Long Questions (Question 7)

Posted on May 17, 2020 by Myhometuition

Question 7:
In a boarding school entry exam, 300 students sat for a mathematics test. The marks obtained follow a normal distribution with a mean of 56 and a standard deviation of 8.

(a) Find the number of students who pass the test if the passing mark is 40.

(b) If 12% of the students pass the test with grade A, find the minimum mark to obtain grade A.

Solution:

\begin{array}{l} Let X = marks obtained by students \\ X ~ N (56, 8^{2}) \\ (a) P (X \geq 40) = P (Z \geq \frac{40 - 56}{8}) \\ = P (Z \geq - 2) \\ = 1 - P (Z \geq 2) \\ = 1 - 0.02275 \\ = 0.9773 \\ Number of students who pass the test \\ = 0.9773 \times 300 \\ = 293 \\ (b) Let the minimum mark to obtain grade A be k \\ P (X \geq k) = 0.12 \\ P (Z \geq \frac{k - 56}{8}) = 0.12 \\ \frac{k - 56}{8} = 1.17 \\ k = (1.17) (8) + 56 \\ = 65.36 \end{array}

Thus, the minimum mark to obtain grade A is 66.

Long Questions (Question 6)

Posted on May 17, 2020 by Myhometuition

Question 6:

The masses of tomatoes in a farm are normally distributed with a mean of 130 g and standard deviation of 16 g. Tomato with weight more than 150 g is classified as grade ‘A’.

(a) A tomato is chosen at random from the farm. Find the probability that the tomato has a weight between 114 g and 150 g.

(b) It is found that 132 tomatoes in this farm are grade ‘A’. Find the total number of tomatoes in the farm.

Solution:

µ = 130

σ = 16

(a)

\begin{array}{l} P (114 < X < 150) \\ = P (\frac{114 - 130}{16} < Z < \frac{150 - 130}{16}) \\ = P (- 1 < Z < 1.25) \\ = 1 - P (Z > 1) - P (Z > 1.25) \\ = 1 - 0.1587 - 0.1056 \\ = 0.7357 \end{array}

(b)

Probability of getting grade ‘A’ tomatoes,

P (X > 150) = P (Z > 1.25)

= 0.1056

\begin{array}{l} Lets total number of tomatoes = N \\ 0.1056 \times N = 132 \\ N = \frac{132}{0.1056} \\ N = 1250 \end{array}

Long Questions (Question 3 & 4)

Posted on May 17, 2020 by Myhometuition

Question 3:

The result of a study shows that 20% of students failed the Form 5 examination in a school. If 8 students from the school are chosen at random, calculate the probability that

(a) exactly 2 of them who failed,

(b) less than 3 of them who failed.

Solution:
(a)

p = 20% = 0.2,

q = 1 – 0.2 = 0.8

X ~ B (8, 0.2)

P (X = 2)

C_{2}

(0.2)² (0.8)⁶

= 0.2936

(b)

P (X < 3)

= P (X = 0) + P (X = 1) + P (X = 2)

C_{0}

(0.2)⁰(0.8)⁸+

C_{1}

(0.2)¹(0.8)⁷ +

C_{2}

(0.2)²(0.8)⁶

= 0.16777 + 0.33554 + 0.29360

= 0.79691

Question 4:
In a survey carried out in a particular district, it is found that three out of five families own a LCD television.
If 10 families are chosen at random from the district, calculate the probability that at least 8 families own a LCD television.

Solution:

\begin{array}{l} Let X be the random variable representing the number of families \\ who own a LCD television . \\ X ~ B (n, p) \\ X ~ B (10, \frac{3}{5}) \\ p = \frac{3}{5} = 0.6 \\ q = 1 - 0.6 = 0.4 \\ P (X = r) = {}^{n}c_{r} . p^{r} . q^{n - r} \\ P (at least 8 families own a LCD television) \\ P (X \geq 8) \\ = P (X = 8) + P (X = 9) + P (X = 10) \\ = {}^{10}C_{8} {(0.6)}^{8} {(0.4)}^{2} + {}^{10}C_{9} {(0.6)}^{9} {(0.4)}^{1} + {}^{10}C_{10} {(0.6)}^{10} {(0.4)}^{0} \\ = 0.1209 + 0.0403 + 0.0060 \\ = 0.1672 \end{array}

Short Questions (Question 4 & 5)

