried out to determine whether the country chosen by students for their university studies was influenced by a person’s

gender. A random sample was taken. The results are shown in the following table.

A test was performed at the 1% significance level.

The critical value for this test is 9.210.

State the null hypothesis.

χ2

MarkschemeCountry chosen and gender are independent. (A1) (C1) Notes: Accept there is no association between country chosen and gender.

Do not accept “not related” or “not correlated” or “influenced”.

[1 mark]

[1 mark]4b. Write down the number of degrees of freedom.


4c. [2 marks]Write down

(i) the statistic;

(ii) the associated p-value.

Markscheme(i) 9.17 (9.16988…) (A1) Notes: Accept 9.169.

(ii) 0.0102 (0.0102043…) (A1) (C2) Notes: Award (A1) for 0.010, but (A0) for 0.01.

[2 marks]

χ2

[2 marks]4d. State, giving a reason, whether the null hypothesis should be accepted.

MarkschemeSince , we accept the null hypothesis. (R1)(A1)(ft)ORSince , we accept the null hypothesis. (R1)(A1)(ft) (C2) Notes: To award (R1) there should be value(s) given in part (c). If a value is given in (c), we do not need it explicitly stated again in

(d).

It is sufficient to state a correct comparison.

e.g. OR

Do not award (R0)(A1). Follow through from part (c).

[2 marks]

0.0102 > 0.01

9.17 < 9.210

p-value > significance level < critical valueχ2calc

[2 marks]5a.

Dumisani has received a scholarship of 5000 US dollars (USD) to study in Singapore.

He has to travel from South Africa and must change USD for his air fare of 6600 South African rand (ZAR).

The exchange rate is 1USD = 8.2421 ZAR.In this question give all answers correct to two decimal places.

Calculate the number of USD that Dumisani must change to pay for his air fare.

Markscheme (M1)

(A1) (C2)[2 marks]

6600 × 18.2421

= 800.77

5b. [2 marks]On arrival in Singapore, Dumisani changes 3000 USD to Singapore dollars (SGD) to pay for his school fees. There is a 2.8%

commission charged on the exchange.

Calculate the value, in USD, of the commission that Dumisani has to pay.

Markscheme (M1)

(accept 84) (A1) (C2)[2 marks]

3000 × 0.028

= 84.00

5c. [2 marks]The exchange rate is .

Calculate the number of SGD Dumisani receives.

Markscheme (M1)

OR (M1)

(A1)(ft) (C2) Notes: Follow through from their answer to part (b).

Note: Do not penalize in part (c) if conversion process has been reversed consistently ie, multiplication by in part (a) and

division by in part (c).

[2 marks]

1 USD = 1.29903 SGD

(3000 − 84) × 1.29903

3000 × 1.29903 × 0.972

= 3787.97

8.2421

1.29903

[1 mark]6a.

The cumulative frequency graph represents the speed, s, in , of 80 cars passing a speed camera.

Write down the number of cars passing the camera with speed of less than or equal to 50 .


km h−1

km h−1

6b. [1 mark]Complete the following grouped frequency table for , the speed of the cars passing the camera.

Markscheme (A1)(ft) (C1)

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

[1 mark]

s

[1 mark]6c. Write down the mid-interval value of the interval.50 < s ⩽ 70

Markscheme60 (A1) (C1) [1 mark]

6d. [3 marks]Use your graphic display calculator to find an estimate of

(i) the mean speed of the cars passing the camera;

(ii) the standard deviation of the speed of the cars passing the camera.

Markscheme(i) (A2)(ft)

Notes: Award (M1) for an attempt to use the formula for the mean with at least two midpoint values consistent with their answer to

part (c). Follow through from their table in part (b).

(ii) (A1)(ft) (C3) Note: Follow through from their table in part (b).

[3 marks]

67.5(km )h−1

18.6(18.6413…)

7a. [3 marks]

The probability that it snows today is 0.2. If it does snow today, the probability that it will snow tomorrow is 0.6. If it does not snow

today, the probability that it will not snow tomorrow is 0.9.

Using the information given, complete the following tree diagram.

Markscheme

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

Note: Award (A1) for each correct pair of probabilities.

[3 marks]

[3 marks]7b. Calculate the probability that it will snow tomorrow.

Markscheme (A1)(ft)(M1)

Note: Award (A1)(ft) for two correct products of probabilities taken from their diagram, (M1) for the addition of their products.

