4.10 SPM Practice (Long Questions)

Posted on May 29, 2020 by Myhometuition

Question 11 (5 marks):
Table 8 shows the information of books purchased by Maslinda.

Table 8

Maslinda purchased x history books and y Science books. The total number of books purchased is 5. The total price of the books purchased is RM17.

(a) Write two linear equations in terms of x and y to represent the above information.

(b) Hence, by using matrix method, calculate the value of x and of y.

Solution:
(a)
x + y = 5
4x + 3y = 17

(b)

$\begin{array}{l} (\begin{array}{l} 1 1 \\ 4 3 \end{array}) (\begin{array}{l} x \\ y \end{array}) = (\begin{array}{l} 5 \\ 17 \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = \frac{1}{1 (3) - 4 (1)} (\begin{array}{l} 3 - 1 \\ - 41 \end{array}) (\begin{array}{l} 5 \\ 17 \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = - 1 (\begin{array}{l} 3 (5) + (- 1) (17) \\ - 4 (5) + (1) (17) \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = - 1 (\begin{array}{l} 15 - 17 \\ - 20 + 17 \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = - 1 (\begin{array}{l} - 2 \\ - 3 \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = (\begin{array}{l} 2 \\ 3 \end{array}) \\ x = 2 and y = 3 \end{array}$

Question 12 (6 marks):

$\begin{array}{l} Given A = (\begin{array}{l} 4 - 2 \\ 3 - 1 \end{array}), B = m (\begin{array}{l} - 1 n \\ - 3 4 \end{array}) \\ and I = (\begin{array}{l} 1 0 \\ 01 \end{array}) . \end{array}$
(a) Find the value of m and of n if AB = I.
(b) Write the following simultaneous linear equation as matrix equation:
4x – 2y = 3
3x – y = 2
Hence, by using matrix method, calculate the value of x and of y.

Solution:
(a)

$\begin{array}{l} If A B = I, then B = A^{- 1} \\ A^{- 1} = \frac{1}{4 (- 1) - (- 2) (3)} (\begin{array}{l} - 1 2 \\ - 34 \end{array}) \\ A^{- 1} = \frac{1}{2} (\begin{array}{l} - 1 2 \\ - 34 \end{array}) \\ By comparison: \\ B = A^{- 1} \\ m (\begin{array}{l} - 1 n \\ - 3 4 \end{array}) = \frac{1}{2} (\begin{array}{l} - 1 2 \\ - 34 \end{array}) \\ m = \frac{1}{2}; n = 2 \end{array}$

(b)

$\begin{array}{l} 4 x - 2 y = 3 \\ 3 x - y = 2 \\ (\begin{array}{l} 4 - 2 \\ 3 - 1 \end{array}) (\begin{array}{l} x \\ y \end{array}) = (\begin{array}{l} 3 \\ 2 \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = \frac{1}{2} (\begin{array}{l} - 1 2 \\ - 34 \end{array}) (\begin{array}{l} 3 \\ 2 \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = \frac{1}{2} (\begin{array}{l} (- 1) (3) + (2) (2) \\ (- 3) (3) + (4) (2) \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = \frac{1}{2} (\begin{array}{l} 1 \\ - 1 \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = (\begin{array}{l} \frac{1}{2} \\ - \frac{1}{2} \end{array}) \\ x = \frac{1}{2} and y = - \frac{1}{2} \end{array}$

4.6 SPM Practice (Long Questions)

Posted on May 29, 2020 by Myhometuition

Question 13 (5 marks):
(a) State whether the following statements are true statement or false statement.

$\begin{array}{l} (i) { } \subset {S, E, T} \\ (ii) { 1 } \subset {1, 2, 3} = {1, 2, 3} \end{array}$

(b) Diagram 7 shows the first three patterns of a sequence of patterns.

Diagram 7

It is given that the diameter of each circle is 20 cm.
(i) Make a general conclusion by induction for the area of the unshaded region.

(ii) Hence, calculate the area of the unshaded region for the 5th pattern.

