Answers to the exam questions extended level includes 2016 papers.

Section 1 Vectors

1 (a)

5

1,

0

1,

3

2RSQRPQ and

5

1RS Make a drawing:

So the vector

2

0PS or:

PS = PQ + QR + RS =

2

0

5

1

0

1

3

2

(b) (i) (a) bbABEA

221

21 (b) abDFDC

222

(c) aababDFADEAEF

23

(d) aababDCADBABC

2232

(ii) (a) Trapezium: aBC

and aAD

3 so AD and BC are parallel (b) Ratio is aaa

3:2: 1 : 2 : 3

2 (a)

10

7

2

3

8

4

2

3

4

222 ba

(b) parallel vectors cak

or

104

2 uk 4k = 10 k = 2.5 so that u = 5.

3 (a) (i) qpAEFAFE

(ii) qpACBABC

24

(b) (i) qkpkqpkpBCkpBDpBDFBFD

2)43()24(333

(ii) F, E and D collinear then they are on one line FDFEm with m .

Using earlier obtained results we get: m )( qpmFE

qkpk

2)43( .

We now get two equations: m = 3 4k or; 4k m = 3 …. because the scalars in front of p

are the same. For the scalars in front of q

we get: and m = 2k…. substitution of into gives k = 1.5.

4 [P1 May/June 2003 Q 23]

(a) (i) qpOQPOPQPE

21

21

21

21 )( (ii) qpqppPEOPOE

21

21

21

21 )(

(b) In the same way: qrOF

21

21 (c) rpqrqpOFEOEF

21

21

21

21

21

21

(d) Since qpPR

and rpEF

21

21 it means that PREF 2 this means EF and PR are parallel and

half the length of RP is the length of EF.

P

Q R

S

1

5 (a)

6 (a) Hexagon

(b) (i) Rotational symmetry of order 6 [Point A can rotate 6 times before it comes back to its own position]

(ii) 6 lines of symmetry

(c) (i) Enlargement, centre T and factor 31 (ii) Reflection in the line through FC.

(d) (i) mOS

3 (ii) nmBA

(iii) mCD

(iv) mnOD

2 (v) mnFC

22

(e) (i) 5n

[△AOB is an equilateral triangle] (ii) 5nm

7 (a)

3

4

0

4

3

8BCABAC

(b) (i) Make a small sketch (see diagram on the RHS)

1

4

hDE

or without a diagram DE = DO + OE =

1

42

1

2

hh using position vectors.

(ii) AB parallel to DE k DE AB with k a constant. So

3

8

1

4

hk then k = 2

Looking at the first component and 2(h + 1) = 3 so that 2h = 3 + 2 or h = 2½.

(iii) ACDE 2222 3)4()1(4 h gives: 16 + h

2 + 2h + 1 = 25 so that h

2 + 2h 8 = 0

Factorize: (h + 4)(h 2) = 0 so h = 2 or h = 4

8 (a) (i)

1

2

2

1

1

3OABOBA (ii)

3

3

1

4

2

1OCAOAC

(b) 521 22OA 2.24 (c)

1

2BA so

1

2AB then

2

4AD

9 (a) (i) AS = ½ b (ii) BS = ASBA = a + ½ b (iii) BT = BC53 = ACBA

53 =

53 a +

53 b

(b) BTABSAST = ½ b + a 53 a +

53 b =

52 a +

101 b =

101 (4a + b)

(c) (i) 3p q + 2r =

17

1

3

12

2

2

3

13 (ii) magnitude of p + q = 101)3( 22 qp

D

E

2 2

h

1

2

b

3

4KN

1

2MN

K L

M

N

10 (a) (i) AC = BCAB = a + b.

(ii) BE = BA + AE = BA + ½ AC = a + ½(a + b) = ½ a + ½ b

(iii) DE = DB + BE = 32 b + ½ a + ½ b = ½ a

61 b.

(b) (i) 2p q =

14

2

4

4

5

32

(ii) p + q 2r =

3

11

2

22

4

4

5

3 so the magnitude is (121+ 9) = 130.

(iii) q and r are parallel and 2│r│ = │q│.

11 [The length AB cannot be 3m since AB is always twice the length of EF= 2m. Think of the enlargement with

centre C and factor 2. It is clear that AB = 4m not 3m]

(a) (i) EC = EF + FC = 2m n

(ii) Here things go wrong: ➀ DE = DB + BE = DB + EC = m + 2m n = m n

➁ DE = DA + AF + FE = 2m n 2m = n

(iii) BD = m

(b) This question can only be answered if you use (a) (ii) ➁.

You can explain that both vectors DE and AC can be expressed in terms of n so they must be parallel.

(c) (i) 2b c =

12

3

4

7

4

22

(ii) a + b =

4

1

4

2

0

1 so |a + b| = √[1

2 + (4)

2] = √17

(d) (i) x

0

18

4

7

4

2y the top line for the x component gives: 2x + 7y =18 and the second or

y component gives: 4x + 4y = 0.

(ii) The second equation tells us 4y = 4x or x = y substitute this in the first equation: 2x + 7x = 18 or 9x = 18

so x = 2 and y = 2.

Section 2 Functions & long division of a polynomial by a binomial.

1 (a) f(k) = k or k = 3k + 4 4 = 2k so that k = 2

(b) f(x) = 3x + 4 or y = 3x + 4 swop: x = 3y + 4 so 3y = x 4 or 3

4

xy this means f

-1(x) =

3

4x

2 (a) f(5.5) = 7.5 × 10 = 75

(b) 5)12(31 x

times 3: 2x 1 = 15 or 2x = 16 x = 8 so g

-1(5) = 8 [No need to work out g

-1(x)]

(a) f(3) = 02

0

2

)3(3

(b)

2

3)(

xxf

or

2

3 xy

swop x and y :

2

3 yx

Make y subject: 2x = 3 + y so that y = 2x 3 this means: f -1

(x) = 2x 3

4 Given f(x) = 1

4

x and g(x) = x

2 3

3

(a) Explain why f(x) is undefined for x = 1 The denominator is 0 for x = 1; hence f(1) does not exist.

(b) Find f(3) f(3) =4 ÷ (3+1) = 4/4 = 1.

(c) fg(x) = f(g(x)) = f(x2 3) =

2

4

13

422

xx

(d) f-1

(x) ? y = 1

4

x swop x and y and make y subject: x = 4)1(

1

4 )1(

yx

y

y; divide both sides by x

you get: x

y4

1 so that y = 1

4

x this means: f

-1(x) = 1

4

x .

(e) f(x) = g(x) 1

4

x = x

2 3 ×(x + 1) both sides: 4 = (x

2 3)( x + 1) work out the brackets on the RHS:

4 = x3 + x

2 3x 3 becomes: x

3 + x

2 3x 7 = 0

(f) x2 + 3x + 3

x 2 x3 + x

2 3x 7

- x3 2x

2

3x2 3x

- 3x2 6x

3x 7

- 3x 6

1

5 (a) (i) f(10) = 2 410

102

aor

6

102

a so that 12 = 10 + a or a = 2

(ii) Solve: 2)4)(4(4

24

)4(by sidesboth mutliply

xx

x

x x 4x + 16 = x + 2 or x = 2.8

Conclusion: f -1

(4) = 2.8

(b) fg(x) = f(g(x)) = f( x + 2) = 2

4

42

22

x

x

x

x

6 x2 + 2x + 7

x 3 x3 x

2 + x 21

- x3 3x

2

2x2 + x

- 2x2 6x n no remainder.

7x 21

- 7x 21

0

7 (c) 2x2 3x + 4

x + 4 2x3 + 5x

2 8x 15

2x3 + 8x

2 3x

2 8x

3x2 12x

4x 15

4x + 16

31

So the quotient is x2 + 3x + 3 and

the remainder is 1.

So the answer is: x2 + 2x + 7 ;

So the quotient is 2x2 3x + 4 and

the remainder is 31.

4

8 (a) g(1) = (2)3 = 8 and f(8) = (8 1)

2 = 81 so fg(1) = 81

(b) gh(x) = g(h(x)) = g(2x + 1) = (2x + 1)3 = (4x

2 + 4x + 1)( 2x + 1) = 8x

3 + 12x

2 + 6x + 1

(c) To find h1

(x) first write y = 2x + 1 swop x and y: x = 2y + 1 then make y subject of the equation:

x 1 = 2y or y = ½ x ½ so h1

(x) = ½ x ½ .

9 Let f(x) = 4x3 + 6x

2 3x + 16 then the remainder is f(3) = 4 27 + 6 9 9 + 16 = 108 + 54 + 7 = 169

The quotient can be found through inspection: 4x3 + 6x

2 3x + 16 = (x 3)(ax

2 + bx + c) → a =4 and c = 5

31

So 4x3 + 6x

2 3x + 16 = (x 3)(4x

2 + bx 5

31 )

3 4 + b = 6 or b = 18 So the quotient is 4x2 + 18x 5

31 .

[This last approach of inspection is tricky but faster; of course you can use the long division as well.]

10 (a) f(2) = 2 4 + 1 = 9 (b) 4 3x = 0 or 3x = 4 so x = 1⅓

(c) y = 4 3x swop x and y: x = 4 3y; make y subject: x 4 = 3y so y = 1⅓ ⅓x so g1

(x) =1⅓ ⅓x

(d) fg(x) = f(4 3x) = 2(4 3x)2 + 1 = 32 48x + 18x

2 + 1 = 18x

2 48x + 33

(e) 2x2 + 1 = 4 3x or 2x

2 + 3x 3 = 0 so

4

333

4

2493

2

42

a

acbbx

So x = 0.69 or x = 2.19

11 Let f(x) = x3 + 3x

2 5 then f(2) = 8 + 12 5 = 15 so R = 15.

Since x3 + 3x

2 5 = (x 2)(ax

2 +bx + c) + 15 or x

3 + 3x

2 10 = (x 2)(ax

2 + bx + c)

One can easily see a = 1; c = 5. For b looking at the coefficient of x2 , valid is 3 = 2 + b so b = 5

Section 3 Linear Programming 1 So the point inside R with integers as coordinates is (2, 2)

2 (a) Inequalities are: x ≥ 2 and x + y ≤ 12½

(b) (i) Point closest to C is: (10, 2) inside ABC; (10, 2) is

closer but outside the triangle.

(ii) Points inside the triangle with x coordinate 7 are:

(7, 2); (7, 3); (7, 4) and (7, 5)

2 4 6 8 10 12 14

x

y

2

4

6

8

10

12

14

A

B

C

x ≥ 2

1 2 3 4 5 6

x

-1 -2

1

2

3

4 y

-1

R

y > x 1

y < 3

x > 1

5

3 (a) The gradient of the line x 3y = 6

3y = x + 6 3

y = ⅓ x 2 so the gradient is ⅓.

(b) Inequality: x + 1 ≥ 0 or x ≥ 1

[The points right of the x = 1]

Inequality: x + y ≤ 2 try point (0, 0): 0 + 0 ≤ 2 is true

So all points below the line x + y = 2 are solutions.

Inequality: x 3y ≤ 6 try again point: (0, 0): 0 3×0 ≤ 6

yes 0 ≤ 6 so all points above the line x 3y = 6 are solutions.

Shaded region is inside the triangle formed by the 3 lines.

4 (a) 400x + 700y ≤ 56000 divide by 100 4x + 7y ≤ 560

(b) y ≤ x (c) y ≥ 10 (d) see the diagram

(e) (40, 40) is a point on the line y = x, so 40 high cost houses can be build.

(f) Two slanted lines meet at 11x = 560 or x = 50.91; so point of intersection is (501110 , 50

1110 ). Try (50, 51) and

(50, 52) .(52, 50) is the point inside the unshaded area that still satisfy all

conditions. 52 × 200 000 + 50 × 500 000 = N$ 35 400 000. The other point just inside is: (122, 10)

building cost = 122 × 200 000 + 10× 500 000 = N$ 29 400 000so highest cost is N$35 400 000.

2 4 6 2 4

2

4

2

4

x = 1

0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150

10

20

30

40

50

60

70

80

90

x

y

6

5 (a) 10 × x + 20 × y ≤ 160 divide by 10: x + 2y ≤ 16

(b) x + y ≤ 12; x ≥ 6 and y ≥ 3

(c) See diagram (Scale is less than 1cm per unit)

(d) Points that satisfies all conditions are: (6, 3); (6, 4);

(6, 5); (7, 3); (7, 4); (8, 3); (8, 4) and (9, 3).

