RA1 Percentages
Further Practice for HKCEE
Classwork
RA1
RA1

Percentages
1
The percentage required  1.2 x  x  100 %
1 .2 x
0 .2

 100 %
1 .2
2
 16 %
3
Further Practice for HKCEE (page RA1.8) Top
1. Let x be the number.
x  39 %  78
x  78  39 %
 200
 The number is 200.
8. Let x be the value of A.
x  20 %  40 15 %
x  40 15 %  20 %
 30
 The value of A is 30.
2. Let x be the number.
x  40 %  72
x  72  40 %
 180
 The number is 180.
9. Let x and y be the original length and width
of the rectangle respectively.
New length  x(1  20 %)
 1. 2 x
New width  y (1  30 %)
 0.7 y
Original area  xy
New area  (1.2 x)(0.7 y)
 0.84 xy
 Percentage change in its area
0.84 xy  xy

100 %
xy
  16 %
3. Let x be the number.
x  (1  20 %)  168
x  1.2  168
x  168  1.2
 140
 The number is 140.
4. (a) New value  250  (1  12 %)
 280
(b) New value  250  (1  16%)
 210
5. (a) Selling price  $200  (1  30 %)
 $260
(b) Selling price  $200  (1  15 %)
 $170
6. (a) New number  n(1  60 %)(1  40 %)
 n(1.6)( 0.6)
 0.96 n
(b) Percentage change of the number
0.96 n  n

100 %
n
  4%
7. Let x be the value of B.
A  x(1  20 %)
 1. 2 x
10. Original number of science books
 120 000  40%
 48 000
New number of science books
 (120 000  20 000)  45%
 63 000
Increase in the number of science books
 63 000  48 000
 15 000
 The percentage required
15 000

100 %
20 000
 75 %
11. Let $x be the cost of the product.
x(1  12 .5%)  423
x  1.125  423
x  423  1.125
 376
 The cost of the product is $376.
 Profit  $423  $376
 $47
2
RA1
Percentages
12. Amount paid  $1 200  (1  10 %)
 $1 080
13. (a) Let $x be the cost of the article.
x(1  20 %)  240
x  0.8  240
x  240  0.8
 300
 The cost of the article is $300.
(b) Selling price  $300  (1  20 %)
 $360
14. (a) Cost of each orange  $300
200
 $1.5
 Selling price of each orange
 Cost of each orange
 The shopkeeper gains.
(b) Profit of each orange  $4  $1.5
 $2.5
$ 2 .5
 100 %
Profit percent 
$ 1 .5
2
 166 %
3
15. (a) Profit  $3 600  $3 000
 $600
$600
 100 %
Profit percent 
$3 000
 20 %
(b) Amount paid  $3 600  (1  20 %)
 $2 880
16. (a) Total profit  $120  5  $85  5  $100 10
 $25
$25
 100 %
$100  10
 2.5%
(b) Profit percent 
17. (a) Total loss  $4 000  8  $3 600  8
 $3 200
$3 200
 100 %
$4 000  8
 10 %
(b) Loss percent 
18. (a) Selling price  $500  (1  30 %)  (1  20 %)
 $500  1.3  0.8
 $520
(b) Profit  $520  $500
 $20
$20
 100 %
Profit percent 
$500
 4%
19. (a) Marked price  $C (1  50 %)
 $1.5C
(b) Selling price  $C (1  35 %)
 $1.35C
(c) 1.5C (1  x%)  1.35C
1  x %  0 .9
x %  0 .1
x  10
20. Let $x be the marked price of the article.
x  (1  20 %)  1 600  (1  30 %)
x  1 600  1.3  0.8
 2 600
 The marked price of the article is $2 600.
21. Simple interest  $25 000  6%  2
 $3 750
1
2
22. (a) Simple interest  $10 000  6%  2
 $1 200
(b) Compound interest
6% 4
 $10 000  (1 
)  $10 000
2
 $1 255 (corr. to the nearest dollar)
23. (a) Let $x be the monthly income of David.
x  70 %  11 200
x  11 200  70 %
 16 000
 The monthly income of David is
$16 000.
 Amount of his savings last month
 $16 000  $11 200
 $4 800
RA1
(b) (i) Amount of his expenditure this month
 $11 200  (1  10 %)
 $12 320
(ii) Amount of his savings this month
 $16 000  $12 320
 $3 680
 Percentage decrease in his savings
$4 800  $3 680

