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1.2.2 Angles and Lines II, PT3 Practice

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Question 6:
In Diagram below, PQRST is a straight line. Find the value of x.

Solution:
Interior angle of  R = ( 180 o − 76 o ) ÷ 2 = 52 o ∠ x = exterior angle of  R   ( corresponding angles ) Hence,  x = 76 o + 52 o = 128 o


Question 7:
In Diagram below, find the value of y.



Solution:


∠ A B C = ∠ B C D    = 108 o  (alternate angle) y o + 130 o + 108 o = 360 o   y o = 360 o − 238 o   y o = 122 o


Question 8:
In Diagram below, PSR and QST are straight lines.
Find the value of x.

Solution:
∠UST+∠STV= 180 o ∠UST= 180 o − 116 o = 64 o ∠PST=∠QSR x o +∠UST= 135 o x o +∠UST= 135 o x o = 71 o x=71


Question 9:
In Diagram below, PWV is a straight line.

(a) Which line is perpendicular to line PWV?
(b) State the value of  ∠ RWU.

Solution:
(a) SW

(b) ∠ RWU = 13o + 29o + 20o = 62o

 


Question 10:
In Diagram below, UVW is a straight line.


(a) Which line is parallel to line TU?
(b) State the value of  ∠ QVS.

Solution:
(a) QV

(b) ∠ QVS = 8o + 18o = 26o

 

12.2.2 Solid Geometry (II), PT3 Focus Practice

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12.2.2 Solid Geometry (II), PT3 Focus Practice

Question 5:
Sphere below has a surface area of 221.76 cm2.


Calculate its radius.
( π = 22 7 )

Solution:
Surface area of sphere = 4πr2
4 π r 2 = 221.76 4 × 22 7 × r 2 = 221.76 r 2 = 221.76 × 7 4 × 22 r 2 = 17.64 r = 17.64 r = 4.2 cm


Question 6:
Diagram below shows a right pyramid with a square base.

Given the height of the pyramid is 4 cm.
Calculate the total surface area, in cm2, of the right pyramid.

Solution:
h2 = 32 + 42
= 9 + 16
= 25
h = √25   = 5 cm2

Total surface area of the right pyramid
= Base area + 4 (Area of triangle)
= (6 × 6) + 4 × 4 (½ × 6 × 5)
= 36 + 60
= 96 cm2


Question 7:
Diagram below shows a prism.

Draw to full scale, the net of the prism on the grid in the answer space. The grid has equal squares with sides of 1 unit.

Solution:


8.2.2 Coordinates, PT3 Focus Practice

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Question 6:
The point M (x, 4), is the midpoint of the line joining straight line Q (-2, -3) and R (14, y).
The value of x and y are

Solution:

x= −2+14 2 x= 12 2 x=6 4= −3+y 2 8=−3+y y=11


Question 7:
In diagram below, PQR is a right-angled triangle. The sides QR and PQ are parallel to the y-axis and the x-axis respectively. The length of QR = 6 units.

Given that M is the midpoint of PR, then the coordinates of M are

Solution:
x-coordinate of R = 3
y-coordinate of R = 1 + 6 = 7
R = (3, 7)

P( 1,1 ),R( 3,7 ) Coordinates of M =( 1+3 2 , 1+7 2 ) =( 2,4 )


Question 8:
Given points P (–2, 8) and Q (10, 8), find the length of PQ.

Solution:
Length of PQ = [ 10−( −2 ) ] 2 + ( 8−8 ) 2 = ( 14 ) 2 +0 =14 units


Question 9:
In diagram below, ABC is an isosceles triangle.

Find
(a) the value of k,
(b) the length of BC.

Solution:
( a ) For an isosceles triangle,  y−coordinate of C is the midpoint of straight line AB. 2+k 2 =−3 2+k=−6   k=−8 ( b ) B=( −2,−8 ) BC= [ 10−( −2 ) ] 2 + [ −3−( −8 ) ] 2  = 12 2 + 5 2  =13 units


Question 10:
Diagram below shows a rhombus PQRS drawn on a Cartesian plane. PS is parallel to x-axis.

