Short Question 10 & 11

Posted on May 18, 2020 by Myhometuition

Question 10 (3 marks):
Diagram 3 shows vectors

\vec{O P}, \vec{O Q} and \vec{O M}

drawn on a square grid.

Diagram

\begin{array}{l} (a) Express \vec{O M} in the form h \underset{˜}{p} + k \underset{˜}{q}, \\ where h and k are constants . \\ (b) On the diagram 3, mark and label \\ the point N such that \vec{M N} + \vec{O Q} = 2 \vec{O P} . \end{array}

Solution:
(a)

\vec{O M} = \underset{˜}{p} + 2 \underset{˜}{q}

(b)

\begin{array}{l} \vec{M N} + \vec{O Q} = 2 \vec{O P} \\ \vec{M N} = 2 \vec{O P} - \vec{O Q} \\ = 2 \underset{˜}{p} - \underset{˜}{q} \end{array}

Question 11 (4 marks):

\begin{array}{l} A (2, 3) and B (- 2, 5) lie on a \\ Cartesian plane . \\ It is given that 3 \vec{O A} = 2 \vec{O B} + \vec{O C} . \\ Find \\ (a) the coordinates of C, \\ (b) | \vec{A C} | \end{array}

Solution:

\begin{array}{l} Given A (2, 3) and B (- 2, 5) \\ Thus, \vec{O A} = 2 \underset{˜}{i} + 3 \underset{˜}{j} and \vec{O B} = - 2 \underset{˜}{i} + 5 \underset{˜}{j} \end{array}

(a)

\begin{array}{l} 3 \vec{O A} = 2 \vec{O B} + \vec{O C} \\ \vec{O C} = 3 \vec{O A} - 2 \vec{O B} \\ = 3 (2 \underset{˜}{i} + 3 \underset{˜}{j}) - 2 (- 2 \underset{˜}{i} + 5 \underset{˜}{j}) \\ = 6 \underset{˜}{i} + 9 \underset{˜}{j} + 4 \underset{˜}{i} - 10 \underset{˜}{j} \\ = 10 \underset{˜}{i} - \underset{˜}{j} \\ Thus, coordinate of C is (10, - 1) \end{array}

(b)

\begin{array}{l} \vec{A C} = \vec{A O} + \vec{O C} \\ = - \vec{O A} + \vec{O C} \\ = - (2 \underset{˜}{i} + 3 \underset{˜}{j}) + 10 \underset{˜}{i} - \underset{˜}{j} \\ = - 2 \underset{˜}{i} + 10 \underset{˜}{i} - 3 \underset{˜}{j} - \underset{˜}{j} \\ = 8 \underset{˜}{i} - 4 \underset{˜}{j} \\ | \vec{A C} | = \sqrt{8^{2} + {(- 4)}^{2}} \\ = \sqrt{80} units \\ = \sqrt{16 \times 5} units \\ = 4 \sqrt{5} units \end{array}

Short Question 8 & 9

Posted on May 18, 2020 by Myhometuition

Question 8 (3 marks):
Diagram shows vectors

\vec{A B}, \vec{A C} and \vec{A D}

drawn on a square grid with sides of 1 unit.

Diagram

\begin{array}{l} (a) Find | - \vec{B A} | . \\ (b) Given \vec{A B} = \underset{˜}{b} and \vec{A C} = \underset{˜}{c}, \\ express in terms of \underset{˜}{b} and \underset{˜}{c} \\ (i) \vec{B C}, \\ (i i) \vec{A D} \end{array}

Solution:
(a)

| - \vec{B A} | = \sqrt{3^{2} + 4^{2}} = 5 units

(b)(i)

\begin{array}{l} \vec{B C} = \vec{B A} + \vec{A C} \\ = - \underset{˜}{b} + \underset{˜}{c} \\ = \underset{˜}{c} - \underset{˜}{b} \end{array}

(b)(ii)

\begin{array}{l} \vec{A D} = \vec{A B} + \vec{B D} \\ = \underset{˜}{b} + 2 \vec{B C} \\ = \underset{˜}{b} + 2 (\underset{˜}{c} - \underset{˜}{b}) \\ = 2 \underset{˜}{c} - \underset{˜}{b} \end{array}

Question 9 (3 marks):
Diagram shows a trapezium ABCD.

