SPM Practice (Short Questions)

Posted on April 23, 2020 by user

Question 1:

In figure above, FAD is a tangent to the circle with centre O. AEB and OECD are straight lines. The value of y is

Solution:

∠OAD = 90^o

∠AOD= 180^o– 90^o – 34^o= 56^o

y = 56^o÷ 2 = 28^o

Question 2:

In figure above, PQR is a tangent to the circle QSTU at Q and TUPV is a straight line. The value of y is

Solution:

∠QTS = ∠RQS = 40^o

∠SQT= ∠QTS = 40^o (isosceles triangle)

∠PQT= 180^o– 40^o – 40^o= 100^o

∠TPQ= 180^o– 115^o = 65^o

y = 180^o– 100^o – 65^o= 15^o

Question 3:

In figure above, ABC is a tangent to the circle BDE with centre O, at B.

Find the value of y.

Solution:

∠BOD= 2 × ∠BED

= 2 × 35^o = 70^o

∠ODB = ∠OBD

= (180^o– 70^o) ÷ 2 = 55^o

∠ EDB = ∠ EBA = 75^o

y^o+ ∠ ODB = 75^o

y^o+ 55^o = 75^o

y= 20^o

8.2 Angle between Tangent and Chord

Posted on April 23, 2020 by user

8.2 Angle between Tangent and Chord

1. In the diagram above, ABC is a tangent to the circle at point B.

2. Chord PB divides the circle into two segments, that is, the minor segment PRB and the major segment PQB.

3. With respect to ∠PBA, ∠PQB is known as the angle subtended by chord BP in the alternate segment.

4. With respect to ∠QBC, ∠BPQ is known as the angle subtended by chord BQ in the alternate segment.

5. The angle formed by the tangent and the chord which passes through the point of contact of the tangent is the same as the angle in the alternate segment which is subtended by the chord.

6. The relationships between the angles are:

Angle ∠ABP = Angle ∠BQP

Angle ∠CBQ = Angle ∠BPQ

SPM Practice (Short Questions)

Posted on April 23, 2020 by user

Question 4:

John has a collection of stamps from Thailand, Indonesia and Singapore. He picks one stamp at random. The probability of picking a Thailand stamp is

\frac{1}{5}

and the probability of picking an Indonesia stamp is

\frac{8}{15} .

John has 20 Singapore stamps. Calculate the total number of stamps in his collection.

Solution:

\begin{array}{l} Probability of picking a Singapore stamp \\ = 1 - \frac{1}{5} - \frac{8}{15} \\ = \frac{4}{15} \\ Given that John has 20 Singapore stamps, \\ ∴ Total number of stamp in John's collection \\ = \frac{15}{4} \times 20 \\ = 75 \end{array}

Question 5:
A box contains 36 green erasers and a number of red erasers.
If an eraser is picked randomly from the box, the probability of picking a red eraser is

\frac{5}{8}

. Find the number of red erasers.

Solution:
Probability of picking a green eraser = 1 –

\frac{5}{8}

\frac{3}{8}

Total number of erasers in the box = 36 ×

\frac{8}{3}

= 96

Hence, number of red erasers in the box = 96 – 36 = 60

SPM Practice (Short Questions)

Posted on April 23, 2020 by user

Question 1:

A box contains 5 pink marbles and 21 yellow marbles. Sharon puts another 4 pink marbles and 1 yellow marble inside the box. A marble is chosen at random from the box.

What is the probability that a pink marble is chosen?

Solution:
Total number of pink marbles = 5 + 4 = 9
Total number of yellow marbles = 21 + 1 = 22
Total number of marbles = 9 + 22 = 31

∴ P (pink marble) = \frac{9}{31}

Question 2:
In a group of 80 prefects, 25 are girls. Then 10 boys leave the group.
If a prefect is chosen at random from the group, state the probability that the prefect chosen is a boy.

Solution:
Number of boys = 80 – 25 = 55

After 10 boys left,
Number of boys = 55 – 10 = 45

Total number of prefects = 25 + 45 = 70

∴ P (b o y) = \frac{45}{70} = \frac{9}{14}

Question 3:
60 people have taken part in a singing contest. If a person is chosen at random from all the contestants, the probability of getting a male contestant is

\frac{8}{15}

. If there are 12 males and 3 females who did not qualify to the second round, find the probability that a male is chosen from the second round contestants.

