munsang college f.3 mathematics revision …data.munsang.edu.hk/~hfchan/f.3/revision exercise (for...

Answers written in the margins will not be marked.

1

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

MUNSANG COLLEGE F.3 MATHEMATICS

Revision Exercise 1 (Answer)

Section A: Multiple Choice Questions

1. Factorize 33 3753 nm .

A. 22 51053 nmnmnm

B. 22 5553 nmnmnm

C. 22 251053 nmnmnm

D. 22 25553 nmnmnm

2. Simplify nn 2223 2)2(2)( .

A. n22

B. 0

C. n22

D. 122 n

3. Amy deposited $ 20 000 in a bank at an interest rate of 4% p.a. compounded half-yearly. Find the

amount after 4 years, correct to the nearest $100.

A. $21 600

B. $23 200

C. $23 400

D. $27 400

4. The following cumulative frequency curve shows the heights of 200 students.

The third quartile of the heights is

A. 150 cm.

B. 155 cm.

C. 160 cm.

D. 165 cm.


2

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

5. Referring to the given figure, which of the following is/are correct?

I. ABC > AHD

II. AE is a median of ABC .

III. CD is an altitude of ABC .

A. III only

B. I and II only

C. II and III only

D. I, II and III

6. In the figure, AB // CD // EF. If BD = DF = 9 cm, AB = 17 cm and

EF = 27 cm, find the length of CD.

A. 22 cm

B. 23 cm

C. 24 cm

D. 25 cm

7. Which of the following shows the respective front view, top view and side view of the given 3-D object?

A.

B.

C.

D.

8. The volume of a pyramid is 450 cm3 and its height is 18 cm. Find the base area of the pyramid.

A. 25 cm2

B. 50 cm2

C. 75 cm2

D. 150 cm2

9. The bearing of Q from P is N60°E. The distance between P and Q is 5 km and R is 7 km due north of Q.

Find the distance of PR correct to 3 significant figures.

A. 15.4 km

B. 12.0 km

C. 11.6 km

Front Top Side

Side

Front

Top


3

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

D. 10.4 km

10. In the figure, which straight line has the most negative slope?

A. L1

B. L2

C. L3

D. L4

Answer for Section A: (20 marks)

Use a pen to mark your answer as follows:

1. 6.

2. 7.

3. 8.

4. 9.

5. 10.

A B C D A B C D

A B C D

L1 L2 L3

L4


4

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

Section B: Short Questions

1. (a) Factorize 23781 ba .

(b) (i) Factorize 1522 yy .

(ii) Hence, factorize 153232

xx .

(a) 23781 ba

baba

ba

379379

37922

(b) (i) 1522 yy

53 yy

(ii) 153232

xx

8

5333

xx

xx

2. Simplify 3xx

xx

and express your answer with positive indices.

x

x

x

x

x

x

x

xx

xx

8

1

8

8

2

2

1

32

33

2

3

2

3


5

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

3. A computer was bought 2 years ago. It is worth $14 000 now and its depreciation rate is 10% per year.

(a) Find the price of the computer 2 years ago, correct to the nearest $10.

(b) After how many years from now will the price of the computer be $11 340?

(a) The price of the computer 2 years ago

$10)nearest the to(Corr. 28017$

%101000142

Or

Let $ x be the price of the computer 2 years ago.

17280

14000%1012

x

x

(correct to the nearest $10)

The price of the computer 2 years ago is $17 280.

(b) Let n years be the required time.

2

9.09.0

81.09.0

34011%10100014

2

n

n

n

n

∴ After 2 years, the price of the computer will be $11 340.

4. (a) Solve 22

511

x, and represent the solution graphically.

(b) If x is a positive integer, write down all the value(s) of x that satisfy the inequality 22

511

x.

3

155

4511

22

511 (a)

x

x

x

x

(b) For x is a positive integer, x = 1, 2 or 3.

x

X ≦ 3


6

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

5. An organization issues 50 000 lucky draw tickets. The prizes are as follows:

Prize item Number of prizes Prize

1st prize 1 $100 000

2nd prize 1 $50 000

3rd prize 3 $10 000

Consolation prize 5 $1 000

(a) Find the expected value of the prizes in each lucky draw.

(b) If Jacky buys a lucky draw ticket of $20, does he gain or suffer a loss on average? Explain your answer.

(a) P(1st prize)00050

1

P(2nd prize)00050

1

P(3rd prize)00050

3

P(Consolation prize)00050

5

00010

1

Expected value of the prize of each of the lucky draw tickets

73$

)106012($

000 10

1000 1

000 50

3000 10

000 50

1000 50

000 50

1000 100$

.

..

(b) Since the expected value of the prize of each of the lucky draw tickets is less than the price of the ticket,

he will suffer a loss on average.


7

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

6. A box has 6 cards numbered from 1 to 6. Nick draws a card from the box and records the number obtained.

He then puts the card back into the box and draws a card again. This process is repeated 30 times. The

following table shows the recorded results.