(e) Check only the vertices towards the right top side:

Profit for (6, 5): 6 × 150 + 5 ×300 = N$2400

Profit for (8, 4): 8 × 150 + 4 ×300 = N$2400

Profit for (9, 3): 9 × 150 + 3 ×300 = N$2250

So maximum profit is N$2400

6

0 100 200 300 400 500 600 700 800

100

200

300

400

500

600

x

y

x + y =500

y = 400

x = 300

R

0 1 2 3 4 5 6 7 8 9 10 11 12 13

1

2

3

4

5

6

7

8

9

10

11

12

13

x

y x = 6

y= 3

x+y = 12

x+2y = 16

7

7 (a) ➀ x 7 ➁ y 8 ➂ x + y 10

(b) & (c)

Section 4 Commercial arithmetic Reverse percentages & percentage increase

1 The short way of solving this question: 800 ÷ 125 × 100 = $640. [In the 800 is the cost price (100%) and the

Profit (25%); go to 1 by dividing by 125 then times 100 to get the cost price]

If you cannot understand/see this you can use a longer method:

Using a three lined table: Cost price $100

Profit $25

Selling price $125 $800

The table above is how you start the calculation. Since the columns are direct proportional one can divide/multiply

each column to form a new column, slowly working yourself towards the final answer.

Cost price $100 $400 $80 $640 So she paid $ 640 for the

painting.

Profit $25 $100 $20 $160

Selling price $125 $500 $100 $800

2 8 × 100150 = 8 × 1.5 = 12 cm; this is the increase so the new length is 8 + 12 = 20 cm.

3 (a) 120 × 2.80 = $336 (b) 224 ÷ 2.80 = £80

4 (a) Total discount is: 1200 × 0.28 × 0.04 = $13.44

(b) Percentage increase =

10028

2835%100

valueold

valuesold- valuenew25%

(c) Discount is: 1200 × 0.35 399 = 420 399 = $21 so the % discount is now: %100420

21 = 5%

×4 ÷5 ×8

0 2 4 6 8 10 x

No of Dobermans

1

2

3

4

5

6

7

8

9

10

No

of

Bu

lld

ogs

y

7

(d) Maximum number of Dobermans is 7

Maximum number of Bulldogs is 8.

(e) Either 2 3500 + 8 2500 = N$27000

or 3 2500 + 7 3500 = N$32000

So the maximum profit is N$32000.

8

(d) In short: cost of posting this parcel in 2005 = 4.60 ÷ (100 + 15) × 100 = $4

Or use a 3 lined table:

So the company had to pay $4 for posting a parcel in 2005.

5 (a) apples cost 5.90 (1.35 + 1.20 + 1.64 + 1.20) = $ 0.51

(b) (i) Draw a line through

(0, 0) and (6.80, 5.90)

(ii) See graph: €1.40

6 (a) 50.48.050.4100

80$3.60

(b) 80% of an amount is $6.20 10% will then be $6.20 ÷ 8 = 0.775 so 100% is then 10 × 0.775 = $7.75

[Check: 0.8 × 7.75 = $ 6.20]

7 (a) $595 + 12 × $171.04 = $2647.48 so he paid $2647.48 $2395 = $252.48 more.

(b) ($3054.20 $395) ÷ 24 = $110.80

(c) Percentage is

%1002395

1595239533.4%

8 (a) Income from selling pens and pencils is: 5x + 3(x 9) = $(8x 27)

(b) 8x 27 < 300 8x < 327 x < 40.875 so x ≤ 40 since x is a natural number.

(c) 40 pens.

9 Pt = P0(1 + 0.01 ×r)n 50000 = P0 × 1.15

4 He has to invest P0 = 50000 ÷ 1.15

4 = N$ 28587.66

Price 2005 $100

Increase $15

Price 2006 $115 $4.60

Price 2005 $100 100÷115 $4

Increase $15

Price 2006 $115 115÷115 $4.60

÷115

Price 2005 $100

Increase $15

Price 2006 $115 $4.60

×4.60

Dollars

Euros 1 2 3 4 5 6

7

1

2

3

4

5

6

7

0

9

10 (a) 7500052 = N$30000.

(b) 15% stands for 30000 so 1% stands for 2000 so 100% stands for N$200 000.

(c) 30000 + 54 × 4550 = N$275700

(d)

%100200000

20000027570037.85%

(e) Interest is 45000 × 1.5 × 0.055 = N$3712.50

11 Cost of oil is 2000 × 33 × 8.20 = N$541200

12 Use the formula Pt = P0(1 + r/100)

t or 1000 = 420 1.08

t or 100 = 42 1.08

t take log:

Log 100 = log(42 1.08t ) or 2 = log 42 + t log 1.08 so t = (2 log 42) log 1.08 = 11.27 years.

13 (a) Percentage girls = (45 29) 45 100% = 35.6%

(b) No of learners passing the test = 0.31 45 = 14 learners

(c) 106 % stands for 18 so 100% stands for 18 106 100 = 17 marks.

(d) The number of learners passing the test decreased from 24 to 14 so % decrease = 10 24 100 = 41.7%

14 Algebraically: 0.9 x = 450 so x = 450 0.9 = N$500; with a table:

15 (a) (i) 0.07 2500 3 = N$525 so he will have 2500 + 525 = N$3025

(ii) Pt = P0(1 + r/100)t or P3 = 2500 1.09

3 = N$ 3237.57

(iii) Increase % = %211002500

525100

valueold

increase [of course easier 3 7 = 21%]

(b) 120% corresponds with 3600 so 1% stands for 30 so the original amount was: N$3000.

16 (a) Taxable income is 210 000 – 10 500 – 30 000 = N$16 9500.

(b) Income tax is 9 000 + 0.25 × 69 500 = N$26 375

(c) %100210000

3200015.2%

(d) Let her salary in 2012 be N$ y then 1.05 × y = 210000 so y = 210000 ÷ 1.05 = N$200 000

(e) Use the formula Pt = P0(1 – r/100)t substitute: Pt = 150000(1 – 0.15)

3 = 150000× 0,85

3 = N$ 92118.75.

Section 5 Accuracy; rounding off; significant figures; standard form.

1 (a) 5000 [correct to one significant figure] (b) 4872 4870 is expressed in 3 significant figures.

2 (a) 0.000 004 cm = 4 × 106

cm.

3 (a) Upper bound is: 50.5 cm (b) 49.5 + 14.5 + 49.5 + 14.5 = 128 cm.

4 (a) 0.000 042 = 4.2 × 105

(b) 7 × 103

× 3 × 109 = 21 × 10

6 = 2.1 × 10

7

5 9080007211037020.0110

6 [P1 Oct/ Nov 2011 Q 17]

% 90 10 100

N$ 450 50 500

10

(a) Upper bound is 57.5 mm (b) 4 × 57.5 mm = 230 mm = 23.0 cm

7 (b) 0.026 m = 26 mm lower bound is 25.5 mm

8 (a) q = 8 × 104

so q2 = 8 × 10

4 × 8 × 10

4 = 64 × 10

8 = 6.4 × 10

7 in standard form.

(b) p ÷ q =

)4(10

41

104

4

11

104108

1032

108

102.3

q

p 4 × 10

14

9 (a) 55 cm ≤ C < 65 cm b (i) max no is 455 ÷ 55 = 8 revolutions (ii) min no is 455 ÷ 65 = 7 revolutions.

10 (a) 62 m2 = 62000000 mm

2 (b) 62000000mm

2 = 6.2 × 10

7

11 Least possible perimeter = 5 × 6.45 cm = 32.25 cm

12 Length of 25 mm means 24.5 mm real length < 25.5 mm; greatest possible area is 25.52 = 650.25 mm

2.

13 (a) 0.00759 = 7.59 103

(b) 0.00759 →0.0076

14 Maximum left over area = biggest possible area cardboard rectangle smallest possible area circle =

30.5 21.5 π 4.52 = 592 cm

2

In the same way the minimum area: 29.5 20.5 π 5.52 = 510 cm

2

15 (a) 200, 100 and 5 (b) (201 99) 4.9 ≈ (200 100) 5 = 4000.

16 (a) 3(2 106 4 10

5) = 6000000 1200000 = 4800000 = 4.8 10

6.

(b) 2 103 + 3 10

2 + 4 10

x + 6 10

y = 2304.06 subtract 2300 from both sides: 4.06 = 4 10

x + 6 10

y

So x = 1 and y = 2

17 Maximum speed: biggest possible distance smallest time used = 850 12.75 = 6632 km/h

18 3× 900152 = 46.4

Section 6 Speed; speed time graph; distance time graph; duration and time difference

1 (a) average speed = 4025.4

170

used timetotal

distance covered total km/h

2 (a) Go back 7 hours in time: in London it is 20 30 the previous

day.

(b) From 6 30 pm to 10 00 pm is 3½ hours, yet the flight took 7½ hours so there is a 4 hour time difference

between Brunei and Dubai. [Brunei is ahead of Dubai since Brunei is more easterly than Dubai]

03 30 03 00 00 00

24 00

21 00 20 30

30 m 3 h 3 h 30 m

11

3 (a) Area under the whole graph is total covered distance:

Area under the graph is area triangle + area rect. + area trap =

½×10×21 +10×21 + ½ × 10 (21 + 16) = 105 + 210 + 185 = 500 m

(b) The cyclist slows down from 21 m/s to 16 m/s in 10 seconds.

That is 5 m/s in 10 sec so the retardation is 0.5 m/s2

4 (a) 1 hour and 12 minutes to reach Tuesday; 7h 30m 1h 12m = 6h 18m so they arrived 06 18 Tuesday morning.

(b) 3

80

15

400

5.7

200

5.7

5.220150

5.7

)55.7(20150

used timetotal

distance covered total

= 26.7 km

5 (a) Speed went down from 50 to 0 in 10 seconds so that is a retardation of 5 m/s2.

As well gradient of the line:

10090

050 5m/s

2 which is a deceleration or retardation of 5 m/s

2.

Retardation or deceleration is the same has a negative acceleration so acceleration of 5m/s2 is correct as well.

(b) Distance travelled is the area under the graph.

Total Area = area trapezium + area triangle =

½ × 90 ×(20 + 50) + ½ × 10 × 50 = 3400 m

6 (a) See clock face: (b) 11 hours ahead from 9.15 am is 20 15.

7 (a) Speed increased from 0 to 11m/s in 10 seconds so acceleration is 1.1m/s2. Or gradient =

010

011

= 1.1m/s

2

(b) Area under the graph: A triangle + A trapezium +A rectangle = 55 + ½ × 30 × 24 + 50 ×13 = 1065 m

(c) Total distance is 65m more than 1 km, yet at a speed of 13m/s it will take 5 sec to cover the 65 m so time

taken 90s 5s = 85s

8 23:05 2h 10 min = 20:55.

9 Area under the graph is the same so ½ × 12 (15 + 9) = ½ ×15 × v or 96 = 5 v v = 19.2 m/s

10 256 km/h = 256000m in 3600 sec = 2560m in 36 sec = 2560 36 m/sec = 71.1m/s

11 (a) (i) from 0 to 30 m/s in 20 sec = 1.5 m/s2

(ii) Area under the graph = travelled distance = ½ 20s 30 m/s = 300 m.

(b) 960 = 300 + 30 (k 20) or 660 = 30k 600 or 1260 = 30 k so that k = 1260 30 = 42 sec.

20m/s

50m/s

90 10

10 s

21 m/s

21 m/s

16 m/s

10 s 10 s

12

Section 7 Variation and substitution

1 (a) p is inversely proportional to the square root of q.

(b) Equation Comment

qp

12

q

124

Substitute p = 4

4 × q = 12 Multiply both sides with q

4

12

4

4

q 3q

Divide both sides by 4

223q q = 9

Square both sides

2 (a) y is inversely proportional to x y

cx with c a constant. Substitute x = 3 and y = 20 you get:

203

c

This makes c = 60. The relation between x and y is now: y

x60

so if x = 4 then y = 15 p = 15

(b) Substitute y = 5 5

60x = 12 so q = 12.

(c) m

n60

n

m60

3 (a) Substitute d = 800 and w = 803 in: 10098

)800803(8003

8

)(3

dwds = 30 cm

(b)

Equation Comment

8

)(3 dwds

8

)(32 dwds

Square both sides.

8s2 = 3dw 3d

2 Multiply both sides with 8 and work out the brackets.

8s2 = 3dw 3d

2 3dw = 8s

2 + 3d

2 Isolate w and flip the equation.

d

dsw

3

38 22

Divide both sides by 3d

Finished.