Diagram

\begin{array}{l} Given \underset{˜}{p} = (\begin{array}{l} 3 \\ 4 \end{array}) and \underset{˜}{q} = (\begin{array}{l} k - 1 \\ 2 \end{array}), \\ where k is a constant, find value of k . \end{array}

Solution:

\begin{array}{l} \underset{˜}{p} = m \underset{˜}{q} \\ (\begin{array}{l} 3 \\ 4 \end{array}) = m (\begin{array}{l} k - 1 \\ 2 \end{array}) \\ (\begin{array}{l} 3 \\ 4 \end{array}) = (\begin{array}{l} m k - m \\ 2 m \end{array}) \\ m k - m = 3 .......... (1) \\ 2 m = 4 ................ (2) \\ From (2) : \\ 2 m = 4 \\ m = 2 \\ Substitute m = 2 into (1) : \\ 2 k - 2 = 3 \\ 2 k = 3 + 2 \\ 2 k = 5 \\ k = \frac{5}{2} \end{array}

Short Questions (Question 7 – 9)

Posted on May 18, 2020 by Myhometuition

Question 7 (2 marks):
Table shows the information about a set of data.

Table

State
(a) the value of p if m = 20,
(b) the value of q if p = 2.5.

Solution:
(a)
New standard deviation = original standard deviation × p
20 = 5 × p
p = 4

(b)
New median = [original median × p] + 1
q = 2p × 1
q = 2(2.5) + 1
q = 5 + 1
q = 6

Question 8 (3 marks):
Table shows the distribution of scores obtained by a group of students in a competition.

Table

(a) State the minimum value of x if the mode score is 4.
(b) Find the mean score of the distribution if x = 1.

Solution:
(a)
Minimum value of x = 8

(b)

\begin{array}{l} Mean \\ = \frac{1 (3) + 2 (6) + 3 (7) + 4 (1) + 5 (1)}{3 + 6 + 7 + 1 + 1} \\ = \frac{45}{18} \\ = 2.5 \end{array}

Question 9 (4 marks):
Table shows the distribution of marks for 40 students in an Additional Mathematics test. The number of students for the class interval 40 – 59 is not stated.

Table

(a) State the modal class.
(b) Puan Zainon, the subject teacher, intends to give a reward to the top ten students. Those students who achieve the minimum mark in the top ten placing will be considered to receive the reward. Elina obtains 74 marks.
Does Elina qualify to be considered to receive the reward? Give your reason.

Solution:
(a)
4 + 10 + x + 8 + 7 = 40
x + 29 = 40
x = 11

Modal class = 40 – 59

(b)

\begin{array}{l} The top ten placings are T_{31}, T_{32}, T_{33}, ... T_{40} \\ T_{31} = 59.5 + \frac{5}{8} (79.5 - 59.5) \\ = 59.5 + 12.5 \\ = 72 \\ A student has to achieve a minimum mark \\ of 72 . \\ Elina qualifies for the reward because her \\ marks > 72 marks . \end{array}

Short Questions (Question 3 & 4)

Posted on May 18, 2020 by Myhometuition

Question 3 (4 marks):
Diagram shows a standard normal distribution graph.

Diagram

The probability represented by the area of the shaded region is 0.2881.
(a) Find the value of h.
(b) X is a continuous random variable which is normally distributed with a mean, μ and a variance of 16.
Find the value of μ if the z-score of X = 58.8 is h.