Number obtained 1 2 3 4 5 6

Frequency 2 5 8 9 4 2

(a) What is the mode of the number obtained?

(b) The median of the number obtained is 3 after the 29th draw. What is the smallest possible

number obtained in the 30th draw?

(c) Nick draws 2 more cards and the mean number of the 32 cards is 3.5.

Suppose the number of 31st card drawn is x and the number of 32nd card drawn is y.

(i) Find the value of x + y.

(ii) Write down 2 possible pairs of the values of x and y.

(a) The mode of the number obtained = 4

(b) The smallest possible number obtained on the 30th draw is 4.

(c) (i) Let x and y be the number of the 2 cards drawn.

Total number obtained when the process repeated 30 times

104

264594835221

8

112104

5.332

104

yx

yx

yx

(ii) Two possible pairs of the 2 cards drawn:

(2 and 6), (3 and 5), (4 and 4), (5 and 3) or (6 and 2)


8

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

7. Figure A is the net of the 3D solid in Figure B.

(a) Where is the projection of point V on the plane ABC?

Hence, find the distance between point V and plane ABC.

(b) Name the angle between plane VAB and plane VAC.

(c) Mark and name the angle between plane VBC and plane ABC.

Figure A Figure B

(a) Point A is the projection of the point V on the plane ABC.

By Pythagoras’ theorem,

VA = 22 513 cm

=12 cm

(b) ∵ VA AC, VA AB

∴∠BAC is the angle between plane VAB and plane VAC, i.e. 90.

(c)

Mark the mid-point N of BC. Join VN and AN.

∵ ∆ACB is an isosceles triangle, AN BC.

∆VCB is an isosceles triangle, VN BC.

∴ ∠VNA is the angle between plane VBC and plane ABC.

B

A C

V

N

B

A C

V

A

13 cm

5 cm

13 cm

5 cm

B

C


9

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

Section C: Long Questions

1. In Figure 1, ABC is an isosceles triangle in which AB = AC, QR // AB and AP = AQ.

(a) Prove that PQRB is a parallelogram.

∵ QR // AB (Given)

∴ PB // QR

AB = AC (Given)

∠ ABC = ∠ ACB (base∠s , isos. Δ)

(or∠PBR =∠QCR)

∠ QRC = ∠ ABC (corr. ∠ s, QR // AB)

(or∠ PBR = ∠ QRC)

∴∠ QRC = ∠ ACB

(or ∠ QRC = ∠ QCR)

∴ QR = QC (sides opp., eq. ∠ s)

AP = AQ (Given)

AB = AC (Given)

AP + PB = AQ + QC

∴ PB= QC

∴ QR = PB

∴PQRB is a parallelogram. (opp. sides eq. and //)

R C B

Q P

A

Figure 1


10

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

(b) Suppose PBRE , PQRF , 62 yPQ cm, yBR 18 cm, PB = 6 cm and RF = 4.2 cm,

Find (i) the value of y and

(ii) the length of RE.

(b) (i)∵ PQ = BR (opp. sides of // gram)

∴ yy 1862

8

243

y

y

(ii) RFPQREPB

2.4626 yRE or 2.4186 yRE

62.4682 RE cm 7RE

Figure 2

P E

4.2 cm

F

(2y – 6) cm

R

(18 – y) cm

C B

Q

A

6 cm


11

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

2. In the figure, ABCD is a square, E and F are points on AD and AB respectively such that ED = FA,

EC and FD intersect at G.

(a) Prove that EDCFAD .

(b) Prove that EC⊥FD.

(c) (i) Prove that EGDFAD ~ .

(ii) If DC = 6 cm and AE = 2 cm,

find the area of BCGF.

(a) In FAD and EDC ,

FA = ED (given)

∠FAD = ∠EDC = 90° (property of square)

AD = DC (property of square)

∴ EDCFAD (SAS)

(b)∵ EDCFAD (proved in (a))

∴∠AFD = ∠DEC (corr. ∠s, s )

i.e. ∠AFD = ∠DEG

In FAD ,

∠AFD +∠ADF +∠FAD = 180° (∠ sum of Δ)

∠AFD +∠ADF + 90° = 180°

∠AFD +∠ADF = 90°

In EDG ,

∠EGF =∠DEG +∠EDG (ext. ∠ of Δ)

=∠AFD +∠ADF

= 90°

∴ EC⊥FD

(c) (i)∵ EDCFAD (proved in (a))

∴ ∠AFD = ∠DEC (corr. ∠s, s )

∠ADF = ∠EDG (common ∠)

In FAD ,

∠ AFD +∠ ADF +∠ FAD = 180° (∠ sum of Δ)

∠ FAD = 180°－∠ AFD +∠ ADF

In EDG ,

∠EGD +∠DEG +∠EDG = 180° (∠ sum of Δ)

