Category: 2nd PUC

Students can Download Maths Chapter 13 Probability Ex 13.4 Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Maths Question Bank Chapter 13 Probability Miscellaneous Exercise

Question 1.
A family has two children. Find the probability that both are boys, if it is known that
(i) At least one of them is boy
(ii) Elder child is a boy (CBSE – 2010)
Answer:
(i) Let A : At least one of them is a boy
{BB,BG,GB,GG}

Question 2.
A bag contains 4 balls
Two balls are drawn at random and are found to be White. What is the probability that all Balls are white.
Answer:
E₁: 2 balls are white, 2- are non white
E₂: 3 balls are white, 1 is non white
E₃: 4 balls are white
A : Two balls are white

Question 3.
Given three identical boxes I, II and III each containing two coins. In box I, both coins are gold, In box II both are silver and in box III there is one gold and one silver coin. A person choose a box and takes out a coin. If the coin is gold, what is the probability that the other coin is also gold  (CBSE – 2011)
Answer:
E₁: Both gold coin
E₂: Both silver coin
E₃: One gold and one silver

Question 4.
Two cards are drawn without replacement from a well shuffled pack of cards. Find the mean and variance of the number of Red cards.  (CBSE – 2012)
Answer:
Let X denote the no. of Red cards

Questions from Competitive Exams

Question 1.
A die is thrown. Let A be the event that the number obtained: is greater than 3. B be the event that the number obtained is less than 5. Find P(A∪B)
Answer:

Question 2.
It is given that, the events A and B are such that
\(\mathbf{P}(\mathbf{A})=\frac{1}{4}, \mathbf{P}(\mathbf{A} | \mathbf{B})=\frac{1}{2}, \mathbf{P}(\mathbf{B} | \mathbf{A})=\frac{2}{3}, \text { Find } P(B)\) [AIEEE – 2008]
Answer:

Question 3.
\(\mathbf{P}(\mathbf{A})=\frac{1}{4}, \mathbf{P}(\mathbf{B})=\frac{1}{2}, \mathbf{P}(\mathbf{A} \cap \mathbf{B})=\frac{1}{8}\),find P(not A, not B) (UPSEE -2008)
Answer:

Question 4.
[Latex]\mathbf{P}(\mathbf{A})=\frac{1}{2}, \mathbf{P}(\mathbf{B})=\frac{7}{12}, \mathbf{P}(\text { not } \mathbf{A} \text { or not } \mathbf{B})=\frac{1}{4}[/latex]
State whether A and B are independent. [AIEEE]
Answer:

Students can Download Maths Chapter 13 Probability Ex 13.5 Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Maths Question Bank Chapter 13 Probability Ex 13.5

2nd PUC Maths Probability NCERT Text Book Questions and Answers Ex 13.5

Ex 13.5 Class 12 Maths Question 1.
A die is thrown 6 times. If ‘getting an odd number’ is a success, what is the probability of
(i) 5 successes?
(ii) at least 5 successes?
(iii) at most 5 successes?
Solution:
There are 3 odd numbers on a die
∴ Probability of getting an odd number on a die = \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\)

Ex 13.5 Class 12 Maths Question 2.
There are 5% defective items in a large bulk of items. What is the probability that a sample of 10 items will include not more than one defective item ?
Solution:
Probability of getting one defective item = 5%
= \(\frac { 5 }{ 100 }\)
= \(\frac { 1 }{ 20 }\)
Probability of getting a good item = \(1-\frac { 1 }{ 20 }\) = \(\frac { 19 }{ 20 }\)
A sample of 10 item include not more than one defective item.
=> sample contains at most (me defective item Its probability = P (0) + P (1)

Ex 13.5 Class 12 Maths Question 3.
A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability, of two successes.
Solution:
n(S) = 36, A = {11,22,33,44,55,66}

Ex 13.5 Class 12 Maths Question 4.
Five cards are drawn successively with replacement from a well- shuffled deck of 52 cards. What is the probability that
(i) all the five cards are spades?
(ii) only 3 cards are spades?
(iii) none is spade?
Solution:
S = {52 cards}, n (S) = 52
Let A denotes the favourable events
A= {13 spade}, n(A)= 13

Ex 13.5 Class 12 Maths Question 5.
The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs.
(i) none
(ii) not more than one
(iii) more than one
(iv) at least one will fuse after 150 days of use
Solution:
Probability that a bulb gets fuse after 150 days of its use = 0.05
Probability that the bulb will not fuse after 150 days of its use = 1 – 0.05 = 0.95
(i) Probability that no bulb will fuse after 150

Ex 13.5 Class 12 Maths Question 6.
A bag consists of 10 balls each marked with one of the digits 0 to 9. If four bails are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0?
Solution:
S = {0,1,2,3,4,5,6,7,8,9},n(S) = 10
Let A represents that the ball is marked with the digit 0.
A = {0}, n(A) = 1

Ex 13.5 Class 12 Maths Question 7.
In an examination, 20 questions of true – false type are asked. Suppose a student tosses fair coin to determine his answer to each question. If the coin falls heads, he answers ‘true,’ if it falls tails, he answers “ false’. Find the probability that he answers at least 12 questions correctly.
Solution:
Probability that student answers a question true = \(\frac { 1 }{ 2 }\)
i.e., when a coin is thrown, probability that a head is obtained = \(\frac { 1 }{ 2 }\)
Probability that his answer is false = \(1-\frac { 1 }{ 2 }\) = \(\frac { 1 }{ 2 }\)
Probability that his answer at least 12 questions correctly = P (12) + P (13) + P (14) +…….. P (20)

Ex 13.5 Class 12 Maths Question 8
Suppose X has a binomial distribution \(B\left( 6,\frac { 1 }{ 2 } \right) \). Show that X = 3 is the most likely outcome.
(Hint: P (X = 3) is the maximum among all P (Xi), xi. = 0,1,2,3,4,5,6)
Solution:
\({ \left( \frac { 1 }{ 2 } +\frac { 1 }{ 2 } \right) }^{ 6 } \)

Ex 13.5 Class 12 Maths Question 9.
On a multiple choice examination with three possible answers for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing?
Solution:
P = \(\frac { 1 }{ 3 }\). q = 1 – P = \(1-\frac { 1 }{ 3 }\) = \(\frac { 2 }{ 3 }\)

Ex 13.5 Class 12 Maths Question 10.
A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is \(\frac { 1 }{ 100 }\) . What is the probability that he will win a prize?
(a) at least once,
(b) exactly once,
(c) at least twice?
Solution:
Probability that the person wins the prize = \(\frac { 1 }{ 100 }\)
Probability of losing = \(1-\frac { 1 }{ 100 }\) = \(\frac { 99 }{ 100 }\)
(a) Probability that he loses in all the loteries

Ex 13.5 Class 12 Maths Question 11.
Find the probability of getting 5 exactly twice in 7 throws of a die.
Solution:
S = {1,2,3,4,5,6},n(S) = 6
A = {5} => n(A) = 1

Ex 13.5 Class 12 Maths Question 12.
Find the probability of throwing at most 2 sixes in 6 throws of a single die.
Solution:
When a die is thrown,
Probabiltiy of getting a six = \(\frac { 1 }{ 6 }\)
Probabiltiy of not getting a six = \(1-\frac { 1 }{ 6 }\) = \(\frac { 5 }{ 6 }\)
Probabiltiy of getting at most 2 sixes in 6 throws of a single die = P (0) + P (1) + P (2)

Ex 13.5 Class 12 Maths Question 13.
It is known that 10% of certain articles manufactured are defective. What is the probability that in a random sample of 12 such articles 9 are defective?
Solution:
p = \(\frac { 1 }{ 100 }\) = \(\frac { 1 }{ 10 }\)
q = 1 – p = \(1-\frac { 1 }{ 10 }\) = \(\frac { 9 }{ 10 }\)

Ex 13.5 Class 12 Maths Question 14.
In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is
(a) \({ 10 }^{ -1 }\)
(b) \({ \left( \frac { 1 }{ 2 } \right) }^{ 5 }\)
(c) \({ \left( \frac { 9 }{ 10 } \right) }^{ 5 }\)
(d) \(\frac { 9 }{ 10 }\)
Solution:
p = \(\frac { 1 }{ 10 }\)
q = \(\frac { 9 }{ 10 }\) n = 5, r = 0, P(X=0) = \({ \left( \frac { 9 }{ 10 } \right) }^{ 5 }\)
Option (c) is correct

Ex 13.5 Class 12 Maths Question 15.
The probability that a student is not a swimmer is \(\frac { 1 }{ 5 }\). Then the probability that out of five students, four are swimmers is:

Solution:
p = \(\frac { 4 }{ 5 }\) , q = \(\frac { 1 }{ 5 }\) , n = 5,r = 4

Option (a) is true

Students can Download Maths Chapter 13 Probability Ex 13.4 Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Maths Question Bank Chapter 13 Probability Ex 13.4

2nd PUC Maths Probability NCERT Text Book Questions and Answers Ex 13.4

Question 1.
State which of the following are not the probability distributions of a random variable. Give reasons for your answer.
(i)

x	0	1	2
P(X)	0.4	0.4	0.2

Answer:

X	0	1	2	T
P(X)	4	4	2	1

(ii)

X	0	1	2	3	4
P(X)	0.1	0.5	0.2	-0.1	0.3

Answer:

X	0	1	2	3	T
P(X)	.1	.5	.2	-.1	1

It is not P.d as P(X) < 0 when x = 3

(iii)

Y	-1	0	1
P(Y)	0.6	0.1	0.2

Answer:

Y	-1	0	1	T
P(Y)	.6	.1	.2	.9

It is not P.d as ΣP(X) = .9 ≠1

(iv)

Z	3	3	1	0	-1
P(Z)	0.3	0.2	0.4	0.1	0.05

Answer:

Z	3	3	1	0	T
P(Z)	.3	.2	.4	.1	1.05

It is not a P.d as ΣP(X) = 1.05 > 1

Question 2.
An urn contains 5 red and 2 black balls.Two balls are randomly drawn. Let X represent the number of black balls. What are the possible values of X ? Is X a random variable ?
Answer:
X can take the values 0, 1, 2 It is a Random variable because the experiment is a random experiment.

Question 3.
Let X represent the difference between the number of heads and the number of tails obtained when a coin is tossed 6 times. What are possible values of X?
Answer:
When m = 6, n = 0  ⇒ |m-n| = 6
m = 5,n = 1                  |m-n| = 4
m = 4, n = 2                 |m-nl=2
m = 3,n = 3                  |m-n| = 0
m = 2, n = 4                 |m-n|=2
m = 1,n = 5                  |m-n| = 4
m = 0, n = 6                 |m-n|=6
∴ hence X can take values 0, 2, 4, 6.

Question 4.
Find the probability distribution of
(i) number of heads in two tosses of a coin.
(ii) number of tails in the simultaneous tosses of three coins.
(iii) number of heads in four tosses of a coin.
Answer:
(i) When a coin is tossed twice, the number of heads may be 0, 1,2
Sample space {HH, HT, TH, TT}

(ii) When a coin is tossed thrice, the number of heads are 0, 1, 2, 3
Sample space {HHH, HHT, HTH, HTT, ITT, TTH, THT, THH

(iii) When coin is tossed four times, the number of heads be 0, 1, 2, 3,4 Sample space {HHHH, HHHT, HHTH, HHTT, HTHT, HTHH, HTTH, TTTT, TTTH, HTTT, TTHT, TTHH, THTH, THTT, THHT, THHH]

Question 5.
Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as
(i) number greater than 4
(ii) six appears on at least one die
Ans:
(i) Let the random variable be X no of successes then X can take the value 0,1,2.

(ii) Let X be random variable, then X can take values
O (no sixes) & 1 (at least 1 six)
P (x = 0) = P (no six)
= [1-P (getting a six)] x [1 – P (getting a six on single throw]

Question 6.
From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with’ replacement. Find the probability distribution of the number of defective bulbs.
Answer:
Let X denote the no. of defective bulbs P denote the probability of obtaining defective bulb

Question 7.
A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails.
Answer:
Let X denote the number of tails in two tosses of a coin
∴ X can assain the values 0, 1,2
Let P be the probability getting head, q the tail

Question 8.
A random variable X has the following probability distribution:

X	0	1	2	3	4	5	6	7
P(X)	0	k	2k	2k	3k	k2	2k2	7k2+k

Determination
(i) K
(ii) P(X<3)
(iii) P(X>6)
(iv) P(0<X<3)
Answer:

Question 9.
The random variable X has a probability distribution P(X) of the following form, where k is some number:
\(P(X)=\left\{\begin{array}{ll}{k,} & {\text { if } \quad x=0} \\{2 k,} & {\text { if } x=1} \\{3 k,} & {\text { if } x=2} \\{0,} & {\text { otherwise }}\end{array}\right.\)
(a) Determine the value of k
(b) Find P (X < 2), P (X≤2), P (X≥2)
Answer:
(a) In P.d. Σ XP(X)= 1
6k = 1 ⇒ \(k=\frac{1}{6}\)

Question 10.
Find the mean number of heads in three tosses of a fair coin.
Answer:
If three coins are tossed, Let X denote the number of heads.
X can take the values 0, 1, 2, 3

Question 11.
Two dice are thrown simultaneously. If X denotes the number of sixes, find the expectation of X.
Answer:
Let X denote the number of sires. Then X can take values 0,1,2

Question 12.
Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained. Find E(X).
Ans:
X can take the values 2, 3, 4, 5, 6

Question 13.
Let X denote the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of X.
Ans:
Let X denote the sum of the numbers obtained
∴  X can take the values 2, 3,4, 5,6,7, 8,9, 10, 11, 12

Question 14.
A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. One student is selected in such a manner that each has the same chance of being chosen and the age X of the selected student is recorded. What is the probability distribution of the random variable X? Find mean, variance and standard deviation of X.
Ans:
Let X denote the age of the students.
Then X can take the values 14,15,16,17, 18,19,20, 21

Question 15.
In a meeting, 70% of the members favour and 30% oppose a certain proposal. A member is selected at random and we take X = 0 if he opposed, and X = 1 if he is in favour. Find E(X) and Var (X).
Answer:

Choose the correct answer in each of the following:

Question 16.
The mean of the numbers obtained on throwing a die having written 1 on three faces, 2 on two faces and 5 on one face is
(A) 1
(B) 2
(C) 5
(D) 1
Answer:

Question 17.
Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained . Then the value of E(X) is
(A) \(\frac{37}{221}\)
(B) \(\frac{5}{13}\)
(C) \(\frac{1}{13}\)
(D) \(\frac{2}{13}\)
Answer:

Students can Download 2nd PUC Basic Maths Previous Year Question Paper June 2017, Karnataka 2nd PUC Maths Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Previous Year Question Paper June 2017

Time: 3.15 hours
Max Marks: 100

Instructions:

• The question paper has 5 parts A, B, C, D and E. Answer all the pats.
• Part-A carries 10 marks), Parts – B carries 20 marks, Part – C carries 3 marks, Part – D carries 30 marks and Part – E carries 1 marks.
• Write the question number properly as indicated in the question paper.

Part – A

Answer all the ten questions: (10 × 1 = 10)

Question 1.
Evaluate \(\left| \begin{matrix} 3200 & 3201 \\ 3202 & 3203 \end{matrix} \right| \)
Answer:

Question 2.
In how many ways can 10 people seated around a table ?
Answer:
(10 – 1)! = 9!

Question 3.
Write the verbal form of the compound proposition pvq where
P:x is an integer; Q: 5 is odd number.
Answer:
PVQ = x is an integer or 5 is odd number.

Question 4.
Find the triplicate ration of 3 : 5.
Answer:
3³ : 5³ = 27:125

Question 5.
A bill was drawn for 3 months was legally due on 18-08-2012. Find the date of drawing of the bill.
Answer:
Drawn date = L.D.D – B.P – 3 days = 15 – 5 – 12

Question 6.
Find the value of cos³ 20° – 3cos 20°.
Answer:
Cos 3(20) = cos60 = \(\frac { 1 }{ 2 }\)

Question 7.
If the length of the latus rectum of the parabola x² = 4k y is 8. Find the value of k.
Answer:
4k = 8 ⇒ k = 2

Question 8.
Evaluate line \(\lim _{x \rightarrow 3}\left(\frac{x^{2}-x}{x-2}\right)\)
Answer:
\(=\frac{3^{2}-3}{3-2}=\frac{9-3}{1}=6\)

Question 9.
Differentiate : 5e^x – log_e x – 3√x w.r.t.x.
Answer:
\(\frac{d y}{d x}=5 e^{x}-\frac{1}{x}-\frac{3}{2 \sqrt{x}}\)

Question 10.
Evaluate ∫sec² (x – 5)dx.
Answer:
Tan (x – 5)

Part – B

Answer any ten questions. (10 × 2 = 20)

Question 11.
If A = \(\left[ \begin{matrix} 2 \\ -1 \\ 3 \end{matrix} \right] \), B = \(\left[ \begin{matrix} 1 & 4 & 2 \end{matrix} \right] \) find AB and BA.
Answer:

Question 12.
In a party each person shakes hands with everyone else. If there are 25 members in the party, calculate the number the number of hand shakes.
Answer:
\(25 C_{2}=\frac{25 \times 24}{2 \times 1}=25 \times 12=300\)

