8.1 Tangents to a Circle (Examples)

Posted on May 27, 2020 by Myhometuition

Example 1:

In the diagram, O is the centre of a circle. ABC and CDE are two tangents to the circle at points B and D respectively. Find the length of OC.

Solution:

OC²= OB² + BC² ← (Pythagoras’ Theorem)

= 6² + 8²

= 100

OC = √100 = 10 cm

Example 2:

In the diagram, AB and BC are two tangents to a circle with centre O. Calculate the values of

(a) x (b) y

Solution:

(a)

AB = BC

7 + x = 12

x = 5

(b)

∠

OBA =

∠

OBC = 21^o

∠

OAB = 90^o ← (OA is perpendicular to AB)

y^o= 180^o – 21^o– 90^o

y = 69

Example 3:

In the diagram, ABC is a tangent to the circle with centre Oat point B. CDE is a straight line. Find the value of x.

Solution:

∠

CBO = 90^o ← (OB is perpendicular to BC)

In ∆ BCE,

x = 180^o – 30^o– 50^o– 90^o

x = 10^o

SPM Practice (Short Questions)

Posted on May 27, 2020 by Myhometuition

Question 6:
A box contains blue marbles and green marbles. A marble is picked at random from the box.
The probability of picking a blue marble is

\frac{6}{11}

. If there are 30 green marbles in the box, how many blue marbles are there in the box?

Solution:
Probability of picking a green marble = 1 –

\frac{6}{11}

\frac{5}{11}

Total number of marbles in the box = 30 ×

\frac{11}{5}

= 66

Hence, number of blue marbles in the box = 66 – 30 = 36

Question 7:

Table below shows the distribution of a group of 90 pupils attending a penalty kick training programme.

	Form Four	Form Five
Girls	33	15
Boys	18	24

A pupil is chosen at random from the group to start the penalty kick training.

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

Solution:

P (a boy from Form Five)

= \frac{24}{90} = \frac{4}{15}

Question 8:

Diagram below shows some number cards.

A card is picked at random. State the probability that a prime number is picked.

Solution:

prime number = {11, 19, 37}
P (a prime number)

= \frac{3}{6} = \frac{1}{2}

6.7 SPM Practice (Long Questions)

Posted on May 26, 2020 by Myhometuition

Question 7:
Diagram below shows the marks obtained by a group of 30 students in a history test.

(a) Using data in diagram above and a class interval of 5 marks, complete the table in the answer space.

(b) Based on the completed table in part (a)
(i) find the modal class,
(ii) calculate the mean mark obtained by the students.

(c) By using a scale of 2 cm to 10 marks on the horizontal axis and 2 cm to 1 student on the vertical axis, draw a frequency polygon for the data.

Answer:

Solution:
(a)

(b)(i)
Modal class = 50 – 54

(b)(ii)

\begin{array}{l} Mean mark = \frac{1615}{30} \\ = 53 .83 \end{array}

(c)

6.7 SPM Practice (Long Questions)

Posted on May 26, 2020 by Myhometuition

Question 6:
The table below shows the frequency distribution of the age of 100 scouts in a camping spot.

(a) Based on table above, complete table in the answer space.

(b) Calculate the estimated age, in years, of a scout in the camping spot.

(c) For this part of the question, use graph paper. You may use a flexible curve rule.
By using a scale of 2 cm to 5 years on the horizontal axis and 2 cm to 10 scouts on the vertical axis, draw an ogive for the data.

(d) Any scout which is older than 18 years old is considered senior scout.
Based on the ogive drawn in part (c), find the percentage of senior scout in the camping spot.

Answer:

Solution:
(a)

(b)

\begin{array}{l} Estimated mean age \\ = \frac{0 \times 2 + 12 \times 7 + 18 \times 12 + 26 \times 17 + 25 \times 22 + 13 \times 27 + 6 \times 32}{100} \\ = \frac{1835}{100} \\ = 18.35 years old \end{array}

(c)

(d)

\begin{array}{l} Number of senior scouts \\ = 100 - 49 \\ = 51 \\ Percentage of senior scouts \\ = \frac{51}{100} \times 100 % \\ = 51 % \end{array}

6.7 SPM Practice (Long Questions)

Posted on May 26, 2020 by Myhometuition

Question 5:

The table below shows the frequency distribution of the time spent by 50 swimmers in the pool in a swimming practice.