4 (a) 2d

cF whereby c is a constant. F = 32 then d = 2cm substitute:

432

c so c = 128. Equation is

2

128

dF

(b) 2

1282

d so 2d

2 = 128 d

2 = 64 so d = 8.

5 (a) x varies inversely to the square root of y → y

cx substitute x = 3 and y = 25 you get:

253

c so c =15

so the equation is y

x15

.

(b) This question has a mistake, obviously x cannot be negative! [The outcome of a square root is always

positive.]

13

Section 8 Mensuration

1 Answer is to be expressed in cm3 so express the area in cm

2: 20 m

2 = 200 000 cm

2

Volume = base area × height = 200 000 × 0.000 004 = 0.8 cm3

2 (a) 3

2

60

40 so 9

32 = 6ℓ (b) 9 ÷ 60 = 0.15 ℓ = 150 ml so 750 ÷ 150 = 5 bottles.

3 37 kg + 40 g = 37 kg + 0.040 kg = 37.040 kg or 37.04 kg

4 (a) Draw the diagonal AC and apply Pythagoras in one of the triangles: ℓ2 = r

2 + r

2 ℓ

2 = 2r

2 take square root

from both sides: ℓ = r 2 ….. *

(b) One can easily see that: perimeter square < circumference

4l < 2r substitute * in this inequality: 4r 2 < 2r divide both sides by 4r: r

r

r

r

4

2

4

24

so 2

2

(c) 2 and are irrational numbers.

5 (a) This is a cuboid so ℓ × ℓ × 2 = V = 500 divide by 2: ℓ2 = 250 so that ℓ 250 15.8 cm

(b) 500150heigtharea base31

31 h or 50h = 500 or h = 10 cm

(c) 500 = 3

34 r 3

4

1500r

so that 3

4

1500

r = 4.92 cm

(d) Let the radius of the inner cylinder be r. 500 = r + 1.5)2 6 r

2 6 work out the brackets:

500 = 6r2 + 18r + 13.5 6r

2 the r

2 term cancels.

500 13.518r so

18

5.13500r = 8.09 cm

(e) Lengths of the two cones are in the ratio 2 : 5 so the volumes are in the ratio 23 : 53

= 8 : 125.

So the small cone has volume = 500133

830.1 cm

3

6(a) O is the centre so OB = OC; AB = DC so OA = OD so △OAD is equilateral. As a result AD is 6cm because

A and D are respective midpoints of OB and OC.

(b) (i) Arc length of BC = 41222360

6061

o

o

r so n = 4.

(ii) Arc length AED = ½ × 2r = ½ × 2 × 3 = 3 so the required ratio is:4

3

4

3

(c) (i) Area BOC = 2412360

60 2

612

o

o

r ≈ 75.4 cm2

(d) Area Shaded part = Area sector OBC Area △OAD area semicircle ADE =

24 ½ OA × OD × sin 60º ½ r2 = 24 ½ × 6 × 6 ×½ 3 ½ × 3

2 = 24 9 3 4.5

3 = 45.7cm2

14

7 (a) (i) Rotational symmetry order 6 [The shape can rotate 6 times, then it falls on itself again.]

(ii) (a) Area large circle is r2 = ×15

2 = 225 cm

2 = 707 cm

2. Radius of the big circle is 3 × 5 = 15 cm.

(b) Area shaded part is 61 (area big circle 7× area small circles) =

61 (707 7 × × 5

2) = 26.2 cm

2

(b) (i) AOB = 60º

(ii) Perimeter shaded area is:

2 half circumferences of the small circle + 61 ×circumference larger circle +

61 ×circumf. small circle =

2× 5 + 61 × 2 ×15 +

61 × 2 × 5 = 10 +

61 (30 = 10

640 52.4 cm.

8 (a) Capacity is 31 r

2 h =

31 × × 6

2 × 18 = 216 = 679 cm

3 [The formula given in the exam paper is wrong!]

(b) The radius of the water is ½ × 6 = 3 cm. Volume is 31 r

2 h =

31 × ×3

2 × 9 = 27 = 84.8 cm

3

(c) (i) Use ratio’s in similar triangles: 18

6

x

r so r =

31 x.

(ii) V(x) = 31 r

2 h =

31 × × (

31 x)

2 × x this simplifies to: V(x) = 3

27x

(iii) V(9) = 84.8 cm3

9 (a) TSA is area hemisphere + area curved part cylinder + area circle [top] = 2r2 + 2rh + r

2 =

3r2 + 2rh = (3 × 15

2 + 2 × 15 × 35) = (675 + 1050) = 1725 ≈ 5420 m

2

(b) (i) Volume = 0.75(32 r

3 + r

2h) = 0.75(

32 × 15

3 + 15

2 × 35) = 0.75(2250 + 7875) = 23856 m

3

(ii) 1000 ℓ = 1 m3 so the time to empty the tank is 23856 seconds = 6.63 hours.

(c) Volume is 34 ×23856 = ½ h (15 + 20) × 72 h = 31808 ÷ (36 × 35) = 25.2 m

10 (a) area sector = 2

1212 5.6

360

30 r = 11.1 cm

2

(b) perimeter sector = 2r + 5.62132360

30121 r = 16.4 cm

11 Volume syringe = π r2 h = π 2.5

2 15 = 295 cm

3

12 (a) Two solutions:

I Drop a perpendicular to BC through M. Now sin 70˚ = ½BC/ BM or BC = 260 sin70˚ = 112.8 mm.

II Using the cosine rule: BC 2

= BM 2

+ MC 2

2 BM MC cos 140˚ or BC = (602 + 60

2 260

2 cos140) =

So BC = 60 2 2cos 140˚) = 112.8 mm

(b) Shaded area = area circle area triangle = πr2 ½ MB MC sin 140˚ =

π602 ½ 60

2 sin140˚ = 10153 mm

2

13 (a) Shape is called a triangular prism.

(b) Volume ramp = area triangle height = ½ b ht hr = ½ 12 9 18 = 972 m3

(c) 8 100 = N$800. [Again something wrong with the question; I think they meant to include the cost of

the soil as well. The question does not make that clear.]

The answer including the cost of the soil is: 8 100 + 972 25 = N$25100.

(d) The lorry contains 3 2.5 4 = 30 m3 of soil. So there are 972 30 = 33 lorries needed to fill up the ramp. 15

14 (i) 300 500 300 10 10 5 = 300 100 3 = 90 000 small boxes

(ii) 90000 13.21 2.33 = N$2 770 137

(iii) 2(3 5 + 3 5 + 3 3) = 78 m2

15 (a) Vsph =3

4π r

3 =

3

4× π × 6

3 = 4 × π × 72 = 288π cm

3 = 905 cm

3 in three significant figures.

(b) TSAsph = 225 = 4 π r2 so r

2 = 225 ÷ 4 π → r = 7.5π

−0.5 =

5.7= 4.23 cm.

Section 9 Inequalities

1 (a) Smallest integer such that 7k ≥ 36 k = 6 [k = 5 won’t work since 7 × 5 is only 35.]

(b) Largest integer so that 3n 1 < 26 3n < 27 n < 9 so n = 8

2 (a) 2(4 x) < x 10 8 2x < x 10 3x < 18 x > 6

(b) 3n > 17 or n > 532 so n = 5

3 (a) 7 y < 9 y < 2 y > 2 [Note when you divide or multiply with a negative number, the sign changes]

(b) z = 3

4 2 ≤ 1 + 2n ≤ 5 subtract 1 from all three sides 3 ≤ 2n ≤ 4 divide by 2 1.5 ≤ n ≤ 2. n is 1; 0; 1 and 2.

5 2 ≤ x < 1

6(a) x – 10 ≤ 2(x −1) < x. work out brackets: x – 10 ≤ 2x −2 < x add 2 to all three sides: x – 8 ≤ 2x < x + 2

Now subtract x from all three sides: −8 ≤ x < 2.

(b) On the number line:

Section 10 Simultaneous equations

1 3x = 7y 3

7yx substitute this in the other equation:12y = 5x 1

2 2x + 3y = 0 2x + 3y = 0………..

x + 4y = 15 2x + 8y = 30…….. gives: 5y = 30 so y = 6 substitute in : 2x 18 = 0

this becomes: 2x = 18 so that x = 9

3 4x y = 9 ……………………………..

2x 3y = 23 4x 6y = 46 …….. gives: 5y = 55 or y = 11 substitute this in :

4x 11 = 9 4x = 20 so that x = 5

4 (b) 3x + 4y = 1 ×2 6x + 8y = 2……..

5x 8y = 9 no change: 5x 8y = 9…….. now + gives: 11x = 11 x = 1 substitute this in :

6 + 8y = 2 or 8y = 4 y = ½

13

7512

yy 1

3

35

3

36

yy

or 13

y

or y = 3. Use the first

equation to get x: 3x = 21 x = 7 × 2

× 2

−9 −8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4

16

5 Substitute y = 2x2 + 2x 6 in y = 3x + 4, you get: 2x

2 + 2x 6 = 3x + 4 or 2x

2 x 10 = 0

Factorize: 2x2 5x + 4x 10 = 0 or x(2x 5) + 2(2x 5) = 0 so that (x + 2)(2x 5) = 0

Solutions are now: ➀ x = 2; substitute this value in y = 3x + 4 and you get y = 2

➁ x = 2.5; substitute this value in y = 3x + 4 and you get y = 11.5.

6 3a – b = 9 3a – b = 9

2a + 2b =14 divide by 2: + a + b = 7 *)

Add the two equations: 4a = 16 so a = 4 then b = 3 [using *)]

7 xy = 20

x – 2y = − 3 → x = 2y – 3 substitute this in the first equation: (2y – 3)y = 20 rewrite to standard form:

2y2

− 3y – 20 = 0 or 2y2

− 8y + 5y – 20 = 0 so that 2y(y – 4) + 5(y – 4) = 0 or (y – 4)(2y + 5) = 0

This means y = 4 → x = 5 [substitute y = 4 in xy = 20]

or y = −2½ → x = −8[substitute y = −2½ in xy = 20]

Section 11 Symmetry

1 6 planes cutting the prism as shown: and one plane of symmetry half way:

Altogether 7 planes of symmetry.

2 (a) Order of rotational symmetry of decagon is 10 [Decagon is a 10 sided shape]

(b) A; M; U

(c) The locus of points is a cylindre with height 10 cm and radius 3 cm, with the line AB as its central line.

The top and bottom of the cylindre is formed by two semispheres with radius 3cm and one with centre A and

the other one centre B.

3 (b) A hexagon can be covered by 4 triangles Sum of interior angles is 4 × 180º = 720º.

In the shape drawn there are 3 angles of 105º and 3 angles of zº so 3 × 105º + 3z = 720º

Solve for z : 3z = 720º 315º = 405º so z = 405 ÷ 3 = 135º

4 (a) Taking the colors in account there is no symmetry. If you see all colors to be identical: one line of symmetry.

(b) Order of rotational symmetry is 4.

5 (a) (b)

6 (a) (i) angle AOB = 360 – 90 – 90 – 70 = 110° [The four angles in a quadrilateral add up to 360°.]

(ii) angle BOC = ½ × 110 = 55°

(iii) C is 90° [given] so the trig relations apply in ∆OCB: cos 55° = OC ÷ OB with OB = 5 cm [radius]

So OC = 5 × cos 55° = 2.9 cm

(b) FB and FA are both tangents to the circle and OB and OA are radii so because of symmetry reasons the must be 90°. 17

(c) One line of symmetry.

(d) Rotational symmetry of order 1.[ It can only turn a full 360° to cover itself again.]

Section 12 Solving Equations

1 (a) x = 6 substitute: 03

36 k or 12 + k = 0 this means k = 12

(b) look for two numbers product 4 [= 2 ×(-2)] sum 3 these are: 4 and 1 so 2y2 3y 2 = 0 becomes:

2y2 4y + y 2 = 0 or 2y(y 2) + 1(y 2) = 0 so that (y 2)(2y + 1) = 0 this means y = 2 or y = ½

2 2y2 + 29y 15 = 0 2y

2 + 30y y 15 = 0 2y(y + 15) 1(y + 15) = 0 (y + 15)(2y 1) = 0\

So that y = 15 or y = ½

3 (x + 1) (3x 2) = 0 a product is zero when one of the factors is zero so x = 1 or x = 32

4 (a) 41

5

x or

1

4

1

5

x use cross multiplication: 5 = 4(x + 1) 5 = 4x + 4 or 4x = 1 so that x =

41

5 (a) 5t(3t + 7) = 0 this equation is already factorized so t = 0 or 3t + 7 = 0 t = 31

37 2

6 (a) Apply cosine rule in △DEF: DF2 = DE

2 + EF

2 2DE × EF cosE . Substitute values given:

122 = x

2 + (5 + x)

2 2x(5 + x) cos 120º using cos 120º = ½ we get:

144 = 2x2 + 10x + 25 + 5x + x

2 3x

2 + 15x 119 = 0

(b) 6

)119(3422515

2

42

b

acbbx This will give: x = 4.28 cm or x = -9.28 cm

The negative value for x is not acceptable.