(A1)(ft) (C3)

Note: Accept any equivalent correct fraction.

Follow through from their tree diagram.

[3 marks]

0.2 × 0.6 + 0.8 × 0.1

= 0.2 ( , 20% )15

[2 marks]8a.

A child’s wooden toy consists of a hemisphere, of radius 9 cm , attached to a cone with the same base radius. O is the centre of the

base of the cone and V is vertically above O.

Angle OVB is .

Diagram not to scale.

Calculate OV, the height of the cone.

27.9∘

Markscheme (M1)

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

(A1) (C2)

[2 marks]

tan =27.9∘ 9OV

OV = 17.0(cm) (16.9980…)

[4 marks]8b. Calculate the volume of wood used to make the toy.

Markscheme (M1)(M1)(M1)

Note: Award (M1) for correctly substituted volume of the cone, (M1) for correctly substituted volume of a sphere divided by two

(hemisphere), (M1) for adding the correctly substituted volume of the cone to either a correctly substituted sphere or hemisphere.

(A1)(ft) (C4)

Note: The answer is , the units are required.

[4 marks]

+ ×π (16.9980…)(9)2

312

4π(9)3

3

= 2970 c (2968.63 … )m3

2970 cm3

[1 mark]9a.

The diagram shows the points M(a, 18) and B(24, 10) . The straight line BM intersects the y-axis at A(0, 26). M is the midpoint of the

line segment AB.

Write down the value of .

Markscheme12 (A1) (C1) Note: Award (A1) for .

[1 mark]

a

(12,18)

[2 marks]9b. Find the gradient of the line AB.

Markscheme (M1)

Note: Accept or (or equivalent).

(A1) (C2)

Note: If either of the alternative fractions is used, follow through from their answer to part (a).

The answer is now (A1)(ft). [2 marks]

26−100−24

26−180−12

18−1012−24

= − (− , − 0.666666…)23

1624

[3 marks]9c. Decide whether triangle OAM is a right-angled triangle. Justify your answer.

Markschemegradient of (A1)(ft)

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

(M1)

Note: Award (M1) for multiplying their gradients.

Since the product is , OAM is a right-angled triangle (R1)(ft) Notes: Award the final (R1) only if their conclusion is consistent with their answer for the product of the gradients.

The statement that OAM is a right-angled triangle without justification is awarded no marks.

OR

and (A1)(ft) (M1)

Note: This method can also be applied to triangle OMB.

Follow through from (a).

Hence a right angled triangle (R1)(ft) Note: Award the final (R1) only if their conclusion is consistent with their (M1) mark.

OR

(cm) an isosceles triangle (A1) Note: Award (A1) for (cm) and (cm).

Line drawn from vertex to midpoint of base is perpendicular to the base (M1)Conclusion (R1) (C3) Note: Award, at most (A1)(M0)(R0) for stating that OAB is an isosceles triangle without any calculations.

[3 marks]

OM = 32

− ×23

32

−1

(26 − 18 +)2 122 +122 182

( + ) + ( + ) =(26 − 18)2 122 122 182 262

OA = OB = 26

OA = 26 OB = 26

[1 mark]10a.

Let .

Write down .

Markscheme (A1) (C1)

[1 mark]

f(x) = x4

(x)f ′

( (x) =)f ′ 4x3

10b. [2 marks]Point lies on the graph of .

Find the gradient of the tangent to the graph of at .

Markscheme (M1)

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

(A1)(ft) (C2)

Note: Follow through from their part (a).

[2 marks]

P(2,6) f

y = f(x) P

4 × 23

= 32

10c. [3 marks]Point lies on the graph of .

Find the equation of the normal to the graph at . Give your answer in the form , where , and are integers.

Markscheme or (M1)(M1)

Note: Award (M1) for their gradient of the normal seen, (M1) for point substituted into equation of a straight line in only and

(with any constant ‘ ’ eliminated).

or any integer multiple (A1)(ft) (C3)

Note: Follow through from their part (b).

[3 marks]

P(2,16) f

P ax + by + d = 0 a b d

y − 16 = − (x − 2)132

y = − x +132

25716

x y

c

x + 32y − 514 = 0

[1 mark]11a.

The amount of electrical charge, C, stored in a mobile phone battery is modelled by , where t, in hours, is the time

for which the battery is being charged.