Time (seconds)	Frequency
35 – 39	5
40 – 44	8
45 – 49	9
50 – 54	15
55 – 59	11
60 – 64	2

(a) State the modal class.

(b) Calculate the estimated mean of the time spent of a swimmer.

(c) Based on table above, complete table below in the answer space by writing down the values of the upper boundary and the cumulative frequency.

Upper Boundary	Cumulative Frequency
34.5	0
39.5




64.5	50

(d) For this part of the question, use graph paper. You may use a flexible curve rule.

By using a scale of 2 cm to 5 seconds on the horizontal axis and 2 cm to 5 swimmers on the vertical axis, draw an ogive for the data.

Solution:
(a) Modal class = time 50 – 54 seconds (highest frequency).

(b)

$\begin{array}{l} Estimated mean \\ = \frac{37 \times 5 + 42 \times 8 + 47 \times 9 + 52 \times 15 + 57 \times 11 + 62 \times 2}{50} \\ = \frac{2475}{50} = 49.5 \end{array}$

(c)

Upper Boundary	Cumulative Frequency
34.5	0
39.5	5
44.5	13
49.5	22
54.5	37
59.5	48
64.5	50

(d)

6.7 SPM Practice (Long Questions)

Posted on May 26, 2020 by Myhometuition

Question 4:
Diagram below shows the marks obtained by a group of 30 students in a Science test.

(a) Based on the data in diagram above, complete the table in the answer space.

(b) Based on the completed table in part (a), calculate the estimated mean mark of a student.

For this part of the question, use graph paper.
(c) By using a scale of 2 cm to 10 marks on the horizontal axis and 2 cm to 1 student on the vertical axis, draw a frequency polygon for the data.

(d) The passing mark for the test is 44. Using the frequency polygon drawn in part (c), find the number of students who passed the test.

Answer:

Solution:
(a)

(b)

\begin{array}{l} Estimated mean mark = \frac{1255}{30} \\ = 41 .83 \end{array}

6.6 Measures of Dispersion (Part 2)

Posted on May 26, 2020 by Myhometuition

(B) Medians and Quartiles

1. The first quartile (Q₁)is a number such that 1 4 of the total number of data that has a value less than the number.

2. The median is the second quartile which is the value that lies at the centre of the data.

3. The third quartile (Q₃) is a number such that 3 4of the total number of data that has a value less than the number.

4. The interquartile range is the difference between the third quartile and the first quartile.

Interquartile range = third quartile – first quartile

Example 2:

The ogive in the diagram shows the distribution of time (to the nearest second) taken by 100 students in a swimming competition. From the ogive, determine

(a) the median,

(b) the first quartile,

(c) the third quartile

(d) the interquartile range of the time taken.

Solution:

\begin{array}{l} (a) \frac{1}{2} of 100 students = \frac{1}{2} \times 100 = 50 \\ From the ogive, median, M = 50.5 second \\ (b) \frac{1}{4} of 100 students = \frac{1}{4} \times 100 = 25 \\ From the ogive, first quartile, Q_{1} = 44.5 second \\ (c) \frac{3}{4} of 100 students = \frac{3}{4} \times 100 = 75 \\ From the ogive, third quartile, Q_{3} = 5 4.5 second \end{array}

(d)

Interquartile range

= Third quartile – First quartile

= 54.5 – 44.5

= 10.0 second

6.2 Mode and Mean of Grouped Data (Examples)

Posted on May 26, 2020 by Myhometuition

Example:

The following frequency table shows the number of magazines sold at a bookshop for 30 days in April 2013.

Number of magazines	Frequency
220 – 229	3
230 – 239	5
240 – 249	11
250 – 259	6
260 – 269	5

Based on the data given,
(a) calculate the size of class,

(b) state the modal class,

(c) calculate the mean number of magazine sold per day.