Solution:
(a)
P(X < h) = 0.5 – 0.2881
P(X < h) = 0.2119
P(X < –0.8) = 0.2119
h = –0.8

(b)

\begin{array}{l} X = 58.8 \\ \frac{X - μ}{σ} = \frac{58.8 - μ}{σ} \\ Z = \frac{58.8 - μ}{4} \\ h = \frac{58.8 - μ}{4} \\ - 0.8 = \frac{58.8 - μ}{4} \\ - 3.2 = 58.8 - μ \\ μ = 58.8 + 3.2 \\ μ = 62 \end{array}

Question 4 (4 marks):
A voluntary body organizes a first aid course 4 times per month, every Saturday from March until September.
[Assume there are four Saturdays in every month]

Salmah intends to join the course but she might need to spare a Saturday per month to accompany her mother to the hospital. The probability that Salmah will attend the course each Saturday is 0.8. Salmah will be given a certificate of monthly attendance if she can attend the course at least 3 times a month.

(a) Find the probability that Salmah will be given the certificate of monthly attendance.

(b) Salmah will qualify to sit for the first aid test if she obtains more than 5 certificates of monthly attendance.
Find the probability that Salmah qualifies to take the first aid test.

Solution:
(a)

\begin{array}{l} P (X = r) = {}^{n}C_{r} p^{r} q^{n - r} \\ p = 0.8, q = 0.2, n = 4, r = 3, 4 \\ P (X \geq 3) \\ = P (X = 3) + P (X = 4) \\ = {}^{4}C_{3} {(0.8)}^{3} {(0.2)}^{1} + {}^{4}C_{4} {(0.8)}^{4} {(0.2)}^{0} \\ = 0.4096 + 0.4096 \\ = 0.8192 \end{array}

(b)

\begin{array}{l} P (X = r) = {}^{n}C_{r} p^{r} q^{n - r} \\ p = 0.8192, q = 0.1808, n = 7, r = 6, 7 \\ P (X > 5) \\ = P (X = 6) + P (X = 7) \\ = {}^{7}C_{6} {(0.8192)}^{6} {(0.1808)}^{1} + \\ {}^{7}C_{7} {(0.8192)}^{7} {(0.1808)}^{0} \\ = 0.3825 + 0.2476 \\ = 0.6301 \end{array}

Short Questions (Question 1 & 2)

Posted on May 18, 2020 by Myhometuition

Question 1 (2 marks):
Diagram shows a probability distribution graph for a random variable X, X ~ N(μ, σ²).

Diagram

It is given that AB is the axis of symmetry of the graph.
(a) State the value of μ.
(b) If the area of the shaded region is 0.38, state the value of P(5 ≤ X ≤ 15).

Solution:
(a)
μ = 0

(b)
P(10 ≤ X ≤ 15)
= 0.5 – 0.38
= 0.12

P(5 ≤ X ≤ 10)
= P(10 ≤ X ≤ 15)
= 0.12

Thus P(5 ≤ X ≤ 15)
= 0.12 + 0.12
= 0.24

Question 2 (3 marks):
Diagram shows the graph of binomial distribution X ~ B(3, p).

Diagram

(a) Express P(X = 0) + P(X > 2) in terms of a and b.
(b) Find the value of p.

Solution:
(a)
P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 1
P(X = 0) + a + b + P(X = 3) = 1
P(X = 0) + P(X = 3) = 1 – a – b
P(X = 0) + P(X > 2) = 1 – a – b

(b)

\begin{array}{l} P (X = 0) = \frac{27}{343} \\ ^{3} C_{0} (p^{0}) {(1 - p)}^{3} = \frac{27}{343} \\ 1 \times 1 \times {(1 - p)}^{3} = {(\frac{3}{7})}^{3} \\ 1 - p = \frac{3}{7} \\ p = \frac{4}{7} \end{array}

(Long Questions) – Question 10

Posted on May 18, 2020 by Myhometuition

Question 11 (10 marks):
Solution by scale drawing is not accepted.
Diagram shows a quadrilateral PQRS on a horizontal plane.
VQSP is a pyramid such that PQ = 12 m and V is 5 m vertically above P.
Find
(a) ∠QSR,
(b) the length, in m, of QS,
(c) the area, in m², of inclined plane QVS.

