Name: ............................................................................. Register no: ............. Class: Sec .........

NGEE ANN SECONDARY SCHOOL

Building CharacterExpanding MindsShaping Lives

Mid-Year Examination 2010

Secondary Two Express
Ngee Ann Secondary School Ngee Ann Secondary School Ngee Ann Secondary School Ngee Ann Secondary School
Ngee Ann Secondary School Ngee Ann Secondary School Ngee Ann Secondary School Ngee Ann Secondary School
Ngee Ann Secondary School Ngee Ann Secondary School Ngee Ann Secondary School Ngee Ann Secondary School
Ngee Ann Secondary School Ngee Ann Secondary School Ngee Ann Secondary School Ngee Ann Secondary School
Ngee Ann Secondary School Ngee Ann Secondary School Ngee Ann Secondary School Ngee Ann Secondary School

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)
 2
1 Simplify
15b 2 4a 3 b [2]
(a) 
8a 45ab 5
4 p  21 [3]
(b)  2
p3 p 9

Solution:

(a)
15b 2 45ab 5

8a 4a 3 b
15b 2 4a 3b
         M1
8a 45ab 5
a
 2          A1
6b

(b)
4 p  21
 2
p3 p 9
4 p  21
 
p  3 ( p  3)( p  3)
4( p  3)  ( p  21)
          M 1
( p  3)( p  3)
4 p  12  p  21

( p  3)( p  3)
3p  9
            M 1
( p  3)( p  3)
3( p  3)

( p  3)( p  3)
3
                   A1
p3

2
3a  5b 2 7a [3]
(a) If  , find the value of .
4a  b 5 b

3 xy  15 y  x 2  5 x [3]
(b) Simplify
2 x 2  11x  5
Solution:
3a  5b 2
(a) 
4a  b 5
53a  5b   24a  b  -------------M1
15a  25b  8a  2b
7 a  27b ---------------M1
 3
7a
 27 -----------------A1
b

(b)

3 xy  15 y  x 2  5 x
2 x 2  11x  5
3 y ( x  5)  x ( x  5)
          M1
( 2 x  1)( x  5)
( x  5)(3 y  x )
            M1
( 2 x  1)( x  5)
3y  x
                  A1
2x 1

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 [2]
to right or right to left. e.g. 121)

Solution:
S = {121, 122, 123, …, 205}

(a) Let E1 be the event the ballot number has ‘7’ as its unit digit.

E1  {127, 137, 147, 157, 167, 177, 187, 197}

8
P( E1 )  [B1]
85

(b) Let E2 be the event the tens digit and unit digit of the ballot number differ
by 1.

E2  {121, 123, 132, 134, 143, 145, 154, 156, 165, 167, 176, 178,
187,189, 198, 201} [M1]

16
P( E2 )  [A1]
85
 4

(c) Let E3 be the event the ballot number is a palindrome.

E3  {121, 131, 141, 151, 161, 171, 181, 191, 202}

9
P( E3 )  [M1]
85

9
P(ballot numbers is not a palindrome)  1 
85
76
 [A1]
85

4 Given that ab  5 and a  b  7 , find the value of (a  b) 2 . [4]

a  b 2  a 2  2ab  b 2
a  b 2  72 ---M1
 49 ----M1

(a  b)2  a 2  b 2  2ab
 49  2 5 ----M1
 49  10
 59----A1

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 [2]
planned, write down an equation in x and show that it reduces to
x 2  6 x  280  0 .

(d) Solve the equation x 2  6 x  280  0 and find the number of hand-made [3]
dolls that Mr Lim finally sold.
 5
Solution
(i) Write down an expression in terms of x for
(a) the number of hand-made dolls Mr Lim originally planned to sell,

2800
---B1
x
(b) the number of hand-made dolls Mr Lim finally sold. [1m]

2730
---B1
x 1
(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 x 2  6 x  280  0 .
[2m]

2800 2730 2800  10 x 2730

 10  =>  ---M1
x x 1 x x 1

( x  1)(2800  10 x)  2730 x => (2800 x  10 x 2  2800  10 x)  2730 x

(2800 x  10 x 2  2800  10 x)  2730 x => 10 x 2  60 x  2800  0

x 2  6 x  280  0 (Shown.)---A1

(iii) Solve the equation x 2  6 x  280  0 and find the number of hand-made
dolls that Mr Lim finally sold. [3m]

x 2  6 x  280  0

Factorise by cross method---M1

( x  14)( x  20)  0

x  14 or x  20 ---M1
No of hand-made dolls = 210---A1
 6
6 (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
2 456
3 00135
4 2267
5 039

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

abcd e
6
5
Mean = a  b  c  d  e  30 ---M1
4  4  5  d  e  30
d  e  17

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
1
(240  250  260  2  300  310  330  350  2  420  460  470  500  530  590)
15
5730 -

15

--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

5 1
(iv)  100%  33 %
15 3

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 [4]
simultaneous equations, and show that it reduces to
6 x  5 y  15
5 x  y  28

(b) Copy and complete the tables provided below, and hence plot the above equations [5]
on a scale of 2cm to 1 unit on the x-axis and 1cm to 2 units on the y-axis
for 0  x  6
 8

(c) Hence, solve the pair of simultaneous equations and determine the area of the triangle. [3]

Solution
Perimeter of triangle = 30 cm
4 x  5 y   6 x  6 y   2 x  y   30 - - - - - - - -[M1]
4 x  6 x  2 x  5 y  6 y  y  30
12 x  10 y  30
26 x  5 y   30
6 x  5 y  15 - - - - - - - -[A1]
Perimeter of rectangle = 86-30
Perimeter of rectangle = 56 cm
23 x  2 y   22 x  3 y   56 - - - - - - - -[M1]
6 x  4 y  4 x  6 y  56
10 x  2 y  56
25 x  y   56
5 x  y  28 - - - - - - - -[A1]
 6 x  5 y  15
5 x  y  28 (shown)

2b)
[A1] [A1]
6x + 5y = 15 5x - y = 28

x 0 5 10 x 0 3 6

y 3 -3 -9 y -28 -13 2

2c) From the graph, the point of intersection is 5,3 .

 x  5, y  3 - - - - - - - -[M1]
Area of triangle  65  6 345  5 3 - - - - - - - -[M1]
1
2
 125
1
2
 30 cm 2 - - - - - - - -[M1]
9