∠EGD = 180°－∠DEG +∠EDG

= 180°－∠AFD +∠ADF

∴∠FAD =∠EGD

∴ EGDFAD ~ (AAA)

1M – any one

∵EC⊥ FD

∴∠ EGD = 90°

∠ FAD = 90°(property of square)

∴∠ EGD =∠ FAD


12

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

(ii) FA = ED

= 6 – 2

=4 cm

In FAD ,

222 AFDADF (Pyth. Thm.)

22 46

52DF

Area of EDC

12

2

64

∵ EDCFAD

∴Area of FAD = 12

∵ EGDFAD ~

∴2

52

4

of Area

of Area

FAD

EGD

13

48

1252

16 of Area

52

16

12

of Area

EGD

EGD

area of BCGF

= area of ABCD － (area of FAD + area of EDC － area of EGD )

= 66 － (12 +12 －13

48)

2cm7.15

or CE


13

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

3. Figure 3 shows a test tube consisting of a hollow cylindrical tube joining a hemispherical bowl of the

same radius. The height of the cylindrical tube is h cm and its radius is r cm. The capacity of the test tube

is 18 cm3 and the capacity of the hemispherical part is

8

1 of the test tube.

(a) (i) Find the values of r and h .

(ii) The test tube is placed upright and water is poured into it until the water level is 2 cm

beneath the rim as shown in Figure 4. Find the volume of water poured into the test tube.

(Leave your answer in terms of .)

(b) The water in the test tube is poured into a conical vessel PAD as shown in Figure 5. The depth

of the water is half the height of the vessel. Find the capacity of the vessel.

(Leave your answer in terms of .)

(a)(i) 188

1

3

4

2

1 3 r

8

273 r

2

3r

188

7

2

32

h

7h

(a)(ii) The volume of water poured in:

=

2

2

3218

=13.5 cm3

2 cm

r cm

h cm

A D

A

B C

P

Figure 3 Figure 4 Figure 5


14

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

(b) Let V1 and V2 be the volume of water poured into the test tube

and the capacity of the vessel respectively.

12

2

1

3

2

1

8

8

1

2

1

VV

V

V

V

V

The capacity of the vessel

3cm108

85.13


15

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

4. The figure shows a building CD of height h m. A and B are two points on the horizontal ground. The

angles of depression of A and B from C are 40° and 75° respectively and AB = 90 m. A helicopter F is

going to land at the top of the building. The angle of elevation of the helicopter F from C is 50°. E is the

point at the same level as C and vertically above B.

(a) Express BD and AD in terms of h.

(b) Find the height of the building CD, correct to 3 significant figures.

(c) Find the height of the helicopter F above the ground, correct to 3 significant figures.

(a) 7590BCD

15

4090ACD

50

In ,BCDCD

BDBCD tan

m 15tan

15tan

hBD

h

BD

In ,ACDCD

ADACD tan

m 50tan

50tan

hAD

h

AD

(b) AD－BD = AB

4.97

42322074.97

15tan50tan

90

9015tan50tan

9015tan50tan

h

h

h

h

hh

(correct to 3 sig. fig.)

The height of the building is 97.4 m.

h m

F

E


16

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

(c) 75CBD (alt ∠s, EC//BD)

15tan50tan75tan

90

75tan

75tan

BD

hBD

BD

h

15tan50tan75tan

90

BDEC

11009986.31

15tan50tan75tan

50tan90

50tan

50tan

EF

ECEF

EC

EF

height of the helicopter F above the ground

= EF + h

15tan50tan

90

75tan

50tan

15tan50tan

90

15tan50tan75tan

50tan90

= 31.11009986 + 97.42322074

=129 m (correct to 3 sig. fig.)


17

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

5. In the figure, the coordinates of A, B and C are (2, 4), (8, 6) and (b, 2) respectively. AB cuts the yaxis

at D (0 , a). CD is perpendicular to AB.

(a) Find the slope of AB.

(b) Find the values of a and b.

(c) Find the height CD of ABC .

(Leave your answer in surd form if necessary.)

(d) Find the ratio AD : DB.

(a) Slope of AB 182

64

(b) Slope of AD = 102

4

a

∴ 2a

CD ⊥ AB

∴ Slope of CD Slope of AB = 1

110

22

b

∴ 4b

(c) By the distance formula,

222204 CD units

24 units

(d). Let the ratio of AD : DB be m : n .

∵ The x-coordinate of D 0

∴ nm

mn

))(8())(2( 0

8m 2n

m : n 1 : 4

∴ The ratio of AD : DB is 1 : 4.

C(b, 2)

y

x

D(0, a)

B(8, 6)

A(2, 4)

O


18

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

An

swer

s w

ritt

en i

n t

he

mar

gin

s w

ill

no

t b

e m

ark

ed.

This is a blank page

End of paper

munsang college f.3 mathematics revision …data.munsang.edu.hk/~hfchan/f.3/revision exercise (for...

Documents