Solution:
(a)(i) True

(a)(ii) False

(b)(i)
Area of unshaded region (first)
= (20 × 20) – π(10)²= 400 – 100π
= 100 (4 – π)

Area of unshaded region (second)
= 4 × 100 (4 – π)
= 400 (4 – π)

Area of unshaded region (third)
= 9 × 100 (4 – π)
= 900 (4 – π)
100 (4 – π), 400 (4 – π), 900 (4 – π), …
10² (4 – π), 20² (4 – π), 30² (4 – π), …

General conclusion = n² (4 – π)
n = 10, 20, 30, …

(b)(ii)
Area of unshaded region (5th)
= 50² (4 – π)
= 2500 (4 – π)

4.6 SPM Practice (Long Questions)

Posted on May 29, 2020 by Myhometuition

Question 11 (5 marks):
(a) State whether the following compound statement is true or false.
2 > 3 and (–2)³ = –8

(b) Write down two implications based on the following statement:
a > b if and only if a – b > 0

(c) Table 1 shows the number of sides and the number of axes of symmetry for some regular polygons.

Table 1

Make a conclusion by induction by completing the statement in the answer space:
The number of axes of symmetry for a regular polygon with n sides is __________.

Solution:
(a)
Palsu. Nilai bagi 2 adalah kurang daripada 3, 2 < 3.

(b)
Implication 1: Jika a > b, maka a – b > 0
Implication 2: Jika a – b > 0, maka a – b > 0

(c)
The number of axes of symmetry for a regular polygon with n sides is
Number of sides of regular polygon = Number of axes of symmetry of regular polygon

Question 12 (5 marks):
(a) State whether the following statement is true or false.

(b) Write down the converse of the following implication:

(c) Write down Premise 2 to complete the following argument:
Premise 1 : If x is an odd number, then x is not divisible by 2.
Premise 2 : ………………………………………………………
Conclusion : 24 is not an odd number.

(d) Based on the information below, make one conclusion by deduction for the surface area of sphere with radius 9 cm.

Solution:
(a)
True

(b)
If an angle is an acute angle, then the angle lies between 0^o and 90^o.

(c)
24 is divisible by 2.

(d)
4π(9)² = 324π
Surface area of the sphere is 324π.

5.7 SPM Practice (Long Questions 6)

Posted on May 29, 2020 by Myhometuition

Question 11 (5 marks):
Diagram 4 shows a parallelogram drawn on a Cartesian plane which represents the locations of Rahman's house, a cinema, a school and a shop.

It is given that the scale is 1 unit : 1 km.
(a) Calculate the distance, in km, between Rahman's house and the school.
(b) Find the equation of the straight line that links the school to the cinema.

Solution:
(a)
2y = 3x + 15
When y = 0
3x + 15 = 0
3x = –15
x = –5

Rahman's house = (–5, 0)
School = (3, 0)

Distance, between Rahman's house and the school
= 3 – (– 5)
= 8 units
= 8 km

(b)

$\begin{array}{l} 2 y = 3 x + 15 \\ y = \frac{3}{2} x + \frac{15}{2} \\ Thus m = \frac{3}{2} \\ At point (3, 0), y_{1} = m x_{1} + c \\ 0 = \frac{3}{2} (3) + c \\ \frac{9}{2} + c = 0 \\ c = - \frac{9}{2} \\ Thus, the linear equation is \\ y = \frac{3}{2} x - \frac{9}{2} \\ 2 y = 3 x - 9 \end{array}$

Question 12 (6 marks):
Diagram 6 shows two parallel straight lines, JK and LM, drawn on a Cartesian plane.
The straight line KM is parallel to the x-axis.

Diagram 6

Find
(a) the equation of the straight line KM,
(b) the equation of the straight line LM,
(c) the value of k.

Solution:
(a)
The equation of the straight line KM is y = 3

(b)

$\begin{array}{l} Given, equation of J K : \\ 2 y = 4 x + 3 \\ y = 2 x + \frac{3}{2} \\ Thus, m_{J K} = 2 \\ m_{L M} = m_{J K} = 2 \\ y = m x + c \\ At M (5, 3) \\ 3 = 2 (5) + c \\ 3 = 10 + c \\ c = - 7 \\ ∴ Equation of the straight line L M \\ is y = 2 x - 7. \end{array}$

(c)

$\begin{array}{l} Substitute (k, 0) into y = 2 x - 7 \\ 0 = 2 (k) - 7 \\ 7 = 2 k \\ k = \frac{7}{2} \end{array}$

Quadratic Equations Long Questions (Question 14 & 15)

Posted on May 29, 2020 by Myhometuition

Question 14 (4 marks):
An aquarium has the length of (x + 7) cm, the width of x cm and the height of 60 cm.
The total volume of the aquarium is 48000 cm³. The aquarium will be filled fully with water.
Calculate the value of x.

