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

Long Question 5

Posted on May 18, 2020 by Myhometuition

Question 5 (10 marks):

\begin{array}{l} (a) Prove \sin (3 x + \frac{π}{6}) - \sin (3 x - \frac{π}{6}) = \cos 3 x \\ (b) Hence, \\ (i) solve the equation s i n (\frac{3 x}{2} + \frac{π}{6}) - \sin (\frac{3 x}{2} - \frac{π}{6}) = \frac{1}{2} for 0 \leq x \leq 2 π \\ and give your answer in the simplest fraction form in terms of π radian . \\ (i i) sketch the graph of y = s i n (3 x + \frac{π}{6}) - \sin (3 x - \frac{π}{6}) - \frac{1}{2} for 0 \leq x \leq π . \end{array}

Solution:

\begin{array}{l} (a) Left hand side, \\ \sin (3 x + \frac{π}{6}) - \sin (3 x - \frac{π}{6}) \\ = [\sin 3 x \cos \frac{π}{6} + \cos 3 x \sin \frac{π}{6}] - [\sin 3 x \cos \frac{π}{6} - \cos 3 x \sin \frac{π}{6}] \\ = 2 [\cos 3 x \sin \frac{π}{6}] \\ = 2 [\cos 3 x (\frac{1}{2})] \\ = \cos 3 x (right hand side) \end{array}

\begin{array}{l} (b) (i) \\ \sin (\frac{3 x}{2} + \frac{π}{6}) - \sin (\frac{3 x}{2} - \frac{π}{6}) = \frac{1}{2}, 0 \leq x \leq 2 π \\ \cos \frac{3 x}{2} = \frac{1}{2} \\ \frac{3 x}{2} = \frac{π}{3}, (2 π - \frac{π}{3}), (2 π + \frac{π}{3}) \\ \frac{3 x}{2} = \frac{π}{3}, \frac{5 π}{3}, \frac{7 π}{3} \\ x = \frac{2 π}{9}, \frac{10 π}{9}, \frac{14 π}{9} \end{array}

\begin{array}{l} (b) (i i) \\ y = s i n (3 x + \frac{π}{6}) - \sin (3 x - \frac{π}{6}) - \frac{1}{2} for 0 \leq x \leq π . \\ y = \cos 3 x - \frac{1}{2} \end{array}

Long Questions (Question 12)

Posted on May 18, 2020 by Myhometuition

Question 12 (10 marks):
(a) The mass of honeydews produced in a plantation is normally distributed with a mean of 0.8 kg and a standard deviation of 0.25 kg. The honeydews are being classified into three grades A, B and C according to their masses:

Grade A > Grade B > Grade C

(i) The minimum mass of a grade A honeydew is 1.2 kg.
If a honeydew is picked at random from the plantation, find the probability that the honeydew is of grade A.

(ii) Find the minimum mass, in kg, of grade B honeydew if 20% of the honeydews are of grade C.

(b) At the Shoot the Duck game booth at an amusement park, the probability of winning is 25%.
Jacky bought tickets to play n games. The probability for Jacky to win once is 10 times the probability of losing all games.

(i) Find the value of n.

(ii) Calculate the standard deviation of the number of wins.

Solution:
μ = 0.8 kg, σ = 0.25 kg

(a)(i)

\begin{array}{l} P (grade A) = P (X > 1.2) \\ = P (Z > \frac{1.2 - 0.8}{0.25}) \\ = P (Z > 1.6) \\ = 0.0548 \end{array}

(a)(ii)

\begin{array}{l} P (grade C) = 0.2 \\ P (X < m) = 0.2 \\ P (Z < \frac{m - 0.8}{0.25}) = 0.2 \\ P (Z < - 0.842) = 0.2 \\ \frac{m - 0.8}{0.25} = - 0.842 \\ m - 0.8 = - 0.2105 \\ m = 0.5895 \\ Minimum mass of grade B honeydew \\ is the same as the maximum mass of \\ grade C honeydew . \\ Minimum mass of grade B = 0.5895 kg \end{array}

(b)

\begin{array}{l} p = 0.25, X = B (n, 0.25) \\ P (X = r) = {}^{n}C_{r} p^{r} q^{n - r} \\ = {}^{n}C_{r} {(0.25)}^{r} {(0.75)}^{n - r} \end{array}

(b)(i)

\begin{array}{l} P (X = 1) = 10 P (X = 0) \\ {}^{n}C_{r} {(0.25)}^{1} {(0.75)}^{n - r} = 10 \times {}^{n}C_{0} {(0.25)}^{0} {(0.75)}^{n} \\ n \times 0.25 \times {(0.75)}^{n - r} = 10 \times 1 \times 1 \times {(0.75)}^{n} \\ 0.25 n \times \frac{{(0.75)}^{n - 1}}{{0.75}^{n}} = 10 \\ 0.25 n \times {0.75}^{- 1} = 10 \\ \frac{1}{4} n (\frac{4}{3}) = 10 \\ \frac{1}{3} n = 10 \\ n = 30 \end{array}

(b)(ii)

\begin{array}{l} n = 30, p = 0.25 \\ Standard deviation \\ = \sqrt{n p (1 - p)} \\ = \sqrt{30 \times 0.25 \times 0.75} \\ = 2.372 \end{array}

Long Questions (Question 11)

Posted on May 18, 2020 by Myhometuition

Question 11 (10 marks):
A study shows that the credit card balance of the customers is normally distributed as shown in Diagram 6.