Posted on May 17, 2020 by Myhometuition

Question 4:
The events A and B are not independent.

\begin{array}{l} Given P (A) = \frac{3}{5}, P (B) = \frac{1}{4} and P (A \cup B) = \frac{1}{5}, find \\ (a) P [(A \cup B)'], \\ (b) P (A \cap B) . \end{array}

Solution:
(a)

\begin{array}{l} P [(A \cup B)'] = 1 - P (A \cup B) \\ = 1 - \frac{1}{5} \\ = \frac{4}{5} \end{array}

(b)

\begin{array}{l} P (A \cap B) = P (A) + P (B) - P (A \cup B) \\ = \frac{3}{5} + \frac{1}{4} - \frac{1}{5} \\ = \frac{13}{20} \end{array}

Question 5 (3 marks):
A biased cube dice is thrown. The probability of getting the number ‘4’ is

\frac{1}{16}

and the probabilities of getting other than ‘4’ are equal to each other.
If the dice is thrown twice, find the probability of getting two different numbers.
Give your answer in the simplest fraction form.

Solution:

\begin{array}{l} P (1) + P (2) + P (3) + P (4) + P (5) + P (6) = 1 \\ Given probabilities of getting numbers are equal . \\ x + x + x + \frac{1}{16} + x + x = 1 \\ 5 x = 1 - \frac{1}{16} \\ 5 x = \frac{15}{16} \\ x = \frac{3}{16} \\ P (Same numbers) \\ = P (1, 1) + P (2, 2) + P (3, 3) \\ + P (4, 4) + P (5, 5) + P (6, 6) \\ = (x \times x) + (x \times x) + (x \times x) + \\ (\frac{1}{16} \times \frac{1}{16}) + (x \times x) + (x \times x) \\ = 5 x^{2} + \frac{1}{256} \\ = 5 {(\frac{3}{16})}^{2} + \frac{1}{256} \\ = \frac{23}{128} \\ P (Two different numbers) \\ = 1 - \frac{23}{128} \\ = \frac{105}{128} \end{array}

Short Question 7 & 8

Posted on May 17, 2020 by Myhometuition

Question 7:
Six members of a committee of a school are to be selected from 6 male teachers, 4 female teachers and a male principal. Find the number of different committees that can be formed if
(a) the principal is the chairman of the committee,
(b) there are exactly 2 females in the committee,
(c) there are not more than 4 males in the committee.

Solution:
(a)
If the principal is the chairman of the committee, the remaining number of committee is 5 members.
Hence, the number of different committees that can be formed from the remaining 6 male teachers and 4 female teachers

\begin{array}{l} =^{10} C_{5} \\ = 252 \end{array}

(b)

\begin{array}{l} Exactly 2 females in the committee \\ =^{4} C_{2} \times^{7} C_{4} \\ = 210 \end{array}

(c)

\begin{array}{l} There are not more than 4 males in the committee \\ = 4 males 2 females + 3 males 3 females + 2 males 4 females \\ =^{7} C_{4} \times^{4} C_{2} +^{7} C_{3} \times^{4} C_{3} +^{7} C_{2} \times^{4} C_{4} \\ = 210 + 140 + 21 \\ = 371 \end{array}

Question 8:

The diagram below shows five cards of different letters.

R E A C T

(a) Calculate the number of arrangements, in a row, of all the cards.

(b) Calculate the number of these arrangements in which the letters E and A are side by side.

Solution:
(a) Number of arrangements = 5! = 120

(b)
Step 1
If the letters ‘E ’ and ‘A’ have to be placed side by side, they will be considered as one item.
Together with the letters ‘R ’, ‘C ’ and ‘T ’, there are altogether 4 items.

EA R C T

Number of arrangements = 4!

Step 2

The letters ‘E ’ and ‘A ’ can also be arranged among themselves in their group.
Number of arrangements = 2!

