Home › View all posts by Myhometuition › Page 7

Author: Myhometuition

7.5 SPM Practice (Long Questions)

Posted on by

7.5.2 Probability (II), SPM Practice (Long Questions)
Question 3:
Diagram below shows two cards labelled with letters in box A and three numbered cards in box B.


A card is picked at random from box A and then a card is picked at random from box B.
By listing the sample of all possible outcomes of the event, find the probability that

(a) a card labelled M and a card with an even number are picked,
(b)   a card labelled Q or a card with a number which is multiple of 2 are picked.

Solution:
Sample space, S
= {(M, 2), (M, 3), (M, 6), (Q, 2), (Q, 3), (Q, 6)}
n(S) = 6

(a)
{(M, 2), (M, 6)}
P( M and even number )= 2 6 = 1 3  
 
(b)
{(Q, 2), (Q, 3), (Q, 6), (M, 2), (M, 6)}
P( Q or multiple of 2 )= 5 6



Question 4:
Table below shows the names of participants from two secondary schools attending a public speaking training programme.
Boys
Girls
School A
Karim
Rosita
Sally
Linda
School B
Ahmad
Billy
Nancy
Two participants are required to give speeches at the end of the programme.
(a)  A participant is chosen at random from School A and then another participant is chosen at random also from School A.
(i) List all the possible outcomes of the event in this sample space.
(ii)Hence, find the probability that a boy and a girl are chosen.

(b)  A participant is chosen at random from the boys group and then another participant is chosen at random from the girls group.
(i) List all the possible outcomes of the event in this sample space.
(ii)Hence, find the probability that both participants chosen are from School B.

Solution:
(a)(i)
Sample space, S
= {(Karim, Rosita), (Karim, Sally), (Karim, Linda), (Rosita, Sally), (Rosita, Linda), (Sally, Linda)}
n(S) = 6

(a)(ii)
{(Karim, Rosita), (Karim, Sally), (Karim, Linda}
P ( a boy and a girl ) = 3 6 = 1 2
 
(b)(i)
Sample space, S
= {(Karim, Rosita), (Karim, Sally), (Karim, Linda), (Karim, Nancy), (Ahmad, Rosita), (Ahmad, Sally), (Ahmad, Linda), (Ahmad, Nancy), (Billy, Rosita), (Billy, Sally), (Billy, Linda), (Billy, Nancy)}
n(S) = 12

(b)(ii)
{(Ahmad, Nancy), (Billy, Nancy)}
P (both participants from School B)
= 2 12 = 1 6

7.4 SPM Practice (Short Questions)

Posted on by

Question 6:
A box contains 18 black cards and 8 blue cards. Fanny puts another 2 black cards and 4 blue cards inside the box. A card is chosen at random from the box.
What is the probability that a black card is chosen?

Solution:
New total number of cards in the box =( 18+2 )+( 8+4 ) =32 Probability a black card is chosen Probability= 18+2 32   = 20 32   = 5 8


Question 7:
A fruits seller wants to check the number of rotten apples in each box. Table below shows the number of rotten apples in each box.
Given that there were 55 boxes of apples.
If a box is selected at random, what is the probability that the box does not contain rotten apples?

Solution:
Number of boxes without rotten apples =55−( 3+9+2+3+1+2 ) =55−20 =35 Probability= 35 55 = 7 11


Question 8:
A form 5 science class has 15 boy students and a number of girl students. A student is chosen at random from the class. The probability of choosing a boy student is 3 5 .
Find the number of girl students in the class.

Solution:
Let x be the number of girl students in the class. P( a girl student )=1− 3 5   = 2 5 x 15+x = 2 5 5x=30+2x 3x=30 x=10

7.4 SPM Practice (Short Questions)

Posted on by

Question 4:
A box contains 5 red cards, 3 yellow cards and a number of green cards. A card is picked at random from the box. Given that the probability of picking a yellow card is 1 6 , find the probability of picking a card that is not green.

Solution:
P( yellow card )= n( yellow card ) n( S )                       1 6 = 3 n( S )                 n( S )=3×6                         =18 n( not green card )=5+3=8 P( not green card )= 8 18                             = 4 9



Question 5:
In an office, 300 workers go to work by bus, 60 workers go by car and the rest go by motorcycle. A worker is chosen at random. The probability of choosing a worker who goes to work by bus is ⅔.
Find the probability of choosing a worker who does not go to work by motorcycle.

