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Mid-Year Examination 2010

Secondary Two Express

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Mathematics Paper 2

Friday Duration: 14 May 2010

1 hr 15 min

Additional Materials: Writing paper, Graph paper Instructions to Candidates:

1. Write your name, register number and class at the top of this page. 2. Answer all questions on the separate answer paper provided. 3. Show all your working on the same page as the rest of the answer. 4. Omission of essential working will result in loss of marks. 5. Calculators should be used where appropriate. 6. At the end of the examination, fasten all your work securely together.

Information for candidates

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 50.

If the degree of accuracy is not specified in the question and if the answer is not exact, the answer should be given to three significant figures. Answers in degrees should be given to one decimal place.

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

After checking of answer scripts

Checked by Signature Date

Student

DO NOT TURN THIS PAGE OVER UNTIL YOU ARE TOLD TO DO SO.

This Question Paper Consists of 4 Pages (including cover page)

21 Simplify

(a) 5

32

45

4

8

15

ab

ba

a

b

[2]

(b)

9

21

3

42

p

p

p

[3]

Solution: (a)

16

145

4

8

15

4

45

8

15

2

5

32

3

52

Ab

a

Mab

ba

a

b

ba

ab

a

b

(b)

13

3

)3)(3(

)3(3

1)3)(3(

93

)3)(3(

21124

1)3)(3(

)21()3(4

)3)(3(

21

3

4

9

21

3

42

Ap

pp

p

Mpp

p

pp

pp

Mpp

pp

pp

p

p

p

p

p

2 (a) If

5

2

4

53

ba

ba, find the value of

b

a7.

[3]

(b) Simplify

5112

51532

2

xx

xxyxy

Solution:

(a) 5

2

4

53

ba

ba

baba 42535 -------------M1 baba 282515

ba 277 ---------------M1

[3]

3

277

b

a-----------------A1

(b)

3 During the annual primary one registration exercise, 85 parents had to ballot

for 20 vacancies at a primary school. Each child is assigned a ballot number starting from 121 in running order.

Find the probability that a child selected at random had a ballot number which

(a) has ‘7’ as its unit digit, [1] (b) has the tens digit and the unit digit differ by 1, [2] (c) is not a palindrome (palindromes are numbers which can be read from left

to right or right to left. e.g. 121) Solution:

S = {121, 122, 123, …, 205} (a) Let 1E be the event the ballot number has ‘7’ as its unit digit.

1E {127, 137, 147, 157, 167, 177, 187, 197}

P( 1E ) 8

85 [B1]

(b) Let 2E be the event the tens digit and unit digit of the ballot number differ

by 1.

2E {121, 123, 132, 134, 143, 145, 154, 156, 165, 167, 176, 178,

187,189, 198, 201} [M1]

P( 2E ) 16

85 [A1]

[2]

112

3

1)5)(12(

)3)(5(

1)5)(12(

)5()5(35112

51532

2

Ax

xy

Mxx

xyx

Mxx

xxxyxx

xxyxy

4 (c) Let 3E be the event the ballot number is a palindrome.

3E {121, 131, 141, 151, 161, 171, 181, 191, 202}

P( 3E ) 9

85 [M1]

P(ballot numbers is not a palindrome) 85

91

76

85 [A1]

4 Given that 5ab and 7 ba , find the value of 2)( ba .

222 2 bababa

22 7 ba ---M1

49 ----M1 (a b)2 abba 222

5249 ----M1 1049

59----A1

[4]

5 Mr Lim planned to sell hand-made dolls at x dollars each. After selling all the dolls, he

hoped to earn $2800 from the sales in total. However, due to poor business, he reduced the price of each hand-made dolls by $1 and finally earned $2730 from the sales.

Write down an expression in terms of x for

(a) the number of hand-made dolls Mr Lim originally planned to sell [1] (b) the number of hand-made dolls Mr Lim finally sold [1] (c) Given that Mr Lim sold 10 more hand-made dolls than he originally

planned, write down an equation in x and show that it reduces to 028062 xx .

