Sec 2 Maths End-of-year exam Paper 2: free PDF with answers

Print it or save it as a PDF, do it under exam conditions, then check the answers and worked solutions on the last page. It's a practice paper in the same style as school exam papers and past year papers, not a copy of one.

In the print window, choose Save as PDF to download it.

MugCornerT-Leng Tuition · mugcornersg.com

Mathematics

Sec 2 Maths · End-of-year exam Paper 2

Primes, HCF and LCM + Integers, rational and real numbers + Approximation and estimation + Basic algebra + Linear equations + Coordinates and linear graphs + Number sequences and patterns + Percentage + Ratio, rate and speed + Angles and parallel lines + Triangles, quadrilaterals and polygons + Perimeter and area + Volume and surface area + Statistics (Sec 1) + Direct and inverse proportion + Linear equations, inequalities and simultaneous equations + Expansion and factorisation of quadratic expressions + Algebraic fractions and formulae + Quadratic equations and graphs + Congruence and similarity + Pythagoras' theorem + Trigonometric ratios + Pyramids, cones and spheres + Statistics (Sec 2) + Probability

Practice paper in the same style as the school and national exam papers. Not a copy of any past year paper.

1 hour 15 minutes

Name

Class

Index number

Candidates answer on the Question Paper.

Read these instructions first

Write your name, class and index number in the spaces at the top of this page.

Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs.

Do not use staples, paper clips, glue or correction fluid.

Answer all questions.

If working is needed for any question it must be shown with the answer. Omission of essential working will result in loss of marks.

The use of an approved scientific calculator is expected, where appropriate.

If the degree of accuracy is not specified in the question, and if the answer is not exact, give the answer to three significant figures. Give answers in degrees to one decimal place.

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

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.

For Examiner's Use

/ 50

(a)(i)The length of a rod is

8.6

cm, correct to 1 decimal place. Given that the actual length of the rod is

l

cm, write down the range of possible values of

l

Answer (a)(i)[1]

(a)(ii)The length of a rod is

8.6

cm, correct to 1 decimal place. Write down the greatest possible difference between the actual length of the rod and

8.6

cm.

Answer (a)(ii)cm[1]

(b)(i)Write

504\,789

correct to 3 significant figures.

Answer (b)(i)[1]

(b)(ii)Write

0.02995

correct to 3 significant figures.

Answer (b)(ii)[1]

(c)By writing each number correct to 1 significant figure, estimate the value of

\dfrac{23.7 \times 0.48}{1.89}

Answer (c)[2]

(a)(i)Write down an algebraic expression for

9

times

n

Answer (a)(i)[1]

(a)(ii)Nurul is

y

years old. Nurul's father is

4

times as old as Nurul. Write down an expression, in terms of

y

, for the father's age in

5

years' time.

Answer (a)(ii)[1]

(b)Factorise

xy - 3x

completely.

Answer (b)[2]

(c)(i)Expand and simplify

-3(4x - 8)

Answer (c)(i)[1]

(c)(ii)Expand and simplify

3(3x - 6) - 6(2x - 8)

Answer (c)(ii)[1]

(a)Solve

\dfrac{x}{5} - 7 = -10

Answer (a)[2]

(b)Solve

2(x + 6) = 4(3x - 7)

Answer (b)[2]

(c)(i)The area of a trapezium is given by

A = \dfrac{1}{2}(a + b)h

. Find the value of

A

when

a = 7

b = 13

and

h = 3

Answer (c)(i)[1]

(c)(ii)The cost, $

C

, of hiring a kayak for

h

hours is given by

C = 15h + 8

. Find the value of

h

when

C = 113

Answer (c)(ii)[2]

(a)The graph shows the height,

y

cm, of a candle

x

hours after it is lit. The candle burns at a steady rate. Find this rate, in cm per hour.

Answer (a)[2]

(b)(i)Find the gradient of the line

y = 3 - x

Answer (b)(i)[1]

(b)(ii)A straight line passes through the points

A(1, -9)

and

B(3, -5)

. Find the equation of the line, in the form

y = mx + c

Answer (b)(ii)[3]

(a)(i)Find

45\%

240

Answer (a)(i)[1]

(a)(ii)A fruit seller had 40 durians. She sold 27 of them by noon. Find the percentage of the durians she sold by noon.