Solution:
(a)

\begin{array}{l} \frac{\sin ∠ Q S R}{20.5} = \frac{\sin 64^{o}}{22} \\ \sin ∠ Q S R = \frac{\sin 64^{o}}{22} \times 20.5 \\ \sin ∠ Q S R = 0.8375 \\ ∠ Q S R = 56^{o} 52' \end{array}

(b)

\begin{array}{l} ∠ Q R S = 180^{o} - 64^{o} - 56^{o} 52' \\ = 59^{o} 8' \\ \frac{Q S}{\sin 59^{o} 8'} = \frac{22}{\sin 64^{o}} \\ Q S = \frac{22}{\sin 64^{o}} \times \sin 59^{o} 8' \\ Q S = 21.01 m \end{array}

(c)

\begin{array}{l} Q V^{2} = P Q^{2} + V P^{2} \\ Q V = \sqrt{12^{2} + 5^{2}} \\ Q V = 13 m \\ S V^{2} = P S^{2} + V P^{2} \\ S V = \sqrt{10^{2} + 5^{2}} \\ S V = \sqrt{125} m \\ Q S = 21.01 m \\ {21.01}^{2} = 13^{2} + {(\sqrt{125})}^{2} - 2 (13) (\sqrt{125}) \cos θ \\ 26 (\sqrt{125}) \cos θ = 169 + 125 - 441.42 \\ \cos θ = \frac{169 + 125 - 441.42}{26 (\sqrt{125})} \\ \cos θ = - 0.5071 \\ θ = 120^{o} 28' \\ Area of Q V S \\ = \frac{1}{2} \times 13 \times \sqrt{125} \times \sin 120^{o} 28' \\ = 62.64 m^{2} \end{array}

(Long Questions) – Question 9

Posted on May 18, 2020 by Myhometuition

Question 9 (10 marks):
Solution by scale drawing is not accepted.
Diagram shows a transparent prism with a rectangular base PQRS. The inclined surface PQUT is a square with sides 12 cm and the inclined surface RSTU is a rectangle. PTS is a uniform cross section of the prism. QST is a green coloured plane in the prism.

It is given that ∠PST = 37^o and ∠TPS = 45^o.
Find
(a) the length, in cm, of ST,
(b) the area, in cm², of the green coloured plane.
(c) the shortest length, in cm, from point T to the straight line QS.

Solution:
(a)

\begin{array}{l} \frac{S T}{\sin 45^{o}} = \frac{12}{\sin 37^{o}} \\ S T = \frac{12}{\sin 37^{o}} \times \sin 45^{o} \\ S T = 14.1 cm \end{array}

(b)

\begin{array}{l} Q T^{2} = Q P^{2} + P T^{2} \\ Q T^{2} = 12^{2} + 12^{2} \\ Q T = \sqrt{12^{2} + 12^{2}} = 16.97 cm \\ ∠ P T S = 180^{o} - 45^{o} - 37^{o} = 98^{o} \\ \frac{P S}{\sin 98^{o}} = \frac{12}{\sin 37^{o}} \\ P S = \frac{12}{\sin 37^{o}} \times \sin 98^{o} \\ P S = 19.75 cm \\ Q S^{2} = Q P^{2} + P S^{2} \\ Q S = \sqrt{Q P^{2} + P S^{2}} \\ Q S = \sqrt{12^{2} + {19.75}^{2}} \\ Q S = 23.11 cm \end{array}

\begin{array}{l} Q S^{2} = Q T^{2} + S T^{2} - 2 (Q T) (S T) \cos ∠ Q T S \\ {23.11}^{2} = {16.97}^{2} + {14.1}^{2} - 2 (16.97) (14.1) \cos ∠ Q T S \\ \cos ∠ Q T S = \frac{{16.97}^{2} + {14.1}^{2} - {23.11}^{2}}{2 (16.97) (14.1)} \\ \cos ∠ Q T S = - \frac{47.28}{478.55} \\ ∠ Q T S = {95.67}^{o} \\ Thus, area of green coloured plane Q T S \\ = \frac{1}{2} (16.97) (14.1) \sin {95.67}^{o} \\ = 119.05 {cm}^{2} \end{array}