Solution:

$\begin{array}{l} Volume of aquarium = 48000 {cm}^{3} \\ So, \\ (x) (x + 7) (60) {cm}^{3} = 48000 {cm}^{3} \\ (x) (x + 7) = \frac{48000}{60} \\ x^{2} + 7 x = 800 \\ x^{2} + 7 x - 800 = 0 \\ (x - 25) (x + 32) = 0 \\ x - 25 = 0 or x + 32 = 0 \\ x = 25 or x = - 32 (not accepted) \\ So the value of x = 25 cm \end{array}$

Question 15 (4 marks):
Diagram 3 shows a garden path with a rectangular shape. There are 8 similar circular stepping stone built in the path.

Given the area of the path is 32 m², find the diameter, in m, of one piece of stepping stone.

Solution:

$\begin{array}{l} Given the area of the path \\ = 32 m^{2} = 320000 {cm}^{2} \\ x (x + 4) = 320000 \\ x^{2} + 4 x - 320000 = 0 \\ a = 1, b = 4, c = - 320000 \\ x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ x = \frac{- 4 \pm \sqrt{4^{2} - 4 (1) (- 320000)}}{2 (1)} \\ x = \frac{- 4 \pm \sqrt{1280016}}{2} \\ x = \frac{- 4 + 1131.38}{2} or \frac{- 4 - 1131.38}{2} \\ x = 563.69 or - 567.69 (ignored) \\ Diameter of one piece of stepping \\ stone = x \\ = 563.69 cm \\ = 5.64 m \end{array}$

2.5 SPM Practice (Long Questions)

Posted on May 29, 2020 by Myhometuition

Question 13 (12 marks):
(a) Complete Table 2 in the answer space, for the equation

$y = \frac{30}{x}$ by writing down the values of y when x = 2 and x = 5.

(b) For this part of the question, use the graph paper. You may use a flexible curve ruler.
Using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, draw the graph of

$y = \frac{30}{x} for 0 \leq x \leq 7.$

(c) From the graph in 12(b), find
(i) the value of y when x = 2.6,
(ii) the value of x when y = 17.5.

(d) Draw a suitable straight line on the graph in 12(b) to find the values of x which satisfy the equation

$\frac{30}{x} = - 5 x + 30 for 0 \leq x \leq 7.$
State the values of x.

Answer:

Table 2

Solution:
(a)

(b)

(c) From the graph
(i) When x = 2.6; y = 11.5
(ii) When y = 17.5; x = 1.7

(d)

$\begin{array}{l} Given, \frac{30}{x} = - 5 x + 30 \\ y = - 5 x + 30 \\ From the graph; x = 1.25 and 4.75 \end{array}$

2.5 SPM Practice (Long Questions)

Posted on May 29, 2020 by Myhometuition

Question 12 (12 marks):
(a) Complete Table 2 in the answer space, for the equation y = –x³ + 4x + 10 by writing down the values of y when x = –2 and x = 1.5.

(b) For this part of the question, use the graph paper. You may use a flexible curve ruler.
Using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 10 units on the y-axis, draw the graph of y = –x³ + 4x + 10 for –3 ≤ x ≤ 4.

(c) From the graph in 12(b), find
(i) the value of y when x = –2.5,
(ii) the positive value of x when y = 4.

(d) Draw a suitable straight line on the graph in 12(b) to find the values of x which satisfy the equation x³ – 14x + 5 = 0 for –3 ≤ x ≤ 4.
State the values of x.

Answer:

Solution:
(a)
y = –x³ + 4x + 10
When x = –2
y = –(–2)³ + 4(–2) + 10
y = 8 – 8 + 10
y = 10

When x = 1.5
y = –(1.5)³ + 4(1.5) + 10
y = –3.375 + 6 + 10
y = 12.625

(b)

(c) From graph
(i) When x = –2.5; y = 16
(ii) When y = 4; x = 2.5

(d)
x³ – 14x + 5 = 0
–x³ + 14x – 5 = 0
–x³ + (4x + 10x) + (10 – 15) = 0
–x³ + 4x + 10 = –10x + 15
Thus, y = –10x + 15

From graph, the values of x are 0.35 and 3.5.

2.5 SPM Practice (Long Questions)

Posted on May 29, 2020 by Myhometuition

Question 11 (12 marks):
(a) Complete Table 12 in the answer space, for the equation y = –x² + 2x + 10 by writing down the values of y when x = –1 and x = 2.

