9.6 SPM Practice (Long Questions)

Posted on April 24, 2020 by user

Question 2:

P(25^o S, 40^o E), Q(θ^o N, 40^o E), R(25^o S, 10^o W) and K are four points on the surface of the earth. PK is the diameter of the earth.

(a) State the location of point K.

(b) Q is 2220 nautical miles from P, measured along the same meridian.

Calculate the value of θ.

(c) Calculate the distance, in nautical mile, from P due west to R, measured along the common parallel of latitude.

(d) An aeroplane took off from Q and flew due south to P. Then, it flew due west to R. The average speed of the aeroplane was 600 knots.

Calculate the total time, in hours, taken for the whole flight.

Solution:

(a)

As PK is the diameter of the earth, therefore latitude of K = 25^o N

Longitude of K= (180^o – 40^o) W = 140^o W

Therefore, location of K = (25^o N, 140^oW).

(b)

Let the centre of the earth be O.

\begin{array}{l} ∠ P O Q = \frac{2220}{60} \\ = 37^{o} \\ θ^{o} = 37^{o} - 25^{o} = 12^{o} \\ ∴ The value of θ is 12 . \end{array}

(c)

Distance from P to R

= (40 + 10) × 60 × cos 25^o

= 50 × 60 × cos 25^o

= 2718.92 n.m.

(d)

Total distance travelled

= distance from Q to P + distance from P to R

= 2220 + 2718.92

= 4938.92 nautical miles

\begin{array}{l} Time taken = \frac{total distance from Q to R}{average speed} \\ = \frac{4938.92}{600} \\ = 8.23 hours \end{array}

9.5 SPM Practice (Short Questions)

Posted on April 24, 2020 by user

Question 1

In diagram below, N is the North Pole and S is the South Pole. The location of point P is (40^o S, 70^o W) and POQ is the diameter of the earth.

Find the longitude of Q.

Solution:

Since PQ is a diameter of the earth and the longitude of P is θ^o W, the longitude of Q is (180^o – θ^o) E.

Longitude of P = 70^o W

Longitude of Q = (180^o – 70^o) E

= 110^oE

Question 2

In diagram below, N is the North Pole and S is the South Pole and NOS is the axis of the earth.

Find the position of point Q.

Solution:

Latitude of Q = (90^o – 42^o) N

= 48^o N

Longitude of Q = (65^o – 30^o) E

= 35^o E

Therefore, position of Q = (48^o N, 35^oE).

9.1 Longitudes

Posted on April 24, 2020 by user

9.1 Longitudes

1. A great circle is a circle with the centre of the Earth as its centre.

2. A meridian is half of a great circle from the North pole to the South pole.

3. The longitude of the Greenwich Meridian is 0^o.

4. The longitude of a meridian is determined by:

(a) The angle between the meridian plane and the Greenwich Meridian.

(b) The position of the meridian to the east or west of the Greenwich Meridian.

Example:

Longitude of P is 55^o W.

Longitude of Q is 30^o E.

5. All points on the same meridian have the same longitude.

Difference between Two Longitudes

1. If both the meridians are on the east (or west) of the Greenwich Meridian, subtract the angles of the longitudes.

2. If both the meridians are on the opposite sides of the Greenwich Meridian, add the angles of the longitudes.

9.2 Latitudes

Posted on April 24, 2020 by user

9.2 Latitudes

1. The great circle which is perpendicular to the earth’s axis is called equator.

2. A latitude is the angle at the centre of the Earth which is subtended by the arc of a meridian starting from the Equator to the parallel of latitude.

3. A circle which is perpendicular to the Earth’s axis and parallel to the equator is called a parallel of latitude.

4. The latitude of the Equator is 0^o.

5. All points on a parallel of latitude have the same latitude.

Example:

Latitude of D is 30^o N.

Latitude of E is 45^o S.

Difference between Two Latitudes

1. If both parallels of latitude are on the north (or south) of the equator, subtract the angles of latitudes.

2. If both parallels of latitude are on the opposite sides of the equator, add the angles of the latitudes.

9.4 Distance on the Surface of the Earth

Posted on April 24, 2020 by user

9.4 Distance on the Surface of the Earth

9.4.1 Distance between Two Points Along a Great Circle

1. A nautical mile is the length of an arc of a great circle that subtends an angle of one minute at the centre of the Earth.