Question 13.
A bag contains 6 white beads and 4 red beads. A bead is drawn at random, what is the probability that the bead drawn is white.
Answer:
P(beaddrawniswhite) = \(\frac{10 C_{1}}{6 C_{1}} \frac{10}{6}=\frac{5}{3}\)

Question 14.
If p,q, r are the propositions with truth values F, T and F repectively. Then find the truth value of the ; compound proposition p → (q → r).
Answer:
P → (q → r)
F → (T → F)
F → F = T

Question 15.
If a : b = 4 : 5. Find \(\frac{3 a+2 b}{3 a-2 b}\)
Answer:

Question 16.
Banker’s gain on a bill due after 6 months at 4% p.a. is 24. Find true discount and Banker’s discount,
Answer:
Given B.G = 24. T = 6 months, = \(\frac { 1 }{ 2 }\) yr, r = 4% = 0.0
BG = TD + r
24 = TD × \(\frac { 1 }{ 2 }\) × 0.04

TD = 1200
BG = BD – TD ⇒ BD = BG + TD = 24 + 1200 = 1224

Question 17.
Prove that \(\frac{\sin 3 A}{1+2 \cos 2 A}\) = sin A.
Answer:

= Sin A = R.H.S

Question 18.
If tan A = \(\frac { 5 }{ 6 }\) tan B = \(\frac { 1 }{ 11 }\) Show that A + B = \(\frac{\pi}{4}\)
Answer:
Consider tan (A+B) = \(\frac{\tan A+\tan B}{1+\tan A \cdot \tan B}\)

Tan (A + B) = 1 ⇒ A + B = \(\frac{\pi}{4}\)

Question 19.
Find the equation of the circle whose centre is at (4,-2) and passing through the origin (0,0)
Answer:
C = (4,-2) & r = \(\sqrt{4^{2}+(-2)^{2}}=\sqrt{16+4}=\sqrt{20}\)
Equation of the circle (x + 4)² + (y + 2)² = (√ 20)
⇒ x² + 16 – 8x + y² + y + 4y – 20 = 0
⇒ x² + y² – 8x + 4y = 0

Question 20.

Answer:
\(\lim _{x \rightarrow 0}\) f(x) = \(\lim _{x \rightarrow 0}\) (1 + 3x)^1/3 = e³ = f(0)
∴ f(x) is continuous at x = 0

Question 21.
If siny = x sin(a+y), prove that \(\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}\)
Answer:

Question 22.
If s = at³ + bt, find a and b given that when t = 3, velocity is 0 and the acceleration is 14 units.
Answer:
If s = at³ + bt velocity = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 3at² + b = 0
Acceleration = \(\frac{\mathrm{dv}}{\mathrm{dt}}\) = 6at = 14
At t = 3sec, 6a × 3 = 14 ⇒ a = \(\frac{14}{18}=\frac{7}{9}\)

Question 23.
Evaluate \(\int \frac{3^{x} \log 3}{\left(3^{x}+1\right)} d x\)
Answer:
= log(3^x + 1) + C ∴ \(\int \frac{f^{1}(x)}{f(x)} d x\) = log(3^x +1)

Question 24.
Evaluate \(\int_{1}^{2} \frac{1}{(2 x+3)} d x\)
Answer:

Part – C

Answer any ten questions: (10 × 3 = 30)

Question 25.
If 2A + B = \(\left[ \begin{matrix} 3 & -1 \\ -2 & 5 \end{matrix} \right] \) and A – 2B = \(\left[ \begin{matrix} 4 & 2 \\ -1 & 5 \end{matrix} \right] \) then find A and B.

Question 26.
Solve \(\left| \begin{matrix} 2+x & 3 & -4 \\ 2 & 3+x & -4 \\ 2 & 3 & -4+x \end{matrix} \right| =0\)
Answer:

Question 27.
Find the number of permutations of the letters of the word ‘ENGINEERING’ How many of these
i. Begin with E and end with E
ii. Have all the 3E’s together?
Answer:
Total-11, N – 3, G – 2, 1 – 2, E = 3
Total permutations \(\frac{11 !}{3 ! \times 3 ! \times 2 ! \times 2 !}\)=
I. Number of permutations which begin with E & end with E is = \(\frac{9 !}{3 ! \times 2 ! \times 2 !}\)
II. Number of Permutations in Which all 3E’s are together (Total 11 – 3 + 1 = 9′) = \(\frac{9 !}{3 ! \times 2 ! \times 2 !}\)

Question 28.
A die is thrown, If E is the event “the number appearing is a multiple of 3” and F be the event “the number appearing is even”. Then find whether E are Fare independent.
Answer:
S = {1,2,3,4,5,6}
E = {3,6), F = {2,4,6}
N(S) = 6, n(E) = 2 n(F) = 3
P(E) = \(\frac{2}{5}=\frac{1}{3}\) P(E) = \(\frac{3}{6}=\frac{1}{3}\)
They are not independent

Question 29.
Two tap can separately fill a tank in 12 minutes and 15 minutes respectively. The tank when full can be emptied by a drain pipe in 20 minutes. When the tank was empty all the three were opened simultaneously in what time will the tank be filled up?
Answer:
A fills tank in 12 mins → In 1 min A will fill 1/12 of tank.
B fills tank in 15 mins → In 1 min B will fill 1/15 of tank.
Lets say C can alone empty the tank in x mins → In 1 min C will empty 1/x of tank.
According to question, if all taps are opened tank will fill in 20 mins.
In each min 1/20 tank is filled which is due to A, B and C. Now A and B fill the tank will C empty the tank.

1/12 + 1/15 – 1/x = 1/20

1/12 + 1/15 – 1/20 = 1/x

(5 + 4 – 3)/60 = 1/x

6/60 = 1/x

x = 10

Question 30.
Bharath bought a shirt for ₹ 336 including 12% sales tax and a neck tie for 110 including 10% sales tax. Find the printed price of shirt and neck tie together.
Answer:
Given S.P = ₹ 336, S.P (neck tie) = 110/
Let C.P of shirt = x, & C.P of neck tie = y
Here, setting price of shirt = C.P + 12% of C.P
336 = x + \(\frac{12 x}{100}\)
336 = \(=\frac{112 x}{100}\) ⇒ x = \(\frac{33600}{112}\)
Selling price of neck tie = C.P + 10% of C.P
110 = y + \(\frac{10 y}{100}\)
110 = \(\frac{10 y}{100}\) ⇒ \(\frac{1100}{11}\) = 100
∴ Printed price of shirt & neck tie = 300 + 100 = ₹ 400

Question 31.
A Banker pays ₹ 4,250 on a bill of ₹ 5,000. 146 days before the legally due date. Find the rate of discount charged by the Banker.
Answer:
Given F = ₹ 5000, Discounted Value = ₹ 4,520
t = 146 days = \(\frac{146}{365}\) year,r = ?
BD = F – Discounted value = 5000 – 4520 = ₹ 480
BD = F/r ⇒ 480 = \(\frac{5000 \times 146 \times r}{365}\)
∴ r = \(\frac{480 \times 365}{5000 \times 146}\) = 0.24 × 100=24

Question 32.
Rakshith decides to invest in TCS shares which are selling at 2020 per share. How many money is required to purchase 10 shares if the brokerage is 0.5%?
Answer:
Selling price of 10 shares at 2020 per share = 20,200
0.5 Brokerage = \(\frac{0.5}{100}\) × 20,200 = ₹ 101
100 Amount required to purchase = 20,200 + 101 = 20,301

Question 33.
Find the equation of the parabola if its focus is (4,0) and directrix is x = -4. Also find the equation of the tangent at the vertex.
Answer:
Given Directrix As x = -4 & Focus = (4,0) ⇒ a = 4
The standard equation is y² = 4ax ⇒ y² = 4 × 4 x X = 16x
∴ Equation of the parabola = 16x

Question 34.
If x = a cos⁴θ, y = a sin⁴6 show that \(\frac{d y}{d x}-\tan ^{2} \theta\)
Answer:
x = a cost⁴θ & y = a sin⁴
Differentiate both w.r.t θ we get
\(\frac{d y}{d \theta}\) = a.4cos³θ.sin θ, \(\frac{d y}{d \theta}\) = 2.4 sin³θ. cosθ

Question 35.
The side of an equilateral triangle is increasing at the rate of √3 cm/s. Find the rate at which its area is including when its side is 200 cms.
Answer:
Given \(\frac{\mathrm{dx}}{\mathrm{dt}}\) = \(\sqrt{3}\) cm/sec., x = 2 meters, \(\frac{\mathrm{dA}}{\mathrm{dt}}\) = ?
Area of equilateral Δ^le = A = \(\frac{\sqrt{3}}{4}\) x²
\(\frac{d A}{d t}=\frac{\sqrt{3}}{4} \cdot 2 x \frac{d x}{d t}\)
= \(\frac{\sqrt{3}}{4}\) . 2. 200. \(\sqrt{3}\) = 300cm² / sec.