(b) For this part of the question, use the graph paper. You may use a flexible curve ruler.
Using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 2 units on the y-axis, draw the graph of y = –x² + 2x + 10 for –3.5 ≤ x ≤ 4.

(c) From the graph in 12(b), find
(i) the value of y when x = –1.5,
(ii) the positive value of x when y = 8.2.
[adinserter block="3"]

(d) Draw a suitable straight line on the graph in 12(b) to find the values of x which satisfy the equation 7 – x = x² for –3.5 ≤ x ≤ 4.
State the values of x.

Answer:

Solution:
(a)
y = –x² + 2x + 10
When x = –1
y = –(–1)² + 2(–1) + 10
y = –1 – 2 + 10
y = 7

when x = 2
y = –(2)² + 2(2) + 10
y = –4 + 4 + 10
y = 10

(b)

(c) From graph
(i) When x = –1.5; y = 4.6
(ii) When y = 8.2; x = 2.7

(d)
y = –x² + 2x + 10 ……. (1)
0 = –x² – x + 7 ………. (2)
(1) – (2): y = 3x + 3

From graph, the values of x are –3.2 and 2.2.

4.10 SPM Practice (Long Questions)

Posted on May 28, 2020 by Myhometuition

Question 9:

$(a) Given \frac{1}{s} (\begin{matrix} - 4 & 2 \\ - 5 & 3 \end{matrix}) (\begin{matrix} t & - 2 \\ 5 & - 4 \end{matrix}) = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}), find the value of s and of t .$

(b) Using matrices, calculate the value of x and of y that satisfy the following matrix equation:

$(\begin{matrix} - 4 & 2 \\ - 5 & 3 \end{matrix}) (\begin{array}{l} x \\ y \end{array}) = (\begin{array}{l} 1 \\ 2 \end{array})$

Solution:

$\begin{array}{l} (a) \\ \frac{1}{s} (\begin{matrix} t & - 2 \\ 5 & - 4 \end{matrix}) = {(\begin{matrix} - 4 & 2 \\ - 5 & 3 \end{matrix})}^{- 1} \\ = \frac{1}{(- 4) (3) - (2) (- 5)} (\begin{matrix} 3 & - 2 \\ 5 & - 4 \end{matrix}) \\ = \frac{1}{- 2} (\begin{matrix} 3 & - 2 \\ 5 & - 4 \end{matrix}) \\ ∴ s = - 2, t = 3 \end{array}$

$\begin{array}{l} (b) \\ (\begin{matrix} - 4 & 2 \\ - 5 & 3 \end{matrix}) (\begin{array}{l} x \\ y \end{array}) = (\begin{array}{l} 1 \\ 2 \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = \frac{1}{- 2} (\begin{matrix} 3 & - 2 \\ 5 & - 4 \end{matrix}) (\begin{array}{l} 1 \\ 2 \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = \frac{1}{- 2} (\begin{array}{l} (3) (1) + (- 2) (2) \\ (5) (1) + (- 4) (2) \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = \frac{1}{- 2} (\begin{array}{l} - 1 \\ - 3 \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = (\begin{array}{l} \frac{1}{2} \\ \frac{3}{2} \end{array}) \\ ∴ x = \frac{1}{2}, y = \frac{3}{2} \end{array}$

Question 10 (5 marks):
During the sport day, students used coupons to buy food and drink. Ali and Larry spent RM31 and RM27 respectively. Ali bought 2 food coupons and 5 drinks coupons while Larry bought 3 coupons and 1 drinks coupon.
Using the matrix method, calculate the price, in RM, of a food coupon and of a drinks coupon.

Solution:
Ali spent RM31. He bought 2 food coupons and 5 drinks coupons.
Larry spent RM27. He bought 3 food coupons and 1 drinks coupons.
x = price of one food coupon
y = price of one drinks coupon

$\begin{array}{l} (\begin{array}{l} 25 \\ 3 1 \end{array}) (\begin{array}{l} x \\ y \end{array}) = (\begin{array}{l} 31 \\ 27 \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = \frac{1}{2 (1) - 5 (3)} (\begin{array}{l} 1 - 5 \\ - 32 \end{array}) (\begin{array}{l} 31 \\ 27 \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = \frac{1}{2 - 15} (\begin{array}{l} 1 (31) + (- 5) (27) \\ - 3 (31) + 2 (27) \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = \frac{1}{- 13} (\begin{array}{l} - 104 \\ - 39 \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = (\begin{array}{l} 8 \\ 3 \end{array}) \\ x = 8 and y = 3 \\ Thus, the price for a food coupon is RM8 \\ and the price for drink coupon is RM3 . \end{array}$

