6.7 SPM Practice (Long Questions)

Posted on April 23, 2020 by user

Question 1:

The data below shows the age of 25 tourists who visited a tourist spot.

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

(b) Based on the completed table above,

(i) State the modal class.

(ii) Calculate the mean age of the tourists.

(c) For this part of the question, use graph paper.

By using a scale of 2 cm to 5 years on the horizontal axis and 2 cm to 1 tourist on the vertical axis, draw a histogram for the data.

Solution:
(a)

Calculation of midpoint for (age 6 – 10) =

\frac{6 + 10}{2} = 8

(b)(i)
Modal class = age 16 – 20 (highest frequency)

(b)(ii)

\begin{array}{l} mean age = \frac{3 \times 2 + 8 \times 5 + 13 \times 3 + 18 \times 8 + 23 \times 3 + 28 \times 4}{25} \\ = 16.4 \end{array}

(c)

6.6 Measures of Dispersion (Part 1)

Posted on April 23, 2020 by user

6.6 Measures of Dispersion

(A) Determine the range of a set of data

1. For an ungrouped data,

Range = largest value – smallest value.

2. For a grouped data,

Range = midpoint of the last class – midpoint of the first class.

Example 1:

Determine the range of the following data.

(a) 720, 840, 610, 980, 900

(b)

Time (minutes)	1 – 6	7 – 12	13 – 18	19 – 24	25 – 30
Frequency	3	5	9	4	4

Solution:

(a)

Largest value of the data = 980

Smallest value of data = 610

Range = 980 – 610 = 370

(b)

Midpoint of the last class

= ½ (25 + 30) minutes

= 27.5 minute

Midpoint of the first class

= ½ (1 + 6) minutes

= 3.5 minute

Range = (27.5 – 3.5) minute = 24 minutes

6.4 Frequency Polygons

Posted on April 23, 2020 by user

6.4 Frequency Polygons

1. A frequency polygon is a line graph that connects the midpoints of each class interval at the top end of each rectangle in a histogram.

2. A frequency polygon can be drawn from a

(a) Histogram,

(b) Frequency table

3. Steps for drawing a frequency polygon:

Step 1: Add a class with 0 frequency before the first class and add also a class with 0 frequency after the last class.

Step 2: Calculate the midpoints or mark the midpoints of the top sides of the rectangular bars including the midpoints of the two additional classes.

Step 3: Joint all the midpoints with straight lines.

Example:

The following frequency table shows the distance travelled by 38 teenagers by motorcycles in one afternoon.

Journey travelled (km)	Frequency
55 – 59	4
60 – 64	4
65 – 69	7
70 – 74	8
75 – 79	9
80 – 84	6

Draw a frequency polygon based on the frequency table.

Solution:

Journey travelled (km)	Frequency	Midpoint
50 – 54	0	52
55 – 59	4	57
60 – 64	4	62
65 – 69	7	67
70 – 74	8	72
75 – 79	9	77
80 – 84	6	82
85 – 89	0	87

6.2 Mode and Mean of Grouped Data

Posted on April 23, 2020 by user

6.2 Mode and Mean of Grouped Data

(A) Modal Class

The modal class of grouped data is the class interval in the frequency table with the highest frequency.

(B) Class Midpoint

The class midpoint is the value of data that lies at the centre of a class.

Class midpoint = \frac{Lower limit + Upper limit}{2}

(C) Calculating the Mean of Grouped Data

The steps to calculate the mean of grouped data are as follows.

Step 1: Calculate the midpoint value of each class.

Step 2: Calculate the value of (frequency × midpoint value) of each class.

Step 3: Calculate the sum of the values of (frequency × midpoint value) of all the classes.

Step 4: Calculate the sum of all the frequencies of all the classes.