Question 36.
Find two positive numbers whose sum is 14 and the sum of whose square is minimum.
Answer:
Let the two numbers be x and y
Given x + y = 14 & S = x² + y² where y = 14 – x
∴ S = x² + (14 – x)² = x² + 14² + x² – 28x = 2x² – 28x + 14²
\(\frac{d s}{d x}\) = 4x – 28 → (1) \(\frac{d s}{d x}\) = 0 ⇒ 4x – 28 = 0 ⇒ x = 7
\(\frac{\mathrm{d}^{2} \mathrm{s}}{\mathrm{d} \mathrm{x}^{2}}\) = 4 > 0, sum is minimum.
∴ y = 14 – x = 14 – 7 = 7
∴ the two positive number are 7 & 7.

Question 37.
Evaluate \(\int \frac{1+\cos 2 x}{1-\cos 2 x} d x\)
Answer:
= \(=\int \frac{2 \cos ^{2} x}{2 \cos } d x\) = ∫cos² dx = ∫(cosec²x – 1)dx = – cos x – x + c

Question 38.
Evaluate \(\int_{0}^{1}(6 x+1) \sqrt{3 x^{2}+x+5 d x}\)
Answer:
Put 3x² + x + 5 = t.
(6x + 1) dx = dt
When x = 0 then t = 5
When x = 1 then t = 9
\(\left.\int_{0}^{1}(6 x+1)(\sqrt{3 x^{2}+x+5}) d x=\int_{0}^{1} \sqrt{t} d t=\frac{2}{3} t^{\frac{3}{2}}\right]_{5}^{7}\)
\(=\frac{2}{3}\left(9^{\frac{3}{2}}-5^{\frac{3}{2}}\right)=\frac{2}{3}(27-5 \sqrt{5})\)

Part-D

Answer any six questions: (6 × 5 = 30)

Question 39.
Simplify (√2 + 1)⁶ – (√2 – 1)⁶ using Binomial theorem.
Answer:

Question 40.
Resolve \(\frac{2 x^{2}-4 x+1}{(x-2)(x-3)^{2}}\) into partial fractions.
Answer:
2x² – 4x + 1 = A[x – 3)² + B(x – 2) (x-3) + C (x – 2)
Put x = 2, 8-8 + 1 = A(-1)² = A = 1
Put x = 3, 18 – 12 + 1 = A(0) + B(0) + C(3 – 2) = 7 = C
Comparing coefficient of x² = 2 = A + B = B = 2-A = 2 -1 =1
∴ \(\frac{2 x^{2}-4 x+1}{(x-2)(x-3)^{2}}=\frac{1}{x-2}+\frac{1}{x-3}+\frac{7}{(x-3)^{2}}\)

Question 41.
Prove that (p ∨q)∧(~P∧q~) is a contradiction.
Answer:

Question 42.
Divide ₹ 1,647 into three parts such that \(\frac { 3 }{ 7 }\) th of the first, \(\frac { 2 }{ 3 }\) rd of second and \(\frac { 4 }{ 5 }\) th of the third are equal..
Answer:
Let the 3 parts be A,B & C
Given \(\frac{3}{7} A=\frac{2}{3} B \Rightarrow \frac{A}{B}=\frac{2}{7} \times \frac{7}{3}=\frac{14}{9}\)
∴ A : B = 14 : 9
Also given \(\frac{2}{3} B=\frac{4}{5} \Rightarrow \frac{B}{C}=\frac{4}{5} \times \frac{3}{2}=\frac{12}{10}=\frac{6}{5}\)
∴ B : C = 6 : 5
Multiple (1) by 2 and (2) by 3 we get A : B = 28 : 18 and B : C = 18 : 5
∴ A : B : C = 28 : 18 : 15
∴ 28x + 18x + 15x = 1647
⇒ 61x = 1647 ⇒ x = \(\frac{1647}{61}\) = 27
∴ A receives = 28 × 27 = ₹ 3 756
B receives = 18 × 27 = ₹ 486
Creceives = 15 × 27 = ₹ 405

Question 43.
A company requires 100 hours to produce the first 10 units at 15/hour. The learning effect is 80%. Find the total labour cost to produce a total of 160 units.
Answer:
Given
a = 100 hrs. 10 units = 1 lot
∴ 160 units = 16 lots

y = ax^b
log y = log a + b log x
= log 100 – 0.3219 log 16
= 2 – 0.3219 1.2041 = 2 – 0.3875 = 1.625
y = A(1.6125) = 40.98 hours/unit
∴ Total time required for 160 units or 16 lots = 40.98 16 = 655.68 hrs.
∴ Total cost to produce 16 units at ₹ 15/hr. = 655.68 × 15 = ₹ 9835.20
Table method

Total time for 16 lots = 655.36 hrs.
∴ Total cost to produce 16 lots at ₹ 15/hr. = 655.36 × 15 = ₹ 19830.4.

Question 44.
Solve the following L.P.P. graphically maximize z = 60x + 15y subject to the constraints x + y ≤ 50, 3x + y ≤ 90, x ≥ 0, y ≥ 0.
Answer:
Consider line x + y = 50
Put x = 0, y = 50, (0,50)
Put y = 0, x = 50 (50,0)
Consider the line 3x + y =- 90
Put x = 0, y = 90 (0,90)
Put y = 0, x = 30 (30,0)
Plot the above lines on the graph we get OABC is the solution region

≥ is maximum at c (30,0)
z_max = 1800

Question 45.
Prove that sin 20° sin 40° sin 60° sin 80°= \(\frac { 3 }{ 16 }\)
Answer:
L.H.S=sin 60° (sin 40° sin 20° )sin 80°

Question 46.
Find the equation of the circle passing through the points (5,1) and (3,4) and has its centre on the x – axis.
Answer:
Let the required equation of the circle is x² + y² + 2gx + 2fy + c = 0
It passes through (5, 1) & (3, 4)
(5, 1) (5)² + (1)² + 2g(5) + 2f (1) +C = 0
10g + 2f + C + 26 = 0 …..(1)
(3,4) 3² + 4² + 2g (3) + 2f (4) +c= 0
68 + 8f + C + 25 = 0 …..(2)
& the centre (- g, – f) lies on x-axis ⇒ f = 0 ….(3)
Solving 1 & 2 we get

⇒ 2x² + 2y² – x – 47 = 0.

Question 47.
If y = \((x+\sqrt{x^{2}+1})\) show that(x² – 1}y₂ + xy₁ – y = 0.
Answer:
y = x + \(\sqrt{x^{2} – 1}\) differentiate w.r.t x

∴ (x² – 1) y₁² = y² Differentiate agin w.r.t x
(x² – 1)2y₁y₂ + y₁² (2x) = 2y .y₁ (÷ 2y₁ we get)
(x² – 1)y₂ + xy₁ – y = 0 Hence proved

Question 48.
Find the area of the region between the parabolas y² = 4x and x² = 4y.
Answer:

Part – E

Answer any one question: (1 × 10 = 10)

Question 49.
(a) A school wants to award its students for the values of punctuality, good behavior and hardwork with a total cash award ₹ 6,000. Three times the award money for hardwork added to that given for punctuality amounts to ₹ 11,000. The award money for punctuality and hardwork together is double the one given for good behavior.

Represent the above situation algebraically and also find the award money for each value using matrix method.