Write down the amount of electrical charge in the battery at .


[1 mark]

C(t) = 2.5 − 2−t

t = 0

1.5

11b. [2 marks]The line is the horizontal asymptote to the graph.

Write down the equation of .

Markscheme (accept ) (A1)(A1) (C2)

Notes: Award (A1) for a positive constant, (A1) for the constant .

Answer must be an equation.

[2 marks]

L

L

C = 2.5 y = 2.5

C (or y) = = 2.5

11c. [3 marks]To download a game to the mobile phone, an electrical charge of 2.4 units is needed.

Find the time taken to reach this charge. Give your answer correct to the nearest minute.

Markscheme (M1)

Note: Award (M1) for setting the equation equal to 2.4 or for a horizontal line drawn at approximately .

Allow instead of .

OR

(M1) (A1)

(A1)(ft) (C3) Note: Award the final (A1)(ft) for correct conversion of their time in hours to the nearest minute.

[3 marks]

2.4 = 2.5 − 2−t

C = 2.4

x t

−t ln(2) = ln(0.1)

t = 3.32192…

t = 3 hours and 19 minutes (199 minutes)

[2 marks]12a.

The average radius of the orbit of the Earth around the Sun is 150 million kilometres.

Write down this radius, in kilometres, in the form , where .


Notes: Award (A2) for the correct answer.

Award (A1)(A0) for 1.5 and an incorrect index.

Award (A0)(A0) for answers of the form .

[2 marks]

a × 10k 1 ⩽ a < 10, k ∈ Z

1.5 × (km)108

15 × 107

12b. [2 marks]

The average radius of the orbit of the Earth around the Sun is 150 million kilometres.

The Earth travels around the Sun in one orbit. It takes one year for the Earth to complete one orbit.Calculate the distance, in kilometres, the Earth travels around the Sun in one orbit, assuming that the orbit is a circle.

Markscheme (M1)

(A1)(ft) (C2) Notes: Award (M1) for correct substitution into correct formula. Follow through from part (a).

Do not accept calculator notation .

Do not accept use of or for .

[2 marks]

2π1.5 × 108

= 942 000 000 (km) (942477 796.1… , 3 × π, 9.42 × )108 108

9.42E8227

3.14 π

12c. [2 marks]Today is Anna’s 17th birthday.

Calculate the total distance that Anna has travelled around the Sun, since she was born.

Markscheme (M1)

(A1)(ft) (C2)

Note: Follow through from part (b).

[2 marks]

17 × 942 000 000

= 1.60 × (km) (1.60221… × , 1.6014 × , 16 022 122 530, (5.1 × )π)1010 1010 1010 109

13a. [2 marks]

Two propositions and are defined as follows

: Eva is on a diet

: Eva is losing weight.

Write down the following statement in words.

MarkschemeIf Eva is losing weight then Eva is on a diet (A1)(A1) (C2) Notes: Award (A1) for If… then…

For Spanish candidates, only accept “Si” and “entonces”.

For French candidates, only accept “Si” and “alors”.

For all 3 languages these words are from the subject guide.

Award (A1) for correct propositions in correct order.

[2 marks]

p q

p

q

q ⇒ p

[2 marks]13b. Write down, in words, the contrapositive statement of .

MarkschemeIf Eva is not on a diet then she is not losing weight (A1)(A1) (C2) Notes: Award (A1) for “not on a diet” and “not losing weight” seen, (A1) for complete correct answer.

No follow through from part (a).

[2 marks]

q ⇒ p

[2 marks]13c. Determine whether your statement in part (a) is logically equivalent to your statement in part (b). Justify your answer.

MarkschemeThe statements are logically equivalent (A1)(ft)The contrapositive is always logically equivalent to the original statement (R1)(ft)ORA correct truth table showing the equivalence (R1)(ft) (C2) Note: Follow through from their answers to part (a) and part (b).

[2 marks]

14a. [1 mark]

The following Venn diagram shows the relationship between the sets of numbers

The number –3 belongs to the set of and , but not , and is placed in the appropriate position on the Venn diagram as an

example.

Write down the following numbers in the appropriate place in the Venn diagram.

4

Markscheme

(A1)(A1)(A1)(A1)(A1)(A1) (C6)

Note: Award (A1) for each number correctly placed.

Award (A0) for any entry in more than one region.

[1 mark]

N, Z, Q and R.