Answer (a)(ii)%[1]

Hire purchase	Television
Cash price	$1,200
Deposit	20% of the cash price
Monthly instalments	30 instalments of $36.80

(b)Ravi buys a television on the hire-purchase terms in the table. The amount still owed after the deposit is the loan, and the interest is simple interest on the loan for the 30 months. Find the rate of simple interest per annum.

Answer (b)%[3]

(a)The diagram shows a wooden box in the shape of a cuboid, with a lid. Its outside, including the lid, is painted with paint that costs $4 per m². Find the cost of the paint needed.

Answer (a)$[2]

(b)A closed cylinder has a radius of

10

m and a total surface area of

280\pi

m². Find its height,

h

Answer (b)m[3]

(a)5 identical pumps can empty a tank in 21 hours. After 6 hours, 2 of the pumps break down. Find the number of hours the pumps take to empty the rest of the tank.

Answer (a)hours[2]

(b)(i)

y

is directly proportional to the square of

x

. Find the percentage decrease in

y

when

x

is decreased by

50\%

Answer (b)(i)%[3]

(b)(ii)

y

is directly proportional to the square of

x

. When

x

increases from

1

6

y

increases by

70

. Find the value of

y

when

x = 10

Answer (b)(ii)[3]

(a)Simplify

\dfrac{3y^2}{4x} \times \dfrac{4}{11x}

, giving your answer in its simplest form.

Answer (a)[2]

(b)Make

x

the subject of

y = \dfrac{-4}{x} + 6

Answer (b)[2]

(c)(i)Simplify

\dfrac{x^{2} + x - 30}{x - 5}

Answer (c)(i)[1]

(c)(ii)Simplify

\dfrac{x^{2} - 9}{x^{2} - 4x + 4} \div \dfrac{x - 3}{x - 2}

Answer (c)(ii)[2]

END OF PAPER

Answers and worked solutions

Question 1

(a)(i)Answer:

8.55 \leq l < 8.65

[1]

Half of

0.1

= 0.05

8.55 \leq l < 8.65

(a)(ii)Answer: Half of

0.1

= 0.05

cm[1]

Half of

0.1

= 0.05

(b)(i)Answer:

504\,789 = 505\,000

(3 s.f.)[1]

The 4th significant figure is

7

.It is 5 or more, so round up.

504\,789 = 505\,000

(3 s.f.)

(b)(ii)Answer:

0.02995 = 0.0300

(3 s.f.)[1]

The 4th significant figure is

5

.It is 5 or more, so round up.

0.02995 = 0.0300

(3 s.f.)

(c)Answer:

= 5

[2]

\dfrac{23.7 \times 0.48}{1.89} \approx \dfrac{20 \times 0.5}{2}

= \dfrac{10}{2}

= 5

Question 2

(a)(i)Answer:

9n

[1]

“Times” means multiply.

9n

(a)(ii)Answer:

4y + 5

[1]

4y

4y + 5

(b)Answer:

x(y - 3) = xy - 3x

✓[2]

\text{HCF of } 1 \text{ and } 3 =

1

Every term has an

x

, so

x

is a common factor too.

xy - 3x = x(y - 3)

x(y - 3) = xy - 3x

✓

(c)(i)Answer:

-3(4x - 8) = -12x + 24

[1]

-3(4x - 8) = -12x + 24

(c)(ii)Answer:

= -3x + 30

[1]

3(3x - 6) - 6(2x - 8) = 9x - 18 - 12x + 48

= -3x + 30

Question 3

(a)Answer:

x = -15

[2]

\dfrac{x}{5} = -10 + 7

\dfrac{x}{5} = -3

x = -15

(b)Answer:

x = 4

[2]

2x + 12 = 12x - 28

2x - 12x = -28 - 12

-10x = -40

x = 4

(c)(i)Answer:

A = 30

[1]