Solution:
Number of male contestants = 60 ×

\frac{8}{15}

= 32

After 12 males and 3 females left,
Total number of contestants left = 60 – 15 = 45

Number of male contestant = 32 – 12 = 20

∴ P (male contestant) = \frac{20}{45} = \frac{4}{9}

7.3 Probability of an Event

Posted on April 23, 2020 by user

(A) Probability of an Event

1. The probability of an event A, P(A) is given by

2. If P(A) = 0, then the event A will certainly not occur.

3. If P(A) = 1, then the event A will certainly to occur.

Example 1:

Table below shows the distribution of a group of 80 pupils playing a game.

	Form Four	Form Five
Girls	28	16
Boys	12	24

A pupil is chosen at random from the group to start the game.

What is the probability that a boy from Form Five will be chosen?

Solution:

Let

A = Event that a boy from Form Five

S = Sample space

n(S) = 28 + 12 + 16+ 24 = 80

n(A) = 24

\begin{array}{l} P (A) = \frac{n (A)}{n (S)} \\ = \frac{24}{80} = \frac{3}{10} \end{array}

(B) Expected Number of Times an Event will Occur

If the probability of an event A and the number of trials are given, then the number of times event A occurs

= P(A) × Number of trials

Example 2:

In a football training session, the probability that Ahmad scores a goal in a trial is ⅝. In 40 trials are chosen randomly, how many times is Ahmad expected to score a goal?

Solution:

Number of times Ahmad is expected to score a goal

= ⅝ × 40

= 25

(C) Solving Problems

Example 3:

Kelvin has 30 white, blue and red handkerchiefs. If a handkerchief is picked at random, the probability of picking a white handkerchief is

\frac{2}{5} .

Calculate

(a) the number of white handkerchiefs.

(b) the probability of picking a blue handkerchief if 8 of the handkerchiefs are red in colour.

Solution:

Let

W = Event that a white handkerchief is picked.

B = Event that a blue handkerchief is picked.

R = Event that a red handkerchief is picked.

S = Sample space

(a)
n(S) = 30

\begin{array}{l} n (W) = P (W) \times n (S) \\ = \frac{2}{5} \times 30 = 12 \end{array}

(b)
Given n(R) = 8

n(B) = 30 – 12 – 8 = 10

\begin{array}{l} P (B) = \frac{n (B)}{n (S)} \\ = \frac{10}{30} = \frac{1}{3} \end{array}

7.1 Sample Space

Posted on April 23, 2020 by user

7.1 Sample Space

(A) Experiment

1. An experiment is a process or an action in making an observation to obtain the required results.

2. The result of the experiment is called the outcome.

(B) Listing all the possible outcomes of an experiment

We can list all the possible outcomes of an experiment by carrying out the experiment or by reasoning.

Example 1:

Six cards as shown in the diagram above are placed in a box. A card is drawn at random from the box. List all the possible outcomes.

Solution:

All the possible outcomes are O, R, A, N, G, E.

(C) Sample space of an experiment

1. A sample space is the set of all the possible outcomes of an experiment.

2. The letter S is used to represent the sample space and all the possible outcomes are written in brackets, { }.

Example 2:

A letter is chosen from the word ‘GARDEN’.

(a) List all the possible outcomes.

(b) Write the sample space, S, using set notation.

Solution:

(a) The possible outcomes are G, A, R, D, E and N

(b) Sample space, S = {G, A, R, D, E, N }

7.2 Events

Posted on April 23, 2020 by user

7.2 Events

(A) Elements of a Sample Space which Satisfy Given Condition

When a specific condition is given, we can list the elements of a sample space which satisfy the given condition.

Example 1:

A two-digit number which is not more than 25 is chosen at random. List the elements of the sample space which satisfy each of the following conditions.

(a) A perfect square is chosen.

(b) A prime number is chosen.

Solution:

Sample space

= {10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25}

(a)

A perfect square is chosen

= {16, 25}

(b)

A prime number is chosen

= {11, 13, 17, 19, 23}

(B) Events for a Sample Space

An event is a set of outcomes which satisfy a specific condition.

Event is a subset is a sample space.