(c) EF = x + 5 = 4.28 + 5 = 9.28 cm = 93 mm

7 (a) )1)(12(

14

)1)(12(

2236

)1)(12(

)1(2

)12)(1(

)12(3

12

2

1

3

xx

x

xx

xx

xx

x

xx

x

xx

(b) (i) x2 + x 6 = (x + 3)(x 2)

(ii) 65

62

2

xx

xx=

3

3

)3)(2(

)2)(3(

x

x

xx

xx

(c) 7x2 = 11 12x or 7x

2 + 12x 11 = 0 use the formula:

14

117414412

2

42

a

acbbx

Use a calculator: x = 0.66 or x = 2.38

8 (a) Area ∆ABC = 12 = ½ AC BC sinACB or 12 =½ x(x + 8) sin 30˚ or 12 = ½x(x + 8) ½ multiply

with 4: 48 = x2 + 8x or x

2 + 8x 48 = 0.

(b) x2 + 8x 48 = 0 → (x + 12)(x 4) = 0 so x = 4 or x = 12 so that BC = 4 cm

(c) BC = 2 cm so x = 2 so AC = 2 + 8 = 10 cm.

Use the cosine rule: AB2 = AC

2 + BC

2 2 AC BC cos C or : AB

2 = 4 + 100 2 2 10 cos 30˚

AB2 = 104 40 ½√3 = 69.4 so AB = 8 cm

18

9 (a) Area original square – area square with reduced sides = 39 cm2 or

x2 – (x – 3)(x – 3) = 39 → x

2 –(x

2 – 6x + 9) = 39 → 6x – 9 = 39 or 6x = 48 so that x = 8 cm

(b) (i) (x + 2)2 = 20 or x

2 + 4x + 4 = 20 → x

2 + 4x − 16 = 0

(ii) 2022

2024

2

804

2

)16(14164

2

42

a

acbbx

So x = −2 +√20 = 2.47 or x = −2 −√20 = −6.47.

(iii) So the length of the square is 2.47 cm.

Section 13 Angles in a circle

1 (a) ADB = 90º 36º = 54º [TD is tangent so BDT = 90º]

(b) ABD = 36º [BAD = 90º and ADB = 54º; the angles in a △ add up to 180º.]

(c) ACD = 36º [ ABD and ACD are both subtended by chord AD, so they are equal.]

2 (a) BOD = 2 × 47º = 94º [Angle at the centre is twice the angle at the circumference]

(b) BCD = 180º 47º = 133º [Opposite angles in a cyclic quadrilateral add up to 180º]

(c) OBD = (180º 94º) ÷ 2 = 43º. [△BDO is an isosceles triangle.]

3 (a) GAE = 38º [AEDC is cyclic quadrilateral so EAC = 180º 128º = 52º and GAC = 90º]

(b) AEB = 180º 65º 52º = 63º [BFC = AFE {Vertically opposite angles} angles in △AEF add up to 180º]

4 (a) (i) APC = 90º because AC is the diameter of the circle. [The intersection of the two diagonals is the centre

of the circle.]

(ii) Draw diagonal AC; CAB = CPB subtended by the same chord BC. But CAB = 45º and APC =

90º. This means APB = PBC = 45º.

(iii) APB = 45º. (b) APD = 135º [3 × 45º]

(b) PRD = 127º as well [vertically opposite angles] so PDR = 180º 127º 45º = 8º.

So PDC = 90º + 8º = 98º.

5 (a) CEA = 146º CEA and AOC are both subtended by chord AC in circle II.

(b) CBA = ½ × 146º = 73º angle at the circumference is half the angle at the centre when they both are

subtended by the same chord.

(c) Quadrilateral AECF is cyclic opposite angles add up to 180º CFA = 180º 146º = 34º

(d) In circle I; CDA is an angle at the circumference subtended by chord AC.

Reflex angle AOC = 360º 146º = 214º is as well subtended by chord AC. CDA = ½ × 214º = 107º.

CDA is as well the exterior angle in △CDF so DCF = 107º 34º = 73º

6 (a) (i) ½ 105 = 52.5˚ Reason: Angle at the circumference is half the angle at the centre if they are subtended by the same chord.

(ii) 90˚. Reason: Tangent is perpendicular to the diameter in the point of touch.

(iii) 90˚ Reason: angle in a semicircle is 90˚. (b) (i) 52.5˚ (ii) 180 105 = 75˚ [AOBC iis as well a cyclic quadrilateral]

(iii) 180 90 52.5 = 37.5˚

19

Section 14 Changing the subject of a formula and substitution

1 (a) Substitute F = 4: )36(9

5)324(

9

5C 20

(b) Equation Comment

)32(95 FC 9C = 5(F 32) Multiply both sides by 9;

9C = 5F 160 Work out the brackets;

5F = 9C + 160 Isolate F and flip the equation

F = 1.8 C + 32. Divide by 5 both sides.

2 Given y = 18 + 3x2 (a) x = 2 y = 18 + 3(2)

2 = 18 + 12 = 30

(b) y = 93 93 = 18 + 3x2 75 = 3x

2 x

2 = 25 so x = 5 or x = 5

(c) y = 18 + 3x2 y 18 = 3x

2 x

2 =

3

18y so that x =

3

18

yor x = 6

31 y

3 (a) x

20 (b)

2

25

x

(c) x

20

2

25

x = 1½ multiply both sides with 2x(x + 2): 20 × 2(x + 2) 25 × 2x = 3 × x(x + 2)

Work out the brackets and take like terms together: 40x + 80 50x = 3x2 + 6x 3x

2 + 16x 80 = 0

(d) 3x2 + 16x 80 = 0 1216

6

1

3

8

6

)80(341616

2

4 22

a

acbbx

x = 3.15 or x = 8.48 not acceptable, x stands for a speed and should have a positive value.

(e) Substitute x = 3.1452 in x

20 +

2

25

x = 11.22h = 11h 13m [x = 3.15 will not give an accurate answer]

4 [P2 Oct/Nov 2007 Q 2 (a)]

(a) 22 baax =

22 84.173.073.0 = 2.71

(b) first isolate the square root: axba 22 now square both sides: a

2 + b

2 = (x a)

2

Then make b2 subject: b

2 = (x a)

2 a

2 take square root: 22)( aaxb or axxb 22

5 (a)When the numerator is zero so a = 2 or a = ½

(b) P = a

aa

3

)12)(2( is undefined for a = 3.

6 x

mp4

4 square both sides: x

mp4

16 22 or x p2 = 64m

2 this means

2

264

p

mx

7

nr

PA

1001 take log from both sides:

log A = log

nr

P100

1 first use law 1 (of the logarithm: log ab = log a + log b)

Log A = log P + log

nr

1001 now use law 3 (of the logarithm: log a

n = n log a)

20

Log A – log P = n log

1001

r divide both sides by log

1001

r:

)1log(

loglog

100r

PAn

.

8 1 xy square both sides: y2 = x + 1 so x = y

2 – 1.

Section 15 Statistics

1 (a) 0.3 (2.3) = 2.6 m (b) order them first: 2.3 1.6 1.2 0.5 0.4 0.1 0.3

So the median is 0.5 m

(c) Mean is: [(2.3) + (1.6) + (1.2) + ( 0.5) + (0.4) + 0.1 + 0.3] ÷ 7 = 5.6 ÷ 7 = 0.8 m

2 (a) (b) 15

14

45

42

180

168

(c) 360º ÷ 45º = 8 so there were 8 × 30 = 240 people in the survey.

3 (a) Frequency density = frequency ÷ class width

2 = k ÷ 2 so k = 4.

4 (a) 0 goals [Highest frequency] (b) 40 teams; the middle team has scored 1 goal.

(c) Mean = (0 × 16 + 1 × 6 + 2 × 6 + 3 × 6 + 4 × 4 + 6 × 2) ÷ 40 = 64 ÷ 40 = 1.6 goals.

5 (a) (i) No of goals scored 0 1 2 3 4 5

No of matches (freq) 4 2 1 2 3 8

(ii) See histogram

(b) Median: 4 goals [The 10th

and 11th

match fall in

the 4 goals column]

(c) Mean = (0 × 4 + 1 × 2 + 2 × 1 + 3 × 2 + 4 × 3 + 5 × 8) ÷ 20 =

(2 + 2 + 6 + 12 + 40) ÷ 20 = 62 ÷ 20 = 3.1 goals

45º

A

B

C

168º

0 1 2 3 4 5 6

Goals scored

2 1

3 4 5

6 7 8 9

10

Frequency

1 2 3 4 5 6 7 8 9

Time (hours)

0 0

1

2

3

4

5

(b)

Frequency

density

21

6 (a) Mode is 0 yellow sweets

(b) Median is 1 yellow sweet. [Packet 13 is in the middle of 25 packets; add freq. until you come across no 13]

(c) Mean = (0×8 + 5×1 + 5×2 + 4×3 + 4×2 + 5×1) ÷ 25 = 40 ÷ 25 = 1.6 yellow sweets.

7 (a) See diagram.

(b) Mass (m grams) m ≤ 35 m ≤ 40 m ≤ 45 m ≤ 50 m ≤ 55 m ≤ 60 m ≤ 70

Cumulative freq. 0 15 35 65 100 128 140

(c) (i) Median mass is 51g

(ii) Interquartile range: 56 45 = 11g

(d) 60 44 = 16g (see the long horizontal arrows)

30 35 40 45 50 55 60 65 70

5

10

15

20

25

30

35

Frequency

density

Mass of eggs in grams

0

20

40

60

80

100

120

140

30 40 50 60 70

Cumulative

frequencies

Mass (m grams)

56 45

70

105

35

30

128

22

8 (a) (i) Midpoints are: 75g; 125g etc.

Total mass is: 4 × 75 + 56 × 125 + 84 × 175 + 76 × 225 + 36 × 275 + 4 × 325 = 50300g = 50.3kg

(ii) Mean mass is: 50300 ÷ 260 = 193 g

(b) (i) Cumulative frequency table:

Mass (m grams) m ≤ 50 m ≤ 100 m ≤ 150 m ≤ 200 m ≤ 250 m ≤ 300 m ≤ 350

Frequency 0 4 60 144 220 256 260

(ii)

9 (a) (i) mean score = (15 + 18 + 3 + 12 + 21 + 18 + 3 + 12 + 6 + 24 + 42 + 18) ÷ 12 = 192 ÷ 12 = 16

(ii) median score order then and take the average of the two that in the middle.

15, 18, 3, 12, 21, 18, 3, 12, 6, 24, 42, 18 3, 3, 6, 12, 12, 15, 18, 18, 18, 21, 24, 42

So the median is (15 + 18) ÷ 2 = 16.5

(iii) Maybe 16 since it is lower.

(b) (i)

Distance (m) 0 ≤ m ≤ 20 20 < m ≤ 40 40< m ≤ 60 60< m ≤ 80 80 < m ≤ 100

Frequency 4 9 15 10 2

Cum freq 4 13 28 38 40

(ii)

50 100 150 200 250 300 350

50

100

150

200

250

(iii) (a) median is 192g

(b) IQR = 230 154 = 76g

(c) There are 260 145=

115 potatoes over 200g in

a bag. 4 bags will give 460

potatoes over 200g; they

have to buy 5 bags to have at

least 500 potatoes of more

than 200g.

130

195

65

Mass in grams

Cumul.

frequency

f

5

0

10

15

20

25

30

35

40

Dist in m

10 20 30 40 50 60 70 80 90 100

(iii) 15% of 40 = 6 and 40 6 = 34.

Draw a horizontal line at 34; it meets

the graph at 68 m.

(iv) Check the dotted lines.

Inter- quartile range = 62 33 = 29 m.

23

10 (a) highest temperature is 1ºC

(b) Range is highest value lowest value = 1 (23) = 24 ºC

11 (a) (9 × 14.5 + 23 × 24.5 + 38 × 34.5 + 30 × 49.5) ÷ 100 = 34.9 email messages per day

(b) 38 + 30 = 68

12 (a) Use the cumulative frequency curve to complete the frequency table below.