A = \dfrac{1}{2}(7 + 13) \times 3

A = 30

(c)(ii)Answer:

h = 7

[2]

113 = 15h + 8

15h = 105

h = 7

Question 4

(a)Answer: Rate:

3

cm per hour[2]

The line goes from

(0, 24)

(8, 0)

.Rise:

0 - 24 = -24

; run:

8 - 0 = 8

\text{Gradient} = \dfrac{-24}{8}

\phantom{\text{Gradient}} = -3

Rate:

3

cm per hour

(b)(i)Answer:

m = -1

[1]

y = -x + 3

m = -1

(b)(ii)Answer:

y = 2x - 11

[3]

m = \dfrac{-5 - (-9)}{3 - 1}

\phantom{m} = \dfrac{4}{2}

\phantom{m} = 2

-9 = 2 \times 1 + c

c = -11

y = 2x - 11

Question 5

(a)(i)Answer:

= 108

[1]

45\% \text{ of } 240 = \dfrac{45}{100} \times 240

= 108

(a)(ii)Answer:

= 67.5\%

[1]

\dfrac{27}{40} \times 100\%

= 67.5\%

(b)Answer: The rate of simple interest is

6\%

per annum.[3]

\text{Deposit} = \dfrac{20}{100} \times 1200

\phantom{\text{Deposit}} = 240

\text{Loan} = 1200 - 240

\phantom{\text{Loan}} = 960

30 \times 36.80 = 1104

\text{Interest} = 1104 - 960

\phantom{\text{Interest}} = 144

T = \dfrac{30}{12}

\phantom{T} = 2.5

R = \dfrac{100 \times 144}{960 \times 2.5}

\phantom{R} = 6

The rate of simple interest is

6\%

per annum.

Question 6

(a)Answer: The paint costs $3.76.[2]

\text{Area} = 2(50 \times 40) + 2(50 \times 30) + 2(40 \times 30)

= 9400

cm²

\dfrac{9400}{10\,000} = 0.94

m²

0.94 \times 4 = 3.76

The paint costs $3.76.

(b)Answer: The height is

4

m.[3]

2\pi r^2 + 2\pi rh = 280\pi

2\pi(10)^2 + 2\pi(10)h = 280\pi

200 + 20h = 280

20h = 80

h = \dfrac{80}{20}

h = 4

The height is

4

Question 7

(a)Answer:

= 25

hours[2]

We assume that every pump empties water at the same rate.

\text{Left} = 5 \times (21 - 6)

= 75

pump-hours

\text{Time} = \dfrac{75}{5 - 2}

= 25

hours

(b)(i)Answer:

= 75\%

[3]

y = kx^2

\dfrac{100 - 50}{100} = 0.5

\text{New } y = k(0.5x)^2

= 0.25kx^2

\text{Percentage decrease} = (1 - 0.25) \times 100\%

= 75\%

(b)(ii)Answer:

= 200

[3]

y = kx^2

When

x = 1

y = k(1)^2

y = 1k

.When

x = 6

y = k(6)^2

y = 36k

36k - 1k = 70

35k = 70

k = \dfrac{70}{35}

= 2

y = 2x^2

y = 2(10)^2

= 200

Question 8

(a)Answer:

= \dfrac{3y^2}{11x^2}

[2]

\dfrac{3y^2}{4x} \times \dfrac{4}{11x} = \dfrac{12 \times y \times y}{44 \times x \times x}

The HCF of

12

and

44

4

\dfrac{12}{44} = \dfrac{3}{11}

= \dfrac{3y^2}{11x^2}

(b)Answer:

x = \dfrac{-4}{y - 6}

[2]

y - 6 = \dfrac{-4}{x}

x(y - 6) = -4

x = \dfrac{-4}{y - 6}

(c)(i)Answer:

= x + 6

[1]

\dfrac{x^{2} + x - 30}{x - 5} = \dfrac{(x - 5)(x + 6)}{(x - 5)}

= x + 6

(c)(ii)Answer:

= \dfrac{x + 3}{x - 2}

[2]

\dfrac{x^{2} - 9}{x^{2} - 4x + 4} \div \dfrac{x - 3}{x - 2} = \dfrac{x^{2} - 9}{x^{2} - 4x + 4} \times \dfrac{x - 2}{x - 3}

= \dfrac{(x - 3)(x + 3)}{(x - 2)^{2}} \times \dfrac{x - 2}{x - 3}

= \dfrac{x + 3}{x - 2}

Try a full paper, free

Make a free account, pick the chapters your exam tests, and start a timed paper in under a minute.