Example 2:

A coin and a die are thrown simultaneously. The events P and Q are defined as follows.

P = Event of obtaining a ‘heads’ from the coin and an odd number from the die.

Q = Event of obtaining a ‘tails’ from the coin and a number more than 2 from the die.

(a) List the sample space, S.

(b) List the elements of

(i) The event P,

(ii) The event Q.

Solution:

(a)

Sample space, S

= { (H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6) }

(b)(i)

P = { (H, 1), (H, 3), (H, 5) }

← (Set of outcomes of Event P: obtaining a ‘heads’ from the coin and an odd number from the die.)

(b)(ii)

Q = { (T, 3), (T, 4), (T, 5), (T, 6) }

← (Set of outcomes of Event Q: obtaining a ‘tails’ from the coin and a number more than 2 from the die.)

6.5 Cumulative Frequency

Posted on April 23, 2020 by user

6.5 Cumulative Frequency

Cumulative Frequency of a data or a class interval in a frequency table is obtained by determining the sum of its frequency with the total frequencies of all its previous data or class interval.

(A) Ogive

Ogive is a cumulative frequency graph which is obtained by plotting the cumulative frequency against the upper boundaries of each class.

Example:

The data below shows the number of books read by a group of 60 students in a year.

Books	Frequency
6-10	3
11-15	7
16-20	11
21-25	16
26-30	11
31-35	8
36-40	4

(a) Construct a cumulative frequency table for the given data.

(b) By using the scales of 2 cm to 5 books on the horizontal axis and 2 cm to 10 students on the vertical axis, draw an ogive for the data.

Solution:

(a)

Add a class with frequency 0 before the first class.
Find the upper boundary of each class interval.

Books	Frequency	Cumulative Frequency	Upper Boundary
1-5	0	0	5.5
6-10	3	0 + 3 = 3	10.5
11-15	7	3 + 7 = 10	15.5
16-20	11	10 + 11 = 21	20.5
21-25	16	21 + 16 = 37	25.5
26-30	11	37 + 11 = 48	30.5
31-35	8	48 + 8 = 56	35.5
36-40	4	56 + 4 = 60	40.5

(b)

6.7 SPM Practice (Long Questions)

Posted on April 23, 2020 by user

Question 3:

The data below shows the mass, in kg, of 50 students.

(a) Copy and complete the table below based on the data above.

(b) Based on the completed table above,

(i) State the size of the class interval used in the table.

(ii) Calculate the estimated mean of the mass of the students.

For this part of the question, use graph paper.
(c) By using a scale of 2 cm to 5 kg on the horizontal axis and 2 cm to 2 students on the vertical axis, draw a frequency polygon for the data.

Solution:
(a)

(b)(i)

Size of class interval

= upper boundary – lower boundary

= 44.5 – 39.5

= 5

(b)(ii)

\begin{array}{l} Estimated mean = \frac{42 \times 4 + 47 \times 9 + 52 \times 7 + 57 \times 9 + 62 \times 18 + 67 \times 3}{50} \\ = \frac{2785}{50} = 55.7 \end{array}

(c)

6.3 Histograms

Posted on April 23, 2020 by user

6.3 Histograms

(A) Draw a histogram based on the frequency table of a grouped data

1. A histogram is a graphical representation of a frequency distribution.

2. A histogram consists of vertical rectangular bars without any spacing between them.

3. Steps for drawing a histogram

(a) Determine the lower boundaries and upper boundaries for each class interval.

(b) Choose suitable scales for the horizontal axis (x-axis) to represent class interval and the vertical axis (y-axis) to represent frequency.

Example:

The following frequency table shows the radii, in cm, of different types of trees in a garden.

Radii (cm)	Frequency
2.0 – 2.4	7
2.5 – 2.9	5
3.0 – 3.4	10
3.5 – 3.9	2
4.0 – 4.4	6
4.5 – 4.9	4

Draw a histogram to represent the above information.

Solution:

Radii (cm)	Frequency	Lower boundary	Upper boundary
2.0 – 2.4	7	1.95	2.45
2.5 – 2.9	5	2.45	2.95
3.0 – 3.4	10	2.95	3.45
3.5 – 3.9	2	3.45	3.95
4.0 – 4.4	6	3.95	4.45
4.5 – 4.9	4	4.45	4.95