Section A (22 marks)

PAGE 1

Preliminary Examination II (2013)Secondary 4 Express/ 5 Normal Academic

Candidate

Name Register No MATHEMATICS PAPER 2 Date: 27 Aug 2013(4016/2)

Duration: 2 h 30 min

For examiners use

/ 100

Additional Materials: Answer sheet Graph paperREAD THESE INSTRUCTIONS FIRST

Setter: Mr Thong Nai KeeThis paper consists of 12 printed pages, INCLUDING the cover page.Mathematical Formulae

Compound Interest

Total amount =

Mensuration

Curved surface area of a cone =

Surface area of a sphere =

Volume of a cone =

Volume of a sphere =

Area of triangle ABC =

Arc length = , where is in radiansSector area = , where is in radians

Trigonometry

Statistics

Mean =

Standard deviation =

1 (a) Express as a single fraction.

[2](b) Solve the following equations

(i)

[2](ii)

[2]

(c) Make the subject of the formula given that

[2]

(d) Given that and , find the value of .[2]2. (a) (i) Express the shaded region in set notation. [1]

(ii) Draw and label clearly the sets P, Q and R on a Venn diagram such that they satisfy all the following conditions: [2]

(b) Given that y varies inversely as the positive square root of x, and y = 3

for a particular value of x.

Find the value of y when this value is reduced by 75%.

[3]

(c) John will be twice as old as Peter in four years time.

The sum of their ages three years ago was 22 years. Find the present age of John and of Peter.

[3]

(d) Given that A = and B = find(i)2A ( B,[1]

(ii)A2.[2]

3. (a) Helen wanted to buy a LCD television set priced at $3000. She paid a downpayment of 25% and the rest was paid over 24 equal monthly instalments. The interest was charged at a flat rate of 6% per annum. Find

(i) the downpayment amount,

[1]

(ii) the hire purchase price,

[2]

(iii) the amount of each instalment.

[1]

(b) May deposits $22000 in a fund that pays compound interest of 1.88% per annum, compounded yearly.

(i) Calculate the total amount of money that May will have in the fund at the end of six years. [2] At the end of six years, May withdraws $18000 from the fund and invests this amount in a bank which offers a simple interest rate of 0.9% per annum.

(ii) Find the minimum number of full years she has to leave the money in the bank in order for the total amount to be more than $20000. [2]4.

(a)The points B and C lie on a circle with centre A.

Given that AB = AC = 10 cm and angle = 0.8 radians. Calculate

(i) the length of arc BGC,

[1]

(ii) the area of sector ABGC,

[1]

(iii) the area of triangle ABC.

[1]

(b)A second circle, centre D, cuts the first circle at B and C.

The radius of the second circle is 8 cm and angle = 1.0 radians.

Calculate (i) the area of triangle BCD,

[1]

(ii) the area of the shaded region,

[2]

(iii) the perimeter of the outline of the shape BECFB.

[2]

5In the diagram below, ABCD and PONM are two identical parallelograms. The points M, N and S are mid-point of AD, BC and MP respectively. The points S and V lie on the lines DC and MN respectively and . The lines BS and VO intersect at point T. VS is parallel to DM.

(a) Prove that triangle BCT is similar to triangle SVT. [2]

(b) Give a reason why . [1]

(c) Find in terms of . [1]

(d) Prove that triangle OVN is congruent to triangle BSC. [2]

6.

In Diagram I, a sphere is placed in a hollow right circular cone for which the centre C of the circular base of the cone is the same as that of the sphere.

The hollow part of the cone is fully filled with liquid.

The diameter AB of the circular base of the cone is 14 cm and its height VC is 24 cm.

(a) Find the length of the slant edge VA.

[1]

(b)By considering the area of the triangle ACV, or otherwise, show that the radius of the sphere is 6.72 cm.

[2]

(c)Calculate, giving your answers in terms of (, the area of the surface of

(i)the cone which is in contact with the liquid,

[1]

(ii)the part of the sphere which is above the liquid level.

[2](d)Show that the volume of the liquid in the cone is about

[2]

The sphere is removed and a lid is used to cover the cone tightly.

The cone is then inverted as shown in Diagram II.

(e)Find the depth of the liquid in the cone.

[3]

7.

Diagram 1 shows a quadrilateral PQRS in which PS = PQ = x cm and

RS = RQ = y cm. PSR = PQR = 90(.