Hence, number of arrangements of all the letters of the word ‘REACT’ in which the letters E and A have to be side by side
= 4! × 2!
= 24 × 2
= 48

Long Question 4

Posted on May 17, 2020 by Myhometuition

Question 4:

Find all the angles between 0° and 360° which satisfy the following equations:

(a) 2 sin ( 2x – 50^o) = –1

(b) 15 sin²x = sin x + 4 sin 30^o

(c) 7 sin x cos x = 1

Solution:
(a)

2 sin ( 2x – 50^o) = –1

sin ( 2x – 50^o) = – ½

basic angle ( 2x – 50^o) = –30^o ← (sin is negative at third and fourth quadrants)

2x – 50^o = –30^o, 180^o+ 30^o, 360^o – 30^o, 360^o + 180^o+ 30^o

← (Take the angles in the range of 0^o ≤ x ≤ 720^o, which in 2 complete revolutions)

2x – 50^o = –30^o, 210^o, 330^o, 570^o

2x = 20^o, 260^o, 380^o, 620^o

x = 10^o, 130^o, 190^o, 310^o

(b)

15 sin²x = sin x + 4 sin 30^o

15 sin²x = sin x + 4 (½) ← (sin 30^o = ½)

15 sin²x = sin x + 2

15 sin²x – sin x – 2 = 0

(5 sin x – 2)(3 sin x + 1) = 0

sin x =

\frac{2}{5}

or sin x = – ⅓

When sin x =

\frac{2}{5}

Basic angle x = 23º 35’

x = 23º 35’, 180º – 23º 35’

x = 23º 35’, 156º 25’

When sin x = – ⅓ ← (sin is negative at third and fourth quadrants)

Basic angle x = 19º 28’

x = 180º + 19º 28’, 360º – 19º 28’

x = 199º 28’, 340º 32’

Hence x = 23º 35’, 156º 25’, 199º 28’, 340º 32’.

(c)

7 sin x cos x = 1

sin x cos x =

\frac{1}{7}

2 sin x cos x =

\frac{2}{7}

← ( × 2 for both sides)

sin 2x =

\frac{2}{7}

sin 2x = 0.2857

Basic angle x = 16º 36’

2x = 16º 36’, 180º – 16º 36’, 360º + 16º 36’, 360º + 180º – 16º 36’

2x = 16º 36’, 163º 24’, 376º 36’, 523º 24’

x = 8º 18’, 81º 42’, 188º 18’, 261º 42’

Hence x = 8º 18’, 81º 42’, 188º 18’, 261º 42’.

Long Question 3

Posted on May 17, 2020 by Myhometuition

Question 3:

(a) Prove that

\frac{2 \tan x}{2 - \sec^{2} x} = \tan 2 x .

(b)(i) Sketch the graph of y = – tan 2x for 0 ≤ x ≤ π .

(b)(ii) Hence, by drawing a suitable straight line on the same axes, find the number of solutions satisfying the equation

\frac{3 x}{π} + \frac{2 \tan x}{2 - \sec^{2} x} = 0

for 0 ≤ x ≤ π .

State the number of solutions.

Solution:

(a)

\begin{array}{l} \frac{2 \tan x}{2 - \sec^{2} x} = \tan 2 x \\ L H S : \\ \frac{2 \tan x}{2 - \sec^{2} x} = \frac{2 \tan x}{2 - (1 + \tan^{2} x)} \\ = \frac{2 \tan x}{2 - \tan^{2} x} \\ = \tan 2 x (RHS) \end{array}

(b)(i)

(b)(ii)

\begin{array}{l} \frac{3 x}{π} + \frac{2 \tan x}{2 - \sec^{2} x} = 0 \\ \frac{3 x}{π} + \tan 2 x = 0 \leftarrow from part (a) \\ - \tan 2 x = \frac{3 x}{π} \\ y = \frac{3 x}{π} \\ The suitable straight line to sketch is y = \frac{3 x}{π} . \end{array}

When x = 0, y = 0.

When x = π, y = 3.

Number of solutions = 3

Long Question 8

Posted on May 16, 2020 by Myhometuition

Question 8:
Diagram below shows quadrilateral OPQR. The straight line PR intersects the straight line OQ at point S.

\begin{array}{l} It is given that \vec{O P} = 7 \underset{˜}{x}, \vec{O R} = 5 \underset{˜}{y}, P S : S R = 3 : 1 and \vec{O R} is parallel to \vec{P Q} . \\ (a) Express in terms of \underset{˜}{x} and \underset{˜}{y}, \\ (i) \vec{P R} \\ (ii) \vec{O S} \\ (b) Using \vec{P Q} = m \vec{O R} and \vec{S Q} = n \vec{O S}, where m and n are constants, \\ Find the value of m and of n . \\ (c) Given that | \underset{˜}{y} | = 4 units and the area of O R S {is 50 cm}^{2}, find the \\ perpendicular distance from point S to O R . \end{array}