Start a free paper

Sec 2 Maths End-of-year exam Paper 2: free PDF with answers

In the print window, choose Save as PDF to download it.

MugCornerT-Leng Tuition · mugcornersg.com

Mathematics

Sec 2 Maths · End-of-year exam Paper 2

Practice paper in the same style as the school and national exam papers. Not a copy of any past year paper.

1 hour 15 minutes

Name

Class

Index number

Candidates answer on the Question Paper.

Read these instructions first

Write your name, class and index number in the spaces at the top of this page.

Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs.

Do not use staples, paper clips, glue or correction fluid.

Answer all questions.

If working is needed for any question it must be shown with the answer. Omission of essential working will result in loss of marks.

The use of an approved scientific calculator is expected, where appropriate.

If the degree of accuracy is not specified in the question, and if the answer is not exact, give the answer to three significant figures. Give answers in degrees to one decimal place.

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

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.

For Examiner's Use

/ 50

(a)(i)The length of a rod is

8.6

cm, correct to 1 decimal place. Given that the actual length of the rod is

l

cm, write down the range of possible values of

l

Answer (a)(i)[1]

(a)(ii)The length of a rod is

8.6

cm, correct to 1 decimal place. Write down the greatest possible difference between the actual length of the rod and

8.6

cm.

Answer (a)(ii)cm[1]

(b)(i)Write

504\,789

correct to 3 significant figures.

Answer (b)(i)[1]

(b)(ii)Write

0.02995

correct to 3 significant figures.

Answer (b)(ii)[1]

(c)By writing each number correct to 1 significant figure, estimate the value of

\dfrac{23.7 \times 0.48}{1.89}

Answer (c)[2]

(a)(i)Write down an algebraic expression for

9

times

n

Answer (a)(i)[1]

(a)(ii)Nurul is

y

years old. Nurul's father is

4

times as old as Nurul. Write down an expression, in terms of

y

, for the father's age in

5

years' time.

Answer (a)(ii)[1]

(b)Factorise

xy - 3x

completely.

Answer (b)[2]

(c)(i)Expand and simplify

-3(4x - 8)

Answer (c)(i)[1]

(c)(ii)Expand and simplify

3(3x - 6) - 6(2x - 8)

Answer (c)(ii)[1]

(a)Solve

\dfrac{x}{5} - 7 = -10

Answer (a)[2]

(b)Solve

2(x + 6) = 4(3x - 7)

Answer (b)[2]

(c)(i)The area of a trapezium is given by

A = \dfrac{1}{2}(a + b)h

. Find the value of

A

when

a = 7

b = 13

and

h = 3

Answer (c)(i)[1]

(c)(ii)The cost, $

C

, of hiring a kayak for

h

hours is given by

C = 15h + 8

. Find the value of

h

when

C = 113

Answer (c)(ii)[2]

(a)The graph shows the height,

y

cm, of a candle

x

hours after it is lit. The candle burns at a steady rate. Find this rate, in cm per hour.

Answer (a)[2]

(b)(i)Find the gradient of the line

y = 3 - x

Answer (b)(i)[1]

(b)(ii)A straight line passes through the points

A(1, -9)

and

B(3, -5)

. Find the equation of the line, in the form

y = mx + c

Answer (b)(ii)[3]

(a)(i)Find

45\%

240

Answer (a)(i)[1]

(a)(ii)A fruit seller had 40 durians. She sold 27 of them by noon. Find the percentage of the durians she sold by noon.