Given the perimeter of PQRS is 40 units, find the coordinates of point R.

Solution:
All sides of rhombus have the same length, therefore length of each side= 40 4 =10 units PQ=10 ( 9− x 1 ) 2 + ( 7−( −1 ) ) 2 = 10 2 81−18 x 1 + x 1 2 +64=100 x 1 2 −18 x 1 +45=0 ( x 1 −3 )( x 1 −15 )=0 x 1 =3,15 x 1 =3 Q=( 3,−1 ),R=( x 2 ,−1 ) QR=10 ( x 2 −3 ) 2 + [ −1−( −1 ) ] 2 = 10 2 x 2 2 −6 x 2 +9+0=100 x 2 2 −6 x 2 −91=0 ( x 2 +7 )( x 2 −13 )=0 x 2 =−7,13 x 2 =13 ∴R=( 13,−1 )



4.2.2 Linear Equations I, PT3 Practice

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Question 6:
Solve each of the following equations:
(a)  5 a − 16 = a (b)  6 + 2 3 ( − 9 p + 12 ) = p


Solution:
(a)
5 a − 16 = a 5 a − a = 16 4 a = 16 a = 4

(b)
6 + 2 3 ( − 9 p + 12 ) = p 6 − 6 p + 8 = p − 6 p − p = − 8 − 6 − 7 p = − 14 p = 2


Question 7::
Solve each of the following equations:
(a)  a − 2 = a 5 (b)  b − 3 4 = 2 + b 5

Solution:
(a)
a − 2 = a 5 5 a − 2 = a 4 a = 2 a = 2

(b)
b − 3 4 = 2 + b 5 5 b − 15 = 8 + 4 b 5 b − 4 b = 8 + 15 b = 23

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Question 8:
Solve each of the following equations:
(a)  x = − 24 − x (b)  y + 5 4 ( 4 − 2 y ) = − 4

Solution:
(a)
x = − 24 − x 2 x = − 24 x = − 12

(b)
y + 5 4 ( 4 − 2 y ) = − 4 4 y + 20 − 10 y = − 16    ( × 4 ) − 6 y = − 16 − 20 − 6 y = − 36 y = 6


Question 9:
Solve each of the following equations:
(a)  5 p = 8 p − 9 (b)  3 q = 20 − 13 q 4

Solution:
(a)
5 p = 8 p − 9 5 p − 8 p = − 9 − 3 p = − 9 p = 3

(b)
3 q = 20 − 13 q 4 12 q = 20 − 13 q 12 q + 13 q = 20 25 q = 20 q = 4 5

2.2.2 Number Patterns and Sequences, PT3 Practice

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Question 6:
How many prime numbers are there between 10 and 40?

Solution:
Prime numbers between 10 and 40
= 11, 13, 17, 19, 23, 29, 31, 37
There are 8 prime numbers between 10 and 40.


Question 7:
Diagram below shows a sequence of prime numbers.
13, 17, x, y, 29 Find the value of x + y.

Solution:
Value of x + y
= 19 + 23
= 42


Question 8:
The following data shows a sequence of prime numbers in ascending order.
 5, 7, x, 13, 17, 19, 23, 29, y, z  Find the value of x, y and of z.

Solution:
x = 11, y = 31, z = 37


Question 9:
The following data shows a sequence of prime numbers in ascending order.
  p, 61, 67, q, 73, r, 83  Find the value of p, q and of r.

Solution:
p = 59, q = 71, r = 79



Question 10:
Diagram below shows several number cards.
    49     39     29         87     97     77          5      15     25     List three prime numbers from the diagram.

Solution:
Prime number card from the diagram = 5, 29 and 97.


2.2.1 Number Patterns and Sequences, PT3 Practice

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Question 1:
Diagram below is part of a number line.
What is the value of P and of Q?

Solution:
Each gradation on the number line represents 4 units. Thus P = –12 and Q = 12.