Solution:
Let y be the number of workers who go to office by motorcycle. P( by bus )= 2 3 Number of workers ( By bus ) Total number of workers = 2 3 300 300+60+y = 2 3 900=720+2y 2y=180 y=90 Probability= 300+60 360+90  = 360 450  = 4 5

6.3 SPM Practice (Long Questions)

Posted on by

Question 9:
Diagram below shows the speed-time graph for the movement of an object for a period of 34 seconds.


(a) State the duration of time, in seconds, for which the object moves with uniform speed.
(b) Calculate the rate of change of speed, in ms-2, of the object for the first 8 seconds.
(c) Calculate the value of u, if the average of speed of the object for the last 26 seconds is 6 ms-1.

Solution:
(a) Duration of time = 26s – 20s = 6s

(b) Rate of change of speed for the first 8 seconds = 10−6 0−8 =− 4 8 =− 1 2  ms −2

(c) Speed= Distance Time ( 1 2 ×12×( 6+u ) )+( 6×u )+( 1 2 ×8×u ) 26 =6   36+6u+6u+4u=156    16u=120    u=7.5  ms −1




Question 10 (6 marks):
Diagram 11 shows the speed-time graph for the movement of a particle for a period of t seconds.

Diagram 11

(a) State the uniform speed, in ms-1, of the particle.

(b) Calculate the rate of change of speed, in ms-2, of the particle in the first 4 seconds.

(c) Calculate the value of t, if the distance travelled in the first 4 seconds is half of the distance travelled from the 6th second to the tth second.


Solution:
(a)
 Uniform speed of the particle = 12 ms-1

(b)
 Rate of change of speed of particle = 12 4 =3  ms −2

(c)
Distance travelled in the first 4 seconds  = 1 2 ( Distance travelled from the  6 th  second to  t th  second ) 1 2 ×4×12= 1 2 [ 1 2 ( 12+20 )( t−6 ) ] 24= 1 2 [ 16( t−6 ) ] 24=8( t−6 ) 24=8t−48 3=t−6 t=9

6.3 SPM Practice (Long Questions)

Posted on by

Question 7:



The diagram above shows the speed-time graph of a moving object for 15 seconds. Find
(a)    the speed of the object at t = 9s.
(b)   the distance travel by the object for the first 12 seconds.

Solution:
(a)
The speed of the object at t = 9s is 6 ms-1

(b)
Distance travel by the object for the first 12 seconds
= Area under the speed-time graph
= Area of triangle
= ½ × 12 × 8
= 48 m


Question 8:


The diagram above shows the speed-time graph of a moving object for 15 seconds.
(a) Find the time interval when the object moves with constant speed.
(b) Find the acceleration from t = 6 s to t = 12 s.
(c) State the time when the object is stationary.

Solution:
(a)
Time interval when the object moves with constant speed
= 12 – 6
= 6 s

(b)
Acceleration from t=6s to t=12s = Gradient of speed-time graph = 6−6 12−6 =0  ms −2 ← The moving object is moving  at a uniform speed

(c)
Time when the object is stationary is at 15 s.


6.2 Quantity Represented by the Area under a Graph (Part 2)

Posted on by

Combination of Graphs
Example 2:

 
The diagram above shows the speed-time graph of a moving object for 15 seconds.
(a) State the length of time, in s, that the particle moves with constant speed.
(b) Calculate the rate of change of speed, in ms-2, in the first 3 seconds.
(c) Calculate the average speed of the object in 15 seconds.

Solution: 
(a)
Length of time that the particle moves with constant speed
= 9 – 3 = 6 s 

(b)
Rate of change of speed in the first 3 seconds
= acceleration = gradient
= y 2 − y 1 x 2 − x 1 = 6 − 3 3 − 0 = 1 m s − 2

(c)
Total distance travelled of the object in 15 seconds
= Area under the graph in the 15 seconds
= Area P + Area Q + Area R
=[ 1 2 ( 3+6 )×3 ]+[ ( 9−3 )×6 ]+[ 1 2 ( 15−9 )×6 ] =13.5+36+18 =67.5 m

Average speed of the object in 15 seconds
= Total distance travelled Total time taken = 67.5 15 =4.5 m s −1

5.1 Direct Variation (Part 3)

Posted on by
(D) Solving problems involving direct variations
1. If   y∝ x n , where n= 1 2 , 2, 3,  then the equation is y = k x n where k is a constant.