Answer (a)(ii)%[1]

Hire purchase	Television
Cash price	$1,200
Deposit	20% of the cash price
Monthly instalments	30 instalments of $36.80

Answer (b)%[3]

(a)The diagram shows a wooden box in the shape of a cuboid, with a lid. Its outside, including the lid, is painted with paint that costs $4 per m². Find the cost of the paint needed.

Answer (a)$[2]

(b)A closed cylinder has a radius of

10

m and a total surface area of

280\pi

m². Find its height,

h

Answer (b)m[3]

(a)5 identical pumps can empty a tank in 21 hours. After 6 hours, 2 of the pumps break down. Find the number of hours the pumps take to empty the rest of the tank.

Answer (a)hours[2]

(b)(i)

y

is directly proportional to the square of

x

. Find the percentage decrease in

y

when

x

is decreased by

50\%

Answer (b)(i)%[3]

(b)(ii)

y

is directly proportional to the square of

x

. When

x

increases from

1

6

y

increases by

70

. Find the value of

y

when

x = 10

Answer (b)(ii)[3]

(a)Simplify

\dfrac{3y^2}{4x} \times \dfrac{4}{11x}

, giving your answer in its simplest form.

Answer (a)[2]

(b)Make

x

the subject of

y = \dfrac{-4}{x} + 6

Answer (b)[2]

(c)(i)Simplify

\dfrac{x^{2} + x - 30}{x - 5}

Answer (c)(i)[1]

(c)(ii)Simplify

\dfrac{x^{2} - 9}{x^{2} - 4x + 4} \div \dfrac{x - 3}{x - 2}

Answer (c)(ii)[2]

END OF PAPER

Answers and worked solutions

Question 1

(a)(i)Answer:

8.55 \leq l < 8.65

[1]

Half of

0.1

= 0.05

8.55 \leq l < 8.65

(a)(ii)Answer: Half of

0.1

= 0.05

cm[1]

Half of

0.1

= 0.05

(b)(i)Answer:

504\,789 = 505\,000

(3 s.f.)[1]

The 4th significant figure is

7

.It is 5 or more, so round up.

504\,789 = 505\,000

(3 s.f.)

(b)(ii)Answer:

0.02995 = 0.0300

(3 s.f.)[1]

The 4th significant figure is

5

.It is 5 or more, so round up.

0.02995 = 0.0300

(3 s.f.)

(c)Answer:

= 5

[2]

\dfrac{23.7 \times 0.48}{1.89} \approx \dfrac{20 \times 0.5}{2}

= \dfrac{10}{2}

= 5

Question 2

(a)(i)Answer:

9n

[1]

“Times” means multiply.

9n

(a)(ii)Answer:

4y + 5

[1]

4y

4y + 5

(b)Answer:

x(y - 3) = xy - 3x

✓[2]

\text{HCF of } 1 \text{ and } 3 =

1

Every term has an

x

, so

x

is a common factor too.

xy - 3x = x(y - 3)

x(y - 3) = xy - 3x

✓

(c)(i)Answer:

-3(4x - 8) = -12x + 24

[1]

-3(4x - 8) = -12x + 24

(c)(ii)Answer:

= -3x + 30

[1]

3(3x - 6) - 6(2x - 8) = 9x - 18 - 12x + 48

= -3x + 30

Question 3

(a)Answer:

x = -15

[2]

\dfrac{x}{5} = -10 + 7

\dfrac{x}{5} = -3

x = -15

(b)Answer:

x = 4

[2]

2x + 12 = 12x - 28

2x - 12x = -28 - 12

-10x = -40

x = 4

(c)(i)Answer:

A = 30

[1]

A = \dfrac{1}{2}(7 + 13) \times 3

A = 30

(c)(ii)Answer:

h = 7

[2]

113 = 15h + 8

15h = 105

h = 7

Question 4

(a)Answer: Rate:

3

cm per hour[2]

The line goes from

(0, 24)

(8, 0)

.Rise:

0 - 24 = -24

; run:

8 - 0 = 8

\text{Gradient} = \dfrac{-24}{8}

\phantom{\text{Gradient}} = -3

Rate:

3

cm per hour

(b)(i)Answer:

m = -1

[1]

y = -x + 3

m = -1

(b)(ii)Answer:

y = 2x - 11

[3]

m = \dfrac{-5 - (-9)}{3 - 1}

\phantom{m} = \dfrac{4}{2}

\phantom{m} = 2

-9 = 2 \times 1 + c

c = -11

y = 2x - 11

Question 5

(a)(i)Answer:

= 108

[1]

45\% \text{ of } 240 = \dfrac{45}{100} \times 240

= 108

(a)(ii)Answer:

= 67.5\%

[1]

\dfrac{27}{40} \times 100\%

= 67.5\%

(b)Answer: The rate of simple interest is

6\%

per annum.[3]

\text{Deposit} = \dfrac{20}{100} \times 1200

\phantom{\text{Deposit}} = 240

\text{Loan} = 1200 - 240

\phantom{\text{Loan}} = 960

30 \times 36.80 = 1104

\text{Interest} = 1104 - 960

\phantom{\text{Interest}} = 144

T = \dfrac{30}{12}

\phantom{T} = 2.5

R = \dfrac{100 \times 144}{960 \times 2.5}

\phantom{R} = 6

The rate of simple interest is

6\%

per annum.

Question 6

(a)Answer: The paint costs $3.76.[2]

\text{Area} = 2(50 \times 40) + 2(50 \times 30) + 2(40 \times 30)

= 9400

cm²

\dfrac{9400}{10\,000} = 0.94

m²

0.94 \times 4 = 3.76

The paint costs $3.76.

(b)Answer: The height is

4

m.[3]

2\pi r^2 + 2\pi rh = 280\pi

2\pi(10)^2 + 2\pi(10)h = 280\pi

200 + 20h = 280

20h = 80

h = \dfrac{80}{20}

h = 4

The height is

4

Question 7

(a)Answer:

= 25

hours[2]

We assume that every pump empties water at the same rate.

\text{Left} = 5 \times (21 - 6)

= 75

pump-hours

\text{Time} = \dfrac{75}{5 - 2}

= 25

hours

(b)(i)Answer:

= 75\%

[3]

y = kx^2

\dfrac{100 - 50}{100} = 0.5

\text{New } y = k(0.5x)^2

= 0.25kx^2

\text{Percentage decrease} = (1 - 0.25) \times 100\%

= 75\%

(b)(ii)Answer:

= 200

[3]

y = kx^2

When

x = 1

y = k(1)^2

y = 1k

.When

x = 6

y = k(6)^2

y = 36k

36k - 1k = 70

35k = 70

k = \dfrac{70}{35}

= 2

y = 2x^2

y = 2(10)^2

= 200

Question 8

(a)Answer:

= \dfrac{3y^2}{11x^2}

[2]

\dfrac{3y^2}{4x} \times \dfrac{4}{11x} = \dfrac{12 \times y \times y}{44 \times x \times x}

The HCF of

12

and

44

4

\dfrac{12}{44} = \dfrac{3}{11}

= \dfrac{3y^2}{11x^2}

(b)Answer:

x = \dfrac{-4}{y - 6}

[2]

y - 6 = \dfrac{-4}{x}

x(y - 6) = -4

x = \dfrac{-4}{y - 6}

(c)(i)Answer:

= x + 6

[1]

\dfrac{x^{2} + x - 30}{x - 5} = \dfrac{(x - 5)(x + 6)}{(x - 5)}

= x + 6

(c)(ii)Answer:

= \dfrac{x + 3}{x - 2}

[2]

\dfrac{x^{2} - 9}{x^{2} - 4x + 4} \div \dfrac{x - 3}{x - 2} = \dfrac{x^{2} - 9}{x^{2} - 4x + 4} \times \dfrac{x - 2}{x - 3}

= \dfrac{(x - 3)(x + 3)}{(x - 2)^{2}} \times \dfrac{x - 2}{x - 3}

= \dfrac{x + 3}{x - 2}

Try a full paper, free

Make a free account, pick the chapters your exam tests, and start a timed paper in under a minute.

Start a free paper