 100 %
$4 800
1
 23 %
3
24. (a) Let $x be the selling price of the computer
2 years ago.
x(1  20 %) 2  4 000
x  0.82  4 000
x  4 000  0.8
 6 250
 2 years ago the selling price of the
2
computer was $6 250.
(b) The value of the computer after 3 years
 $4 000  (1  20 %) 3
 $4 000  0.83
 $2 048
25. (a) Let $x and y% be the cost of the brand Y
VCD player and the profit percent on
this VCD player respectively, then
cost of the brand X VCD player
 $ x(1  25 %)
 $1.25 x
selling price of the brand X VCD player
 $1.25 x(1  60 %)
 $2 x
selling price of the brand Y VCD player
 $x(1  y%)
total profit  [2 x  x(1  y%)]  (1.25 x  x)
 (3  y%) x  2.25 x
 x(0.75  y%)
 Total profit percent  50%
x(0.75  y %)

 100 %  50 %
1.25 x  x
0.75  y %
 0.5
2.25
y %  0.375
y  37 .5
 The profit percent on the brand Y
VCD player is 37.5%.
Percentages
3
(b) If the cost of the brand Y VCD player is
$310, then
cost of the brand X VCD player
 $(1.25  310 )
 $387 .5
total profit on the two VCD players
 ($387 .5  $310 )  50 %
 $348 .75
26. Final volume of water in the bottle
 [( 20  4)  80 %  4] litres
 16.8 litres
16 .8 litres
 The percentage required 
 100 %
20 litres
 84 %
2
27. (a) Amount obtained  $20 000  (1  6%)
 $22 472
20 000
 (1  8%) 2 ]
12 .5
 £1 866 .24
(b) Amount obtained  £[
(c) According to Plan B, amount obtained
by Joseph after 2 years
 $(1 866 .24  12 )
 $22 394 .88
 Plan A gives a better return.
 Joseph will adopt Plan A.
28. (a) Let x be the number of students in that
school.
108 
x
 360
360 
x  1 200
 There are 1 200 students in that school.
(b) The number of students in each item in
the year 2003 are as follows:
Number of students who went to school
by bus
72 
 1 200 
 (1  10 %)
360 
 216
Number of students who went to school
by private car
12 
 1 200 
360 
 40
Number of students who went to school
by school bus
 360  (1  15%)
 414
4
RA1
Percentages
Number of students who went to school
by minibus
36 
 1 200 
360 
 120
Number of students who went to school
by MTR
42 
 1 200 
 (1  15 %)
360 
 161
 Number of students who went to
school on foot
 1 200  216  40  414  120  161
 249
Number of students who went to school
on foot in year 2002
90 
 1 200 
360 
 300
 Percentage change in the number of
students who went to school on foot
249  300

100 %
300
 17 %
 x  17%
$66 000
12
 $5 500
29. (a) Monthly basic salary 
(b) Bonus  $372 000 1.5%
 $5 580
 Total income  $66 000  $5 580
 $71 580
(d) Let y% be the bonus percentage in 2003.
72 000  568 000  y %  84 760
568 000  y %  12 760
12 760
y% 
568 000
y  2.2
(corr. to 1 d.p.)
 The bonus percentage in 2003 was
2.2%.
30. (a) (i)
Month
Loan
interest
($)
Loan
repaid
($)
Outstanding
balance
($)
1
400.00
5 600.00
34 400.00
2
344.00
5 656.00
28 744.00
3
287.44
5 712.56
23 031.44
4
230.31
5 769.69
17 261.75
5
172.62
5 827.38
11 434.37
6
114.34
5 885.66
5 548.71
7
55.49
5 548.71
0.00
(ii) Amount of his last instalment
 $5 548 .71  $55 .49
 $5 604.20 (corr. to 2 d.p.)
(iii) Total instalment
 $6 000  6  $5 604 .20
 $41 604 .20 (corr. to 2 d.p.)
Interest earned by the bank
 $41 604 .20  $40 000
 $1 604.20 (corr. to 2 d.p.)
Loan
Instalment
interest
(b) Month
($)
($)
Loan
repaid
($)
Outstanding
balance
($)
(c) (i) Percentage increase in his annual
basic salary
$72 000  $66 000