2. The distance between two points on the surface of the earth is the arc of a circle that joins the two points.

3. To find the distance between two points along a meridian or the equator:

Distance of P and Q

= ( θ x 60 ) nautical miles

9.4.2 Distance between Two Points along a Parallel of Latitude

Posted on April 24, 2020 by user

9.4.2 Distance between Two Points along a Parallel of Latitude

1. To find the distance along a parallel of latitude:

Distance between points P and Q

= ( different in longitude in minutes ) × cos latitude

= ( θ × 60 × cos x^o ) nautical miles

Example:

In the diagram above, calculate

(a) The distance of PQ, measured along the parallel of latitude 34^o N,

(b) The distance of DM, measured along the parallel of latitude 54^o S.

Solution:

(a)

Difference in longitude between P and Q

= 40^o+ 58^o= 98^o

Hence, the distance of PQ, measured along the parallel of latitude 34^o N

= 98 × 60 × cos 34^o← [θ × 60 × cos (latitude)]

= 4874.74 n.m.

(b)

Difference in longitude between D and M

= 82^o– 40^o= 42^o

Hence, the distance of DM, measured along the parallel of latitude 54^o S

= 42 × 60 × cos 54^o← [θ × 60 × cos (latitude)]

= 1481.22 n.m.

7.2 Probability of the Complement of an Event

Posted on April 24, 2020 by user

7.2 Probability of the Complement of an Event

1. The complement of an event A is the set of all the outcomes in the sample space that are not included in the outcomes of event A.

2. The probability of the complement of event A is:

P (A') = 1 - P (A)

Example:

A number is chosen at random from a set of whole number from 1 to 40. Calculate the probability that the chosen number is not a perfect square.

Solution:

Let

A = Event of choosing a perfect square.

A’ = Event that the number is not a perfect square.

A = {1, 4, 9, 16, 25, 36}

n(A) = 6

\begin{array}{l} P (A) = \frac{n (A)}{n (S)} \\ = \frac{6}{40} = \frac{3}{20} \\ ∴ P (A') = 1 - P (A) \\ = 1 - \frac{3}{20} = \frac{17}{20} \\ Hence, the probability that the number chosen \\ is not a perfect square is \frac{17}{20} . \end{array}

7.4 SPM Practice (Short Questions)

Posted on April 24, 2020 by user

Question 1:
A bag contains 36 marbles which are black and white. It is given that the probability for a black marble being picked at random from the bag is

\frac{5}{9}

.
Calculate the number of white marbles to be taken out from the bag so that the probability of picking a black marble is

\frac{5}{8}

Solution:

\begin{array}{l} Number of black marbles in the bag \\ = \frac{5}{9} \times 36 = 20 \\ Let y is the total number of marbles left in the bag . \\ y \times \frac{5}{8} = 20 \\ y = 20 \times \frac{8}{5} = 32 \\ Number of white marbles to be taken out from the bag \\ = 36 - 32 \\ = 4 \end{array}

Question 2:

Table below shows the number of different coloured balls in three bags.

	Green	Brown	Purple
Bag A	3	1	6
Bag B	5	3	4
Bag C	4	6	2

If a bag is picked at random and then a ball is drawn randomly from that bag, what is the probability that a purple ball is drawn?

Solution:
Probability of picking a bag =

\frac{1}{3}

Probability of picking purple ball from bag A =

\frac{6}{10} = \frac{3}{5}

Probability of picking purple ball from bag B =

\frac{4}{12} = \frac{1}{3}

Probability of picking purple ball from bag C =

\frac{2}{12} = \frac{1}{6}

\begin{array}{l} P (purple ball) = (\frac{1}{3} \times \frac{3}{5}) + (\frac{1}{3} \times \frac{1}{3}) + (\frac{1}{3} \times \frac{1}{6}) \\ = \frac{1}{5} + \frac{1}{9} + \frac{1}{18} \\ = \frac{11}{30} \end{array}

Question 3:
A box contains 48 marbles. There are red marbles and green marbles. A marble is chosen at random from the box. The probability that a red marble is chosen is

\frac{1}{6} .

How many red marbles need to be added to the box so that the probability that a red marble is chosen is

\frac{1}{2} .