(a) Show that the area of this quadrilateral is xy cm2.[2]

Five of these quadrilaterals are joined together to form Diagram 2.

The total area of Diagram 2 is 840cm2.

(b) Show that the perimeter, P of Diagram 2 is P = 10x + cm.[2]

(c) When P = 118, show that 5x2 59x + 168 = 0.[2]

(d) Find x. [2]

8.

A, B, C and D are points on level ground of a safari, with A due north of C and D. (BAD = 75(, (BDC = 130(, AB = 55 m and CD = 30 m.

Calculate the

(a) bearing of D from B,

[1]

(b) length of BD,

[2]

(c) length of CB,

[2]

(d) area of (ABD.

[2]One night, a wild bear was spotted from the bushes at B and it ran along the path BA towards the gate of the town at A at a speed of 8 m/s. A ranger at the top of a 20-metre guard tower at D spotted the bear at B. The ranger fired a shot from the tranquilizer gun which hit the bear when it was closest to the guard tower. Find the (e) angle of depression of the bear from the ranger when the shot was fired,

[3](f) time that elapsed from the instant the ranger spotted the bear at B to the instant he fired the shot. Assume that the time taken by the shot to hit the bear from the time it was fired was negligible.

[2]

9 (a) The vectors and are defined by = and = .(i) Evaluate .

[1]

(ii) If + 2 = 6, express as a column vector.

[2](b) In the diagram, ABCD is a parallelogram, M is the midpoint of DC and N is a point on AB such that 4NB = AB. Given that = , and = , express as simply as possible, in terms of and/or ,

(i) ,

[1]

(ii) ,

[1]

(iii) .

[1](iv) The point E lies on AM such that 3AE = AM.

Show that DE is parallel to MN.

[2]

Answer the whole of this question on a piece of graph paper.

Variables x and y are connected by the equation y = . The table shows some values of x and y.

x

3

2

1

0

1

2

3

4

y

2.5

p

13.5

10

4.5

0

0.5

6

X

-0.5

3.5

2.5-2.5-1.5y

12.18751.6875-0.93758.437513.5625

(a)Calculate the value of p. [1]

(b)Draw a horizontal x-axis of 2 cm to 1 unit for 3 ( x ( 4. [3]Draw a vertical y-axis of 1 cm to 1 unit for 2 ( y ( 14.

Plot the points (x, y) and join them with a smooth curve

From your graph,

(c)determine the other values of x that give equal value of y as x = 4,

[2]

(d)draw a tangent line to calculate the gradient of the curve at x = , [2]

(e)solve the equation for = 0 for 3 ( x ( 4 by drawing a

suitable straight line. [3]

11

The diagram shows the first four sequence of arranging triangles. The table below shows the number of edges (E) and vertices (V) as the number of triangles (T) increases.

Figure no.123456

No. of triangles (T)123456

No. of edges (E)35810p15

No. of vertices (V)3467q10

(a) State the value of p and of q.

[2]

(b) Write an equation connecting T, E and V .

[1]

(b) When T is odd, E = . Use the answer to (b) and E = , find the number of vertices, V, in Figure 21.

[2]

(d)Express E in terms of T when T is even.

[1]

**************End of Paper ****************2013 CCHY 4E5N Prelim2 EMaths Paper 2 Marking Scheme

1 (a) 3/(x-2) + 1/(2-x) (x+6) = [3(x+6) 1]/((x-2)(x+6)M1 = (3x + 19) / (x-2) (x+6)A1

(b) (i) x ( x+3) = 54 x2+ 3x 54 = 0 M1 (x +9 ) (x 6) = 0

x = 9 , 6A1(b) (ii) 9y2 = 4y

y (9y2 4)= 0M1 y (3y+2)(3y2) = 0

y = 0, 2/3 or 2/3A1(c)

r3=

A2, [-1] for each error(d) x + y = 9; x2 y2= 72

(x+y)(x-y) = 72

9 (x y) = 72

x y = 8 M1

3 (x-y) 2 = 3x82 = 192 A1

2(a) (i) A( ( B or B onlyA12(a)(ii)

A2, [-1] for each error(b) y = k/ x

M1

3 = k / x

y1 = k/(x/4)

M1

= 6

A1

(c) Let Johns age at the present time be J and Peters age be p.