 100 %
$66 000
 9.1% (corr. to 1 d.p.)
1
6 000.00
400.00
5 600.00
34 400.00
2
6 600.00
344.00
6 256.00
28 144.00
3
7 260.00
281.44
6 978.56
21 165.44
4
7 986.00
211.65
7 774.35
13 391.09
5
8 784.60
133.91
8 650.69
4 740.40
(ii) Let $x be the sales made by the
salesman in 2002.
72 000  x  1.5%  79 320
x  1.5%  7 320
x  7 320  1.5%
 488 000
 The sales made by him in 2002
6
4 787.80
47.40
4 740.40
0.00
was $488 000.
(c) Total instalment of the first 5 months
 $6 000 .00  $6 600 .00  $7 260 .00
 $7 986 .00  $8 784 .60
 $36 630 .60
Loan repaid by Mr. Chan in the first
5 months
 $7 000  5
 $35 000
 Mr. Chan does not have enough
money to repay his loan in full.
RA1
Percentages

T (12)  P
10 000  1.0112  1.0111 P
10
 1.01 P    1.01P  P
P  1.01P
10
11
12
   1.01 P  1.01 P  10 000  1.01
10
11
P (1  1.01    1.01  1.01 )  10 000  1.0112
12
1.01  1
12
P(
)  10 000  1.01
1.01  1
 P  888 (corr. to the nearest integer)
31. (a) Amount obtained
 $30 000 (1  8%)16
 $102 778.28 (corr. to 2 d.p.)
(b) Amount obtained
 $2 000  (1  8%)16  $2 000  (1  8%)15
   $2 000  (1  8%)
 $2 000  1.0816  $2 000  1.0815
   $2 000  1.08
 $2 000  1.08  (1.0815  1.0814    1)
 $2 160  (1  1.08    1.0815 )
1.08  1
)
1.08  1
 $65 500.45 (corr. to 2 d.p.)
 $2 160  (
16
(c) Method in (a) would give a better return.
Difference
 $102 778 .279  $65 500 .451
 $37 277.83 (corr. to 2 d.p.)
32. (a) Interest rate for each month  12 %
12
 1%
 Outstanding balance at the end of
the 1st month before the 1st
instalment
 $10 000  (1  1%)
 $10 100
(b) Principal at the beginning of the 2nd
month  $(10 100  P)
 Outstanding balance at the end of
the 2nd month before the 2nd
instalment
 $(10 100  P)  (1  1%)
 $(10 201  1.01P)
(c) Let $T(n) be the outstanding balance at
the end of the nth month before the nth
instalment.
T (1)  10 000  1.01
T (2)  (10 000  1.01  P)  1.01
 10 000  1.01 2  1.01P
33. Let $x be the cost of 2 000 kg of sugar.
x(1  20 %)  3 000
x  1.2  3 000
x  3 000  1.2
 2 500
 The cost of 2 000 kg of sugar is $2 500.
In order to make a profit of 44%,
selling price of 2 000 kg of sugar
 $2 500  (1  44 %)
 $3 600
 Selling price of each kg of sugar
$3 600

2 000
 $1.8
 The answer is C.
34. Selling price  $24  (1  20 %)
 $28 .8
Loss  $30  $28 .8
 $1.2
$ 1 .2
 100 %
Loss percent 
$30
 4%
 The answer is D.
12 % 12
)  10 000
12
P(1.01)12  10 000
10 000
P
1.0112
 8 874 .49 (corr. to 2 d.p.)
The answer is C.
35. P(1 

T (3)  (10 000  1.01  1.01P  P)  1.01
2
 10 000  1.013  1.01 2 P  1.01P
 T (12 )  10 000 1.01  1.01 P
12
11
 1.0110 P    1.01P
 Outstanding balance at the end of
the 12th month before the 12th
instalment  The 12th instalment
5
36. Cost of the mixture per kg
2
3
 $12 
 $17 
23
23
 $15
6
RA1
Percentages
Profit obtained by selling each kg of the
mixture
 $24  $15
 $9
$9
 100 %
Profit percent 
$15
 60 %
 The answer is D.
37. Let x kg be the original consumption of
pork.
New consumption of pork
 x(1  30%) kg
 0.7x kg
 Percentage change in Mrs. Lau’s
expenditure on pork
$35 (0.7 x)  $28 x

 100 %
$28 x
24 .5 x  28 x

 100 %
28 x
 12 .5%
 Her expenditure on pork reduces by
12.5%.
 The answer is A.
38. Let x be the number of students in the
school, then
number of boys in the school  x  60 %  0.6 x
number of girls in the school  x  40 %  0.4 x
0.6 x(70 %)  0.4 x(40 %)  696
0.58 x  696
x  1 200
 There are 1 200 students in the school.
 The answer is A.
39. Interest after 5 years  $ P  4%  5
 Amount after 5 years  $ P  4%  5  $ P
 The answer is D.
12 % 124
)
12
1 48
 $ P(1 
)
100
40. Amount after 4 years  $ P(1 