(b) The total revenue function is given by R = 400x – 2x² and the total cost function given by C = 2x² + 40x + 4000 find
i. The marginal revenue and marginal cost function.
ii. The average revenue and average cost.
Answer:
(a) Let the values of punctuality, Good Behaviour and Hard work be denoted by x,y & z receptively
Given x + y +z = 6000 ———1
X + 3z = 11000 ———-2
X + z = 2y ⇒ x – 2y + z = 0 ——- 3
Solve these equations by matrix method

Thus x = ₹ 500, y = ₹ 2000 & z = ₹ 3500

(b) (i) MR = \(\frac{d}{d x}\) (TR) = \(\frac{d}{d x}\) (400x – 2x²) = 400 – 4x
MC = \(\frac{d}{d x}\) (TC) = \(\frac{d}{d x}\)(2x² + 40x + 400 )= 4x – 40

(ii) \(\mathrm{AR}=\frac{\mathrm{TR}}{\mathrm{x}}=\frac{400 \mathrm{x}-2 \mathrm{x}^{2}}{\mathrm{x}}=400-2 \mathrm{x}\)
\(A C=\frac{T R}{x}=\frac{2 x^{2}+40 x+4000}{x}=2 x+40+4000\)

Question 50.
(a) Prove that if \(\lim _{x \rightarrow a}\left(\frac{x^{n}-a^{n}}{x-a}\right)\) = na^n-1 for all values of’n’.
Answer:

Case 1: Let n be a positive integer.
xⁿ – aⁿ = (x-a)(x^n-1 + x^n-2. a + x^n-3. a² + ———- + 2n-1)

Case 2: Let n be a negative integer
put n = -m, m > 0

Case 3: Let n = p/q where p and q are integers and q ≠ 0

Hence it proved for al the values of n

(b) A person is at the top of a tower 75 feet high. From there he observes a vertical pole and finds the angles of depressions of top and bottom of the pole which are 30° and 60° respectively. Find the height of the pole.
Answer:

width=”278″ height=”194″ />
Let AB = Tower, CD = Pole = h
From triangle ABC we get
tan 60° = \(\frac{A B}{A C} \Rightarrow \frac{75}{A C}=\sqrt{3}\)
⇒ AC = \(\frac{75}{\sqrt{3}} ….. (1)\) …. (1)

From triangle BDE

⇒ AC = \(\sqrt{3}(75-h)\)

From 1 and 2 we get
\(\frac{75}{\sqrt{3}}=\sqrt{3}(75-h) \Rightarrow 75=(\sqrt{3})^{2}(75-h)\)
75 = 225 – 3h 3h ⇒ 150 ⇒ h = 50
∴ Height of the pole -50ft

Students can Download 2nd PUC Basic Maths Previous Year Question Paper June 2016, Karnataka 2nd PUC Maths Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Previous Year Question Paper June 2016

Time: 3.15 hours
Max Marks: 100

Instructions:

• The question paper has 5 parts A, B, C, D and E. Answer al the parts
• Part-A carries 10 marks\, Parts – B carries 20 marks, Part – C carries 3 marks, Part – D carries 30 marks and Part – E carries 1 marks.
• Write the question number properly as indicated in the question paper.

Part – A

Answer All the ten questions: (10 × 1 = 10)

Question 1.
Find x such that \(\left[ \begin{matrix} 3 & x \\ 4 & 7 \end{matrix} \right] \) is symmetric
Answer:
X = 4

Question 2.
Find n if ⁿC₈ = ⁿC₇
Answer:
ⁿC₈ = ⁿC₇ ⇒ ⁿC₈ = ⁿC₇ ⇒ 8 = n – 7 ⇒ n = 15

Question 3.
Negate the proposition “4 is an even integer or 7 is a prime number”.
Answer:
4 is not an even integer and 7 is not a prime number.

Question 4.
Find the value of x if 32 : x = 75 : 50.
Answer:
\(x=\frac{32 \times 50}{75}=\frac{64}{3}\)

Question 5.
Find the income obtained by investing ₹ 3,600 in 5% stock at 90.
Answer:
Income = \(\frac { 5 }{ 100 }\) × 3600 = 180

Question 6.
Find the value of 3 sin 10° – 4 sin 10°.
Answer:
3 sin 10° – 4 sin 10° = sin 3.10° = sin 30° = \(\frac { 1 }{ 2 }\)

Question 7.
If x² + y² – 4x – 8y + k = 0 represents a point circle find k.
Answer:
R = \(\sqrt{g^{2}+f^{2}-c}\), g = -2, f = -4,c = k, r = 0.
∴ \(\sqrt{4+16-k}=0\)
20 – k = 0 ⇒ k = 20

Question 8.
Evaluate \(\lim _{x \rightarrow 0}(1+3 x)^{\frac{1}{x}}\)
Answer:
\(\lim _{x \rightarrow 0}(1+3 x)^{\frac{1}{x}}=e^{3}\)

Question 9.
If y = cos (x³) find \(\frac{\mathrm{d} \mathrm{y}}{\mathrm{d} \mathrm{x}}\)
Answer:
Let y = Cos(x³)
∴ \(\frac{\mathrm{d} \mathrm{y}}{\mathrm{d} \mathrm{x}}\) = sin(x³).3x²

Question 10.
Evaluate : ∫(x^e + e^x – log a)dx
Answer:
\(\frac{x^{e+1}}{e+1}+e^{x}-\log a \cdot x+C\)

Part – B

Answer any ten questions. (10 × 2 = 20)

Question 11.
Solve using Cramer’r rule:
5x + 7y = 3
7x + 5y = 9
Answer:
5x + 7y = 3
7x + 5y = 9
Δ = \(\left| \begin{matrix} 5 & 7 \\ 7 & 5 \end{matrix} \right| \) = 25 – 69 = -24 Δ_x = \(\left| \begin{matrix} 3 & 7 \\ 9 & 5 \end{matrix} \right| \) = 15 – 63 = -48 Δ_y = \(\left| \begin{matrix} 5 & 3 \\ 7 & 9 \end{matrix} \right| \) = 45 – 21 = 24
∴ x = \(\frac{\Delta_{x}}{\Delta}=\frac{-48}{-24}\) = 2 y = \(\frac{\Delta_{y}}{\Delta}=\frac{24}{-24}\) = -1
∴ x = 2, y = -1

Question 12.
Find the number of parallelograms that can be formed form a set of 6 parallel lines intersecting another set of 4 parallel lines.
Answer:

Question 13.
What is the probability that there will be 53 Mondays in a randomly selected leap year?
Answer:
Every year has 52 complete weeks 8 = 364 days (52 × 7) A leap year has 366 days, in the remaining two days we have to see what is the probability I getting Monday, the two days may occur in any one of the following (Sunday, Monday) (Mon,Tue), (Wed, Thur) (Thur,Fri) (Fri, Satur) (Satur, Sun) Out of the 7 Possibilities, the first two cases have Monday
∴ P(53Mondays in a leap) = \(\frac { 2 }{ 7 }\)

Question 14.
Write the converse and income of the statement “If Maths is easy then child is brave”.
Answer:
Converse = (q → P) = If the child is brave then maths is easy
Inverse = (~ P → ~q) = If maths is not easy then the child is not brave.

Question 15.
Divide 6,000 in the ratio 3 : 4 : 5.
Answer:

II part = \(\frac{4}{12}\) × 6000 = 2000

III part = \(\frac{5}{12}\) × 6000 = 2500

Question 16.
A banker pays 2,380 on a bill of 2,500,73 days before the legally due date. Find the rate of discount charged by the banker.
Answer:
F = ₹ 2,500, discounted value = 2380, t = 73 days, t = \(\frac{73}{365}=\frac{1}{5}\) yr
Discounted value = F (1 – tr}
2380 = 2500 \(\left(1-\frac{1}{5} \times r\right)\) ⇒ \(\frac{2380}{2500}\) = 1 – \(\frac{r}{5}\) ⇒ 1 – 0.952
0.2r = 0.048
r = \(\frac{0.048}{0.2}\) = 24 = 24%

Question 17.
If A + B + C = 180° and tan A = 1, tan B=2, show that tan C = 3,
Answer:
Given A + B + C = 180° ⇒ A + B = 180 – C ∴ tan(A + B) = tan (180 – C)

Question 18.
Find the focus and equation of the directrix of the parabol x² = – 16y
Answer:
Given x² = – 16y
Compare x² = -4ay ⇒ 4a = 16 ⇒ a = 4
The curve turns down words ∴ focus S = (0,-4)
Equation of directrix is y = 4 or y – 4 = 0

Question 19.

Answer:
Given the function is continuous \(\lim _{x \rightarrow 4}\) f(x) = f(4)
\(\lim _{x \rightarrow 4} \frac{x^{4}-4^{4}}{x-4}=K\) ∴ k = 4.4^+-1 = 4.4³ = 4⁴ = 256

Question 20.
If y = x^x find \(\frac{d y}{d x}\)
dx
Answer:
Let y = x^x
log y = x.log x
\(\frac{1}{y} \frac{d y}{d x}=x, \frac{1}{x}+\log x, 1 \Rightarrow \frac{d y}{d x}\) = y (1 + log x) = x^x(1 + logx)

Question 21.
When brakes are applied to a moving car, the car travels a distance ‘s’ feet in ‘t’ secs is given by s = 20t – 40t². When does the car stop?
Answer:
Given s = 20t – 40t²
v = \(\frac{d s}{d t}\) = 20 – 80t
The car stops when the velocity = 0 ⇒ 20 – 80t = 0 ⇒ t = \(\frac{1}{4}\) sec

Question 22.
The radius of a sphere is increasing at the rate of 0.5 mts / sec. Find the rate of Increase of its volume when r=1.5 mts.
Answer:
Given \(\frac{d r}{d t}\) = 0.5 mts / sec. \(\frac{d v}{d t}\) = ? when r =1.5mts v = \(\frac{4}{3}\)πr³

