1.2.1 Directed Numbers, PT3 Practice

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1.2.1 Directed Numbers, PT3 Practice 1
Question 1:
Complete the following calculation.
− 4.6 × ( 3 5 − 1 3 4 ) = − 4.6 × ( 0.6 − x ) = − 4.6 × x = x

Solution:
− 4.6 × ( 3 5 − 1 3 4 ) = − 4.6 × ( 0.6 − 1.75 ) = − 4.6 × − 1.15 = 5.29


Question 2:
Complete the following calculation.
− 43 + 35 ÷ 1 1 6 = − 43 + 35 ÷ 7 6 = − 43 + 35 × x = − 43 + x = − 13

Solution:
− 43 + 35 ÷ 1 1 6 = − 43 + 35 ÷ 7 6 = − 43 + 35 × 6 7 = − 43 + 30 = − 13



Question 3:
Calculate each of the following.
  (a) (–9) × 14
  (b)   6.7 – 3.2 × (–0.5)  
  (c) –144 ÷ (–8) ÷ (–2)   

Solution:
 (a) (–9) × 14 = –126
 (b)   6.7 – 3.2 × (–0.5) = 6.7 – (–1.6)
= 6.7 + 1.6
 = 8.3
 (c) –144 ÷ (–8) ÷ (–2) = –144 ÷ 4
= 36


Question 4:
Calculate the value of
0.6 + ( − 1 2 3 ) ÷ ( 4 15 ) − 2

Solution:
0.6 + ( − 1 2 3 ) ÷ ( 4 15 ) − 2 = 0.6 + [ − 5 3 × 15 5 4 ] − 2 = 0.6 + [ − 25 4 ] − 2 = 0.6 − 6 1 4 − 2 = 0.6 − 6.25 − 2 = − 7.65



Question 5:
Calculate the value of
( − 0.6 × 1 1 4 ) − [ − 1.6 × ( − 1.5 ) ] ÷ 0.4

Solution:
( − 0.6 × 1 1 4 ) − [ − 1.6 × ( − 1.5 ) ] ÷ 0.4 = ( − 0.6 − 0.3 × 5 4 2 ) − [ − 1.6 − 4 × ( − 1.5 ) 0.4 1 ] = − 1.5 2 − 6 = − 0.75 − 6 = − 6.75


1.1 Directed Numbers

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1.1 Directed Numbers

1.1.1 Multiplication and Division of Integers
1. Multiplication and division of like signs gives (+)

( + ) × ( + ) = + ( + ) ÷ ( + ) = + ( − ) × ( − ) = + ( − ) ÷ ( − ) = +


2. Multiplication and division of unlike signs gives (–)

( + ) × ( − ) = − ( + ) ÷ ( − ) = − ( − ) × ( + ) = − ( − ) ÷ ( + ) = −

Example:
  (a) –25 ÷ 5 = –5
  (b)   8 × (–5) = –40
 

3. Multiplication of 3 integers.

 ( + ) × ( + ) × ( + ) = ( + ) ( + ) × ( + ) × ( − ) = ( − ) ( + ) × ( − ) × ( − ) = ( + )


4. Division of 3 integers.

( + ) ÷ ( + ) ÷ ( + ) = ( + ) ( + ) ÷ ( + ) ÷ ( − ) = ( − ) ( + ) ÷ ( − ) ÷ ( − ) = ( + )


5. The product of an integer and zero is always zero.
Example:
–5 × 0 = 0


6. When zero is divided by any integer except zero, the quotient is zero. Any integer divided by zero is undefined.
Example:
  (a) 0 ÷ 9 = 0
  (b)   –6 ÷ 0 is undefined


1.1.2 Combined Operations of Integers
 
1. BODMAS → (Brackets of Division, Multiplication, Addition and Subtraction) 

 1. Operations in the brackets should be carried out first.
 2. Followed by × or ÷ from left to right.
 3. Followed by + or – from left to right.

Example 1:
(a) –52 ÷ 13 – 15 × 4
(b)   63 ÷ (16 – 7) × (–2)
(c) –30 + 9 × 7 – 16

Solution:
(a)
–52 ÷ 13 – 15 × 4
= (–52 ÷ 13) – (15 × 4) ← (calculate from left to right; ÷ and × are done first)
= –4 – 60
= –64

(b)
63 ÷ (16 – 7) × (–2)
= 63 ÷ 9 × (–2) ← (bracket is done first, then work from left to right)
= 7 × (–2)
= –14

(c)
–30 + 9 × 7 – 16
= –30 + (9 × 7) – 16 ← ( multiply first)
= –30 + 63 – 16
= 17


1.1.3 Fractions
1. Positive fractions are  fractions with the positive sign (+), and their values are greater than 0.
2. Negative fractions are  fractions with the negative sign (–), and their values are less than 0.