Solution:
(a)(i)

\begin{array}{l} \vec{P R} = \vec{P O} + \vec{O R} \\ = - 7 \underset{˜}{x} + 5 \underset{˜}{y} \end{array}

(a)(ii)

\begin{array}{l} \vec{O S} = \vec{O P} + \vec{P S} \\ = 7 \underset{˜}{x} + \frac{3}{4} \vec{P R} \\ = 7 \underset{˜}{x} + \frac{3}{4} (- 7 \underset{˜}{x} + 5 \underset{˜}{y}) \\ = 7 \underset{˜}{x} - \frac{21}{4} \underset{˜}{x} + \frac{15}{4} \underset{˜}{y} \\ = \frac{7}{4} \underset{˜}{x} + \frac{15}{4} \underset{˜}{y} \end{array}

(b)

\begin{array}{l} \vec{P S} = \vec{P Q} - \vec{S Q} \\ \frac{3}{4} \vec{P R} = m \vec{O R} - n \vec{O S} \\ \frac{3}{4} (- 7 \underset{˜}{x} + 5 \underset{˜}{y}) = m (5 \underset{˜}{y}) - n (\frac{7}{4} \underset{˜}{x} + \frac{15}{4} \underset{˜}{y}) \\ - \frac{21}{4} \underset{˜}{x} + \frac{15}{4} \underset{˜}{y} = 5 m \underset{˜}{y} - \frac{7}{4} n \underset{˜}{x} - \frac{15}{4} m \underset{˜}{y} \\ - \frac{21}{4} \underset{˜}{x} + \frac{15}{4} \underset{˜}{y} = - \frac{7}{4} n \underset{˜}{x} + \frac{5}{4} m \underset{˜}{y} \\ - 21 \underset{˜}{x} + 15 \underset{˜}{y} = - 7 n \underset{˜}{x} + 5 m \underset{˜}{y} \\ 7 n = 21 \\ n = 3 \\ 5 m = 15 \\ m = 3 \end{array}

(c)

\begin{array}{l} Area of Δ O R S = 50 \\ \frac{1}{2} \times (5 \underset{˜}{y}) \times t = 50 \\ \frac{1}{2} \times 5 (4) \times t = 50 \\ 10 t = 50 \\ t = 5 \\ ∴ Perpendicular distance from point S to O R = 5 units . \end{array}

Long Question 7

Posted on May 16, 2020 by Myhometuition

Question 7:
Diagram below shows quadrilateral OPBC. The straight line AC intersects the straight line PQ at point B.

\begin{array}{l} It is given that \vec{O P} = \underset{˜}{a}, \vec{O Q} = \underset{˜}{b}, \vec{O A} = 4 \vec{A P}, \vec{O C} = 3 \vec{O Q}, \vec{P B} = h \vec{P Q} and \\ \vec{A B} = k \vec{A C} . \\ (a) Express \vec{O B} in terms of h, \underset{˜}{a} and \underset{˜}{b} . \\ (b) Express \vec{O B} in terms of k, \underset{˜}{a} and \underset{˜}{b} . \\ (c)(i) Find the value of h and of k . \\ (ii) Hence, state \vec{O B} in terms of \underset{˜}{a} and \underset{˜}{b} . \end{array}

Solution:
(a)

\begin{array}{l} \vec{O B} = \vec{O P} + \vec{P B} \\ = \underset{˜}{a} + h \vec{P Q} \\ = \underset{˜}{a} + h (\vec{P O} + \vec{O Q}) \\ = \underset{˜}{a} + h (- \underset{˜}{a} + \underset{˜}{b}) \\ = \underset{˜}{a} - h \underset{˜}{a} + h \underset{˜}{b} \\ \vec{O B} = (1 - h) \underset{˜}{a} + h \underset{˜}{b} \end{array}

