HKDSE Common Question Types

New value Original value Original value

-

Cost Cost Selling price -

Cost Selling price Cost -

Marked price Selling price Marked price -

1 2

Actual value Absolute error

Measured value Maximum absolute error

f b c ma

f b c ma

4

b b ac

a 2

! 2

- -

a b

a c

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HKDSE Common Question Types

Pan Lloyds Publishers Ltd

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Conventional Questions Section A(1)

1. The cost of a machine is $20 000. It is sold at a profit percentage of 5%.

(a) Find the selling price of the machine.

(b) If the marked price is $24 000, find the discount percentage.

Reference: HKDSE 15 I Q6 (4 marks) Solution:

(a) Selling price = $20 000 × (1 + 5%) [1M]

= $21 000 [1A]

(b) Discount percentage = $24 000 – $21 000$24 000 ×100% [1M]

= 12.5% [1A]

Let’s Practice

The cost of a cup is $40. It is sold at a profit percentage of 20%.

(a) Find the selling price of the cup.

(b) If the marked price is $80, find the discount percentage.

(4 marks)

2. In a bank, there are 450 staff and the number of male staff is 25% more than the number of female staff. Find the difference of the numbers of male and female staff. Reference: HKDSE 16 I Q5 (4 marks) Solution:

Let the number of female staff be x.

Then the number of male staff = (1 + 25%)x = 1.25x

x + 1.25x = 450 [1M]

2.25x = 450

x = 200 [1A]

The number of female staff is 200 and the number of male staff is 250.

The difference = 250 – 200 [1M]

= 50 [1A]

Let’s Practice

There are 1200 Chinese and English books in a library. The number of English books is 40% less than the number of Chinese books. Find the difference of the numbers of Chinese and English books. (4 marks)

3. (a) Round up 56 834.62 to the nearest integer.

(b) Round down 56 834.62 to 1 decimal place.

(c) Round off 56 834.62 to 4 significant figures.

Reference: HKDSE 14 I Q3 (3 marks)

For (b), we use the formula Discount percentage

= Marked price – Selling priceMarked price

× 100%

Solving Strategy

First express the number of male staff in terms of the number of female staff. Then set up an equation to solve the unknown.

Solving Strategy

HKDSE Exam Series – Mathematics Mock Exam Papers (Compulsory Part) 2018/19 Edition

2

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Solution:

(a) Let the final volume of the water be x cm³. By the property of similar solids,

x

x – 95 = 15 10



 



3 [1M]+[1A]

xx – 95 = 27 8

8x = 27x – 2565 19x = 2565

x = 135 [1A]

The final volume of the water is 135 cm³.

(b) Let the side length of the base of the water surface be a cm.

13 a² × 15 = 135 a² = 27

a = 27 or – 27 (rejected) [1A]

For the vessel, the lateral faces are isosceles triangles.

Let the slant height be s cm.

( 27 ÷ 2)² + 15² = s² (Pyth. theorem)

s = 231.75

Area of the wet surface

= 4 × 1

2× 27 × 231.75

 

 cm²

≈ 158.21 cm² [1A]

> 150 cm²

` Sam is correct. [1A]

Let’s Practice

An inverted right conical vessel contains some water. The vessel is placed vertically and the depth of the water is 12 cm. Bowie pours 370π cm³ of water into the vessel without overflowing. The depth of water is now 16 cm.

(a) Express the final volume of water in the vessel in terms of π.

(3 marks) (b) Bowie claims that the final area of the wet surface of the vessel is

less than 600 cm². Do you agree? Explain your answer. (3 marks)

Consider two similar pyramids with heights 10 cm and 15 cm.

Then find the required volume by the property of similar solids.

Solving Strategy

15 cm

s cm 27 cm

Wet surface consists of 4 identical triangles.

HKDSE Exam Series – Mathematics Mock Exam Papers (Compulsory Part) 2018/19 Edition

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A B

9:00 9:20 9:47 10:12 11:00 0

2 12

Time Kenny

Ryan

Distance travelled (km)

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a b a b 3 7

3 -

+

a b c

2 3

2

5 3

= 1

= b c

a b - +

11 16

16 11

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( ) a b a b^{3 5 2}

2 -5

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x x

x 1

1 4

4

2-

+ +

- =

x x

x

1 4

3

- -

^ h^ h

x x

x

1 4

3 2

- -

^ h+^ h

x x

x

1 4

3 2

- - -

^ h+^ h

1

x x

x 4 - 3

- -

^ h^ h

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HKDSE Exam Series – Mathematics Mock Exam Papers (Compulsory Part) Solution Guide

Preparation for HKDSE Section A Percentages

Paper 1

1. (a) Selling price = $480 # (1 - 25%) 1M

= $360 1A

(b) Let $x be the cost of the vase.

x

360 - # 100% = 20% x 1M 360 - x = 0.2x

1.2x = 360

x = 300

∴ The cost is $300 1A

Common Mistakes

Candidates may wrongly write the selling price as the denominator.