Mark x 0 x < 20 20 x < 30 30 x < 40 40 x <50 50 x ≤ 60

No of students 58 110 58 = 52 159 110 = 49 190 159 = 31 200 190 = 10

(b) Median mark is at 28. {start a horizontal line at 100; it cut the graph at (28, 100)

(c) 40% of 200 is 80 students. So the pass mark is 24.

13 Use the equation: Frequency density class width = frequency.

Time in days Number of seeds germinated

0 < t 5 50

5 < t 1 50

15 < t 20 75

14 (a) a = 40; b =27 [The sum is 94 this is caused by 4 students taller than 190 cm and the frequency for the

group 180 < h 190 is 11 not 9.]

(b) Based on the table the mean is (40 155 + 27 165 + 18 175 + 9 185) 94 = 165 cm

15 (a) 71 left in the first 10 minutes so answer is 80 71 = 9 people.

(b) 52 people left in the first 8 m; 10 left in the first 3 minutes so answer 52 10 = 42 people.

24

(c) about 7 minutes or just a bit less than 7 minutes. [Correct is as well 6.9 minutes]

16 (a)

Mass (m) in kg Midpoints m Frequency f m f Mean = 2850 50 = 57 kg

40 < m 50 45 20 900

50 < m 60 55 15 775

60 < m 80 70 5 350

80 < m 85 82.5 10 825

Sum: 50 Sum: 2850

(b) Median is the mass of the 25th

boy which is in the interval: 50 < m 60.

0

10

20

30

40

50

60

70

80

Cu

mu

lati

ve

freq

uen

cy

2 4 6 8 10 12 14 16

time (minutes)

(a)

(b)

(b)

(c)

25

(c) Frequency = freq density class interval so first interval: freq density = 20 10 = 2 etc.

(d)10

1

50

5

(e) P(First boy 60m 80; second boy 80< m 85) + P(First boy 80m 85; second boy 60< m 80) =

25

1

50

2

50

5

50

10

50

10

10

1

(f) P = 560

13

48

13

7

2

10

3

48

13

49

14

50

15

17 (a)

Time (t) in minutes Number of learners Time in minutes Cumulative Frequency

0 < t ≤ 10 8 t ≤ 0 0

10 < t ≤ 20 12 t ≤ 10 8

20 < t ≤ 30 34 t ≤ 20 20

30 < t ≤ 40 46 t ≤ 30 54

40 < t ≤ 50 42 t ≤ 40 100

50 < t ≤ 60 38 t ≤ 50 142

60 < t ≤ 70 20 t ≤ 60 180

t ≤ 70 200

35 40 45 50 55 60 65 70 75 80 85 90

0.5

1

Freq

dens

mass

1.5

2

26

18 (a) clothes → 180 – 10 = 170° (b) 180360

10N$5 (c)

%100

360

901027.8%

19 (a) Height (h cm) 130 < h ≤ 150 150 < h ≤ 155 155 < h ≤ 160 160 < h ≤ 165 165 < h ≤ 180

Midpoints m 140 152.5 157.5 162.5 172.5

Frequency f 3 7 20 25 5

m × f 420 1067.5 3150 4062.5 862.5

Mean height is (420 + 1067.5 + 3150 + 4062.5 + 862.5) ÷ 60 = 159.4 cm.

(b) All measurements are taken as the midpoint of their respective classes so one does not get an exact number

but an estimate of the mean.

20

40

60

80

100

120

140

Cum

ula

tiv

e fr

equ

ency

10

Time in minutes

160

180

20 30 40 50 60 70

200

(c)

(i) Median is 40 min.

(ii) Lower Quartile

is 29 minutes.

(iii) IQR is 52 – 29 =

23 minutes.

(iv) Point of the graph

with x coordinate 45

is (45,122). This

means there are 122

learners spending 45

minutes or less on

Facebook.

Conclusion: there are

200 – 122 = 78

learners spending

more than 45 minutes

on Facebook.

27

Section 16 Probability

1 (a) (i)

(ii) (a) 6 prime numbers: 5, 7, 13 (each twice) so P = 6/25 = 0.24

(b) 5 are perfect squares: 4; 9 and 16, so P = 5/25 = 0.2

(b) P = 0 [by definition of a prime number]

2 (a) p = 51 ; q = 1 and r = 0

(b) (i) 75 ×

64 ×

53 =

72 (ii)

75 ×

62 ×

54 =

214

(iii) P(R,R,B) + P(R,B,R) + P(B,R,R) = 75 ×

64 ×

52 +

75 ×

62 ×

54

+

72 ×

65 ×

54 =

74

3 (a)

(b) 0.4 × 0.3 × 0.8 = 0.096

(c) Fifth attempt [probability for failing is then 0.]

4 (a) (i) 131

524 (ii)

41

5213

(b) (i) See diagram

(ii) P(Exactly one queen) =

P( Queen, No Queen) + P(No Queen, Queen)

22132

22116

22116

514

1312

1716

131

(iii) 502

513

524 =

55251

5 (a) 6/12 = ½

(b)

6 (a) 41

164 [1, 3, 5, 15] (b) ½ [All odd numbers] (c) 1 [1-16] (d) 0

7 (a) 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

(b) (i) digits that are prime numbers are: 2, 3, 5, 7 so the probability is 4/10 = 0.4

(ii) digits that are square numbers are 1; 4 and 9 so the probability is 3/10 = 0.3 (What about 0?)

(iii) digits that are not less or equal to 6 are: 7, 8, 9 so the probability is 3/10 = 0.3

(c) The full registration number is: ABC 219 NA.

+ 2 3 5 7 11

2 4 5 7 9 13

3 5 6 8 10 14

5 7 8 10 12 16

7 9 10 12 14 18

11 13 14 16 18 22

Queen

No Queen

52

4

52

48

Queen

No Queen

51

3

51

48

Queen

No Queen

51

4

51

47

Red

Pink

Whi

Red

Pink

Whi

Red

Pink

Whi

Red

Pink

Whi

124

122

126

113

112

116

114

111

116

114

112

115

(c) (i) 661

111

122

(ii) P(W,R) + P(R,W) =

114

116

124

114

126

Pass

Fail

0.8

0.2

Pass

Fail

0.7

0.3

Pass

Fail

0.6

0.4

First att. Second att. third att.

28

8 (a) P = 0.05 and Q = 0.95

(b) (i) Probability = 0.05 0.05 = 0.0025

(ii) P(exactly one is faulty) = P(first one is faulty; the second not) + P(first one is not faulty; the second is F) =

0.05 0.95 + 0.95 0.05 = 0.095

(c) P(3 are faulty) = 0.053 = 0.000125

9 (a) 4 pairs have the same number so 12 are different so P = 43

1612

(b) Only prime numbers appearing here are 2 and 3 so there are only 4 favourable outcome so P = 41

164

(c) Only 1 and 4 are perfect squares so there are 4 favourable outcome: (1,1); (1,4); (4,1) and (4,4) P = 41

(d) Favourable outcome are now: (1,1); (2,1); (3,1); (4,1); (1,2); (1,3); (1,4) so only those with a 1 in one of

the two spins qualify. P = 167

10 (a)

Section 17 Bearings and angles

1

(a)

2 (a) t = 180º (54º + 57º) = 69º [ ABC and BCD are co-interior angles or supplementary angles]

(b) u = 57º [EBC and AEB are alternate angles.]

(c) x = 180º 2 × 54º = 72º [ Angles in a triangle add up to 180º]

(d) y = 69º 54º = 15º [Opposite angles in a parallelogram are equal]

A 14 cm B

40º 7 cm

C

(b) (ii) cos 40º = 7/AC AC = o40cos

7= 9.14 cm

Actual size of AC = 5 × 9.14 = 45.7 m.

(iii) Reflex angle BAC = 360º 40º = 320º.

Jersey

No Jersey

Scarf

No Scarf

Scarf

No Scarf

8

7

3

2

6

2

8

1

3

1

3

2

6

4

(b) (i) 121

32

81

(ii) 241

31

81

(iii) P(Jersey, No Scarf) + P(no Jersey, Scarf)=

31

248

31

81

31

87

(c) 120

1101

32

81 ?

[To me this question is not well set. There is only

something known about the probability of wearing

gloves under the condition that he wears a jersey

and a scarf. That does not give the probability for

wearing gloves when he has not put on a jersey

and a scarf.]

29

3 (a) PSR = 60º because PSR and QPS

are alternating angles. [PQ is parallel to RS]

(b) (i) 60º + 70º = 130º (ii) 130º + 180º = 310º (iii) 70º + 180º = 250º

(c) △PQT is similar to △SRT because QPS = PSR see (a).

As well PTQ = RTS (vertically opposite angles)

The triangles have two equal angles so even the third angle will be equal

(180º sum rule.) Conclusion the two triangles are similar.

Similar triangles, so the corresponding sides have the same ratio:

TR

TQ

TS

PT or

TQ

TQ

8536

54 after simplifying the first fraction we get:

TQ

TQ

852

3

apply cross multiplication: 3(85 TQ) = 2TQ 255 3TQ = 2TQ 5TQ = 255 TQ = 51 m

4 (a) speed × time = distance or time = dist ÷ speed 0.75 ÷ 5 = 0.15 h = 0.15 × 60 = 9 minutes so the boat arrived

in S at 23 05

(b) First find S: sin S = 300 ÷ 750 = 2 ÷ 5 = 0.4 so R = sin-1

0.4 = 24º. Bearing is now: 270 + 24 = 294º.

5 (a) (i) PQM and LMQ are co-interior angles; the add up to 180º x = 56º

(ii) PQM and MQR are supplementary angles so MQR = 56º. Angles in △QRM add up to 180º:

y = 180º 2 × 56º = 68º [△QRM is an isosceles triangle]

(b) (i) V = Y = 90º and WXV = YXZ (vertically opposite angles) so the two triangles are similar.

(ii) It is easy to see that the big triangle is 4 times bigger than the small triangle YZ = 4 × 25 = 100m

Or with ratios: XY

YZ

VX

WV or

16040

25 YZ so that m100

40

16025

YZ

6 (a) (i) x = 60˚ (ii) y = 60˚ (iii) z = 60 + 35 = 95˚ [Draw a line through F, parallel to AC]

(b) Equilateral triangle.

Section 18 Factorization; simplification of expressions and completion of square.

1 (a) (i) x(3x + 2) (2x + 4) = 3x2 + 2x 2x 4 = 3x

2 4 (ii)

x

x

ax

ax

xax

xax 2222

)1(

)1(x

(b) 7x2 63 = 7(x

2 9) = 7(x + 3)(x 3)

2 9x2 1 = (3x + 1)(3x 1)

3 Look for two numbers product 12 sum 7; these are 3 and 4 so x2 7x + 12 = (x 3)(x 4)

4 (a) Expand and simplify: (x 1)(x2 + x + 1) = x

3 + x

2 + x x

2 x 1 = x

3 1

(b) Factorise: ax bx 3ay + 3by = x(a b) 3y(a b) = (a b)(x 3y)

5 (a) (i) 4(3 2p) 3(1 p) = 12 8p 3 + 3p = 9 5p

(ii) (3q r)(q + 2r) = 3q2 + 6qr qr 2r

2 = 3q

2 + 5qr 2r

2

(b) Factorize completely: 18t2 2 = 2(9t

2 1) = 2(3t + 1)(3t 1)

P

Q

R

S

T

N

N

60º

70º

N

30

6 (a) Express as a single fraction:

)32)(2(

236

)32)(2(

841510

)2)(32(

)2(4

)32()2(

)32(5

32

4

2

5

pp

p

pp

pp

pp

p

pp

p

pp

(b) Simplify: 12

1

)1)(12(

)1)(1(

132

12

2

q

q

qq

qq

qq

q

7 x2 + 6x + 1 = x

2 + 6x + 9 9 + 1 ……….. the 9 is needed because (x + 3)

2 = x

2 + 6x + 9

= (x + 3)2 8 so p = 3 and q = 8.

8 (a) (x 2)2 = x

2 4x + 4 (b) x

2 4x 3 = x

2 4x + 4 4 3 = (x 2)

2 7

9 3(x 2)2 (3x 2) = 3x

2 12x + 12 3x + 2 = 3x

2 15x +14

10 (x + 2)2 = x

2 + 4x + 4 so 5x has to be added to x

2 x + 4.