1.1.4 Decimals
1. There are positive decimals and negative decimals.
2. Positive decimals are greater than zero and negative decimals are less than zero.

1.1.5 Directed Numbers
1. Integers, fractions and decimals are directed numbers.
2.The computations of directed numbers is the same as that for whole numbers.

Example 2:
(a)  − 1 2 + ( − 0.37 ) − ( − 5 )
(b)   [(–28) – (–4)] ÷ (–5.147 – 0.853)

Solution:
(a)
− 1 2 + ( − 0.37 ) − ( − 5 ) = − 0.5 − 0.37 + 5 = − 0.87 + 5 = 4.13

(b)
[(–28) – (–4)] ÷ (–5.147 – 0.853)
= [–28 + 4] ÷ (–6)
= (–24) ÷ (–6)
= 4

11.2.3 Perimeter and Area, PT3 Practice

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Question 9:
In diagram below, ADB is a right-angled triangle and DBFE is a square. C is the midpoint of DB and CH = CD.
Calculate the area, in cm2, of the coloured region.

Solution:
Area of △ ABC = 1 2 ×6×8 =24  cm 2 Area of trapezium BCHF = 1 2 ×( 12+6 )×6 =54  cm 2 Area of CDEFH =( 12×12 )−54 =144−54 =90  cm 2 Area of coloured region =24+90 =114  cm 2


Question 10:
Diagram below shows a rectangle ACDE.

Calculate the area, in cm2, of the coloured region.

Solution:
Using Pythagoras' theorem (Refer Form 2 Chapter 6) F E 2 =D F 2 −D E 2         = 13 2 − 12 2         =169−144         =25 FE= 25 =5 cm AF=8−5=3 cm AB=12−8=4 cm Area of rectangle ACDE =8×12 =96  cm 2 Area of △ ABF = 1 2 ×3×4 =6  cm 2 Area of △ DEF = 1 2 ×5×12 =30  cm 2 Area of coloured region =96−30−6 =60  cm 2


Question 11:
Diagram below shows a sketch of parallelogram shaped garden, PQRS that consists of flower beds and a playground.

Calculate the area, in m2, of the flower beds.

Solution:
Area flower bed =( 12×14 )−( 1 2 ×12×7 ) =168−42 =126  m 2



11.2.2 Perimeter and Area, PT3 Practice

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Question 5:
In diagram below, PQUV is a square, QRTU is a rectangle and RST is an equilateral triangle.


The perimeter of the whole diagram is 310 cm.
Calculate the length, in cm, of PV.

Solution:
PV=VU=TS=SR=QP Given perimeter of the whole diagram=310 cm PV+VU+UT+TS+SR+RQ+QP=310 PV+PV+50+PV+PV+50+PV=310 5PV+100=310  5PV=210    PV=42 cm


Question 6:
In diagram below, ABCD and CGFE are rectangles. M, G, E and N are midpoints of AB, BC, CD and DA respectively.


Calculate the perimeter, in cm, of the coloured region.

Solution:


Using Pythagoras' theorem M G 2 =M F 2 +F G 2 = 5 2 + 12 2 =25+144 =169 MG=13 cm Perimeter of the coloured region =13+13+5+12+5+12 =60 cm


Question 7:
Diagram below shows a trapezium BCDE and a parallelogram ABEF. ABC and FED are straight lines.

The area of ABEF is 72 cm2.
Calculate the area, in cm2, of trapezium BCDE.

Solution:
Base × Height =Area of ABFE 9 cm × Height=72  cm 2  Height= 72 9 =8 cm CD=8 cm BC=23−9  =14 cm Area of trapezium BCDE = 1 2 ×( BC+ED )×CD = 1 2 ×( 14+9 )×8 =92  cm 2


Question 8:
In diagram below, ACEF is a trapezium and BCDG is a square.


Calculate the area, in cm2, of the coloured region.

Solution:
Area of trapezium ACEF = 1 2 ×( 8+15 )×10 =115  cm 2 Area of square BCDG =4×4 =16  cm 2 Area of trapezium GDEF = 1 2 ×( 4+15 )×6 =57  cm 2 Area of coloured region =115−16−57 =42  cm 2

11.2.1 Perimeter and Area, PT3 Practice

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Question 1:
In the diagram, ABCD is a trapezium and ABEF is a parallelogram.

Calculate the area, in cm2, of the coloured region.

Solution:
Area of trapezium ABCD = 1 2 ×( 8+14 )×10 =110  cm 2 Area of parallelogram ABEF =8×6 =48  cm 2 Area of the shaded region =110−48 =62  cm 2


Question 2:
Diagram below shows a rectangle ABCD.


Calculate the area, in cm2, of the coloured region.