Z, Q R N

14b. 13

Markscheme

(A1)(A1)(A1)(A1)(A1)(A1) (C6)



[1 mark]

3

[1 mark]14c.

Markscheme

(A1)(A1)(A1)(A1)(A1)(A1) (C6)



[1 mark]

π

[1 mark]14d. 0.38

Markscheme

(A1)(A1)(A1)(A1)(A1)(A1) (C6)



[1 mark]

[1 mark]14e.

Markscheme

(A1)(A1)(A1)(A1)(A1)(A1) (C6)



[1 mark]

5√

[1 mark]14f. −0.25

Markscheme

(A1)(A1)(A1)(A1)(A1)(A1) (C6)



[1 mark]

[2 marks]15a.

Chocolates in the shape of spheres are sold in boxes of 20.

Each chocolate has a radius of 1 cm.

Find the volume of 1 chocolate.

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

(M1)

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

(A1) (C2)

[2 marks]

π(143

)3

= 4.19 (4.18879… , π) c43

m3

[1 mark]15b. Write down the volume of 20 chocolates.


(A1)(ft) (C1)

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

[1 mark]

83.8 (83.7758… , π) c803

m3

15c. [2 marks]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.


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

(A1) (C2)

[2 marks]

10 × 8 × 2

= 160 cm3

15d. [1 mark]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.


(A1)(ft) (C1)

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

[1 mark]

76.2 (76.2241… , (160 − π)) c803

m3

[1 mark]16a.

Ludmila takes a loan of 320 000 Brazilian Real (BRL) from a bank for two years at a nominal annual interest rate of 10%,

compounded half yearly.

Write down the number of times interest is added to the loan in the two years.


[3 marks]16b. Calculate the exact amount of money that Ludmila must repay at the end of the two years.

Markscheme (M1)(A1)

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

OR

(A1)(M1) Note: Award (A1) for seen, (M1) for correctly substituted values from the question into the finance application.

OR

(A1)(M1) Note: Award (A1) for seen, (M1) for correctly substituted values from the question into the finance application.

amount to repay (A1) (C3) Note: Award (C2) for final answer if not seen previously.

[3 marks]

320 000(1 + )102×100

2×2

N = 2

I% = 10

PV = −320000

P / Y = 1

C / Y = 2

C / Y = 2

N = 4

I% = 10

PV = −320000

P / Y = 2

C / Y = 2

C / Y = 2

= 388962

389000 388962

16c. [2 marks]Ludmila estimates that she will have to repay BRL at the end of the two years.

Calculate the percentage error in her estimate.

Markscheme (M1)

Note: Award (M1) for correctly substituted percentage error formula.

(A1)(ft) (C2)

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

[2 marks]

360 000

× 100∣∣360 000−388 962

388 962∣∣

= 7.45 (% ) (7.44597…)

[1 mark]17a.

The graph of the quadratic function intersects the y-axis at point A (0, 5) and has its vertex at point B (4, 13).

Write down the value of .


f(x) = a + bx + cx2

c

[3 marks]17b. By using the coordinates of the vertex, B, or otherwise, write down two equations in and .

Markschemeat least one of the following equations required

(A2)(A1) (C3) Note: Award (A2)(A0) for one correct equation, or its equivalent, and (C3) for any two correct equations.

Follow through from part (a).

The equation earns no marks.

[3 marks]

a b

a(4 + 4b + 5 = 13)2

4 = − b

2a

a(8 + 8b + 5 = 5)2

a(0 + b(0) = 5)2

[2 marks]17c. Find the value of and of .

Markscheme (A1)(ft)(A1)(ft) (C2)

Note: Follow through from their equations in part (b), but only if their equations lead to unique solutions for and .

[2 marks]

a b

a = − , b = 412

a b

18a. [2 marks]

Two propositions are defined as follows:

Quadrilateral ABCD has two diagonals that are equal in length.

Quadrilateral ABCD is a rectangle.

Express the following in symbolic form.

“A rectangle always has two diagonals that are equal in length.”

p :

q :

Markscheme (A1)(A1) (C2)

Note: Award the first (A1) for seeing the implication sign, the second (A1) is for a correct answer only. Not using the implication

earns no marks.

[2 marks]

q ⇒ p

[1 mark]18b. Write down in symbolic form the converse of the statement in (a).