J + 4 = 2 (P+4)

J 3 + p 3 = 22

M1

J = 28 P 28 P = 2 P + 4

3P = 24

P = 8

A1

J = 20

A1(d) (i) 2AB = 2 M1

= A1(ii) A2=A2, [-1] for each error3(a) (i)0.75 ($3000) = $750 A1 (ii)Interest = $(2250x6x2) /100 = $270

M1

hire purchase price = $3250

A1

(iii)instalment = $(2250+270) / 24 = $105A1

(b)(i)A = $22000 (1+1.88/100)6 = $24601.20M1,A1

To exceed $20000, the interest > 20000 18000 = $2000M1

18000 X 0.9 X t/100 > 2000

t > 12.3457

time is 13 yearsA14(a)(i)arc BGC = r = 10 (0.8)cm = 8 cmA1

(ii)area of sector ABGC = r2 = (10) 2 (0.8) cm2 = 40 cm2 A1 (iii)area of triangle ABC = (10) 2 sin 0.8 = 35.86780454 cm 2 = 35.9 cm 2A1

(b)(i)Area of (BCD = r2 = (8) 2 (sin 1.0) cm 2 = 26.92707151 ~ 26.9 cm 2A1(ii) Shaded area = area of sector ABGC area of triangle ABC + area of sector DBC area of triangle BDCM1

= 40 35.86780454 + (8) 2(1) 26.92707151 cm 2

= 9.205883086 cm 2

= 9.21 cm 2 A1(iii) perimeter = 8 (2 1) + 10 (2 0.8) cmM1

= 97.09733553 cm

= 97.1 cm A1

5(a) (vertically opposite s)

or (alternate s, OM //VS, VS//BC)Therefore, is similar to (2 pairs of corresponding angles are equal).

(b) (corresponding s of identical parallelogram)

(c) (sum of interior s, MP // NO)

(d) BC = ON (ABCD & PONM are identical parallelograms)

VN = SC (VM = DS, DM // SV)

Therefore, (SAS)6(a)By Pythagoras Theorem in right-angled (VCA,

VA = cm

= 25 cm(b)CP = radius of circle

(CPV = 90( (radius ( tangent)

Area of (VAC = EQ \f(1,2) x AC x CV = EQ \f(1,2) x AV x rM1

EQ \f(1,2) x 7 x 24 = EQ \f(1,2) x 25 x r

r = EQ \f(7 x 24,25)

= 6.72( the radius is 6.72 cm [shown]A1c(i)surface area of cone = ( r l = ( (7)(25) = 175( cm2A1c(ii)surface area of hemisphere = 2( (6.72)2 = 90.3168( cm2M1,A1(c) Volume of liquid = (1/3) ( (72) (24) (2/3) ( (6.72)3 cm3M1

= 189.690368 ( cm3

~ 190 ( cm3 [shown]A1(d) let the height of V about the liquid surface in the inverted cone be h cm and its radius be r.by similar triangles,

EQ \f(h,24) = EQ \f(r,7)

r = EQ \f(7h,24) ---------------- (1)vol of water = 1/3 (()(72) 24 1/3( (r)2 h = 189.890368( ----------- (2) M1subst (1) into (2),M1

1/3 (49)(24) 189.890368 = 1/3 (7/24h)2 h

h3=

= 7127.458043 h =19.24471822 depth of liquid = 24 19.24471822 cm

= 4.75528178 ~ 4.76 cmA17aarea of (PSR = cm2M1 Area PQRS = 2 x area of (PSR

= xy cm2 (shown)A1

7b5xy = 840

y = ------------(1)M1Perimeter = 10x+2y ---------------- (2)Subst (1) into (2),

( perimeter = 10x+2

Hence P = 10x + (shown) A17cGiven 10x +=118M1( 10x2 + 336 = 118x (2,5x2 59x + 168 = 0 (shown) A1

7d5x2 59x + 168 = 0

( 5x 24 )( x 7 ) = 0

x = 4.8 or x = 7 A2

8(a) bearing of D from B = 360( 130(

= 230(A1

(b) (ADB = 180( 130( ((s on a straight line)

= 50(By Sine Rule,

=

BD=

mM1

= 69.3509638 m

~ 69.4 mA1(c)By Cosine Rule

BC2= 302 + 69.35096382 2 (30)(69.3509638) cos 130(M1BC = 91.56545525 m

~ 91.6 mA1

(d) (ABD = 180( 75( 50( ((s sum of ()