Step 5: Calculate the value of the mean using the formula below.

\begin{array}{l} \begin{array}{l} Mean of grouped data, \bar{x} \\ = \frac{Sum of (frequency \times midpoint)}{Sum of frequencies} \\ = \frac{\sum f x}{\sum f} \\ Where Σ is the notation of summation, \\ x is the midpoint of a class and f is its frequency . \end{array} \end{array}

6.7 SPM Practice (Long Questions)

Posted on April 23, 2020 by user

Question 2:
Diagram below shows the marks, obtained by a group of 24 students in a mathematics quiz.

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

(b) State the modal class.

(c) Calculate the estimated marks obtained by a student.

(d) For this part of the question, use graph paper.
By using a scale of 2 cm to 5 marks on the horizontal axis and 2 cm to 1 student on the vertical axis, draw a histogram for the data.

(e) Based on the histogram drawn in (d), state the number of students who obtain less than 32 marks in the quiz.

Answer:

Solution:
(a)

(b) Modal class = 27 – 31 (highest frequency)

\begin{array}{l} (c) \\ Estimated mean \\ = \frac{4 \times 24 + 7 \times 29 + 6 \times 34 + 4 \times 39 + 2 \times 44 + 1 \times 49}{24} \\ = \frac{796}{24} \\ = 33.17 marks \end{array}

(d)

(e)
Number of students who obtained less than 32 marks
= 4 + 7
= 11

6.1 Class Intervals

Posted on April 23, 2020 by user

6.1 Class Interval

1. Data that consist of the measurement of a quantity can be grouped into few classes and the range of each class in known as the class interval.

(A) Class Limits and Boundaries

Lower Limit and Upper Limit

2. For class interval, for example 30 – 39, the smaller value (30) is known as the lower limit while the larger value (39) is known as the upper limit.

Lower Boundary and Upper Boundary

3. The lower boundary of a class interval is the middle value between the lower limit of the class interval and the upper limitof the class before it.

4. The upper boundary of a class interval is the middle value between the upper limit of the class interval and the lower limitof the class after it.

Example:

20 – 29

30 – 39

40 – 49

\begin{array}{l} Lower boundary of the class 30 - 39 \\ = \frac{29 + 30}{2} = 29.5 \\ Upper boundary of the class 30 - 39 \\ = \frac{39 + 40}{2} = 39.5 \end{array}

(B) Class size

5. The class size is the difference between the upper boundary and the lower boundary of the class.

Example:

Size of class interval 30 – 39

= Upper boundary – Lower boundary

= 39.5 – 29.5

= 10

5.7 SPM Practice (Long Questions 2)

Posted on April 23, 2020 by user

Question 3:

Diagram below shows a straight line JK and a straight line ST drawn on a Cartesian plane. JK is parallel to ST.

Find

(a) the equation of the straight ST,

(b) the x-intercept of the straight line ST.

Solution:

(a)
JK is parallel to ST, therefore gradient of JK = gradient of ST.

\begin{array}{l} = \frac{8 - 0}{0 - 4} \\ = - 2 \end{array}

Substitute m = –2 and S (5, 6) into y = mx + c

6 = –2 (5) + c

c = 16

Therefore equation of ST: y = –2x + 16

(b)
For x-intercept, y = 0

0 = –2x + 16

2x = 16

x = 8

Therefore x-intercept of ST = 8

Question 4:

In the diagram above, PQRS is a parallelogram. Find

(a) the gradient of SR,

(b) the equation of QR,

(c) the x-intercept of QR.

Solution:

(a)
PQ is parallel to SR, gradient of PQ = gradient of SR.

Gradient of S R = - \frac{6}{- 3} = 2

(b)

\begin{array}{l} Gradient of Q R = \frac{8 - 6}{5 - 0} = \frac{2}{5} \\ Substitute m = \frac{2}{5} and R (5, 8) into y = m x + c \\ 8 = \frac{2}{5} (5) + c \\ c = 6 \\ Therefore equation of Q R : y = \frac{2}{5} x + 6 \end{array}

(c)

\begin{array}{l} For x-intercept, y = 0 \\ 0 = \frac{2}{5} x + 6 \\ x = - 15 \end{array}

Therefore x-intercept of QR = –15.