Note: Award (A1)(ft) where the propositions in the implication in part (a) are exchanged.

[1 mark]

p ⇒ q

[2 marks]18c. Determine, without using a truth table, whether the statements in (a) and (b) are logically equivalent.

MarkschemeNot equivalent; a kite or an isosceles trapezium (for example) can have diagonals that are equal in length. (A1)(R1) (C2) Notes: Accept a valid sketch as reasoning.

If the reason given is that a square has diagonals of equal length, but is not a rectangle, then award (R1)(A0). Do not award (A1)(R0). Do not accept solutions based on truth tables.

[2 marks]

[1 mark]18d. Write down the name of the statement that is logically equivalent to the converse.

MarkschemeInverse (A1) (C1) Note: Do not accept symbolic notation.

[1 mark]

[1 mark]19a.

Consider the curve .

Write down .


[1 mark]

y = + kxx3

dy

dx

3 + kx2

19b. [3 marks]The curve has a local minimum at the point where .

Find the value of .

Markscheme (A1)(ft)(M1)

Note: Award (A1)(ft) for substituting 2 in their , (M1) for setting their .

(A1)(ft) (C3)

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

[3 marks]

x = 2

k

3(2 + k = 0)2

dy

dx= 0dy

dx

k = −12

19c. [2 marks]The curve has a local minimum at the point where .

Find the value of at this local minimum.

Markscheme (M1)

Note: Award (M1) for substituting 2 and their –12 into equation of the curve.

(A1)(ft) (C12)

Note: Follow through from their value of found in part (b).

[2 marks]

x = 2

y

− 12 × 223

= −16

k

[1 mark]20a.

The distribution of rainfall in a town over 80 days is displayed on the following box-and-whisker diagram.

Write down the median rainfall.


[1 mark]43 (mm)

20b.


[1 mark]10 (mm)

[2 marks]20c. Find the interquartile range.

Markscheme (A1)

(A1) (C2) Note: Award (A1) for identifying correct quartiles, (A1) for correct subtraction of the quartiles.

[2 marks]

48 − 20

= 28

20d. [2 marks]Write down the number of days the rainfall will be

(i) between 43 mm and 48 mm;

(ii) between 20 mm and 59 mm.

Markscheme(i) 20 (days) (A1)(ii) 60 (days) (A1) (C2)[2 marks]

[1 mark]21a.

The surface of a red carpet is shown below. The dimensions of the carpet are in metres.

Write down an expression for the area, , in , of the carpet.

Markscheme or (A1) (C1)

Note: Award (A0) for .

[1 mark]

A m2

2x(x − 4) 2 − 8xx2

x − 4 × 2x

21b. [3 marks]The area of the carpet is .

Calculate the value of .

10 m2

x

Markscheme (M1)

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

(M1)

ORSketch of and (M1)ORUsing GDC solver and (M1)OR

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

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

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

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

[3 marks]

2x(x − 4) = 10

10

− 4x − 5 = 0x2

y = 2 − 8xx2 y = 10

x = 5 x = −1

2(x + 1)(x − 5)

x = 5 (m)

5 −1

21c. [2 marks]The area of the carpet is .

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

Markscheme (A1)(ft)

(A1)(ft) (C2) Note: Follow through from their answer to part (b).

Do not accept negative answers.

[2 marks]

10 m2

2 × 5 = 10 (m)

5 − 4 = 1 (m)

[1 mark]22a.

The diagram shows a wheelchair ramp, , designed to descend from a height of .

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


[1 mark]

A 80 cm

− (−0.0851,−0.085106… ,− )80940

447

22b. [1 mark]The gradient for a safe descending wheelchair ramp is .

Using your answer to part (a), comment on why wheelchair ramp is not safe.


Notes: Accept “less than” in place of inequality. Award (A0) if incorrect inequality seen. Follow through from part (a). [1 mark]

− 112

A

−0.0851 (−0.085106…) < − (−0.083333…)112

22c. [4 marks]The equation of a second wheelchair ramp, B, is .

(i) Determine whether wheelchair ramp is safe or not. Justify your answer.

(ii) Find the horizontal distance of wheelchair ramp .

Markscheme(i) ramp is safe (A1)the gradient of ramp is (R1)

Notes: Award (R1) for “the gradient of ramp is ” seen.

Do not award (A1)(R0). (ii) (M1) Note: Accept alternative methods.