Solution:

(a) Size of the class

= upper boundary – lower boundary

= 229.5 – 219.5

= 10

(b) Modal class = 240 – 249 (Highest frequency)

(c)

Number of magazines	Frequency (f)	Class midpoint (x)
220 – 229	3	224.5
230 – 239	5	234.5
240 – 249	11	244.5
250 – 259	6	254.5
260 – 269	5	264.5

\begin{array}{l} Midpoint = \frac{220 + 229}{2} = 224.5 \\ mean, \bar{x} = \frac{\sum f x}{\sum f} \\ = \frac{3 \times 224.5 + 5 \times 234.5 + 11 \times 244.5 + 6 \times 254.5 + 5 \times 264.5}{30} \\ = \frac{7385}{30} \\ = 246.2 \end{array}

5.7 SPM Practice (Long Questions 5)

Posted on May 26, 2020 by Myhometuition

Question 9:

The diagram above shows that the straight line PMQ intersects with PNR at N. Given that OQ = OR and M is the midpoint of PQ. Find
(a) the coordinate of P
(b) the value of m and n.

Solution:

\begin{array}{l} (a) \\ Given M is midpoint of P Q \\ x -coordinate of P = 0 \\ For y -coordinate of P, \\ \frac{y + 0}{2} = 3 \\ y = 6 \\ Coordinates of P = (0, 6) . \end{array}

\begin{array}{l} (b) \\ \frac{0 + 4}{2} = m \\ 2 m = 4 \\ m = 2 \\ Gradient of P R \\ = \frac{6 - 0}{0 - (- 4)} \\ = \frac{6}{4} \\ = \frac{3}{2} \\ At point N (n, 4), \\ using y_{1} = m x_{1} + c \\ 4 = \frac{3}{2} n + 6 \\ 8 = 3 n + 12 \\ 3 n = - 4 \\ n = - \frac{4}{3} \end{array}

Question 10 (5 marks):
Diagram 6 shows two parallel straight lines, POQ and RMS drawn on a Cartesian plane.

Diagram 6

Find
(a) the equation of the straight line RMS,
(b) the x-intercept of the straight line MN.

Solution:
(a)

\begin{array}{l} m_{R M S} = m_{P O Q} \\ = \frac{1 - 0}{- 2 - 0} \\ = - \frac{1}{2} \\ At point (4, 6) \\ y_{1} = m x_{1} + c \\ 6 = - \frac{1}{2} (4) + c \\ 6 = - 2 + c \\ c = 8 \\ The equation of straight line R M S is \\ y = - \frac{1}{2} x + 8 \end{array}

(b)

\begin{array}{l} When y = 0, \\ - \frac{1}{2} x + 8 = 0 \\ \frac{1}{2} x = 8 \\ x = 16 \\ x -intercept of straight line M N is 16. \end{array}

5.7 SPM Practice (Long Questions 4)

Posted on May 26, 2020 by Myhometuition

Question 7:

The diagram above shows a parallelogram on a Cartesian plane. MP and NO are parallel to the y-axis. Given that the distance of MZ is 4 units. Find
(a) the value of p and q.
(b) the equation of the straight line MN.

Solution:

\begin{array}{l} (a) \\ Line N O is parallel to y -axis, \\ p = 2 \\ M P = \sqrt{3^{2} + 4^{2}} \\ = \sqrt{9 + 16} \\ = \sqrt{25} \\ = 5 \\ N O = M P = 5 units \\ q = 7 - 5 = 2 \end{array}

\begin{array}{l} (b) \\ Point O = (2, 2) \\ Gradient P O = \frac{2 - 0}{2 - 0} = 1 \\ Gradient M N = g radient P O = 1 \\ y_{1} = m x_{1} + c \\ 7 = 1 (2) + c \\ c = 5 \\ Equation of line M N is \\ y = x + 5 \end{array}

Question 8:

The diagram above shows that two straight line intersect at point (0 , -2). Find
(a) the value of b
(b) the x-intercept of the straight line XY if the gradient of XY is equal to 2.
(c) the equation of XY.

Solution:
(a)
Value of b
= 2 units + 3 units
= 5 units
= –5

\begin{array}{l} (b) \\ Given m = 2, c = - 2 \\ For x -intercept, y = 0 \\ 0 = 2 x + (- 2) \\ 0 = 2 x - 2 \\ 2 x = 2 \\ x = 1 \\ Therefore x intercept of X Y = 1. \end{array}

\begin{array}{l} (c) \\ Substitute m = 2 and (0, - 2) into y = m x + c \\ y = 2 x + (- 2) \\ y = 2 x - 2 \\ Therefore equation of X Y : y = 2 x - 2 \end{array}