2. (a) Number of books

= 120 # (1 + 25%) 1M

= 150 1A (b) Number of books on bookshelf C

= 1 25% 120

- 1M

= 160 > 150

∴ Bookshelf C has the most number of books

on it. 1A

Common Mistakes

Some candidates may wrongly think that the number of books on bookshelves A and C are the same.

Paper 2

1. Amount

= $120 000 # 1 2 4% ⁴

c + m

= $129 891.8592

1

Interest

= $(129 891.8592 - 120 000)

= $9892 (cor. to the nearest dollar) The answer is C.

2. Let $x be the selling price of the flat.

x % 4 000 000 4 000 000 1-30 -

^ h # 100% = 20%

0.7x = 4 800 000

x = 6 857 000

(cor. to the nearest thousand) The selling price of the flat is $6 857 000.

The answer is A.

3. Let $x be the original salary of Susan.

New salary = $x(1 + 40%)(1 - 30%) = $0.98x

Percentage change = . x x x

0 98 - # 100%

= -2%

The answer is B.

4. Let a and b be the height and base of the original triangle respectively.

% %

% %

a b

x a b a b

2 2

1 1 80

2 100 116

#

# #

#

+ +

-

=

^ h ^ h

. %

ab

ab x

2 1 2

1 1 8 1 1

100

# + - 116

^ h =

6 @

1.8(1 + x%) - 1 = 100 116

1.8 + 1.8x% - 1 = 100 116

0.8 + .1001 8 x = 100 116

.

1001 8 x = 100 36

x = 20

The answer is A.

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4

HKDSE Exam Series – Mathematics Mock Exam Papers (Compulsory Part) Solution Guide The percentage error

= .

.

363 23 2 # 100%

= 0.88% (cor. to the nearest 0.01%) The answer is B.

Inequalities

Paper 1

1. (a) x

7 - 5 11 < 2x - 1 and 7 - 3x ≤ 138

x

7 - 2x < 5 11 - 1 and 7 - 13 ≤ 3x8

- x

3 < 5 83 and -6 ≤ 3x x > -85 and x ≥ -2

1A + 1A

Combining the above inequalities,

x > -85 1A

(b) The least integer is 0. 1A Guidelines

Remember that zero is also an integer.

2. x – 2 > 0 and 2[8 + (x – 2)] ≤ 50 1M x > 2 and 2(x + 6) ≤ 50

x > 2 and x + 6 ≤ 25

x > 2 and x ≤ 19 1A Combine the above inequalities,

2 < x ≤ 19 1A

Paper 2

1. 4x + 9 ≥ -7 and -2x + 7 < x - 2 4x ≥ -16 and 9 < 3x x ≥ -4 and x > 3

Combining the two inequalities, we have x > 3.

The answer is B.

3

2. 31 (y + 5) ≥ 1 or 4y + 21 > -3 y + 5 ≥ 3 or 4y > -24 y ≥ -2 or y > -6

Combining the two inequalities, we have y > -6.

Since y is a non-negative integer, y ≥ 0.

Thus, the least value of y is 0.

The answer is C.

3. 3x > x - 6 or 9x < 12x + 15 2x > -6 or -3x < 15 x > -3 or x > -5

Combining the two inequalities, we have x > -5.

The answer is A.

Formulas and Polynomials

Paper 1

1. a b b

3 4 2 5

^ -h

= a b b⁵

6 -8 1M

= a b^{5 8}

6 - -^ h

1M

= a b

6 13

1A

2. (a) 4(5p - 2q) = 2p - 3

20p - 8q = 2p - 3 1M 18p = 8q - 3

p = q

18 8 -3

1A (b) Let pʹ and qʹ be the new values of p and q

respectively.

pʹ = q 18 8 l-3

= q 5

18 8_^ + _h-3

1M

= q 40

18 8 3

+ 18 -

= p + 9 20

∴ The value of p is increased by 20 . 1A9

4

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33 Mock Exam 2 Paper 1

Mock Exam 2 Paper 1

Section A(1)