11 (a) (i) 8x 4(x + 3) = 8x 4x 12 = 4x 12

(ii) 7)7)(3(

)3(

214

32

2

a

a

aa

aa

aa

aa

(b) (i) 2a2 + 2ab b

2 ab = 2a(a + b) b(b + a) = (a + b)(2a b)

(ii) 6x2 x 1 = 6x

2 3x + 2x 1 = 3x(2x 1) + 1(2x 1) = (3x + 1) (2x 1)

12 (a) 2x 3(x + 2) = 2x 3x 6 = x 6

(b) (2x + 3)(x 5) = 2x2 10x + 3x 15 = 2x

2 7x 15

13 (a) (i) c

ab

bc

ab

5

6

15

18 23

(ii) 4

3

)4)(4(

)3)(4(

16

122

2

x

x

xx

xx

x

xx

(iii) )1(2

11

4

82

x

xx

= 644331

44

331

)1(22

131

222

22

)2(

)2(2

xxx

x

xx

x

xx

= 64

(b) x3 + 4x

2 9x 36 (x

3 5x

2 6x 2) = 9x

2 3x 34

(c) (i) Substitute x = 2 2y in x2 + 4y

2 = 8 you get: (2 2y)

2 + 4y

2 = 8 or 4 8y + 4y

2 + 4y

2= 8 or

8y2 8y 4 = 0 divide by 4: 2y

2 2y 1 = 0

(ii) 2y2 2y 1 = 0 → 2[y

2 y +

41

41 ] = 0 →2[(y ½)

2

41 ] = 0 → 2(y ½)

2 = ½ or (y ½)

2 =

41

So y ½ = ±½ so y = 0 or y = 1

(iii) Points of intersection are (2, 0) or (4, 1) by substituting y = 0 or y = 1in x = 2 2y.

14 First work out a(x + p)2 + q = a(x

2 + 2xp + p

2) + q = ax

2 + 2axp + ap

2 + q now 2x

2 + 3x – 2 is equal to

2axp + ap2 + q for all values of x. So 2x

2 + 3x – 2 ≡ ax

2 + 2axp + ap

2 + q.

So 2x2 = ax

2 → a = 2; 3x = 2axp or 3x = 4xp → 3 = 4p → p =

43 = 0.75.

Lastly we have −2 = ap2 + q using what we have found already it changes into: −2 = 2 × (

43 )

2 + q or

−2 = 181 + q so that q = −3

81 .

31

15 (a) (i) 2a – 3b – 4xa + 6xb =1(2a – 3b) – 2x(2a – 3b) = (2a – 3b)(1 – 2x)

(ii) x2y

2 + x

2 + y

2 + 1. = x

2(y

2 + 1) + 1(y

2 + 1) = (x

2 + 1)(y

2 + 1)

(b) )3)(3(

135

)3)(3(

1552

)3)(3(

)3(5

)3)(3(

2

3

5

9

22

xx

x

xx

x

xx

x

xxxx

Section 19 Coordinate geometry

1 (a) Midpoint AB

2

62,

2

41½

(b) AB = 222614 = 25169 5

(c) Gradient of AB = 311

3

4

14

26

(d) y = cx311 A(1, 2) on the line so 2 = c11

31

c = 32 so the equation is: y =

32

311 x

(e) C(7, 2)

(f) Gradient AB is 311 so BAC = tan

-1

311 = 53º DAB = 180º 53º = 127º cos DAB = 0.6

One can as well use the cosine rule in △DAB.

2 [P2 Oct/Nov 2007 Q 11 (a)]

(a) PQ = 1394)96()24( 22 3.61

(b) PQ has gradient 2

3

42

69

= 1½. So the equation of the line is y = 1.5 x + c; Q(4, 6) on the line so

6 = 6 + c ; this means c = 12 so that the requested equation is: y = 1.5 x + 12.

3 (a) Make a sketch on your answer sheet!

Midpoint AC:

2

15,

2

51 (2, 3)

(b) In a parallelogram the diagonals bisect each other so diagonal

BD has to go through the midpoint of AC.

Parallel to y = 2 5x so the gradient of BD is 5 as well.

Equation of BD is y = 5x + c M(2, 3) on the line: 3 = 10 + c

c = 13; the equation of BD is y = 5x + 13

(c) Gradient AC is 4 /6 = 32 BC has gradient

23

Equation BC is y = 23 x + c. C(5, 1) on the line so 1 =

23 × 5 + c c = 6.5; equation BC is y =

23 x 6.5.

(d) B is where BC and BD meet. Equate: y = 23 x 6.5 and y = 5x + 13

23 x 6.5 = 5x + 13 or

6.5 x = 19.5 or x = 3. Substitute this in y = 5x + 13 y = 15 + 13 = 2. So B( 3,2) indeed.

4 (a) midpoint AB:

2

15,

2

24= (1, 3)

D(-5, 2) C(7, 2) A(1, 2)

B(4, 6)

x = 4

x

y

A

C

1

1

2

2

3

3

3

3

3

3

4

3

3

4

3

3

5

3

3

6

3

3

5

3

3

-1

-2

-1

-3

-3 -2 -4

-4

-5

-5 x

y

32

(b) AB has gradient: 32 see diagram, as well:

32

6

4

)4(2

51

So equation of L1 is y = cx 32

C(1, -4) on the line 4 = c 132

This means equation L1 is:

(c) perpendicular to line AB that means gradient is 1½ (product gradients 1)

so the equation is y = 1.5 x + c since D is on the line

it means 0 = 1.5 × (5) + c c = 7.5 so L2 has equation: y = 1.5 x + 7.5

(d) Equate y = 1.5 x + 7.5 and 31

32 3 xy or 1.5x + 7.5 =

31

32 3 x ×6 9x + 45 = 4x 20 or

13x = 65 so that x = 5 and y = 0 [substitute x = 5 in y = 1.5 x + 7.5] Point of intersection: (5, 0)

5 (a) OC = {(5 0)2 + (5 0)

2} = 50 ≈ 7.07.

(b) gradient is 1 [ (5 0) ÷ (5 0)] = 1

(c) y = x + c and the line passes through A(12,0) so 0 = 12 + c c = 12. Eq. through AB is y = x 12

(d) Substitute y = 5 in y = x 12 5 = x 12 so x = 17 so that B(17, 5)

6 (a) (16 4) ÷ (11 2) = 12 ÷ 9 = 1 1/3

(b) so the gradient of BC = 3/4

(c) Line through BC has equation y = ¾ x + c substitute B(11, 16) in it: 16 = ¾ × 11 + c so c = 23 ¾

So the equation through BC is y = ¾ x + 23 ¾ substitute y = 19 in it: 19 = ¾ x + 23 ¾ x = 6 1/3.

7 (a) A has y co-ordinate 4 so x + 2 4 = 6 so x = 6 8 = 2. Conclusion: A(2, 4)

(b) At B the y co-ordinate is zero so 3x = 12 → x = 4 so B(4, 0).

(c) (i) M has x co-ordinate ½(2 + 4) = 1 and y co-ordinate ½(4 + 0) = 2 so M(1, 2)

(ii) rewrite 3x + y = 9 → y = 3x + 9 so the gradient is 3. The perpendicular has now a gradient of

So the equation is y = x + c ; so M(1, 2) is on this line so 2 = + c so c = 2

The equation is y = x + 2

8 (a) The line can be written as y = mx + c or y = 31

x +c. (9, 4) is on the line so 4 = 31 9 + c so c = 7.

The line is y = 31

x 7…*. In the requested format: 31

x y 7= 0 or x 3y 21 = 0

(b) Equation line CD is y = 2x. Equate this equation to * and you get: 2x = 31

x 7 multiply with 3:

6x = x 21 or 7x = 21 so x = 3; use y = 2x to find y gives y = 6 so E(3, 6).

9 (a) The formula to remember is MAB

2,

2

BABA yyxx so here: M(

2

53

2

42,

) = M(−1, −4)

(b) Use: 22 )()( BABAAB yyxxd here: distance = √[{2−(−4)}2 + {(−3) – (−5)}

2] = √[36+ 4] = √40

[You may leave √40 as an exact answer, if you round off you have to do it correctly: √40 = 6.32]

(c) Gradient through line AB is AB

ABAB

xx

yym

here:

3

1

6

2

24

)3(5

ABm so the perpendicular has gradient −3.

The requested equation is now: y = −3x + c and the line passes through M(−1, −4) so −4 = 3 + c so c = −7.

The required equation is : y = −3x – 7.

So c = 331

A

C

1

1

2

2

3

3

3

3

3

3

4

3

3

4

3

3

5

3

3

6

3

3

5

3

3

-1

-2

-1

-3

-3 -2 -4

-4

-5

-5

x

y

B

D

31

32 3 xy

31

31

31

31

31

31

33

Section 20 Ratio & Direct and inverse proportion

1 $1.20 ÷ 0.75 = 1.6 kg

2 Answer: 18$30$5

3

3 (a) (b) 450 : 550 = 45 : 55 = 9 : 11 (c)

4 (a) 22500 : 37500 = 225 : 375 = 9 : 15 = 3 : 5.

(b) Ada’s profit is $3600 so total profit is 8 × (3600 ÷ 3) = $9600.

(c) Use a three lined table:

or: 112.5∼22500 so

answ: 1005.112

22500 = $20 000

5 (a) (i) 753

175

6

2 = 25ℓ

(ii) 5ℓ 2 ÷ 5 = $0.40 per ℓ; 25ℓ 8.75 ÷ 25 = $0.35 per ℓ and 27 ÷ 75 = $ 0.36 per ℓ.

Conclusion: 25ℓ bag is the cheapest.

(b) (i) Volume plant pot = volume large cone volume small cone = 31 × R

2 ×H 31 × r

2 ×h =

= 31 × ×10

2 × 24

31 × × 5

2 × 12 = 800 100 = 700 = 2199 cm

3 = 2.20 ℓ

(ii) 75 ÷ 2.20 = 34 pots

(iii) Ratio of the lengths is length large pot : length small pot = 2 : 1 volumes have ratio 8 : 1 so

8 times more pots can be filled. So 8 × 34 = 272 small pots can be filled.

6 (a) 31

155

5375

(b) 153 2250 = 450 tonnes = 450 000 kg

(c) 231 times more sp 8 2

31 = 19 tonnes.

Section 21 Order of numbers; number properties and calculator skills

1 6500 s = 6500 ÷ 60 = 108.3 m and 431 h = 105 min, so order is:

431 h 6500 s 110 m.

2 (a) 99 = 3 × 33 = 3 × 3 × 11

(b) 24 = 23 × 3 so 99n has already a factor 3 but it still needs an 8 so n = 8

3 (a) 73 37 = 36

(b) 64 46 = 18 81 18 = 63 72 27 = 45

(c) All results are multiples of 9.

(d) (i) 10x + y (ii) 10x + y 10y x = 9x 9y

4 2312323

)2(63 3

3

= 3.56 in 3 significant figures. [Use the brackets on your calculator!]

5 Put in descending order: 81 = 0.125; 12.56% = 0.1256;

250631 = 1 +

25063 = 1 +

1000252 = 1.252 and 0.1251

So order is: 250631 > 12.56% > 0.1251 >

81 [Use the original numbers!]

Invest 2007 100 100 ÷ 112.5 (100 ÷ 112.5) × 22500 = $ 20000

Increase 12.5 12.5 ÷ 112.5

Invest 2008 112.5 1 1 × 22500 = 22500

g1502505

3 %48%100

1250

450150

34

6 Given: 4

9, 3 729 ,

2

9, 9 Rational are:

4

9; 3 729 and 9

7 (a) Prime numbers: 5; 17 (b) Square numbers: 1; 9 and 121 (c) Irrational no: √2;

8 (a) Modern calculators show 2 options: 0.1571348403 or 9

2 (b) 0.1571

9 No of students that take German = 135 31

135 53

135 = 135 45 81 = 9.

10 (a) (i) 2 (ii) 18 (iii) 8 (iv) 2

(b) (i) 0.2 < (ii) 2 < 0.2

(c) (i) 3.74 (ii) 2.41

11 (a) 167 (b) 169 (c) 168 (d) 165

12 First use calculator to find: cos 80°= 0.174 tan 275° = − 11.4 sin 122° = 0.848 sin 195 = −0.259 so

sin 122° > cos 80° > sin 195 > tan 275°.

13 (i) √5 (ii) −4 (iii) 8 (iv) 2 (v) 0.3

Section 22 Indices

1 (a) 40 4

2 = 1

1615

161 (b) (2x

2)3 = 2x

2 × 2x

2 × 2x

2 = 8x

6

2 (a) k = 2 [Since 50 = 1 and indices add up.]