(A1) (C4)

[4 marks]

2x + 24y − 1920 = 0

B

B

B

B − 112

B − 112

2x = 1920

960 (cm)

[2 marks]23a.

A group of 100 students gave the following responses to the question of how they get to school.

A test for independence was conducted at the significance level. The null hypothesis was defined as

: Method of getting to school is independent of gender.

Find the expected frequency for the females who use public transport to get to school.

χ2 5%

H0

Markscheme OR (M1)

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

(A1) (C2)

[2 marks]

× × 10030100

48100

30×48100

= 14.4 ( )725

[2 marks]23b. Find the statistic.


Note: Award (A1)(A0) for .

[2 marks]

χ2

13.0 (12.9554…)

12.9

23c. [2 marks]The critical value is at the significance level.

State whether or not the null hypothesis is accepted. Give a reason for your answer.

Markschemethe null hypothesis is not accepted (A1)(ft)

OR (R1)

ORthe null hypothesis is not accepted (A1)(ft)p-value (R1) (C2) Notes: Follow through from their answer to part (b).

Do not award (A1)(ft)(R0). [2 marks]

χ2 7.815 5%

>χ2calc

χ2crit 13.0 > 7.82

(0.0047) (0.00473391 … ) < 0.05

24a. [2 marks]

In this question give all answers correct to the nearest whole number.

Fumie is going for a holiday to Great Britain. She changes Japanese Yen (JPY) into British Pounds (GBP) with no

commission charged.

The exchange rate between GBP and JPY is

1 GBP = 129 JPY.

Calculate the value of JPY in GBP.

Markscheme (M1)

(A1) (C2)[2 marks]

100 000

100 000

100 000129

= 775 (GBP)

24b. [4 marks]At the end of Fumie’s holiday in Great Britain she has 239 GBP. She converts this back to JPY at a bank, which does not

charge commission, and receives 30 200 JPY

(i) Find the exchange rate of this second transaction.

(ii) Determine, when changing GBP back to JPY, whether the exchange rate found in part (b)(i) is better value for Fumie than the

exchange rate in part (a). Justify your answer.

Markscheme(i) (M1)

(A1) Note: Accept ( ).

Award (M1) for .

Award (A0) for

(ii) No, the part (b)(i) rate is not better value than the part (a) rate. (A1)(ft)

(R1)ORNo, the part (b)(i) rate is not better value than the part (a) rate. (A1)(ft)

(R1) (C4) Note: Accept “part (a) rate is better” for the (A1)(ft). Follow through from part (b)(i).

A numerical comparison must be seen to award (R1).

[4 marks]

30 200239

1 GBP = 126 JPY

126 JPY239

30 300

1 JPY = 0 GBP

30 200 < 30 831

129 > 126

[3 marks]25a.

Günter is at Berlin Tegel Airport watching the planes take off. He observes a plane that is at an angle of elevation of from where

he is standing at point . The plane is at a height of 350 metres. This information is shown in the following diagram.

Calculate the horizontal distance, , of the plane from Günter. Give your answer to the nearest metre.

Markscheme (M1)

(A1) (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 without working.

[3 marks]

20∘

G

GH

350tan 20∘

= 961.617…

= 962 (m)

961

25b. [3 marks]The plane took off from a point , which is metres from where Günter is standing, as shown in the following diagram.

Using your answer from part (a), calculate the angle , the takeoff angle of the plane.

Markscheme (A1)(ft)

(M1)

(A1)(ft) (C3) Notes: Accept from use of 3 sf answer from part (a). Follow through from their answer to part (a).

Accept alternative methods.

[3 marks]

T 250

ATH

961.617… − 250 = 711.617…

( )tan−1 350711.617…

= (26.1896 … )26.2∘

26.1774… 962

[1 mark]26a.

In a trial for a new drug, scientists found that the amount of the drug in the bloodstream decreased over time, according to the model

where is the amount of the drug in the bloodstream in mg per litre and is the time in hours.

Write down the amount of the drug in the bloodstream at .


[1 mark]

D(t) = 1.2 × (0.87 , t ⩾ 0)t

D (mg )l−1 t

t = 0

1.2 (mg )l−1

[2 marks]26b. Calculate the amount of the drug in the bloodstream after 3 hours.