(b)

\begin{array}{l} \vec{O B} = \vec{O P} + \vec{P B} \\ = \underset{˜}{a} + \vec{P A} + \vec{A B} \\ = \underset{˜}{a} + (- \frac{1}{5} \vec{O P}) + k \vec{A C} \\ = \underset{˜}{a} + (- \frac{1}{5} \underset{˜}{a}) + k (\vec{A O} + \vec{O C}) \\ = \frac{4}{5} \underset{˜}{a} + k (- \frac{4}{5} \vec{O P} + 3 \vec{O Q}) \\ = \frac{4}{5} \underset{˜}{a} + k (- \frac{4}{5} \underset{˜}{a} + 3 \underset{˜}{b}) \\ = \frac{4}{5} \underset{˜}{a} - \frac{4}{5} k \underset{˜}{a} + 3 k \underset{˜}{b} \\ \vec{O B} = \frac{4}{5} (1 - k) \underset{˜}{a} + 3 k \underset{˜}{b} \end{array}

(c)(i)

\begin{array}{l} (1 - h) \underset{˜}{a} + h \underset{˜}{b} = \frac{4}{5} (1 - k) \underset{˜}{a} + 3 k \underset{˜}{b} \\ 1 - h = \frac{4}{5} - \frac{4}{5} k .......... (1) \\ h = 3 k .......... (2) \\ Substitute (2) into the (1) \\ 1 - 3 k = \frac{4}{5} - \frac{4}{5} k \\ 5 - 15 k = 4 - 4 k \\ 11 k = 1 \\ k = \frac{1}{11} \\ Substitute k = \frac{1}{11} into (2) \\ h = 3 (\frac{1}{11}) \\ = \frac{3}{11} \end{array}

(c)(ii)

\begin{array}{l} \vec{O B} = (1 - h) \underset{˜}{a} + h \underset{˜}{b} \\ when h = \frac{3}{11} \\ = (1 - \frac{3}{11}) \underset{˜}{a} + (\frac{3}{11}) \underset{˜}{b} \\ = \frac{8}{11} \underset{˜}{a} + \frac{3}{11} \underset{˜}{b} \end{array}

Long Question 6

Posted on May 16, 2020 by Myhometuition

Question 6:
Diagram below shows a trapezium OABC and point D lies on AC.

\begin{array}{l} It is given that \vec{O C} = 18 \underset{˜}{b}, \vec{O A} = 6 \underset{˜}{a} and \vec{O C} = 2 \vec{A B} . \\ (a) Express in terms of \underset{˜}{a} and \underset{˜}{b}, \\ (i) \vec{A C} \\ (ii) \vec{O B} \\ (b) It is given that \vec{A D} = k \vec{A C}, where k is a constant . \\ Find the value of k if the points O, D and B are collinear . \end{array}

Solution:
(a)(i)

\begin{array}{l} \vec{A C} = \vec{A O} + \vec{O C} \\ = - 6 \underset{˜}{a} + 18 \underset{˜}{b} \\ = 18 \underset{˜}{b} - 6 \underset{˜}{a} \end{array}

(a)(ii)

\begin{array}{l} \vec{O C} = 2 \vec{A B} \\ 18 \underset{˜}{b} = 2 (\vec{A O} + \vec{O B}) \\ 18 \underset{˜}{b} = 2 (- 6 \underset{˜}{a} + \vec{O B}) \\ 18 \underset{˜}{b} = - 12 \underset{˜}{a} + 2 \vec{O B} \\ \vec{O B} = 6 \underset{˜}{a} + 9 \underset{˜}{b} \end{array}

(b)

\begin{array}{l} \vec{O D} = h \vec{O B} \\ = h (6 \underset{˜}{a} + 9 \underset{˜}{b}) \\ = 6 h \underset{˜}{a} + 9 h \underset{˜}{b} \\ \vec{A D} = \vec{O D} - \vec{O A} \\ = 6 h \underset{˜}{a} + 9 h \underset{˜}{b} - 6 \underset{˜}{a} \\ = \underset{˜}{a} (6 h - 6) + 9 h \underset{˜}{b} \\ \vec{A D} = k \vec{A C} \\ \underset{˜}{a} (6 h - 6) + 9 h \underset{˜}{b} = k (18 \underset{˜}{b} - 6 \underset{˜}{a}) \\ \underset{˜}{a} (6 h - 6) + 9 h \underset{˜}{b} = - 6 k \underset{˜}{a} + 18 k \underset{˜}{b} \\ 6 h - 6 = - 6 k \\ h - 1 = - k \\ h = 1 - k .......... (1) \\ 9 h = 18 k \\ h = 2 k \\ From (1), \\ 1 - k = 2 k \\ 3 k = 1 \\ k = \frac{1}{3} \end{array}