NAN HUA HIGH SCHOOL 2010 COMMON TEST 1 Subject

:

Mathematics

Paper

:

4016

Level

:

Secondary Three Special/ Express

Date

:

23 February 2010

Duration

:

1 hour

SOLUTIONS READ THESE INSTRUCTIONS FIRST Write your class, index number and name on all the work you hand in. Write in dark blue or black pen. You may use a pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correcting fluid. Answer all questions. If working is needed for any question it must be shown with the answer. Omission of essential working will result in loss of marks. Calculators should be used where appropriate. If the degree of accuracy is not specified in the question, and if the answer is not exact, give the answer to 3 significant figures. Give answers in degrees to one decimal place.

Ď€ , use either your calculator value or 3.142, unless the question requires the answer in terms of Ď€ . For

At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [

] at the end of each question or part question.

The total number of marks for this paper is 40.

This paper consists of 7 printed pages. Page 1 of 7

Nan Hua High School 2010 Common Test 1

1.

Mathematics 4016

Factorise 4ax +16 y − 4ay −16 x completely.

(a) [3]

4ax +16 y − 4ay −16 x = 4ax − 4ay +16 y −16 x = 4a ( x − y ) +16( y − x ) = 4a ( x − y ) −16( x − y ) = ( 4a −16)( x − y ) = 4( a − 4)( x − y )

(b)

Simplify

1 x −1 5x + + . 2 − x ( x − 2) 2 x − 2

[3]

1 x −1 5x 1 x −1 5x + + =− + + 2 2 2 − x ( x − 2) x−2 x − 2 ( x − 2) x−2 − ( x − 2) + x − 1 + 5 x( x − 2) = ( x − 2) 2

(c)

2x +3y

=

− x + 2 + x − 1 + 5 x 2 − 10 x ( x − 2) 2

=

5 x 2 − 10 x + 1 ( x − 2) 2

7

x

Given that 2 x − 3 y = 3 , find the value of y . 2x +3 y 2x −3 y 3( 2 x + 3 y ) 6x +9 y 30 y x y

[3]

7 3 = 7( 2 x − 3 y ) = 14 x − 21 y = 8x 30 = 8 15 = 4 3 =3 4 =

Page 2 of 7

Nan Hua High School 2010 Common Test 1

1.

(d)

Make

Mathematics 4016

x the subject of the formula y =

x x + a2 2

.

[3]

x

y=

x + a2 x2 y2 = 2 2 x +a 2 2 2 y x + a = x2

(

2

)

y x + y a = x2 2 2

2 2

x 2 (1 − y 2 ) = y 2 a 2 x2 =

y 2a2 (1 − y 2 )

x= ± =±

2.

y 2a 2 (1 − y 2 ) ay (1 − y 2 )

Showing your working clearly, evaluate each of the following, giving your answers in the standard form. (i)

3.2 ×10 6 × 5.0 ×10 7

[2]

(ii)

7.2 ×105 2.4 ×10 2

[2]

(i)

3.2 ×10 6 ×5.0 ×10 7 = (3.2 ×5.0) ×10 6+7 =16 ×1013 =1.6 ×1014

(ii)

7.2 ×10 5 7.2 5−2 = ×10 2 2.4 ×10 2.4 = 3.0 ×10 3

Page 3 of 7

Nan Hua High School 2010 Common Test 1

3.

(a)

Mathematics 4016

Solve the equation 2 k × 4 k +1 = 82 k −3 .

[3]

2 k × 4 k +1 = 8 2 k −3

( )

2k × 22

k +1

( )

= 23

2 k −3

2 k × 2 2 k +2 = 2 6 k −9

⇒

(b)

Simplify

2 k +2 k +2 = 2 6 k −9 3k + 2 = 6k − 9 3k =11 11 k = 3 2 =3 3

1 3 1 − 12 x + x 2 x 2 − x 2 2 3 1 2 x − x 2

1 3 1 − 12 x + x 2 x 2 − x 2 2 1 32 x − x 2

.

[3]

x 2 − x + x − x0 = x3 −2 x 2 + x

=

x 2 −1 x ( x 2 − 2 x +1)

x 2 −1 2 x( x −1) ( x +1)( x −1) = 2 x( x −1) ( x +1) = x( x −1) =

Page 4 of 7

Nan Hua High School 2010 Common Test 1

4.

(a)

Mathematics 4016

Find the integer values of x for which 2 x + 4 < 3x + 6 and 3 x ≤ x + 6 . 2 x + 4 < 3x + 6 and 3x − 2 x > 4 − 6 x > −2

[2]

3x ≤ x + 6

3x − x ≤ 6 2x ≤6 x ≤3

∴ −2 < x ≤ 3

Integer values of x are −1, 0, 1, 2 and 3. (b)

Given that 3 ≤ a ≤ 5 and − 4 ≤ b ≤ −1 , find (ii)

a , b the smallest possible value of a 2 + b 2 .

(i)

The largest possible value of

(ii)

The smallest possible value of ( a 2 + b 2 ) = ( 3) 2 + (−1) 2

(i)

the largest possible value of

(

[1]

)

[1]

a 3 = b ( −4) 3 =− 4

=10

Page 5 of 7

Nan Hua High School 2010 Common Test 1

5.

Mathematics 4016

(a)

Total commissions = $ 4000 + $ 2,500 + $ 4,300 = $ 10, 800

(b)

Net earnings = $ 10,800 − $ 1,350 − $ 470 = $ 8, 980

(c)

ABC Bank : Interest =

$8,980 ×2.1 ×1.5 = $ 282.87 100

OCB Bank : Interest = $8,9801 +

18

1.85 1200

− $8,980

= $ 9,232.4875 − $ 8,980 = $ 252.49 As ABC Bank gives more interest, Mr Hassan should choose ABC Bank. 6

(a)

£1 = S$ 2.35 £ 3,400 = S$ 7,990 Alan needs to pay S$ 7,990 to change to £ 3,400.

(b)

100 ×476 = £ 425 112

Amount the shop paid for the phone is £ 425 (c)

S$ 2.35 = £ 1 S$ 1,081 = £ 460 Percentage savings =

(d)

16 ×100% = 3.36% 476

£ 1 = S$ 2.2580 £ 315 = S$ 711.27 Alan could get S$ 711.27 after changing his money

7

(a) (b)

Total tax reliefs = $ 24, 300 Chargeable income = ( $ 5,500

× 12) − $ 24,300

= $ 41, 700 (c)(i) Tax payable for the first $ 40,000 and the next $ 40,000 is $ 900 and $ 3,400 respectively. Hence, tax payable for the first $ 80,000 is $ 900 + $ 3,400 = $ 4,300.

Page 6 of 7

Nan Hua High School 2010 Common Test 1

Mathematics 4016

(ii) For the first $ 40,000, Mr Lim has to pay $ 900. For the subsequent $ 1,700, he has to pay

8.5 Ă—1,700 = $144.50 . 100

So, total amount of income tax is $ 1,044.50

Page 7 of 7