2. The graph of y against xn is a straight line passing through the origin.

3.  If   y ∝ x n and sufficient information is given, the values of variable x or variable y can be determined.

Example
y varies directly to x3 and if y = 54 when x = 3, find
(a) The value of x when y = 16
(b) The value of y when x = 4

Solution:
Given y α x3, y = kx3
When y = 54, x = 3,
54 = k(3)3
54 = 27k
k = 2
Therefore y = 2x3

(a) When y = 16
 16 = 2x³
  x³ = 8
x = 2

(b) When x = 4
y = 2(4)³ = 128
where k is a constant.

SPM Practice (Long Questions)

Posted on by

Question 8:
(a) Diagram 8.1 shows point A and point B marked on a Cartesian plane.

Diagram 8.1

Transformation R is a rotation of 90o, clockwise about the centre B.

Transformation T is a translation  (  5 −2 )

State the coordinates of the image of point A under each of the following transformations:
(i) RT,
(ii) R2.

(b) Diagram 8.2 shows three trapeziums ABCD, PQRS and TUVS, drawn on a Cartesian plane.

Diagram 8.2

(i) Trapezium PQRS is the image of trapezium ABCD under the combined transformation MN.
Describe in full, the transformation:
(a) N,  
(b) M.

(ii) It is given that trapezium ABCD represents a region of area 30 m2.
Calculate the area, in m2, of the shaded region.


Solution:
(a)


(i) A (–1, 6) → T → (4, 4 ) → R → (3, 1)
(ii) A (–1, 6) → R → (5, 6) → R → (5, 0)

(b)(i)(a)
N is a reflection in the line y = 4.

(b)(ii)(b)
M is enlargement of scale factor 2 with centre S (1, 5).

(b)(ii)
Area of PQRS = (scale factor)2 x Area of object ABCD
  = 22 x 30
 = 120 m2

Therefore,
Area of shaded region
= Area PQRS – area ABCD
= 120 – 30
= 90 m2



SPM Practice (Long Questions)

Posted on by

Question 7:
Diagram shows the point J(1, 2) and quadrilaterals ABCD and EFGH, drawn on a Cartesian plane.




(a) Transformation U is a rotation of 90o, clockwise about the centre O.
Transformation T is a translation  ( 2 3 )  
Transformation R is a reflection at the line x = 3.

State the coordinates of the image of point J under each of the following transformations:
(i) RU,
(ii) TR.

(b) EFGH is the image of ABCD under the combined transformation MN.
Describe in full, the transformation:
(i) N,   
(ii) M.

(c) It is given that quadrilateral ABCD represents a region of area 18 m2.
Calculate the area, in m2, of the shaded region.


Solution:




(a)
(i) J (1, 2) → U → (2, –1 ) → R → (4, –1)
(ii) J (1, 2) → R → (5, 2) → T → (7, 5)

(b)(i)
N is a reflection in the line x = 6.

(b)(ii)
M is enlargement of scale factor 3 with centre (8, 7).

(c)
Area EFGH = (scale factor)2 x Area of object ABCD
  = 32 x 18
 = 162 m2

Therefore,
Area of shaded region
= Area of EFGH – Area of ABCD
= 162 – 18
= 144 m2


SPM Practice (Long Questions)

Posted on by

Question 6:
(a) Diagram below shows point M marked on a Cartesian plane.

Transformation T is a translation ( 2 3 ) and transformation R is an anticlockwise rotation of 90o about the centre O.
State the coordinates of the image of point M under each of the following transformations:
(i) RT,
(ii) TR,

(b) Diagram below shows two hexagons, ABCDEF and JKLANO, drawn on square grids.



(i) JKLANO is the image of ABCDEF under the combined transformation VW.
Describe in full the transformation:
(a) W   (b) V

(ii) It is given that quadrilateral ABCDEF represents a region of area 45 m2.
Calculate the area, in m2, of the region represented by the shaded region.


Solution:
(a)



(b)



(i)(a)
V: A reflection in the line EC.

(i)(b)
 Scale factor= KL BC = 6 2 =3 W: An enlargement of scale factor 3 at centre J.

(ii)
Area of JKLANO
= 32 x area of ABCDEF
= 9 x 45
= 405 m2

Area of the coloured region
= area of JKLANO – area of ABCDEF
= 405 – 45
= 360 m2