= 55( Area of (ABD = (1/2) (BD)(AB) sin 55( m2M1 = 0.5 (69.3509638)(55) sin 55( m2 = 1562.247054 m2

= 1560 m2 (3sf)A1Accept 1562 m2 (4sf), 1562.2 (5sf)

Let be the angle of depression and the shortest distance of bear to path BA be x m.

sin 55( = x/BD

x = 69.3509638 sin 55( = 56.80898377 mM1

tan = 20/56.80898377M1

= 19.39497392( ~ 19.4(A1( angle of depression = 19.4((e) Time taken = distance run by bear / speed of bear = BW / 8

M1

= (BD cos55( ) / 8

= ( 69.3509638 cos 55( ) / 8= 4.972259834

~ 4.97 Hence, the time taken = 4.97 sA19(a)(i)

= units

= 2units A1Or 4.47 units

(ii)

=

M1

=

=

A1

b(i)

A1

(ii) 4 NB = AB

( EQ \f(NB,AB) = EQ \f(1,4) ( AN = EQ \f(3,4) AB

( = = A1

(iii)

+ =

M1

=

= + = A1

Or EQ \f(1,2) (2 )

(c)3AE = AM

EQ \f(AE,AM) = EQ \f(1,3) ( AE = EQ \f(1,3) AM

( = = EQ \f(1,3) ( + )

+ =

=

= EQ \f(1,3) ( + )

= M1

= EQ \f(1,3) ( 2 ) ---------- (1)

From (iii), = EQ \f(1,2) (2 )

( 2 = 2 ------------ (2)

subst (2) into (1), = EQ \f(1,3) x 2

= M1

Since ED is a scalar multiple of NM, ED is parallel to NM (shown)

(a) 10p = 12 A1(b) axes : [1 ], points : [1], graph : [1 ](c) x = 0.6 to 0.8 or x = 2.8 to 2.65

(d) drawn tangent line [1]

gradient from 3.7 to 3.1 (e) x3 2x2 8x + 12 = 0

(2, EQ \f(1,2) x3 x2 4x + 6 = 0

EQ \f(1,2) x3 x2 = 4x 6 EQ \f(1,2) x3 x2 5x + 10 = 4x 6 5x + 10

y = x + 4Draw straight line y = x + 4 [1]

From the graph, x = 2.8 to 2.6, 1.3 to 1.45, 3.4 to 3.5 [A2]

11(a)p = 13 , q = 9 [A2](b)

[A1]

(c)Figure 21

So

(

-----------------(1)M1And

-----------------(2)

Subst (2) into (1), V = 53 20

= 33Therefore, there are 33 vertices in Figure 21.

A1(d) When T is even, E = or E = 2.5T

3

Class

Write your name, register number and class in the space provided above.

Write in dark blue or black pen.

You may use a pencil for any diagrams or graphs.

Do not use paper clips, highlighters, glue or correction fluid / tape.

Answer all questions.

If working is needed for any question, it must be shown with the answer.

Omission of essential workings and units of measurement will result in loss of marks.

Calculators should be used where appropriate.

Simplify your answers to their simplest form. If the answer is not exact, give the answer correct to three significant figures or in fraction where applicable. Give answers in degrees correct to one decimal place.

For EMBED Equation.3 , use your calculator value, unless the question requires the answer in terms of EMBED Equation.3 .

At the end of the examination, staple the answers with the graph paper.

The number of marks is given in brackets [ ] at the end of each question or part question.

The total of the marks for this paper is 100.

B

(

A

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

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EMBED Equation.3

EMBED Equation.3

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EMBED Equation.3

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Diagram not drawn to scale

Diagram I

Diagram II

14 cm

24 cm

14 cm

A

B

V

C

(

A

B

V

(

C

P

x

x

Q

S

y

y

R

Diagram 2

Diagram 1

North

A

D

C

B

30 m

55 m

75(

130(

EMBED Equation.3

EMBED Equation.3

Figure 1

Figure 2

Figure 3

Figure

4

Q

(

P

R

r

C

A

V

P

r

7

24

h

8

5x 24 24x

x 7 35x

5x2 168 59x

D

(

(

R

W

A

B

20

x

55(

69.3509638

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