Tuen Mun Government Secondary School
S.1 Mathematics
Easter Holiday Revision Exercise
Suggested Solution
Chapter 4: Linear Equations in One Unknown
1.
(a) 11  5x  4
 5x  4  11
 5x  15
x3
(d)
 3 y  16  6 y  2
(b)
(e)
16  2  6 y  3 y
y2
2y 1
3
5
(h)
2y 1
 5  3 5
5
2 y  1  15
(f)
2  2y
y 1
9  a  4(a  6)
9  a  4a  24
9  24  4a  a
 15  5a
a  3
(i)
4x
26
7
x  8
3z 9

8 32
3z 8 9 8
 

8 3 32 3
7
4
z
3
4
x  14
y7
(a)
4( z  3)  5 z  17
4 y  21  6 y  19
21  19  6 y  4 y
4x
8
7
2 y  14
2.
(c)
4z  12  5z  17
 12  17  5z  4z
z 5
18  9 y
(g)
15x  3x  24
15x  3x  24
12x  24
x2
8( x  3)  7(5 x  39)
(b)
8x  24  35x  273
24  273  35x  8x
297  27 x
x  11
3(4 x  17)  4(2 x  7)  7
(c)
12x  51  8x  28  7
4x  23  7
4x  16
x4
b  8 2b  5

10
6
b8
2b  5
 30 
 30
10
6
3(b  8)  5(2b  5)
3b  24  10b  25
24  25  10b  3b
49  7b
b7
(d)
(e)
2
(4 z  1)  10
5
4 z  1  10 
4z  1  25
4 z  24
x6
5
2
 10  18a
2
13
 10  18a
 13  2  13
13
 10  18a  26
 18a  26  10
 18a  36
a  2
(f)
1 x7

0
2
6
1 x7
( 
)6  06
2
6
3  ( x  7)  0
 x4  0
x  4
(g)
4  10 
6 
z2
3
(h)
z2
3
 63 
4y 
3y  1
 10
5
3y  1
)  10  5
5
20 y  (3 y  1)  50
5  (4 y 
z2
3
3
17 y  1  50
17 y  49
18  z  2
z  20
3.
y
Let x be the required number.
4.
x6
 15
5
Let x be the current age of Billy. Then the
6.
current age of Billy’s mother is x + 25.
1
( x  25  7)
6
Let $x be Romeo’s amount.
x
18  5x  1  24x  3
19  3  24x  5x
16  29x
x
16
29
Let x be the number of postcards that Peter has
x
.
3
x
 1200
3
4x
 1200
3
4x 3
3
  1200 
3 4
4
x  900
 Romeo’s amount is $900.
Let x be the current ages of Anthony. Then the
4x  4
x 1
 The current ages of Anthony and Lawrence
are 1 and 3 respectively.
6 x  42  x  18
5x  60
x  12
 The current ages of Billy and his mother
are 12 and 37 respectively.
Then Juliet’s amount is $
5x  1 8x  1
)
6
6
2
18  (5 x  1)  3(8 x  1)
6  (3 
current age of Lawrence is 3x.
( x  15)  (3x  15)  34
6( x  7)  x  18
7.
5x  1 8x  1

6
2
5x  20  x  20  4
5x  x  16  20
4x  36
x9
 Peter has 9 postcards and Mary has 45
postcards originally.
x  6  15  5
x  6  75
x  69
 The required number is 69.
x7 
3
originally. Then Mary has 5x postcards
originally.
x6
 5  15  5
5
5.
49
17
(i)
8.
Perimeter of ABCD = 2  Perimeter of  EFG
2  [(3x  2)  (4  x)]  2  3x
2(2 x  2)  6 x
4x  4  6x
4  6x  4x
4  2x
x2
 The perimeter of  EFG  2 3
 6cm
9.
Let $x be the entrance fee for an adult.
Then the entrance fee for a child is $
2x  3 
2x 
x
.
2
x
 875
2
3x
 875
2
7x
 875
2
10. Let $x be the price of a ruler.
Then the price of a pen is $(x+2)
5 x  7  ( x  2)  56
5x  7 x  14  56
12x  14  56
12x  42
x  3.5
 The price of a ruler is $3.5
7x 2
2
  875 
2 7
7
x  250
 The entrance fee for an adult is $250.
11. Let $x the original price of each packet of
salt.
1500 x  (1500  1250)  ( x  10)
1500 x  250  ( x  10)
1500 x  250 x  2500
1250 x  2500
x2
 The original price of a packet of salt is
$2.
13. Let $x be Paul’s monthly salary.
3 2 1
x  (1    )  2500
8 7 4
5x
 2500
56
5 x 56
56