Solution:
The area of the coloured region =Area of rectangle−Area of trapezium =( 12×8 )− 1 2 ×( 4+6 )×4 =96−20 =76  cm 2


Question 3:
In diagram below, AEC is a right-angled triangle with an area of 54 cm2 and BCDF is a rectangle.
Calculate
(a) the perimeter, in cm, of the coloured region.
(b) the area, in cm2, of the coloured region.

Solution:
(a) Given area of △ ACE 1 2 ×AC×9=54    AC=54× 2 9    AC=12 cm Using Pythagoras' theorem: AE= 9 2 + 12 2   =15 cm Perimeter of coloured region =6+4.5+6+4.5+15 =36 cm

(b) Area of the coloured region =Area of △ ACE−Area of rectangle BCDF =54−( 6×4.5 ) =54−27 =27  cm 2


Question 4:
Diagram below shows a trapezium ABCDE. ABGF is a square with an area of 36 cm2.


Calculate
(a) the perimeter, in cm, of the coloured region.
(b) the area, in cm2, of the coloured region.

Solution:

(a) Using Pythagoras' theorem: In △ CDH, CD= 8 2 + 6 2  =10 cm AB=BG=GF=FA= 36 =6 cm Perimeter of coloured region =6+10+18+2+6+6 =48 cm

(b) Area of the coloured region =Area of trapezium ABCDE−Area of square ABGF =[ 1 2 ( 12+18 )×8 ]−36 =[ 1 2 ×30×8 ]−36 =120−36 =84  cm 2

2.4.3 The Nitrogen Cycle and Its Importance (Structured Questions)

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Question 1:
Diagram 1 shows part of nitrogen cycle.


(a) What is represented by W? [1 mark]

(b) Name bacteria X and bacteria Y. [2 marks]

(c) State one importance of the nitrogen cycle in Diagram 1.1. [1 mark]

(d) Bacteria X can be found in the root nodules of a plant as shown in Diagram 1.2.[1 mark]


Name the type of plant shown in Diagram 1.2.
(e) The nitrogen cycle consists of several processes.
In Table below, mark (\/) the process involved in the nitrogen cycle. [1 mark]




Answer:
(a) Animal proteins

(b)
X: Nitrogen-fixing bacteria
Y: Nitrifying bacteria

(c)
1. Ensures that nitrogen and compounds of nitrogen are always present around us.
2. Maintains relatively steady levels of nitrogen and compounds of nitrogen around us.
(any one)

(d) Leguminous plant, pea plant, beans (any one)

(e)

8.2.3 Basic Measurements, PT3 Practice

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Question 11:
George departed for Kuantan from Kuala Lumpur at 9.30 a.m. He stopped at a rest station for 45 minutes. Then, he continued his journey and arrived at his destination at 3.45 p.m. How long did he travel on the road.

Solution:
Time taken on the road
= (1545 – 0930) – 0045 ← [3.45 p.m. change to 24-hour system = 1545 (1200 + 0345)]
= 0615 – 0045
= 0530
= 5 hours 30 minutes


Question 12:
Rita began doing her homework at 8.25 in the morning. She took a break for 45 minutes to have some exercises and ½ hour to have lunch. If she finished her homework at 4.45 in the evening on the same day, how long did she take to do her homework?

Solution:
Time taken for Rita to finish her homework
= (1645 – 0825) – (0030 + 0045) ← [4.45 p.m. = 1645 (1200 + 0445)]
= 0820 – 0115
= 0705
= 7 hours 5 minutes


Question 13:
Roslan attended a tuition class for 2½ hours. The class started at 7.15 p.m. The class finished 35 minutes earlier due to power failure.
At what time did the class finish?

Solution:
Class finish at
= (1915 + 0230) – 0035 ← [7.15 p.m. = 1915 (1200 + 0715)]
= 2145 – 0035
= 2110 → (9.10 p.m.)


Question 14:
An aeroplane took off from Kuala Lumpur to Singapore at 2.45 p.m. The flight normally takes 55 minutes. The aeroplane was delayed and landed at Singapore at 4.50 p.m.
How long, in minutes, was the aeroplane delayed?

Solution:
Expected arrival time
= 2.45 p.m. + 55 minutes
= 3.40 p.m.

Time delayed
= 4.50 p.m. – 3.40 p.m.
= 1 hour 10 minutes
= 70 minutes

Question 15:
The flight from Kuala Lumpur to Kota Kinabalu takes 1 hour 45 minutes.
Jacky’s flight should leave at 2.15 p.m. but is delayed by 50 minutes.
At what time, in the 24-hour system, he will reach Kota Kinabalu?

Solution:
2.15 p.m. = 1415 hours
Actual departure time
= 1415 hours + 50 minutes
= 1505 hours

Arrival time at Kota Kinabalu
= 1505 + 0145
= 1650


8.2.2 Basic Measurements, PT3 Practice

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Question 6:
The mass of a box containing 6 papayas is 21.32 kg. The mass of the box when it is empty is 1.46 kg.
Calculate the average mass, in g, of a papaya.