4.10 SPM Practice (Long Questions)

Posted on May 28, 2020 by Myhometuition

Question 7:

$(a) Find the inverse matrix of (\begin{matrix} 3 & 2 \\ 5 & 4 \end{matrix}) .$
(b) Ethan and Rahman went to the supermarket to buy cucumbers and carrots. Ethan bought 3 cucumbers and 2 carrots for RM9. Rahman bought 5 cucumbers and 4 carrots for RM16.
By using matrix method, find the price, in RM, of a cucumber and the price of a carrot.

Solution:

$\begin{array}{l} (a) \\ Inverse matrix of (\begin{matrix} 3 & 2 \\ 5 & 4 \end{matrix}) \\ = \frac{1}{12 - 10} (\begin{matrix} 4 & - 2 \\ - 5 & 3 \end{matrix}) \\ = \frac{1}{2} (\begin{matrix} 4 & - 2 \\ - 5 & 3 \end{matrix}) \\ = (\begin{matrix} 2 & - 1 \\ - \frac{5}{2} & \frac{3}{2} \end{matrix}) \end{array}$

$\begin{array}{l} (b) \\ 3 x + 2 y = 9................. (1) \\ 5 x + 4 y = 16............... (2) \\ (\begin{matrix} 3 & 2 \\ 5 & 4 \end{matrix}) (\begin{array}{l} x \\ y \end{array}) = (\begin{array}{l} 9 \\ 16 \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = (\begin{matrix} 2 & - 1 \\ - \frac{5}{2} & \frac{3}{2} \end{matrix}) (\begin{array}{l} 9 \\ 16 \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = (\begin{array}{l} (2) (9) + (- 1) (16) \\ (- \frac{5}{2}) (9) + (\frac{3}{2}) (16) \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = (\begin{array}{l} 18 - 16 \\ - \frac{45}{2} + 24 \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = (\begin{array}{l} 2 \\ \frac{3}{2} \end{array}) \\ x = 2, y = \frac{3}{2} \\ ∴ Price of a cucumber = RM2 \\ Price of a carrot = RM1 .50 \end{array}$

Question 8:
The inverse matrix of

$(\begin{matrix} 4 & - 1 \\ 2 & 5 \end{matrix}) is t (\begin{matrix} 5 & 1 \\ - 2 & n \end{matrix}) .$
(a) Find the value of n and of t.
(b) Write the following simultaneous linear equations as matrix equation:
4x – y = 7
2x + 5y = –2
Hence, using matrix method, calculate the value of x and of y.

Solution:

$\begin{array}{l} (a) \\ t (\begin{matrix} 5 & 1 \\ - 2 & n \end{matrix}) = {(\begin{matrix} 4 & - 1 \\ 2 & 5 \end{matrix})}^{- 1} \\ = \frac{1}{(4) (5) - (- 1) (2)} (\begin{matrix} 5 & 1 \\ - 2 & 4 \end{matrix}) \\ = \frac{1}{22} (\begin{matrix} 5 & 1 \\ - 2 & 4 \end{matrix}) \\ ∴ t = \frac{1}{22}, n = 4 \end{array}$

$\begin{array}{l} (b) \\ (\begin{matrix} 4 & - 1 \\ 2 & 5 \end{matrix}) (\begin{array}{l} x \\ y \end{array}) = (\begin{array}{l} 7 \\ - 2 \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = \frac{1}{22} (\begin{matrix} 5 & 1 \\ - 2 & 4 \end{matrix}) (\begin{array}{l} 7 \\ - 2 \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = \frac{1}{22} (\begin{array}{l} (5) (7) + 1 (- 2) \\ (- 2) (7) + (4) (- 2) \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = \frac{1}{22} (\begin{array}{l} 35 - 2 \\ - 14 - 8 \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = \frac{1}{22} (\begin{array}{l} 33 \\ - 22 \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = (\begin{array}{l} \frac{3}{2} \\ - 1 \end{array}) \\ ∴ x = \frac{3}{2}, y = - 1 \end{array}$