 2500 
56 5
5
x  28000
 Paul’s monthly salary is $28000.
12. Let $x be May’s amount.
Then Ken and Leona have $4x and $(x+18)
respectively.
x  4 x  ( x  18)  108
6x  18  108
6x  90
x  15
 May has $15
Ken has $ 15 4 =$60
Leona has $(15+18) = $33
14. Let xcm be the length of the square. Then the
length and the width of the rectangle is (x+2)cm
and xcm respectively.
4 x  2  [ x  ( x  2)]  68
4 x  2  (2 x  2)  68
4x  4x  4  68
8x  4  68
8x  64
x 8
 The length of the shorter wire is 4  8  32 cm
and the length of the longer wire is
68-32 = 36cm.
Chapter 5: Percentages
1.
Percentage of buses and mini-buses
2.
27  23
 100%
150
1
 33 %
3
Percentage of money that he saves
140

 100%
175
 80%

3.
Number of boys  1200  (1  55%)
 1200 
4.
Number of the elderly
 4300  (1  65%  30%)
5
 4300 
100
 215
6.
(a) Volume of milk = 180  30%  120  40%
45
100
 540
5.
Amount that he saves = $(175-15-20) = $140
Let x be the number of adults in the city.
x  (10%  15%)  5550
 102mL
x  5550  25%
x  5550 
100
25
(b) Percentage of milk 
x  22200
 There are 22200 adults in the city.
7.
(a)
102
 100%
180  120
 34%
Number of pears remaining
 500  (1  48%  27%)  75
25
 500 
 75
100
 125  75
 50
(b)
Total number of fruits remaining
 500  140  85  75
 200
Percentage of pears
50

 100%
200
 25%
(c) Number of apples  500  48%  140
 240  140
 100
Number of oranges  500  27%  85
 135  85
 50
100  50

 100%
50
Required percentage
 100%
8.
Percentage increase 
15000  12000
 100%
12000
3000
 100%
12000
 25%

9.
375  355
 100%
375
1
5 %
3
Percentage decrease 
10.
Christine’s salary in June = $(10000-2000)
= $8000
11.
Number of eggs  30  (1  20%)
 24
2000
 100%
Percentage increase 
8000
= 25%
12.
Let x be the population of the city in 2007.
x  (1  35%)  3712500
13.
New area  240 260
x  3712500  135%
x  3712500 
(a) New length  300  (1  20%)  240 m
New width  200  (1  30%)  260 m
100
135
 62400m 2
x  2750000
New perimeter  2  (240  260)
 The population was 2750000 in 2007.
13.
 1000m
(c) Original perimeter  2(300  200)
(b) Original area  300 200
 10000m
There is no change in perimeter,
i.e. The percentage change in perimeter is
0%.
 60000m
Percentage increase in area
2
64000  60000
 100%
60000
2
6 %
3

14.
Cost of the pen = $(8 + 2) = $10
Percentage loss 
16.
 80  (1  15%)
17.
Let $x be the cost of the flat.
x  (1  20%)  1380000
x  1380000  120%
 80  85%
 $68
18.
Percentage profit 
2
 100%
10
= 20%
Selling price of the book
2000  1500
 100%
1500
1
 33 %
3
15.
Selling price of his car  80000  (1  120%)
220
100
 $176000
 80000 
Selling price of his ship  320000  (1  30%)
 320000 
70
100
 $224000
Overall, he received $176000 + $224000
= $400000
which is the same amount as the total cost of
his car and his ship.
 The percentage profit/loss is 0%.
x  1150000
 The cost of the flat is $1150000
19. Amount he received for selling all the eggs
 1800  8%  0.1  1800  (1  8%) 
 $2774.4
Percentage profit
2774.4  1500

 100%
1500
 84.96%
5
3
20.
(a) Let $x be the cost of the table.
x  (1  25%)  1625
21.
x  1625  125%
x  1625 
100
125
x  1300
Discount  450 25%
25
 450 
100
 $112.5
Selling price = $(450 - 112.5)
= $337.5
 The cost of the table is $1300.
(b) Selling price  1300  (1  70%)
 1300 
170
100
= $2210
22.
Let $x be the marked price.
x  (1  15%)  2550
23.
 $x
x  2550  85%
x  2550 
Let $x be the marked price of the each book.
Discount for buying 5 book in Bookshop A
100
85
Discount for buying 5 book in Bookshop B
x  3000
 $ x  5  25%
 The marked price is $3000
The discount = $(3000 – 2550)
= $450
25
)x
100
 $1.25 x
 $(5 
 Bookshop B offers a larger discount for
buying 5 books.
24.
(a) Marked price  800  (1  50%)
 800 
150
100
 $1200
(b) To make a profit of 20%,
selling price  800  (1  20%)
 800 
120
100
 $960
Percentage discount 
1200  960
 100%
1200
= 20%