Question 23.
Evaluate : \(\int \frac{e^{x}}{e^{x}-1} d x\)
Answer:
\(\int \frac{e^{x}}{e^{x}-1} d x\) = log (e^x+ 1) + C ∴ \(\int \frac{f^{\prime}(x)}{f(x)}\) dx = log(f(x))

Question 24.
Evaluate \( \int_{1}^{2}\left(x+e^{x}\right) d x\)
Answer:

Part – C

Answer any ten questions: (10 × 3 = 30)

Question 25.
If A = \(\left[ \begin{matrix} 2 & -1 \\ 1 & 4 \end{matrix} \right] \) and B = \(\left[ \begin{matrix} -3 & 2 \\ -1 & 4 \end{matrix} \right] \) show that (AB)’ = B’ A’.
Answer:

From 1 and 2 we get (AB)’ = B’A’

Question 26.
Solve for \(\left[ \begin{matrix} x+2 & 3 & 4 \\ 2 & x+3 & 4 \\ 2 & 3 & x+4 \end{matrix} \right] =\quad 0\)
Answer:

Question 27.
Find the number of permutations of the letters of the word Engineering How many of these
(a) Begin with GRIN
(b) Have all 3 E’s together?
Answer:
The word has 11 letters, E = 3, N = 3, G = 2, I = 2
Total permutations \(\frac{11 !}{3 !, 3 !, 2 ! 2 !}\)
(a) Permutations word which begin with GRIN = \(\frac{7 !}{3 ! 2 !}\)
(b) All 3E’s are together Consider 3E as one unit remaining letters = 8 + 1 = 9
N = 3,G = 2, 1 – 2 No of permutation = \(\frac{9 !}{3 ! 2 ! 2 !}\)

Question 28.
One Card is randomly drawn from a pack of 52 cards, Find the probability that
(a) Card is either black or jack
(b) Card is red
(c) Card is diamond.
Answer:
Totally there are 52 cards and any one can be drawn
∴ Total no cases = n = 52

(a) There are 26 black cards and 2 jack totally = 28 cards ∴ no of favourable case = m = 28
∴ P(black or jack) = \(\frac{28}{52}\)

(b) There are 26 red cards
∴ m = 26,

(C) There are 13 diamonds cards
∴ m = 13,
P(getting diamond) = \(\frac{13}{52}=\frac{1}{4}\)

Question 29.
Walking 4 kmph a student reaches his college 5 minutes late and if he walks at 5 kmph, he reaches \(2 \frac{1}{2}\) minutes early. What is the distance from his house to the college?
Answer:
Let the distance from his house to the college = x km
Time taken to Cover x kns at 4km ph = \(\frac{x}{4}\) hr⁵
Time to cover 5 km ph = \(\frac{x}{5}\) hrs

Question 30.
A bill for ₹ 12,900 was drawn on 3 Feb, 2014 at 6 months and discounted on 13 March 2014 at 8% p.a. For what sum was the bill discounted ?
Answer:
F = 12,900, bill period = 6 months, Drawn date = 03 – 2 – 14
∴ LDD = DD + BP + Grace period

Date of discount 13-3-2014. This means, the no of days from 13-2-2014 to 06-08-14 becomes the unexpired period

∴ t = 145 days = \(\frac{146}{365}\) year, r = 80% = 0.08, DV = ?
BD = Ftr = 12,900 × \(\frac{146}{365}\) × 0.08 = 412.8
∴ Discounted value = 12,900 – 412.8 = 12,487.20

Question 31.
Prathik sells out ₹ 6,000 of 7.5% stock at 108 and reinvests the proceeds in 9% stock. If Prathik’s income increase by ₹ 270, at what price did Prathik buy 9% stock?
Answer:
To calculate income

Income = \(\frac{6000 \times 7.5}{10}\) = 459.0 = ₹ 450

For 9% stock given income increase by ₹ 270
∴ New income = 450 + 270 = ₹ 270.

Question 32.
A furniture dealer sold furniture for ₹ 21,000 and added 5% sales tax to the quoted price. The customers agree to buy it for ₹ 21,000 including sales tax. Find the discount he received.
Answer:
Discount = 5% 21,00 = \(\frac{5}{10}\) × 21,000 = 1050

Question 33.
Prove that \(\frac{\cos 75^{\circ}+\cos 15^{\circ}}{\sin 75^{\circ}-\sin 15^{\circ}}=\sqrt{3}\)
Answer:

Question 34.
Find equation of the parabola given that its vertex is (0,0) axis is Y-axis and passed through (-1,-3).
Answer:
Given v = (0,0) and axis is y-axis
The parabola is of the form x² = 4ay or x² = -4ay
But this equation passes through (-1,-0) which is in III qhadrant.
∴ The curve turn down words and is of the form x² = – 4ay, But it passes through (-1,-3)
∴ (-1)² = -4a(-3) ⇒ 1 = 12a ⇒ a = \(\frac{1}{12}\)
∴ The required equation is x² = -4 . \(\frac{1}{12}\)y ⇒ x² = \(\frac{-1}{3}\) or 3x² + y = 0

Question 35.
If x = \(\frac{1-t^{2}}{1+t^{2}}\) y = \(\frac{2 t}{1+t^{2}}\) find \(\frac{\mathrm{dy}}{\mathrm{dt}}\)
Answer:
Differentiate both w.r.tt


If the total cost function is given by C(x) = 350 + 12x + \(\frac{x^{2}}{4}\) and revenue function is given by R(x) = 75x – 2x², find the level of output at which profit is maximum.
Profit function = TR – TC = 75x – 2x – 350 – 12x – \(\frac{x^{2}}{4}\) p(x) = -2x² – x² + 63x – 350
For maximum profit differentiate w.r.t x \(\frac{d}{d x}\) P(x) = 0

126 = 9x ⇒ x = \(\frac{126}{9}\) = 14 ∴ x = 14

Question 37.
Evaluate ∫x² log xdx
Answer:

Question 38.
Evaluate \(\int \frac{1+\cos x}{1-\cos x} d x\)
Answer:

Part – D

Answer any six questions: (6 × 5 = 30)

Question 39.
Find the term independent of x in \(\left(x^{3}-\frac{3}{x^{2}}\right)^{15}\)
Answer:
T_r+1 = ⁿC_r x^n-r a^r
Here x → x³a → \(\frac{-3}{x^{2}}\) n → 15
T_r+1 = ¹⁵C_r (x³)^{15 – r} \(\left(\frac{-3}{x^{2}}\right)\) = ^{15C_{r .}} x^{45 – 3r} (-3)^r . x^-2r
T_r+1 = ¹⁵C_r(-3)^r . x^{45 – 5r}
To find the term independent of x equate the power of x = 0
⇒ 45 – 5r = 0 ⇒ 45 = 5r = 0 ⇒ r = 9
T₉₊₁ = ¹⁵C₉(-3)⁹ . x⁰ = ¹⁵C₉(-3)⁹
T₁₀ = 3⁹, ¹⁵C₉ is the term independent of x

Question 40.
Resolve into partial fractions \(\frac{3 x+2}{(x-2)(x+3)^{2}}\)
Answer:

3x + 2 = A(x +3 )² + B(x – 2)(x – 3) + C(x – 2)
Put x = 2,6 + 2 = A (2 + 3))² ⇒ A = \(\frac { 8 }{ 25 }\)
Put x = -3 -9 + 2 = 0 + 0 + C(-5)
-7 = -5C ⇒ C = 715
Put x = 0 we get B = \(-\frac { 8 }{ 25 }\)

Question 41.
Verify whether the proposition ~(p ∨ q) → (~P ∧ ~q) is a tautology contradiction or neither.
Answer:

Question 42.
Two taps can separately fill a tank in 12 minutes and 15 minutes respectively. The tank when full can be emptied by a drain tap in 20 minutes. When the tank was empty all the three taps were opened simultaneously. In what time will the tank be filled up?
Answer:
1^st tap can fill \(\frac{1}{12}^{\text {th }}\) tank in 1 min 2nd tap can fill \(\frac{1}{15}^{\text {th }}\) tank in 1 min
Drain pipe drain out = \(\frac{1}{20}^{\text {th }}\) tank in 1 min
∴ In 1 min = \(\frac{1}{12}+\frac{1}{15}-\frac{1}{20}=\frac{15+12-9}{180}=\frac{18}{180}=\frac{1}{10}\) of tank will get filled
∴ The tank will get filled in 10 minutes.