The answer is B.
41. Let  and w be the original length and width
of the rectangle respectively.
New length  (1  30 %)
 1.3
New width  w(1  20 %)
 1.2 w
Original area  w
New area  (1.3)(1.2 w)
 1.56 w
 Percentage increase in its area
1.56 w  w

100 %
w
 56%
 The answer is C.
42. Let r and h be the original base radius and
height of the cone respectively.
New base radius  r (1  10 %)
 1.1r

Original volume  r 2 h
3

New volume  (1.1r ) 2 h
3
 Percentage increase in its volume

(1.1r ) 2 h  3 r 2 h
3

 100 %
 2
r h
3

 21 %
The answer is C.
Classwork
Top
Classwork RA1.1 (page RA1.3)
(a) 24% of 150  150  24%
 36
(b) 120% of 25  25 120 %
 30
Classwork RA1.2 (page RA1.4)
(a) Let x be the number.
x  40 %  18
x  18  40 %
 45
 The number is 45.
(b) Let x be the number.
x 150 %  12
x  12  150 %
8
 The number is 8.
RA1
Percentages
Classwork RA1.3 (page RA1.4)
Classwork RA1.6 (page RA1.5)
(a) Let x be the number.
x  (1  40 %)  108
x  0.6  108
x  108  0.6
 180
 The number is 180.
(a) Let $y be the cost of the reference book.
280  y(1  x%) .......... (1)
190  y(1  2 x%) ........ (2)

(b) Let x be the number.
x  (1  150 %)  210
x  2.5  210
x  210  2.5
 84
 The number is 84.
Classwork RA1.4 (page RA1.4)
(a) Let x be the value of B.
A  x(1  25 %)
 0.75 x
 The percentage required
x  0.75 x

 100 %
0.75 x
0.25

 100 %
0.75
1
 33 %
3
7
280 1  x%

190 1  2 x%
2x
x
280  280 
 190  190 
100
100
750 x
90 
100
x  12
(1)
,
( 2)
(b) Substitute x  12 into (1),
280  y (1  12 %)
y  1.12  280
y  250

The cost of the reference book is $250.
Classwork RA1.7 (page RA1.6)
Selling price of the set of jewellery
 $3 500  (1  20 %)
 $2 800
 Cost of the set of jewellery  $2 800  $800
 $2 000
(b) Let x be the value of B.
A  x(1  25 %)
 1.25 x
 The percentage required
1.25 x  x

 100 %
1.25 x
0.25

 100 %
1.25
 20 %
Classwork RA1.8 (page RA1.6)
Classwork RA1.5 (page RA1.5)
Classwork RA1.9 (page RA1.7)
Let x and y be the original length and width of
the rectangle respectively.
New length  x(1  30 %)
 1.3x
New width  y (1  20 %)
 0.8 y
Original area  xy
New area  (1.3x)( 0.8 y )
 1.04 xy
 Percentage change in its area
1.04 xy  xy

100 %
xy
  4%
(a) Interest rate for each month 
Simple interest  $50 000  6%  4
 $12 000
5% 8
)  $50 000
2
 $10 920 .145 (corr. to 3 d.p.)
Difference  $12 000  $10 920 .145
 $1 079.86 (corr. to 2 d.p.)
Compound interest  $50 000  (1 


12 %
 1%
12
Outstanding balance at the end of the 1st
month before the 1st instalment
 $60 000  (1  1%)
 $60 600
(b) (i) Principal at the beginning of the 2nd
month  $( 60 600  P )
(ii) Outstanding balance at the end of the
2nd month before the 2nd instalment
 $( 60 600  P)  (1  1%)
 $( 61 206  1.01P)
8
RA1
Percentages
(c) Principal at the beginning of the 3rd month
 $( 61 206  1.01P  P)
 The 3rd instalment
 Outstanding balance at the end of the
3rd month before the 3rd instalment

$ P  $( 61 206  1.01P  P)  (1  1%)
P  (61 206  2.01P)  1.01
P  61 818 .06  2.030 1P
3.030 1P  61 818 .06
P  20 401
(corr. to the nearest integer)
Classwork RA1.10 (page RA1.8)
Let x be the population of the city ten years ago.
6 000 000  x(1  0.5%)10
6 000 000
x
(1  0.5%)10
 5 710 000 (corr. to 3 sig. fig.)
 The population of the city was 5 710 000 ten
years ago.