5.7 SPM Practice (Long Questions 3)

Posted on April 23, 2020 by user

Question 5:

In the diagram above, a straight line 5x + 7y + 35 = 0 intersects with the x-axis and y-axis at R and S respectively. Determine

(a) the gradient of the straight line RS.

(b) the x-intercept of the straight line RS.

(c) the distance of RS.

Solution:
(a)

\begin{array}{l} 5 x + 7 y + 35 = 0 \\ 7 y = - 5 x - 35 \\ y = - \frac{5}{7} x - 5 \\ ∴ The gradient of the straight line R S = - \frac{5}{7} . \end{array}

(b)

\begin{array}{l} At x-intercept, y = 0 \\ 0 = - \frac{5}{7} x - 5 \\ \frac{5}{7} x = - 5 \\ x = - 7 \\ ∴ x-intercept of the straight line R S = - 7. \end{array}

(c)

\begin{array}{l} Point R = (- 7, 0) and point S = (0, - 5) \\ Distance of R S = \sqrt{{(- 7 - 0)}^{2} + {(0 - (- 5))}^{2}} \\ Distance of R S = \sqrt{49 + 25} \\ Distance of R S = \sqrt{74} units \end{array}

Question 6:

In the diagram above, O is the origin of the Cartesian plane, AOB is a straight line and OA = AC. Find

(a) the coordinates of C.

(b) the value of h.

(c) the equation of BC.

Solution:
(a)

x-coordinate of C = –3 × 2 = –6

Therefore coordinates of C = (–6, 0).

(b)

\begin{array}{l} Gradient of A O = Gradient of O B \\ \frac{0 - (- 4)}{0 - (- 3)} = \frac{h - 0}{6 - 0} \\ \frac{4}{3} = \frac{h}{6} \\ h = 8 \end{array}

(c)

\begin{array}{l} Gradient of B C = \frac{8 - 0}{6 - (- 6)} = \frac{8}{12} = \frac{2}{3} \\ At point C (- 6, 0), \\ 0 = \frac{2}{3} (- 6) + c \\ c = 4 \\ The equation of B C is, \\ y = \frac{2}{3} x + 4 \end{array}

5.2 Gradient of a Straight Line in Cartesian Coordinates

Posted on April 23, 2020 by user

5.2 Finding the Gradient of a Straight Line

The gradient, m, of a straight line which passes through P (x₁, y₁) and Q (x₂, y₂) is given by,

m_PQ =

\frac{y_{2} - y_{1}}{x_{2} - x_{1}}

Example 1:

Find the gradient of the straight line joining two points P and Q in the above diagram.

Solution:

P = (x₁, y₁) = (4, 3), Q = (x₂, y₂) = (10, 5)

Gradient of the straight line PQ

\begin{array}{l} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ = \frac{5 - 3}{10 - 4} \\ = \frac{2}{6} = \frac{1}{3} \end{array}

Example 2:

Calculate the gradient of a straight line which passes through point A (7, -3) and point B (-3, 6).

Solution:

A = (x₁, y₁) = (7, -3), B = (x₂, y₂) = (-3, 6)

Gradient of the straight line AB

\begin{array}{l} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ = \frac{6 - (- 3)}{- 3 - 7} \\ = - \frac{9}{10} \end{array}

5.3 Intercepts

Posted on April 23, 2020 by user

5.3 Intercepts

1. The x-intercept is the point of intersection of a straight line with the x-axis.

2. The y-intercept is the point of intersection of a straight line with the y-axis.

3. In the above diagram, the x-intercept of the straight line PQ is 6 and the y-intercept of PQ is 5.

4. If the x-intercept and y-intercept of a straight line are given,

Gradient, m = - \frac{y - intercept}{x - intercept}