1. (m n ) n

2 3 4 6

-

= m n n

8 6

12

- 1M

= n^{6 – (–12)}m^-8

= n¹⁸m^-8 1M

= m n

8

18 1A

2. (a) x7 + 3y = 2(x - y + 2)2 7x + 6y = 4x - 4y + 8

7x - 4x = -4y + 8 - 6y 1M 3x = 8 - 10y

x = ( y)

3 2 4-5

1A

(b) Z

[

\ ]]]

]]]

x = 4 5y 3 2_^ - _h

…… (1) 3x

2 + 4y = 11 …… (2) Put (1) into (2),

( y)

3 2 4 5 2

3 -

; E + 4y = 11 1M

4 - 5y + 4y = 11

-y = 7

y = -7

Put y = -7 into (1), x =

3 2 4₆ -5_^-7_h@ x = 26

So, the values of x and y are 26 and -7 respectively. 1A 3. Use tabular form to show the result:

1st card

2nd card 2 3 5 7

1 2 3 5 7

2 4 6 10 14

4 8 12 20 28

6 12 18 30 42 1M

The required probability = 1016 1M

= 58 1A

4. (a) The selling price of the product A

= $800 × (1 - 25%) 1M

= $600

The cost of the product A = $(600 - 120)

= $480 1A (b) The cost of the product B

= $ 1 75%

700

c + m 1M

= $400

< $480

Thus, the cost of B is not higher than A. 1A 5. Let x be the number of English books.

Then the number of Chinese books is 1.25x.

x + 1.25x = 1080 1M

2.25x = 1080

x = 480 1A

The number of English books is 480 and the number of Chinese books is 600.

The difference = 600 – 480 1M

= 120 1A

Common Mistakes

Some candidates may forget to calculate the difference of the number of books.

6. (a) Maximum absolute error = 12 # 1 g

= 0.5g 1M

The upper limit of the weight = (200 + 0.5)g

= 200.5 g 1A (b) Upper limit of the weight of 45 standard packs = (45 # 200.5) g

= 9022.5 g 1A

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35 Mock Exam 2 Paper 1 Area of wet surface

= 4 cm

2

6 333 ³ 2

#_c # _m

≈ 218.98 cm² < 220 cm² 1M May is not correct. 1A 12. (a) Range = (74 - 55) kg

= 19 kg 1A

Upper quartile = 66 kg Lower quartile = 58 kg

Inter-quartile range = (66 - 58) kg 1M

= 8 kg 1A

(b) New sum of the weights

= (64 # 35 + 55 + 62 + 63 + 68 + 76) kg = 2564 kg

New mean = 35 5 2564

+ kg 1M

= 64.1 kg 1A

Since 55 and 62 < 63, 68 and 76 > 63, the median will remain unchanged. 1M New median = 63 kg 1A

13. (a) Let f(x) = k1x² + k2x, where k1 and k2 are non-zero constants.

f(-1) = k1(-1)² + k2(-1) 10 = k1 - k2 …… (1) f(3) = k1(3)² + k2(3)

-6 = 9k1 + 3k2 …… (2) 1M (1) # 3 + (2): 24 = 12k1 1M

k₁ = 2

Substituting k₁ = 2 into (1), we have 10 = 2 - k2

k2 = -8 1A

∴ f(x) = 2x² - 8x f(-2) = 2(-2)² - 8(-2)

= 24 1A (b) Substituting x = 6, we have

f(6) = 2(6)² - 8(6) = 72 - 48

= 24 1A

The coordinates of A and B are (-2, 24) and (6, 24).

Since the y-coordinates of A and B are both 24, AB is a horizontal line.

The height of △ABC regarding the base AB is

24 units. 1A

Area = 1

2 # 24 # [6 - (-2)]sq. units 1M

= 96 sq. units 1A

Guidelines

By (a), we know f(-2) = 24.

∴ m = 24

14. (a) By remainder theorem,

f(1) = 6

(1 + h)²(1 - 1) + k = 6 1M

k = 6 1A

By factor theorem,

f(-5) = 0

(-5 + h)²(-5 - 1) + 6 = 0 1M (h - 5)² = 1

h - 5 = -1 or h - 5 = 1 1M h = 4 or 6 1A

(b) When h = 4,

f(x) = 0

(x + 4)²(x - 1) + 6 = 0

x³ + 7x² + 8x - 10 = 0 1M (x + 5)(x² + 2x - 2) = 0 1M x = -5, -1 ! 3 1A Therefore, the student is not correct. 1A

Section B

15. (a) The required probability

C

C

7 22

7 9

= 1M

170 544

= 36

14 212

= 3 (or ≈ 0.000 211) 1A

Guidelines

The number of combination of choosing r objects from n objects (n ≥ r) = C ⁿr.

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