(b) m773 or m77 3

1

so m = 31

3 Simplify: 22)1(3)1(3

11

33

1

33

1

33

323232

32

)32(

32

6

32

nnnn

nn

nn

n

nn

n

nn

= 36

4 163

2

x 323

43

2

x x2 = 4

6 x = 4

3 so x = 64.

5 Simplify (a) 2

41

5.045.045.0

4

41616x

xxx

(b) 32 y3

23y2

= yy

y

y

y

8

9

8

9

8

93

2

2

3

6 Take log from both sides: log 4x + 1

= log 18 → (x + 1) log 4 = log 18 so x = (log 18 log 4) 1 = 0.244

7 (a)

1

2

2

2

3

=

2

2

3

2 =

9

4 (b)

9

4= 0.4.

43

35

Section 23 Sequences

1 (a) 57 25 = 57 32 = 25 (b) 56 55 52 45 so the n

th term is 57 2

n + n

2 (a) T3 = 11 a + 2d = 11 ….. T12 = 7 a + 11d = 7 ….. then gives: 9d = 18 so d = 2.

Substitute this in a + 2 ×(2) = 11 so a = 15

(b) Sum of the first 24 terms Sn = ½ n[2a + (n 1)d] so S24 = ½ × 24[2 × 15 + 23 × (2)] =

12 × (16) = 192

3 (a) (i) 50c + N$3 = N$3.50

(ii) This is an arithmetic progression with a = ½ and d = 3 so Tn = a + (n 1)d or T8 = 50c + 7 × 3 = N$21.50

(iii) Apply the formula Sn = ½ n[2a + (n 1)d] : S8 = ½ × 8[2 ×½ + (8 1) ×3] = N$88

(b) (i) Week1: N$0.50; week 2: N$1; week 3: N$2 etc.

(ii) This is a geometric progression because the pockets money for the next week comes from the previous week

through a multiplication.

(c) Ben: T8 = a × rn 1

= 0.50 × 27

= N$64. So the difference = 64 21.50 = N$42.50.

4 (a) (i) 50 60 N$70

(ii) a = 50 and d = 10 so Tn = a + (n 1)d in this case: T7 = 50 + 6 × 10 = N$110

(iii) Sum formula is Sn = ½ n[2a + (n 1)d] here: S12 = ½ × 12[100 + 11 × 50] = N$3900

(b) (i) Month 2: N$10; month 3: N$20; month 4: N$40.

(ii) Use: 1

)1(

r

raS

n

n here: 12

)12(5 12

12

S = 5 × 4095 = N$20475.

5 T1 = 3 so 3 = a + b….. T2 = 10 so 10 = 4a + 2b ….. These form two simultaneous equations:

2 × gives: 2a + 2b = 6….. and gives 2a = 4 so that a = 2. Substitute a = 2 in you get b = 1

6 (a) (i) Next two terms are: 4729

2918 ,

(ii) Next two terms are: ba

ba

ba

ba

32

2,

2

(b) (i) 4

5,

3

4

aa (ii)

49

50

a (iii)

2

1

9

10

a cross multiplication: 20 = a + 9 a = 11

7 (a) Following sequence can be formed: 2, 1 ,½ ,…. Etc being a geometric sequence with a = 2 and r = ½

T6 = 2 × (½)5 =

161 m

(b) Tn = 2 × (½)n 1

(c) 2 × (½)n 1

= 512

1 n = 11 [remember 210

=1024]

8 (a) phase 1 → 1 = 20 cell (t1) ;

phase 2 → 2 = 21 cells (t2);

phase 3 → 4 = 22 cells (t3);

phase 4 → 8 = 23 cells (t4);

(b) This is a geometrical sequence because the next term is found by multiplying the previous term with 2.

Compare the two sequences

Phase 20 has 219

= 524288 cells.

36

9 (a) (i) 25 = 5 5; 30 = 6 5 etc until 300 = 60 5 so there are 60 4 = 56 terms.

(ii) Use the formula or faster: ½ 56 (25 + 300) = 28 325 = 14 650 = 7 1300 = 9100

(b) (i) arithmetic so t3 t2 = t2 t1 or log 3x log 3

3 = log 3

3 log 3

or x log 3 3log 3 = 3 log 3 log 3

Take log 3 outside brackets: log 3(x 3) = log 3 (3 1) divide by log 3: x 3 = 2 so x = 5

(ii) geometric then constant ratio: 1

2

2

3

t

t

t

t or

3log

3log

3log

3log 3

3

x

rewrite: 3log

3log3

3log3

3log

x cancel log 3

That makes x = 9

10 Geometric progression → constant ratio so QP

QP 288.12

3 these are 3 equations on which one can apply

Cross multiplication: ➀: P2 = 3 Q; ➁ Q

2 = 12.288P and ➂ PQ = 12.288 × 3

From ➀ we get P = √[3Q] substitute this in ➁: Q2 = 12.288 √[3Q] or Q

2 ÷ √Q =12.288 √3 so that

Q1½

= 12.288 √3 square both sides: Q3 = 3 × 12.288

2 so Q = 3 2288.123 = 7.68

P = √[3 × 7.68] = 4.8

11 (i) Term 1 2 3 4 5 n

Sequence A 7 11 15 19 23 4n + 3

Sequence B 1 8 27 64 125 n3

Sequence C −5 0 5 10 15 5n −10

Sequence D 6 14 24 36 50 n2 + 5n

(ii) n2 + 5n = 500 or n

2 + 5n – 500 = 0 (n + 25)(n – 20) = 0 so n = 20 so the 20

th term is 500.

Section 24 Polygons

1 (a) Let x be the interior angle and y the exterior angle. Valid is now x = py … Valid is as well: x + y = 180º ….

Substitute x from in to you get py + y = 180º so that y(p + 1) = 180º and y = 1

180

p

o

(b) 360º ÷ exterior angle is the number of sides no of sides is: 360º ÷ 1

180

p

o

= 360º ×180

1p= 2p + 2

2 There are 3 exterior angles of 180º 150º = 30º, the other interior angles are 45º.

The sum of all exterior angles is 360º. Let there be n exterior angle of 135º so 3 × 30º + 45n = 360º so 45n = 270º

so n = 6. So this polygon has 9 sides.

3 Exterior angle is 360º ÷ 10 = 36º. So the interior angle is 180º 36º = 144º.

4 (a) Exterior angle is 180º 140º = 40º. 360º ÷ 40º = 9 9 sided regular polygon.

(b) 360º 2 × 140º = 360º 280º = 80º. [Congruent polygons have the same interior angle.]

5 (a) false [All equilateral triangles are similar, not congruent]

(b) true

(c) false

6 In a heptagon (7-sided polygon) one can draw 5 triangles. So the sum of the internal angles is 5 180˚ = 900˚. So one of the remaining angles is [900˚ 4 120˚] 3 = 420˚ 3 = 140˚

37

7 (a) (i) By drawing two diagonals from one vertex of a pentagon you can see it consists of 3 triangles→

sum of the angles is 3 × 180 = 540 so 540 = x + 3x + 2x + 104 + 85 or 6x = 351 so x = 58.5°so the smallest

angle is 58.5°.

(ii) Largest exterior angle is 180 – 58.5 = 121.5°.

(b) [This question is not properly set; omitted is the word regular polygon]

(i) Hexagon

(ii) The area of the hexagon is 6 × area of an equilateral triangle with side 8 cm.

The height of an equilateral ∆ = ½ × base × height = ½ × 8 ×√[82 −4

2] = 4√32

Shaded area is area circle – 6 × area eql∆ = π×64 – 6 × 4√32 = 65.3 cm2

Section 25 Similar & congruent shapes

1 (a) ABC = 180º xº yº

(b) △ABC is similar to △XYZ so YZ

BC

XZ

AC

XY

AB fill in what is given:

54

3 BC

XZ

AC we can ignore the

middle term; using cross multiplication we get: 4 × BC = 3 × 5 so that BC = 15 ÷ 4 = 3.75 cm

(c) The length are in the ratio 3 : 4 the areas will be in the ratio 9 : 16

2 △ PQR is similar to △ PSQ SQ

QR

PQ

PR

PS

PQ * [The ratio’s are obtained by selecting the first two letters

from both triangles. The sequence of the letters in the triangle is not at random but RPQ = SPQ etc.]

Replace all known distances in * 21

14

18

18

PR

PS.

(a) PR = cm 1221

1418 (b)

18

1218

PSso 12PS = 18

2 or PS = 27. Then RS = 27 12 = 15cm

3 (a) △APR; △CQR and △PBQ are congruent because AP =RC=BQ; A = B = C = 60º and

AR = QC = BP (△ABC is an equilateral triangle) [case: SAS]

(b) (i) ABC

PQR

triangleof Area

triangleofArea =

25

16

5

42

2

(The sides are in the ratio 4 : 5 so areas have ratio 42 : 5

2)

(ii) ABC

APR

triangleof Area

triangleofArea=

25

3 (25 16 = 9 and there are three such triangles 9 ÷ 3 = 3)

4 Mass in the ratio 135 : 320 = 27 : 64 length in the ratio 4:364:27 33 ;

so the smaller statue has height 6.143 = 1.2m. [The volume and mass are direct proportional]

5 Corresponding lengths are in the ratio 6 : 12 : 18 = 1 : 2 : 3 areas are in the ration 12

: 22 : 3

2 = 1 : 4 : 9

X : 500 = 1 : 4 so X = 125 cm2. Lengths are in the 1 : 2 : 3 volumes are in the ratio 1

3 : 2

3 : 3

3 = 1 : 8 : 27

So 27

1400

Y so Y = 400 × 27 = 10800 cm

3

[NB this question is not well set up; the rule Volume is base area × height does not work out here for the small

cylinder: (6 × X 400); very confusing for candidates who know this often applied rule]

6 (a) Angle AEB = 56º as well; reason: They are corresponding angles.

(b) Triangle ABE is similar to triangle ACD.

8

4

h

38

(c) Since they are similar triangles the following ratio is valid: CD

AC

BE

AB substitute:

12

15

8

AB

So that 12 AB = 120 [Cross-multiplication] or AB = 10 cm.

(d) The sides in triangle ACD are 1.5 times bigger than the corresponding sides in triangle ABE.

This means the area of triangle ACD is 1.52 times bigger than the area of triangle ABE.

Area triangle ACD = 1.52 × 36 = 81 cm

2

7 Since PQ is parallel to RS it follows that P = S and Q = R. So ∆PQO is similar to ∆SRO.

In addition PQ = RS (given) so ∆PQO is congruent to ∆SRO based on the case ASA (angle-side-angle).

8 (a) Lengths small pot : Lengths Large pot = 6 : 18 = 1 : 3 so the diameter sp : diameter lp = 1 : 3 and area = x : 9 so

93

1 x or 3x = 9 so x = 3 so the small pot has a diameter of 3 cm.

(b) lengths in the ratio 1 : 3 so areas in the ratio 12 : 3

2 = 1 : 9

Section 26 Constructions & locus 1 (a) See diagram

(b) (i) See diagram: the two thick

lines parallel to AC.

(ii) Dotted line through B.

(c) PQ = 5.7 cm

6 cm 8 cm

D

A

B

C

2.5 cm 2.5 cm

2.5 cm 2.5 cm

P

Q

39

2 see diagram on the right.

3

4 (a) & (b)

A

B

C 1.5 cm 1.5 cm

T

7 cm

A B

C D

6 m

8 m

40

13 cm

5 cm

A

B C

D

T

5

6 (a) 120 ÷ 20 = 6 cm; 180 ÷ 20 = 9 cm and 260 ÷ 20 = 13 cm

Section 27 Transformations

1 (a) (i) Angle is 270º (ii) Centre (2, 0)

(b) (i) Scale factor enlargement: 2

(ii) Centre of enlargement: (1, 1)

2 (a) Reflection in the line y = 1

(b) (i) Rotation of 90º about (3, 3) anticlockwise (ii) Reflection in the line y = x

(c) P′ (2, 10)

1 2 3 4 5 6 0 1 2 x

3

1

2

3

4

5

y

A

B

C

(a) 12cm : 360 m = 12 cm : 36000 cm

= 1 : 3000

(b) 360˚ 60˚ = 300˚ (c) See diagram

(d) (i) see diagram

(ii) 6.1 cm

N

60˚

X

Y

12 cm

Z

A B

C

41

3 (a) & (b) see diagram

Translation

2

7

4 (a) Reflection in the line x = 3 (b) Translation

1

4 (c) Rotation 90º anti clockwise about (0, 2)

5 (a) Translation

2

4

(b) A''(4, 6)

(c) (i) Triangle ABC ≡ to triangle A'B'C' { ≡ means is congruent to}

(ii) Triangle ABC ||| to triangle A''B''C'' { ||| means is similar to}

6 (a) Enlargement of factor 2; centre (1, 1)

(b) Reflection in the x-axis. (c) Translation

1

3 (d) Rotation about the origin of 180˚.

