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}

5.5 Parallel Lines (Part 2)

Posted on May 26, 2020 by Myhometuition

(B) Equation of Parallel Lines

To find the equation of the straight line which passes through a given point and parallel to another straight line, follow the steps below:

Step 1 : Let the equation of the straight line take the form y = mx + c.

Step 2 : Find the gradient of the straight line from the equation of the straight line parallel to it.

Step 3 : Substitute the value of gradient, m, the x-coordinate and y-coordinate of the given point into y = mx + c to find the value of the y-intercept, c.

Step 4 : Write down the equation of the straight line in the form y = mx + c.

Example 2:

Find the equation of the straight line that passes through the point (–8, 2) and is parallel to the straight line 4y + 3x = 12.

Solution:

\begin{array}{l} 4 y + 3 x = 12 \\ 4 y = - 3 x + 12 \\ y = - \frac{3}{4} x + 3 \\ ∴ m = - \frac{3}{4} \\ At (- 8, 2), substitute m = - \frac{3}{4}, x = - 8, y = 2 into: \\ y = m x + c \\ 2 = - \frac{3}{4} (- 8) + c \\ c = 2 - 6 \\ c = - 4 \\ ∴ The equation of the staright line is y = - \frac{3}{4} x - 4. \end{array}

4.6 SPM Practice (Long Questions)

Posted on May 26, 2020 by Myhometuition

Question 9:
(a) State whether the following statements are true statement or false statement.

\begin{array}{l} (i) {} \subset {H,O,T} \\ (i i) {2} \subset {2, 3, 4} = {2, 3, 4} \end{array}

(b) Complete the following statement to form a true statement by using the quantifier ‘all’ or ‘some’.
……… factor of 24 are factor of 40

(c) Write down two implications based on the following compound statement.

The perimeter of square ABCD is 60 cm if and only if the side of square ABCD is 15 cm.

(d) It is given that the volume of the sphere is

\frac{4}{3} π r^{3}

where r is the radius.
Make one conclusion by deduction for the volume of the sphere with radius 3 cm.

Solution:
(a)(i) True

(a)(ii) False

(b) …Some… factor of 24 are factor of 40

(c)(i) If the perimeter of square ABCD is 60 cm, then the side of square ABCD is 15 cm.

(c)(ii) If the side of square ABCD is 15 cm, then the perimeter of square ABCD is 60 cm.

(d)

\begin{array}{l} \frac{4}{3} π \times 3^{3} = 36 π \\ Volume of the sphere is 36 π . \end{array}

Question 10:
(a)(i) State whether the following compound statement is true or false.

All straight lines have positive gradients.

(ii) Write down the converse of the following implication.

If x = 5, then x² = 25.

(b) Write down Premise 2 to complete the following argument:
Premise 1: If Q is an odd number, then 2 × Q is an even number.
Premise 2: _____________________
Conclusion: 2 × 3 is an even number.

(c) Make a general conclusion by induction for the sequence of numbers 4, 18, 48, 100, … which follows the following pattern.
4 = 1 (2)²18 = 2 (3)²48 = 3 (4)²100 = 4 (5)² .
.
.

Solution:
(a)(i) False

(a)(ii) If x² = 25, then x = 5

(b) Premise 2: 3 is an odd number

(c) n (n + 1)², where n = 1, 2, 3, …

4.5.1 Argument (Sample Questions)

Posted on May 25, 2020 by Myhometuition

Example 1:

Complete the conclusion in the following argument:

Premise 1: All regular polygons have equal sides.
Premise 2: ABCD is a regular polygon.
Conclusion:

Solution:

Conclusion: ABCD has equal sides.

Example 2:

Complete the conclusion in the following argument:

Premise 1: If m > 4, then 2m > 8.

Premise 2: 2m < 8

Conclusion:

Solution:

Conclusion: m < 4.

Example 3:

Complete the premise in the following argument:

Premise 1:

Premise 2: m ×n is not an even number.

Conclusion: m and n are not even numbers.

Solution:

Premise 1: If m and n are even numbers, then m ×n is an even number.

Example 4:

Complete the premise in the following argument:

Premise 1: If x = 3, then x² = 9.

Premise 2:

Conclusion: x ≠ 3

Solution:

Premise 2: x² ≠ 9.

4.4.1 Implications (Sample Questions)

Posted on May 25, 2020 by Myhometuition

Example 1:

Write down two implications based on the following statement.

y³ = -125 if and only if y = -5.

Solution:

Implication 1: If y³ = -125, then y = -5.

Implication 2: If y = -5, then y³ = -125.

Example 2:

Write down two implications based on the following statement.

8 is a factor of 24 if and only if 24 can be divided exactly by 8.

Solution:

Implication 1: 8 is a factor of 24 if 24 can be divided exactly by 8.

Implication 2: 24 can be divided exactly by 8 if 8 is a factor of 24.

Example 3:

State the converse of the following statement and hence, determine whether its converse is true or false.

(a) If 2x > 8, then x > 4.

(b) If x is a multiple of 6, then it is a multiple of 3.

Solution:

(a) Converse implication: If x > 4, then 2x > 8.

The converse is true.

(b) Converse implication: If x is a multiple of 3, then it is multiple of 6.

The converse is false. (9 is a multiple of 3 but it is not a multiple of 6)