Solution:
Mass of 6 papayas
= 21.32 – 1.46
= 19.86 kg

Average mass of a papaya
= 19.86 ÷ 6
= 3.31 kg
= 3.31 × 1000 g
= 3310 g



Question 7:
Louis bought 600 g of cookies. Dennis bought twice the mass of cookies that Jackson bought. They bought 1.35 kg of cookies altogether. Calculate the mass, in g, of cookies bought by Jackson.

Solution:
Let Jackson bought w g of cookies.
600 g + 2 × w + w = 1.35 kg
600 g + 3w = 1.35 kg
600 g + 3w = 1350 g
3w = 1350 g – 600 g
w = 750 g ÷ 3
w = 250 g



Question 8:
If the mass of 4 packets of candies is 2.6 kg, what is the mass of 9 packets of the same candies, in kg?

Solution:
Mass of 4 packets of candies = 2.6 kg Mass of 9 packets of candies= 2.6 4 ×9                                              =5.85 kg

Question 9:
It is given that 1 4 of fruits is supplied to Juice Stall A and 2 7 to Juice Stall B. The remaining 133.25 kg is sold to a fruit stall.
Calculate the mass of fruits that has been supplied to Fruit Stall B.

Solution:
1 4 + 2 7 = 7 28 + 8 28          = 15 28 Remaining fruits sold to fruit stall =1− 15 28 = 13 28 Total mass of fruits = 28 13 ×133.25 =287 kg Mass of fruits supplied to Juice Stall B = 2 7 ×287 =82 kg

Question 10:
The mixture of metal to produce a piece of 50 sen coin are 2 3  zinc,  1 5  nickel and the rest is copper.
If the mass of copper is 1.6 g, find the total mass, in g, of zinc and nickel.

Solution:
Fraction of copper =1− 2 3 − 1 5 = 15 15 − 10 15 − 3 15 = 2 15 Portion of copper = 2 g Portion of zinc and nickel = 13 g 2↔1.6 g Total mass of zinc and nickel, 13↔ 13×1.6 2 =10.4 g

8.2.1 Basic Measurements, PT3 Practice

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[adinserter name="Block 3"]
Question 1:
Karen has 18 m of ribbon to tie 16 bars of chocolate and 28 boxes of sweets. A bar of chocolate needs 24 cm of ribbon and a box of sweets needs 38 cm of ribbon.
Calculate the length, in m, of the remaining ribbon.

Solution:
Length of ribbon needed
= (16 × 24 cm) + (28 × 38 cm)
= 384 cm + 1064 cm
= 1448 cm
= 14.48 m

Length of remaining ribbon
= 18 m – 14.48 m
= 3.52 m


[adinserter name="Block 3"]
Question 2:
The length of a red string is 4 m 80 cm. The length of a brown string is ⅔ of the length of the red string. The length of a black string is twice the length of the brown string.
Calculate the total length, in m, of the brown and the black strings.

Solution:
Length of brown string = 2 3 ×4 m 80 cm = 2 3 ×4.8 =3.2 m Length of black string =2×3.2 m =6.4 m Total length=3.2 m+6.4 m =9.6 m


[adinserter name="Block 3"]
Question 3:
Kelly has 13 m of cloth. She uses it to make 6 curtains and 4 tablecloths. Each curtain requires 1.25 m of cloth and each tablecloth requires 90 cm of cloth.
What is the length, in m, of the remaining cloth?

Solution:
Length of cloth used to make curtains
= 6 × 1.25 m
= 7.5 m

Length of cloth used to make tablecloth
= 4 × 90 cm
= 360 cm
= 3.6 m

Length of the remaining cloth
= 13 – 7.5 – 3.6
= 1.9 m



[adinserter name="Block 3"]
Question 4:
A rope measuring 6 m 30 cm is cut into 1.6 m, 96 cm and the remainder into two equal parts. Find the length of each of the remaining parts, in cm.

Solution:
(6 m 30 cm – 1.6 m – 96 cm) ÷ 2
= (630 cm – 160 cm – 96 cm) ÷ 2
= 374 cm ÷ 2
= 187 cm

 

6.2.1 Integers, PT3 Practice

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Question 1:
Evaluate 5 + (–7)   

Solution:

Therefore,
5 + (–7) = 5 – 7
= –2


Question 2:
Evaluate –6 – (–5)   

Solution:

Therefore,
–6 – (–5) = –6 + 5   
= –1


Question 3:
Simplify –3 + 7 + (–2)
  
Solution:


Therefore,
–3 + 7 + (–2) = 4 + (–2)
= 4 – 2
= 2


Question 4:
Simplify 5 + (–4) + (–3)  

Solution:


Therefore,
5 + (–4) + (–3)
= 5 – 4 – 3
= –2