Markscheme (M1)

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

(A1) (C2)

[2 marks]

1.2 × (0.87)3

= 0.790 (mg ) (0.790203 … )l−1

0.333 mg1−1

Markscheme (M1)

Note: Award (M1) for setting up the equation.

(M1) Notes: Some indication of scale is to be shown, for example the window used on the calculator.

Accept alternative methods.

(hours) ( , 9 hours 12 minutes, 9:12) (A1) (C3)

[3 marks]

0.333 mg

1.2 × = 0.3330.87t

9.21 9.20519…

[2 marks]27a.

A survey investigated the relationship between the number of cleaners, , and the amount of time, , it takes them to clean a school.

Use your graphic display calculator to write down the equation of the regression line on .

Markscheme

(A1)(A1) (C2) Notes: Award (A1) for and seen,

(A1) for an equation involving and .

[2 marks]

n t

t n

t = −20.1n + 205

t = (−20.1046…)n + (204.755…)

−20.1 205

t n

[2 marks]27b. Write down the value of the Pearson’s product–moment correlation coefficient, .r


Notes: Award (A0)(A1) for .

[2 marks]

−0.941(−0.941366…)

+0.941

[2 marks]27c. Use your regression equation to find the amount of time 4 cleaners take to clean the school.

Markscheme (M1)

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

(minutes) ( ) (A1)(ft) (C2)

Notes: Follow through from their regression equation found in part (a). Accept (minutes) ( ).

[2 marks]

−20.1046… × 4 + 204.755…

124 124.337…

125 124.6

[1 mark]28a.

Consider the graph of the function .

Label the local maximum as on the graph.

f(x) = + 2 − 5x3 x2

A

Markscheme

correct label on graph (A1) (C1)[1 mark]

[1 mark]28b. Label the local minimum as B on the graph.

Markscheme correct label on graph (A1) (C1)[1 mark]

[1 mark]28c. Write down the interval where .


[1 mark]

(x) < 0f ′

−1.33 < x < 0 (− < x < 0)43

[1 mark]28d. Draw the tangent to the curve at on the graph.x = 1

Markscheme

tangent drawn at on graph (A1) (C1)[1 mark]

x = 1

[2 marks]28e. Write down the equation of the tangent at .


Notes: Award (A1) for , (A1) for .

If answer not given as an equation award at most (A1)(A0). [2 marks]

x = 1

y = 7x − 9

7 −9

[1 mark]29a.

The heights of apple trees in an orchard are normally distributed with a mean of and a standard deviation of .

Write down the probability that a randomly chosen tree has a height greater than .


[1 mark]

3.42 m 0.21 m

3.42 m

0.5 (50%, , )50100

12

[1 mark]29b. Write down the probability that a randomly chosen tree will be within 2 standard deviations of the mean of .3.42 m


Note: Accept or .

[1 mark]

0.954(0.954499… ,95.4%,95.4499… %)

95% 0.95

29c. [2 marks]Use your graphic display calculator to calculate the probability that a randomly chosen tree will have a height greater than

.

Markscheme

(M1)

Note: Accept alternative methods.

(A1) (C2)

[2 marks]

3.35 m

0.631(0.630558… ,63.1%,63.0558… %)

[2 marks]29d. The probability that a particular tree is less than metres high is . Find the value of .

Markscheme

(M1)

Note: Accept alternative methods.

(A1) (C2)

[2 marks]

x 0.65 x

3.50 (3.50091...)

[4 marks]30a.

A function is given as .

Write down the derivative of the function.

f(x) = 2 − 5x + + 3, − 5 ⩽ x ⩽ 10, x ≠ 0x3 4x

Printed for Victoria Shanghai Academy

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

Markscheme (A1)(A1)(A1)(A1) (C4)

Note: Award (A1) for , (A1) for , (A1) for , (A1) for or .

Award at most (A1)(A1)(A1)(A0) if additional terms are seen.

[4 marks]

6 − 5 −x2 4x2

6x2 – 5 – 4 x−2 1x2

[2 marks]30b. Use your graphic display calculator to find the coordinates of the local minimum point of in the given domain.


Notes: Award (A1)(A1) for “ and ”.

Award at most (A0)(A1)(ft) if parentheses are omitted.

[2 marks]

f(x)

(1.15, 3.77) ((1.15469..., 3.76980...))

x = 1.15 y = 3.77

2014 paper 1 marking scheme - kennso - home · pdf file · 2015-01-292014 paper 1...

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