1 2 3 4 5 6 7 8 0 -1 -2 -3 -4 -5 -6 -7 -8

-1

-2

-3

-4

-5

-6

1

2

3

4

5

6

7

8

8

x

y

P

Q

R

A

B

C

y= 1

P1

Q1

R1

P2

Q2

R2

42

7

Section 28 Trigonometry

1 Area triangle is ½ × side × side × sin(angle in between) = ½ × 5 × 6 × sin 150º = 7.5 cm2

43

(a) (i) Reflect

triangle A in the

line y = 4.

Label it B [1]

(ii) Rotate triangle

A through 180°

around the origin.

Label it C. [2]

(b) Describe fully

the single

transformation

that maps

(i) Translation

5

8

(ii) Enlargement,

Centre (−6, 0) and

factor ⅓.

E

A

1 2 3 4 6 7 8 0 -1 -2 -4 -5 -6 -7 -8

-2

-1

-3

1

2

3

4

8

x

y

5

-4

5

8

6

8

7

8

8

8

D

y=4

B

C

-3

2 (a) Use cos in △ABC: cos 55º = BC

20 or

oBC

55cos

20 = 34.9 cm

(b) AB′ = 20 + 34.9 = 54.9 cm

(c) A′CB′ = 55º as well so sin 55º = 20

h so the requested height h = 20 sin 55º = 16.4 cm

3 (a) (i) Use sine rule in △ABD: 600

ˆsin

950

118sin BDAo

BDA ˆ = sin-1

950

118sin600 o

= 34º

so ABD = 180º 118º 34º = 28º

(ii) Use cosine rule in △BDC: DC2 = BD

2 + BC

2 2BD × BC cos B ; replace all known values:

DC2 = 950

2 + 1040

2 2 × 950 × 1040 cos 42º DC = 718 m

(iii) sin 42º = h/1040 so the height is 1040 × sin 42º = 696 m (draw a line through C perpendicular to BD)

(b) tan-1

(500/ 696) = 35.7º (imagine a line straight down from the helicopter to BD and a line connecting the

helicopter to C. In this right angled triangle the basic trig ratio’s can be used.)

4 (a) In △PCB tan PCB = BC

PB so tan 20º =

10

PB so that PB = 10×tan 20º = 3.64 cm

(b) In △ABC tan BAC = AB

BC so tan 40º =

AB

10 this means AB =

o40tan

10= 11.9 cm. This makes AP= 8.28 cm

(c) In △PCB cos PCB = PC

BC so cos 20º =

PC

10 this means PC =

o20cos

10= 10.6 cm.

The perimeter is now 10 + 10.642 + 3.6397= 24.3 cm [This is a better answer than 24.2, the 24.3 is obtained by

using numbers that have not been rounded off. Always round off at the final stage of the calculation. If you use

values based on earlier calculations, do not use the rounded off values but values with a higher accuracy.]

5 (a) Apply sine rule in △ABC: AC

CBA

BC

CAB ˆsinˆsin or

11

ˆsin

5.5

25sin o CBA

5.5

25sin11ˆsin

o

CBA

The calculator gives 57.7º for sin-1

5.5

25sin11

o

this means CBA ˆ = 180º 57.7º = 122.3º

6 (a) tan CAB = 200 ÷ 65 so CAB = tan-1

(200 ÷ 65) = 72.0º

7 (a) (i) Use the cosine rule in ∆ABD: AD2 = AB

2 + BD

2 2AB×BD × cosB or

AD2 = 16

2 + 24

2 2 × 16 × 24 × cos 112º so AD = 33.5 cm

(ii) DBC = 180º 112º = 68º. In △BDC the sine rule can be used: DC

CBD

BD

DCB ˆsinˆsin or

20

68sin

16

ˆsin o

DCB

20

68sin16ˆsin

o

DCB = 0.741747 angle BCD = sin-1

0.741747 = 47.9º

(iii) Area ∆ABD = ½ AB.BD sin 112º = ½ × 24 × 16 × sin 112º = 178 cm2

(b) Scale 1 : 250 000 so 1cm on the map is 250 000 cm = 2500 m = 2.5 km so AB is 24 × 2.5 km = 60 km.

44

8 (a) Use cosine rule in triangle DCF: FCDCFDCCFDCDF ˆcos2222

DF = 82cos12402031 22 = 34.473545724359698496390947489216 ≈ 34.5 m

(b) This has to be done with the sine rule: DF

FCD

CF

FDC ˆsinˆsin

5.34

82sin

20

ˆsin

FDC angle CDF = 35.0º

So = 90 35.0 = 55.0º

9 (a) DB = (202 + 10

2) = 500; use the cosine rule in ∆BCD: DC

2 = BC

2 + DB

2 2 BC DB cos 40˚ →

DC = (169 + 500 2 500 13 cos 40˚) = 15.0 m

(b) (i) tan ADB = 10/20 → angle ADB = 26.6˚

(ii) Angle BDC with the sine rule: BC

BDC

DC

DBC sinsin or

13

sin

15

40sin BDC → angle BDC = 33.9˚

10 (i) Use cosine rule: AB2 = AC

2 + BC

2 – 2AC BC cosACB substitute the lengths:

2602 = 120

2 + 180

2 – 2 × 120 × 180 cosACB so ACB = cos

−1

1801202

260180120 222

= 118.8°

(ii) Area park is ½ AC × BC × sin 118.8° = ½ × 120 × 180 × sin 118.8 = 9470 m2

(iii) Use sin rule: BC

BAC

AB

ACB sinsin substitute what is known:

180

sin

260

8.118sin BAC

So BAC = sin−1

260

8.118sin18037.3°

Section 29 Graphs of functions

1 (a)

0 1 2 3 4 5 6 7 8 9 t

y

1

2

3

4

45

(b) (i) gradient is 46.013

6

5.6

3

[Use the numbers on the axes for both distances]

(ii) The water level is declining with a rate of 0.46 m/h.

(c) (i) 4m [replace t = 0 in the relation: y = 4 ½ t]

(ii) 80 cm (See the vertical double arrow at t = 4 [noon])

(iii) t = 5.75 (See graph where the curves meets the straight line)

2 (a) See graph in the Cartesian diagram.

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

10

20

30

40

50

x

y

46

3 (a) x2 + 10x + 24 = 0 (multiply both sides with 1) x

2 10x 24 = 0 or (x 12)(x + 2) = 0 so x = 12 or x = 2

Whereby x = 2 most likely not acceptable.

(b) (i) See graph (ii) See graph (iii) 45 m (iv) About 8.8 m [ = 9.4m 0.6m]

(c) (i) x2 + 10x + 24 = p (x 5)

2 work out the square x

2 + 10x + 24 = p x

2 + 10x 25 or 24 = p 25 so p = 49

(ii) (a) greatest height is: 49m (b) 5 m

y

32

(b) gradient is

02

)6.6(4.6

= 6.5

(c) (i) see graph

(ii) (1.3, 0.7)

(iii) Equate:

y = 2 x and

x

xy2

52

2 x = 2x x2

5 or

2 3x + x2

5 = 0

Multiply both sides

with (2x) :

6x2 4x 5 =0

So B = 4 and C = 5.

(d) (i) f(a) = b or:

aab

2

52 given.

Then

f(a) = a

a2

52 =

b indeed.

(ii) see graph.[first

reflect in the x-axis,

then reflect in the y-

axis. Or a rotation of

180º with centre the

origin.

1 2 -1 -2

-2

-4

-6

-8

-10

2

4

6

8

10

0

P

47

5 (a) Figure 1 (b) Figure 3 (c) Figure 2.

4.(a) see graph on the right hand side:

(b) f(x) = 0 for x = 1.2 or x = 2 or

x = 3.4

(c) Any value between 0 < k ≤ 14

will do.

(d) Rotational symmetry of order 2

(e) (i) (10 + 8) ÷ 1.6 = 11.25

(ii) At point (3.7, 4)

10

2 1 1 2 3 4

5

5

10

15

20

x

48

6 (a) s = 1.3 t = 12 u = 3

Section 30 Logarithms and exponential equations

1 Use logarithms to solve for x: 0.2x = 20 take log from both sides: log 0.2

x = log 20 or x log 0.2 = log 20

2.0log

20logx -1.86

2 [P4 NSSC 2011 Q 11]

(a) 3p =

27

1 or 3

p =

33

1 or 3

p =3

3 so p= 3.

(b) 4log (x + 2) ½ log x = log (x + 2)4 log √x =

x

x 4)2(log

(c) 5x+1

= 6 take log from both sides: log 5x+1

= log 6 or (x + 1)log 5 = log6 so that x log 5 + log 5 = log6

x log 5 = log 6 log5 so that x =

5log

2.1log

5log

5log6log0.113

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

2

4

6

8

10

12

14

16

18

y = 8

(c) (i) x = −0.7 or

x = + 0.7

(ii) Draw the graph

of y = 2x + 6

x = −0.9 or x = 0.7

(d) m = −1.4 ÷0.85

= − 1.6

y = 2x + 6

49

(d) Work out (x3 7x + 6) ÷ (x + 2) x

2 2x 3

x + 2 x3 + 0x

2 7x + 6

- x3 +2x

2

2x2 7x

- 2x2 4x

3x + 6

- 3x 6

12

So the quotient is x2 2x 3 and the remainder is 12

3 5x = 4 take log from both sides log 5

x = log 4 x log 5 = log 4 so x =

5log

4log0.861

4 4 10log 3 1 = 10log 34 10log 10 = 10log

10

34

= 10log 8.1 or simply log 8.1.

5 2 2 10log 5 = 10log 100 10log 52 = 10log

25

100 = 10log 4 or simply log 4.

6 (a) 18000 + 18000 × 100

5.4 × 5 = N$22050

(b) (i) Initial amount is N$15000

(ii) 15000 × 1.0455 = N$18693

(c) (i) 25000 = 15000 × 1.045t 1.045

t =

500015000

500025000

gives: 1.045

t =

3

5

(ii) log 1.045t = log

3

5 t =

045.1log

log35

= 11.6 years.

7 (a) Simplify log p + 2 log q 3 log r = log

3

2

r

pq

(b) 3x 1

= 11 take log on both sides: log 3x 1

= log 11 or (x 1) log 3 = log 11 or x = 13log

11log = 3.18

8 Three different solutions.

I 22log

log

x

xcross multiplication: log x = 2 log 2x → log x = log (2x)

2 or x = 4x

2 so that 4x

2 x = 0

Factorize: x(4x 1) = 0 so that x = 0 (Not allowed) or x = ¼

II 22log

log

x

x…* let p = log x then * can be rewritten as 2

log2log

log

x

xor 2

2log

p

puse

cross multiplication: 2(log 2 + p) = p or 2log 2 + 2p = p or p =2 log 2 so that p = log 22

or p = log ¼.

Yet p = log x so we get log ¼ = log x this means x = ¼.

III The fourth law of logs, the one for change of base is: a

bb

c

ca

log

loglog whereby c is any positive

number (not 1)

Use law 4 on 22log

log

x

xand you get: 2log

2log

log2 x

x

xx rewrite in exponential form: (2x)

2 = x or

4x2 x = 0 so that x = ¼.

50

9 (a) 2 log x 5 log y + 1 = log x2 + log 10 log y

5 =

5

210log

y

x

(b) 32x

= 5; take log from both sides: 2x log 3 = log 5 so x = 9log

5log

3log2

5log0.7

10 (c) (i) 2 × 3x + 5

= 43.74 2bydivide3

x + 5 = 21.87 logtake

(x + 5) log 3 = log 21.87 3log

x + 5 = log 21.87 ÷ log 3 so that x = log 21.87 ÷ log 3 – 5 = −2.19.

(ii)

0)2(4

)2(12)2)(1(4203

4

1

2

2

:34

1

2

2

rdenominato same termsall give

x

xxxx

x

rewritex

x

A fraction cannot be made zero by its denominator so 8 + x2 – x − 2 – 12x + 24 = 0 or x

2 −13x + 30 = 0

This equation can be factorized: (x – 3)(x – 10) = 0 so x = 3 or x = 10

(d) log 2 = a → 2log 2 = 2a or log 22 = 2a (law 3) this means log 4 = 2a. Now consider log 12 = log (4 × 3) = log 4 + log 3.

(law 1 of logs) so log 12 = 2a + b

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