(c)

\begin{array}{l} Let the shortest length from point T \\ to the straight line Q S is h . \\ Area of Δ Q T S = 119.05 \\ \frac{1}{2} (h) (23.11) = 119.05 \\ h = 10.3 cm \end{array}

(Long Questions) – Question 6

Posted on May 18, 2020 by Myhometuition

Question 6 (10 marks):
Table 2 shows the information related to four ingredients, J, K, L and M, used in the production of a type of energy drinks.

The production cost for this energy drinks is RM47600 in the year 2017.
(a) If the price of ingredient K in the year 2013 is RM4.20, find its price in the year 2017.

(b) Percentage of usage for several ingredients was given in Table 2.
Calculate the corresponding production cost in the year 2013.

(c) The production cost is expected to increase by 50% from the year 2017 to the year 2019.
Calculate the percentage of changes in production cost from the year 2013 to the year 2019.

Solution:
(a)

\begin{array}{l} \frac{P_{2017}}{P_{2013}} \times 100 = 120 \\ \frac{P_{2017}}{4.20} \times 100 = 120 \\ P_{2017} = \frac{4.20}{100} \times 120 \\ P_{2017} = RM 5.04 \end{array}

(b)

\begin{array}{l} Percentage of usage of ingredient L \\ = (100 - 10 - 10 - 50) % \\ = 30 % \\ Composite index for the production cost in \\ the year 2017 based on the year 2013, \bar{I} \\ \bar{I} = \frac{\sum I W}{\sum W} \\ \bar{I} = \frac{(140) (10) + (120) (10) + (160) (30) + (90) (50)}{10 + 10 + 30 + 50} \\ \bar{I} = \frac{11900}{100} \\ \bar{I} = 119 \\ \frac{P_{2017}}{P_{2013}} \times 100 = 119 \\ \frac{47600}{P_{2013}} \times 100 = 119 \\ P_{2013} = 47600 \times \frac{100}{119} \\ P_{2013} = RM40000 \\ The production cost in the year 2013 \\ is RM40000 . \end{array}

(c)

\begin{array}{l} year 2013 \overset{+ 19 %}{\to} year 2017 \overset{+ 50 %}{\to} year 2019 \\ ∴ Expected composite index for the year 2019 \\ based on the year 2013, \\ \bar{I} = 119 \times \frac{150}{100} = 178.5 \\ Percentage of changes \\ = [178.5 - 100] % \\ = 78.5 % \end{array}

(Long Questions) – Question 5

Posted on May 18, 2020 by Myhometuition

Question 5 (10 marks):
Table 2 shows the prices and the price indices of three types of ingredients P, Q and R, used in the production of a type of instant noodle.

(a) The price of ingredient Q increased by 20% from the year 2014 to the year 2016.
(i) State the value of x.
(ii) Find the value of y.

(b) Calculate the composite index for the cost of making the instant noodles for the year 2016 based on the year 2014.

(c) It is given that the composite index for the cost of making the instant noodles increased by 40% from the year 2012 to the year 2016.

(i) Calculate the composite index for the cost of making the instant noodles in the year 2014 based on the year 2012.

(ii) The cost of making a packet of instant noodle is 10 sen in the year 2012.
Find the maximum number of packet of instant noodles that can be produced using an allocation of RM80 in the year 2016.