Question 2:
Diagram below shows a sequence of numbers. K and M represent two numbers.
    19     13     K      1      M     What are the values of K and M?

Solution:



K = 13 – 6 = 7
M = 1 – 6 = –5



Question 3:
Diagram below shows five integers.


Find the sum of the largest integer and the smallest integer.

Solution:
The largest integer = 1
The smallest integer = –10
Sum of the largest integer and the smallest integer
= 1 + (–10) 
= –9


1.2.3 Whole Numbers, PT3 Practice

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Question 11:
Britney rents a stall to sell cakes. The rental of the stall is RM500 per month. The cost for baking a cake is RM5 per box and she sells it at RM10 per box. What is the minimum box of cakes must Britney sell each month to make a profit?

Solution:
Rental of the stall = RM500 monthly
Cost of baking a cake per box = RM5
Selling price per box = RM10
Profit per box
= RM10 – RM5
= RM5

The number of boxes of cakes to accommodate the cost
= RM500 ÷ RM5
= 100

Hence, the minimum box of cakes must Britney sell each month to make a profit = 101


1.2.2 Whole Numbers, PT3 Practice

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Question 6:
Solve 4 (8 + 14)(37 – 18) + 46

Solution:
4 (8 + 14)(37 – 18) + 46
= 4(22)(19) + 46
= 4 × 22 × 19 + 46
= 1672 + 46
= 1718


Question 7:
Calculate the value of
113 + 6(47 – 4 × 9)

Solution:
113 + 6(47 – 4 × 9)
= 113 + 6[47 – (4 × 9)]
= 113 + 6(47 – 36)
= 113 + 6(11)
= 113 + 66
= 179


Question 8:
Solve 128 + (9 × 4) – 318 ÷ 6

Solution:
128 + (9 × 4) – 318 ÷ 6
= 128 + 36 – 53
= 164 – 53
= 111


Question 9:
Calculate the value of
62 × 15 – 40 + 68 ÷ 17.

Solution:
(62 × 15) – 40 + (68 ÷ 17)
= 930 – 40 + 4
= 894


Question 10:
402 – ( ) = 78 – 6 × 8 ÷ 16.
Find the missing number in the box.

Solution:
402 – ( ) = 78 – 6 × 8 ÷ 16
402 – ( ) = 78 – 48 ÷ 16
402 – ( ) = 78 – 3
402 – ( ) = 75
( ) = 402 – 75
( ) = 327

1.2.1 Whole Numbers, PT3 Practice

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Question 1:
Solve 8 × 2 ÷ 4 + 12.

Solution:
8 × 2 ÷ 4 + 12
= 16 ÷ 4 + 12
= 4 + 12
= 16


Question 2:
Solve 7(6 + 5) – 3(13 – 9).

Solution:
7(6 + 5) – 3(13 – 9)
= 7(11) – 3(4)
= 77 – 12
= 65


Question 3:
Solve 64 + (136 – 87) × (28 ÷ 4).

Solution:
64 + (136 – 87) × (28 ÷ 4)
= 64 + 49 × 7
= 64 + 343
= 407


Question 4:
Solve 84 ÷ 4 – 9 + 6 × 7

Solution:
84 ÷ 4 – 9 + 6 × 7
= 21 – 9 + 42
= 12 + 42
= 54


Question 5:
Solve 92 + 13 × 5 – 42

Solution:
92 + 13 × 5 – 42
= 92 + (13 × 5) – 42
= 92 + 65 – 42
= 157 – 42
= 115


15.2.3 Trigonometry, PT3 Focus Practice

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Question 10:
A technician needs to climb a ladder to repair a street lamp as shown in Diagram below.
(a) What is the length, in m, of the ladder?
(b) Find the height, in m, of the lamp post.

Solution:


(a)
cos 70 o = 1 h h= 1 cos 70 o h= 1 0.342 h=2.924 m Hence, the length of the ladder =2.924 m

(b)
tan 70 o = T 1 T=tan 70 o ×1    =2.747×1    =2.747 m The height of the lamp post =2.747+1.2 =3.947 m