Solution:

\begin{array}{l} Number of red marbles in the box \\ = \frac{1}{6} \times 48 \\ = 8 \\ Let the number of red marbles needed to be added be y . \\ P (red marble) = \frac{1}{2} \\ \frac{8 + y}{48 + y} = \frac{1}{2} \\ 16 + 2 y = 48 + y \\ 2 y - y = 48 - 16 \\ y = 32 \\ ∴ Number of red marbles need to be added = 32 \end{array}

7.3 Probability of a Combined Event

Posted on April 24, 2020 by user

7.3 Probability of a Combined Event

7.3b Finding the Probability of Combined Events (a) A or B (b) A and B

1. The probability of a combined event ‘A or B’ is given by the formula below.

\begin{array}{l} P (A or B) = P (A \cup B) \\ = \frac{n (A \cup B)}{n (S)} \end{array}

2. The probability of a combined event ‘A and B’ is given by the formula below.

\begin{array}{l} P (A and B) = P (A \cap B) \\ = \frac{n (A \cap B)}{n (S)} \end{array}

Example:

The probabilities that two Form 5 students, Fiona and Wendy will pass the English oral test are

\frac{1}{3} and \frac{2}{5}

respectively. Calculate the probability that

(a) both Fiona and Wendy past the English oral test,
(b) both Fiona and Wendy fail the English oral test,

(c) either one of them passes the English oral test,
(d) at least one of them passes the English oral test.

Solution:

Let

F = Event that Fiona passes the English oral test

W = Event that Wendy passes the English oral test

Therefore,

F’ = Event that Fiona fails the English oral test

W’ = Event that Wendy fails the English oral test

\begin{array}{l} P (F) = \frac{1}{3}, P (F') = \frac{2}{3} \\ P (W) = \frac{2}{5}, P (W') = \frac{3}{5} \end{array}

(a)

P (both Fiona and Wendy past the English oral test)

= P (F ∩ W)

= P (F) x P (W)

\begin{array}{l} = \frac{1}{3} \times \frac{2}{5} \\ = \frac{2}{15} \end{array}

(b)

P (both Fiona and Wendy fail the English oral test)

= P (F’ ∩ W’)
= P (F’) x P (W’)

\begin{array}{l} = \frac{2}{3} \times \frac{3}{5} \\ = \frac{2}{5} \end{array}

(c)

P (either one of them passes the English oral test)

= P (F ∩ W’) + P (F’ ∩ W)
= (P (F) x P (W’)) + (P (F’) x P (W))

\begin{array}{l} = (\frac{1}{3} \times \frac{3}{5}) + (\frac{2}{3} \times \frac{2}{5}) \\ = \frac{7}{15} \end{array}

(d)

P (at least one of them passes the English oral test)

= 1 – P (Both of them fail) ← (concept of complement event)

= 1 – P (F’) x P (W’)

\begin{array}{l} = 1 - \frac{2}{5} \\ = \frac{3}{5} \end{array}

7.3 Probability of a Combined Event

Posted on April 24, 2020 by user

7.3 Probability of a Combined Event

7.3a Finding the Probability of a Combined Event by Listing the Outcomes

1. A combined event is an event resulting from the union or intersection of two or more events.

2. The union of combined event ‘A or B’ = A υ B

3. The intersection of combined event ‘A and B’ = A ∩ B

Example:

Diagram below shows five cards labelled with letters.

All these cards are put into a box. A two-letter code is to be formed by using any two of these cards. Two cards are picked at random, one after another, without replacement.

(a) List all sample space

(b) List all the outcomes of the events and find the probability that

(i) The code begins with the letter P.

(ii) The code consists of two vowel or two consonants.

Solution:

(a)

Sample space, S

= {(G, R), (G, A), (G, P), (G, E), (R, G), (R, A), (R, P), (R, E), (A, G), (A, R),

(A, P), (A, E), (P, G), (P, R), (P, A), (P, E), (E, G), (E, R), (E, A), (E, P)}

(b)

n(S) = 20

Let

A = Event of choosing a code begins with the letter P

B = Event of choosing the code consists of two vowel or two consonants.

(i)

A = {(P, G), (P, R), (P, A), (P, E)}

n(A) = 4

P (A) = \frac{4}{20} = \frac{1}{5}

(ii)

B = {(G, R), (G, P), (R, G), (R, P), (A, E), (P, G), (P, R), (E, A)}

n(B) = 8

P (B) = \frac{8}{20} = \frac{2}{5}