Solution:
(a)(i)
x = 120

(a)(ii)

\begin{array}{l} \frac{P_{2016}}{P_{2014}} \times 100 = 120 \\ \frac{3.00}{y} \times 100 = 120 \\ y = \frac{3.00}{120} \times 100 \\ y = 2.50 \end{array}

(b)

\begin{array}{l} Composite index for the cost of making the instant noodles \\ in the year 2016 based on the year 2014, \bar{I} \\ \bar{I} = \frac{\sum I W}{\sum W} \\ \bar{I} = \frac{(132.8) (50) + (120) (20) + (190) (1)}{50 + 20 + 1} \\ \bar{I} = \frac{9230}{71} \\ \bar{I} = 130 \end{array}

(c)(i)

\begin{array}{l} Given \frac{I_{2016}}{I_{2012}} \times 100 = 140 \to \frac{I_{2016}}{I_{2012}} = \frac{140}{100} \\ and \frac{I_{2016}}{I_{2014}} \times 100 = 130 \to \frac{I_{2016}}{I_{2014}} = \frac{130}{100} \\ Composite index for the cost of making the instant noodle \\ in the year 2014 based on the year 2012, \\ \bar{I} = \frac{I_{2014}}{I_{2012}} \times 100 \\ \bar{I} = \frac{I_{2014}}{I_{2016}} \times \frac{I_{2016}}{I_{2012}} \times 100 \\ \bar{I} = \frac{100}{130} \times \frac{140}{100} \times 100 \\ \bar{I} = 107.69 \end{array}

(c)(ii)

\begin{array}{l} \frac{P_{2016}}{P_{2012}} \times 100 = 140 \\ \frac{P_{2016}}{10} \times 100 = 140 \\ P_{2016} = \frac{140 \times 10}{100} \\ P_{2016} = 14 sen \\ Number of pack of instant noodles \\ = \frac{RM80}{RM 0.14} \\ = 571.4 \\ Maximum number of pack of instant noodles = 571. \end{array}

Long Question 6

Posted on May 18, 2020 by Myhometuition

Question 6 (10 marks):
(a) Prove that 2 tan x cos² x = sin 2x.

(b) Hence, solve the equation 4 tan x cos² x = 1 for 0 ≤ x ≤ 2π.

(c)(i) Sketch the graph of y = sin 2x for 0 ≤ x ≤ 2π.

(c)(ii) Hence, using the same axes, sketch a suitable straight line to find the number of solutions for the equation 4π tan x cos² x = x – 2π for 0 ≤ x ≤ 2π.
State the number of solutions.

Solution:
(a)

\begin{array}{l} 2 \tan x \cos^{2} x = \sin 2 x \\ Left hand side \\ = 2 \tan x \cos^{2} x \\ = 2 \times \frac{\sin x}{\cos x} \times \cos^{2} x \\ = 2 \sin x \cos x \\ = \sin 2 x \\ = Right hand side (Proven) \end{array}

(b)

\begin{array}{l} 4 \tan x \cos^{2} x = 1, 0 \leq x \leq 2 π \\ 2 (2 \tan x \cos^{2} x) = 1 \\ 2 \sin 2 x = 1 \\ \sin 2 x = \frac{1}{2} \\ Basic angle = \frac{π}{6} \\ 2 x = \frac{π}{6}, (π - \frac{π}{6}), (2 π + \frac{π}{6}), (3 π - \frac{π}{6}) \\ 2 x = \frac{π}{6}, \frac{5 π}{6}, \frac{13 π}{6}, \frac{17 π}{6} \\ x = \frac{π}{12}, \frac{5 π}{12}, \frac{13 π}{12}, \frac{17 π}{12} \end{array}

(c)(i)
y = sin 2x, 0 ≤ x ≤ 2π.

(c)(ii)

\begin{array}{l} 4 π \tan x \cos^{2} x = x - 2 π \\ 2 π (2 \tan x \cos^{2} x) = x - 2 π \\ 2 π \sin 2 x = x - 2 π \\ \sin 2 x = \frac{x}{2 π} - \frac{2 π}{2 π} \\ \sin 2 x = \frac{x}{2 π} - 1 \\ y = \frac{x}{2 π} - 1 \end{array}

Number of solutions = 4