(a)(i) Find the standard deviation.
(ii) If 30 customers are chosen at random, find the number of customers who have a credit card balance between RM1800 and RM3000.

(b) It is found that 25% of the customers have a credit card balance less than RM y.
Find the value of y.

Solution:
(a)(i)

\begin{array}{l} μ = 2870, x = 3770 \\ P (X > 3770) = 15.87 % \\ P (Z > \frac{3770 - 2870}{σ}) = 0.1587 \\ P (Z > 1.0) = 0.1587 \\ \frac{3770 - 2870}{σ} = 1.0 \\ σ = 900 \end{array}

(a)(ii)

\begin{array}{l} P (1800 < X < 3000) \\ = P (\frac{1800 - 2870}{900} < Z < \frac{3000 - 2870}{900}) \\ = P (- 1.189 < Z < 0.144) \\ = 1 - P (Z \leq - 1.189) - P (Z \geq 0.144) \\ = 1 - 0.1172 - 0.4427 \\ = 0.4401 \\ Number of customers = 0.4401 \times 30 \\ = 14 \end{array}

(b)

\begin{array}{l} μ = 2870, x = y \\ P (x < y) = 25 % \\ P (Z < \frac{y - 2870}{900}) = 0.25 \\ \frac{y - 2870}{900} = - 0.674 \\ y = 2263.40 \end{array}

Long Question 10

Posted on May 17, 2020 by Myhometuition

Question 10 (10 marks):
Diagram shows a curve y = 2x² – 18 and the straight line PQ which is a tangent to the curve at point K.

It is given that the gradient of the straight line PQ is 4.
(a) Find the coordinates of point K
(b) Calculate the area of the shaded region.

(c) When the region bounded by the curve, the x-axis and the straight line y = h is rotated through 180^o about the y-axis, the volume generated is 65π unit³.
Find the value of h.

Solution:
(a)

\begin{array}{l} y = 2 x^{2} - 18 \\ \frac{d y}{d x} = 4 x \\ Gradient of straight line P Q = 4 \\ 4 x = 4 \\ x = 1 \\ When x = 1, \\ y = 2 {(1)}^{2} - 18 = - 16 \\ Coordinates of K = (1, - 16) . \end{array}

(b)

\begin{array}{l} At x -axis, y = 0 \\ ∴ 2 x^{2} - 18 = 0 \\ 2 x^{2} = 18 \\ x^{2} = 9 \\ x = \pm 3 \\ The curve cuts the x -axis at (- 3, 0) and (3, 0) . \\ Area of shaded region \\ = Area of triangle - Area under the curve \\ = \frac{1}{2} (5 - 1) (16) - \int_{1}^{3} y d x \\ = 32 - \int_{1}^{3} (2 x^{2} - 18) d x \\ = 32 - | {[\frac{2 x^{3}}{3} - 18 x]}_{1}^{3} | \\ = 32 - | (\frac{2 {(3)}^{3}}{3} - 18 (3)) - (\frac{2 {(1)}^{3}}{3} - 18 (1)) | \\ = 32 - | (18 - 54 - \frac{2}{3} + 18) | \\ = 32 - | - 18 \frac{2}{3} | \\ = 32 - 18 \frac{2}{3} \\ = 13 \frac{1}{3} {units}^{2} \end{array}

(c)

\begin{array}{l} Volume generated = 65 π \\ π \int_{h}^{0} x^{2} d y = 65 π \\ π \int_{h}^{0} (\frac{y}{2} + 9) d x = 65 π \leftarrow \begin{array}{l} y = 2 x^{2} - 18 \\ x^{2} = \frac{y}{2} + 9 \end{array} \\ {[\frac{y^{2}}{4} + 9 y]}_{h}^{0} = 65 \\ 0 - (\frac{h^{2}}{4} + 9 h) = 65 \\ - \frac{h^{2}}{4} - 9 h = 65 \\ h^{2} + 36 h + 260 = 0 \\ (h + 10) (h + 26) = 0 \\ h = - 10 or h = - 26 (rejected) \\ Thus, h = - 10 \end{array}