[2]

(d) Solve the equation 028062 xx and find the number of hand-made

dolls that Mr Lim finally sold.

[3]

5Solution

(i) Write down an expression in terms of x for (a) the number of hand-made dolls Mr Lim originally planned to sell,

x

2800 ---B1

(b) the number of hand-made dolls Mr Lim finally sold. [1m]

1

2730

x ---B1

(ii) Given that Mr Lim sold 10 more hand-made dolls than he originally planned, write down an equation in x and show that it reduces to 028062 xx . [2m]

1

273010

2800

xx =>

1

2730102800

xx

x ---M1

xxx 2730)102800)(1( => xxxx 2730)102800102800( 2

xxxx 2730)102800102800( 2 => 028006010 2 xx

028062 xx (Shown.)---A1

(iii) Solve the equation 028062 xx and find the number of hand-made dolls that Mr Lim finally sold. [3m]

028062 xx

Factorise by cross method---M1

0)20)(14( xx

14x or 20x ---M1

No of hand-made dolls = 210---A1

66 (a) Five boys participated in a fishing competition. None of them caught more than 9

fishes. The mean, median and mode of the numbers of fish caught by them are 6, 5 and 4 respectively. Find the number of fishes caught by each boy. (b) The stem-and-leaf diagram below shows the prices of some digital cameras. Stem-and-leaf diagram for the prices of digital cameras

Stem Leaf 2 4 5 6

3 0 0 1 3 5 4 2 2 6 7

5 0 3 9 Key: 2 4 means $240.

(a) Find the mean of the distribution. (b) Find the median of the distribution. (c) Can you explain a possible reason for the significantt difference between the mean

and median value?

(d) Find the percentage of cameras with prices more than $450.

Solution (a) Let a,b, c, d and e be the number of fishes caught in ascending order by the boys. Median = c = 5 Mode = a = b = 4 ---M1

Mean =

17

30544

30

65

ed

ed

edcba

edcba

---M1

As d and e are integers less or equal to 9, d= 8 and e = 9. ---M1 The number of fishes caught by each boy are 4,4,5,8 and 9 respectively. ---A1 (b) (i) Mean price =

7

15

5730

)59053050047046042023503303103002260250240(15

1

-

--M1 =$382---A1 (ii)Median price =$350---B1 (iii) When the mean is higher that the median, it implies that some of the cameras are exceptional high in price, skewing the mean to the higher end.---M2

(iv) %3

133%100

15

5

M1 A1

7 A wire of length 86cm is cut into two parts to form a right-angle triangle and a rectangle.

The dimensions of the shapes are shown in the figures below:

(a) Given that the perimeter of the right-angle triangle is 30 cm, form a pair of simultaneous equations, and show that it reduces to

285

1556

yx

yx

[4]

(b) Copy and complete the tables provided below, and hence plot the above equations

on a scale of 2cm to 1 unit on the x-axis and 1cm to 2 units on the y-axis for 60 x

[5]

8

(c) Hence, solve the pair of simultaneous equations and determine the area of the triangle.

Solution Perimeter of triangle = 30 cm

-[A1]------- 1556

30562

301012

3065264

-[M1]------- 3026654

yx

yx

yx

yyyxxx

yxyxyx

Perimeter of rectangle = 86-30 Perimeter of rectangle = 56 cm

-[A1]------- 285

5652

56210

566446

-[M1]------- 56322232

yx

yx

yx

yxyx

yxyx

(shown) 285

1556

yx

yx

2b) [A1] [A1] 2c) From the graph, the point of intersection is 3,5 . -[M1]------- 3 ,5 yx

Area of triangle -[M1]------- 355436562

1

-[M1]------- cm 30

5122

1

2

5x - y = 28

x 0 3 6

y -28 -13 2

6x + 5y = 15

x 0 5 10

y 3 -3 -9

[3]

9