Question 43.
ABC company required 1000 hours to produce first 30 engines. If the learning effect is 90%, find the total labour cost at ₹ 20 per hour to produce total of 120 engines.
Answer:
Assume 1 lot = 30 engines
∴ 120 engines = 4 lots

The total no of hrs to produce 4 lots = 3240
The lab cost at 20 per hour = 3240 × 20 = ₹ 64,800

Question 44.
Solve the L.P.P graphically
Maximize Z = 2x + 3y
Subject to constraints x + y ≤ 400, 2x + y ≤ 400, x,y ≥ 0.
Answer:
Consider
X + y = 400 2x + y = 600

Point of intersection of x + y = 400 & 2x + y = 600

Function is maximum at the point A (0.400) and maximum value = 1200

Question 45.
The angles of elevation of the top of a tower the base and the top of a building are 60° and 30°. 3x + 5y ≤ 15, 5x + 2y ≤ 10 and x ≥ 0, y ≥ 0.
Answer:
Let AB = Building & CD is the tower. From the rt angled ∆ ACD
We have tan 60 = \(\frac { h }{ x }\)
√3 = \(\frac { h }{ x }\) ⇒ x = √3(h -20)
From 1 and 2 wet √3(h – 20) = \(\frac{\mathrm{h}}{\sqrt{3}}\) ⇒ 3(h – 20) = h ⇒ 3h – 60 = h ⇒ 3h – 60 = h ⇒ 2h = 60 ⇒ h = 30 mts
∴ Height of the tower = 30mts

Question 46.
Find the equation of circle passing through points (5,3) (1,5) (3,-1)
Answer:
Let the required equation of the circle is
x² + y² + 2gx + 2fg + c = 0
But this eqn passes through the points
(5,3) 10g + 6f + C + 34 = 0
(1,5) 2g + 10 f + C + 26 = 0
(3,-1) 6g – 2f + C + 10 = 0
Solving 2,3 and 4 we get
2 – 3 ⇒ 2g – f + 2 = 0
3 – 4 ⇒ g – 3f – 4 = 0
Again solving we get y = -2, f = -2, c = -2
This the required egn of the circle is x² + y² – 4x – 4y -2 = 0

Question 47.
If y = \((x+\sqrt{1+x^{2}})^{m}\) prove that (1 + x²)y₂ + xy₁ – m²y=0
Answer:
Given y = \((x+\sqrt{1+x^{2}})^{m}\)
Diff. w.r.tx

\(\sqrt{1+x^{2}} y_{1}\) = my S.BS
(1 + x²)y²₁ = m²y²
Again differentiate w.r.t.x (1 + x²).2y₁y₂ + y²₁ (2x) = m².2yy₁ ÷ 2y₁ ∴ (1 + x²).y₂ + xy₁ – m²y = 0

Question 48.
Find the area enclosed between parabola y² = x and the line x + y = 2.
Answer:
Given y² = x and x + y = 2
⇒ (2 – x)² = x ∴ y = 2 – x
⇒ 4 + x² – 4x – x = 0 ⇒ x² – 5x – 4 = 0 ⇒ x = 4 or

Part – E

Answer any one questions: (1 × 10 = 10)

Question 49.
(a) A school wants to award to award its students for the values of leadership, good behavior and hard work with a total cash award of ₹ 6,000. Three times the award money for hard work added to that given for leadership amounts to 11,000. The award money for leadership and hard work together is double the one given for good behavior. Calculate the award money given for each value, using matrix method.
Answer:
Let the values of punctuality, good behavior and hard work be denoted by x,y & z respectively
x + y + z = 6000
x + oy + 3z = 11,000
x + z = 2y or x – 2y + z = 0
The given equations in matrix form is \(\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 0 & 3 \\ 1 & -2 & 1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \left[ \begin{matrix} 6000 \\ 11,000 \\ 0 \end{matrix} \right] \)
A x = B
⇒ X = A^-1B
Where A = \(\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 0 & 3 \\ 1 & -2 & 1 \end{matrix} \right] \)

X = ₹ 500, y = ₹ 2000, z = ₹ 3500.

(b). A company produces two products P and Q. Each P require 4 hours of grinding and 2 hours of polishing and each Q requires 2 hours of grinding and 5 hours of polishing. The total available hours for grinding is 20 and for polishing is 24. Profit per unit of P is 6 and that of Q is ₹ 8. Formulate the L.P.P
Answer:
Let x be the units of P manufactured and y be the units of Q manufactured


Maximize z = 6x + 8y
Subject to constraints

Question 50.
(a). Prove that \(\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\), where θ is in radius.
Answer:
Let ‘O’ the centre of unit circle, assume that is measured in positive radians.

From the figure, we have area of ∆AOB = area of sector AOB = area of ∆AOD

∴ Area of sector = \(\frac { 1 }{ 2 }\) (radius)² × angle in radians
From the triangle DOA, tan θ = \(\frac{\mathrm{DA}}{\mathrm{OA}}\) [∴ DA = r tanθ]
From the triangle BOC, sin θ = \(\frac{\mathrm{BC}}{\mathrm{OB}}\) [∴ BC = r sin θ]
(1) becomes
⇒ BC ≤ θ ≤ DA
⇒ sin θ ≤ θ ≤ tan θ

Dividing by sin θ

(b). Find the value of (1.1)⁵ using binomial theorem upto 4 decimal places.
Answer:
(1.1)⁵ = (1 + 0.1)⁵
= 1⁵ + ⁵C₁ (0.1) + ⁵C₂(0.1)² + ⁵C₃ (0.1)³ + ⁵C₄(0.1)⁴ + ⁵C₅(0.1)⁵
= 1 + 5(0.1) + 10(0.1) + 10(0.001) + 1(0.00001)
= 1 + 0.5 + 0.1 + 0.01 + 0.005 + 0.00001
(1.1)⁵ = 1.61051

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Karnataka 2nd PUC Maths Question Bank Chapter 13 Probability Ex 13.3

2nd PUC Maths Probability NCERT Text Book Questions and Answers Ex 13.3

Question 1.
An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red ?
Answer:
In the first draw the ball may be red or black

(1) If red ball is drawn, 2 more red balls are put and then draw a red ball
(2) If black ball is drawn 2 more black ball is put and then again draw a red ball
∴ Required probability is
\(\frac{5}{10} \times \frac{7}{12}+\frac{5}{10} \times \frac{5}{12}=\frac{60}{120}=\frac{1}{2}\)

Question 2.
A bag contains 4 red and 4 black balls , another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first
Answer:
Let E₁ : The first bag is selected
E₂ : The second bag is selected
A : The ball is red

Question 3.
Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year result report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hostlier ?
Answer:
E₁ : Student is a hostler
E₂ : Student is a day scholar
A : student has an A grade

Question 4.
In answering a question on a multiple r. choice test, a student either knows the answer or guesses. Let \(\frac{3}{4}\) be the probability that he knows the answer and \(\frac{1}{4}\) be the probability that he guesses.Assuming that a student who guesses at the answer will be correct with probability \(\frac{1}{4}\) . What is the probability that the student knows the answer given that he answered it correctly ?
Answer:
E₁: student knows the answer
E₂ : student guesses the answer
A : Answer is correct

Question 5.
A laboratory blood test is 99% effective in detecting a certain disease when it is in
fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e. if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive ?
Answer:
E₁: person has the disease
E₂: person is healthy
A : Test is positive

Question 6.
There are three coins. One is a two headed coin (having head on both faces), another is a based coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?
Answer:
E₁ : 1^st coin is chosen
E₂ 2^nd coin is chosen
E₂: 3^rd coin is chosen
Let ‘A’ be the even that shows head

Question 7.
An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?
Answer:
E₁: Insured person is scooter driver
E₂: Insured person is car driver
E₃: Insured person is truck driver
A : Meets with an accident

Question 8.
A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B?
Answer:
E₁ : Item produce by machine A
E₂: Item produce by machine B
A : Item is defective

Question 9.
Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.
Answer:
E₁ : 1^st group will win
E₂ : 2^nd group will win
A : New product is introduced

Question 10.
Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?
Answer:
E₁ : 5 or 6 shows on the die     HHH
E₂ : 1, 2, 3,4 shows on the die  HHT
A : Shows Head exactly once   HTH

Question 11.
A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, where as the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on the job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that it was produced by A?
Answer:
E₁ : item produce by A
E₂ : item produce by B
E₃ : item produce by C
A : item produced is defective

Question 12.
A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are . drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.
Answer:
Let E₁ : Lost card is diamond.
E₂ : Lost card is not diamond
A : Two cards drawn are diamond

Question 13.
Probability that A speaks truth is \(\frac{4}{5}\) A coins is tossed. A reports that a head appears. The probability that actually there was head is
(A) \(\frac{4}{5}\)
(B) \(\frac{1}{2}\)
(C )\(\frac{1}{5}\)
(D) \(\frac{2}{5}\)
Answer:
Let E₁ : Head appers
E₂ : Tail appers
A : Reports that head appears

Question 14.
If A and B are two events such that A⊂B and P(B) ≠ 0,then which of the following is correct?
(A) \(\mathbf{P}(\mathbf{A} | \mathbf{B})=\frac{\mathbf{P}(\mathbf{B})}{\mathbf{P}(\mathbf{A})}\)
(B) P(A|B) < p(A)
(C) P(A|B) ≥ p(A)
(D) None of these
Answer:

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Karnataka 2nd PUC Maths Question Bank Chapter 13 Probability Ex 13.2

2nd PUC Maths Probability NCERT Text Book Questions and Answers Ex 13.2

Question 1.
If \(\mathbf{P}(\mathbf{A})=\frac{3}{5} \text { and } \mathbf{P}(\mathbf{B})=\frac{1}{5}, \text { find } \mathbf{P}(\mathbf{A} \cap \mathbf{B})\) ,find P(A∩B) if A and B are independent events.
Answer:
If two events A and B are independents P (A∩B) = P(A). P(B)
\((A \cap B)=\frac{3}{5} \times \frac{1}{5}=\frac{3}{25}\)

Question 2.
Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.
Answer:
There are 26 black cards out 52 cards. When cards are drawn without replacement,
\(P(\text { both black cards })=\frac{26 \mathrm{C}_{2}}{52 \mathrm{C}_{2}}=\frac{26 \times 25}{52 \times 51}=\frac{25}{102}\)

Question 3.
A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.
Answer:
Since selection is without replacement

Question 4.
A fair coin and an unbiased die are tossed.Let A be the event ‘head appears on the coin’ and B be the event ‘3 on the die’.Check whether A and B are independent events or not.
Answer:

Question 5.
A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, ‘the number is event, and B be the event, ‘the number is red’. Are A and B independent?
Answer:
A : number is even, B : The number is red

Question 6.
Let E and F be events with P(E) \({ P }({ E })=\frac { 3 }{ 5 } ,{ P }({ F })=\frac { 3 }{ 10 } { \quad and\quad }{ P }({ E }\cap { F })=\frac { 1 }{ 5 } \). Are E and F independent?
Answer:

hence they are not independent.

Question 7.
Given that the events A and B are such that \(P(A)=\frac { 1 }{ 2 } ,P(A\cup B)=\frac { 3 }{ 5 } \quad { and }\quad P(B)=p\) Find p if they are
(i) mutually exclusive
(ii) independent.
Answer:
(i) If the events are mutually exclusive

Question 8.
Let A and B be independent events with
P(A) = 0.3 and P (B) = 0.4. Find
(i) P(A∩B)
(ii)P(A∪B)
(iii) P (A|B)
(iv) P (B|A)
Answer:

Question 9.
If A and B are two events such that
\(P(A)=\frac { 1 }{ 4 } ,{ P }({ B })=\frac { 1 }{ 2 } { and }{ P }({ A }\cap { B })=\frac { 1 }{ 8 } \)
Answer:

Question 10.
Events A and B are such that \({ P }({ E })=\frac { 3 }{ 5 } ,{ P }({ F })(B)=\frac { 7 }{ 12 } \) P (not A or not B) = \(\frac{1}{4}\). State whether A and B are independent ?
Answer:

Question 11.
Given two independent events A and B such that P (A) = 0.3, P (B) = 0.6. Find
(i) P (A and B)
Answer:
i.e. P (A∩B) = P (A) x P (B) = .3 x .6 = .18

(ii) P (A and not B)
Answer:
P(A and not B) = \(P(A \cap \bar{B})\)
= P(A) x \(P(\bar{B})\) ⇒ P (A) x [1-P(B)]
= .3 x (1 – 0.6) = .3 x .4 = .12

(iii) P (A or B)
Answer:
P (A or B) = P (A∪B)
= P (A) + P (B) – P (A∩B)
= .3+ .6-.18 = .9 -.18 = .72

(iv) P (neither A nor B)
Answer:
P (neither A nor B) = \(P(\bar{A} \cap \bar{B})\)
= \(P(\overline{A \cup B})\)
= 1 -P(A∪B)= 1 -.72 = .28

Question 12.
A die is tossed thrice. Find the probability of getting an odd number at least once.
Answer:
E : an odd number at least once
\(\overline{\mathrm{E}}\) : not an odd number
\(\overline{\mathrm{E}}\) : an even number on all times
we get even all time 3 x 3 x 3 = 27
\(P(E)=1-P(\bar{E})=1-\frac{27}{216}=\frac{7}{8}\)

Question 13.
Two balls are drawn at random With replacement from a box containing 10 black and 8 red balls. Find the probability that
(i) both balls are red.
(ii) first ball is black and second is red
(iii) one of them is black and other is red
Answer:

Question 14.
Probability of solving specific problem independently by A and B are \(\frac{1}{2} \text { and } \frac{1}{3}\) respectively. If both try to solve the problem independently, find the probability that
(i) the problem is solved
(ii) exactly one of them solves the problem.
Answer:

Question 15.
One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent ?
(i) E : ‘the card drawn is a spade’
F : ‘the card drawn is a ace’
(ii) E : ‘the card drawn is black’
F : ‘the card drawn is a king’
(iii) E : ‘the card drawn is a king or queen’
F : ‘the card drawn is a queen or jack’.
Answer:

Question 16.
In a hostel, 60% of the students read Hindi news paper, 40% read English news paper and 20% read both Hindi^and English news papers. A student is selected at random.
(a) Find the probability that she reads neither Hindi nor English news papers.
(b) If she reads Hindi news paper, find the probability that she reads English news paper.
(c) If she reads English news paper, find the probability that she reads Hindi news paper.
Answer:
(a) 20% read neither Hindi nor english

Choose the correct answer in Exercises 17 and 18.

Question 17.
The probability of obtaining an even prime number on each die, when a pair of dice is rolled is
(A) 0
(B) \(\frac{1}{3}\)
(C) \(\frac{1}{12}\)
(D) \(\frac{1}{36}\)
Answer:
Even Prime no is (2,2)
P (even prime no) = \(\frac{1}{36}\)
Option ‘D’

Question 18.
Two events A and B will be independent, if
(A) A and B are mutually exclusive
(B) P (A’B’) = [1 – P(A)] [1 – P(B)]
(C) P (A) = P (B)
(D) P (A) + P (B) = 1
Answer:
P(A’B’) = P(A’∩B’)
= \(P\bar { A\cup B } \)
= 1 – P (A∪B)
= 1 – P(A) – P(B) – P(A) – P(B)
= 1 – P (A) – P (B) (1 – P (A)]
= [1 – P (A)] [1 -P(B)
Option ‘B’

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Karnataka 2nd PUC Maths Question Bank Chapter 13 Probability Ex 13.1

2nd PUC Maths Probability NCERT Text Book Questions and Answers Ex 13.1

Question 1.
Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E∩F) =0.2, find p(E|f) and P(F|E).
Answer:

Question 2.
Compute P(A|B), if P(B) = 0.5 and P (A∩B) = 0.32
Answer:

Question 3.
If P(A) = 0.8, P(B) = 0.5 and P(B|A) = 0.4 Find.
(i) P(A∩B)
(ii) P(A|B)
(iii) P(A∪B)
Answer:

Question 4.
Evaluate P(A∪B), if 2P(A) = \(P(B)=\frac { 5 }{ 13 } { and }{ \quad P }({ A }|{ B })=\frac { 2 }{ 5 } \)
Answer:

Question 5.
\(\text { If } P(A)=\frac{6}{11}, P(B)=\frac{5}{11} \text { and } P(A \cup B)=\frac{7}{11}\) find
(i) P(A∩B)
(ii) P(A|B)
(iii) P(B|A)
Answer:

Question 6.
A coin is tossed three times, where
(i) E : head on third toss, F : heads on first two tosses
(ii) E : at least two heads, F : at most two heads
(iii) E : at most two tails, F : at least one tail
Answer:
sample space {HHH, HHT, HTT, TTT, THH, TTH, THT, HTH}


Question 7.
Two coins are tossed once, where
(i) E : tail appears on one coin, F : one coin shows head
(ii) E: no tail appears, F: no head appears
Answer:

Question 8.
A die is thrown three times,
E : 4 appears on the third toss,
F : 6 and 5 appears respectively on first two tosses
Answer:

Question 9.
Mother, father and son line up at random for a family picture
E : son on one end,
F : father in middle
Answer:
sample space [MFS, MSF, SMF, SFM, FMS, FSM]

Question 10.
A black and a red dice are rolled.
(a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.
(b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.
Answer:
(a) E : Sum gr than 9 = {(4,6), (6,4), (5,5), (6,5), (5,6), (6,6)}
F : Black resulted in 5 = {(5,1), (5,2), (5,3), (5,4),(5,5), (5,6)}
E∩F = sum gr than 9 and black die resulted in 5

Question 11.
A fair die is rolled. Consider events E = {1,3, 5}, F = (2,3} and G = {2, 3,4,5} Find
(i) P (E|F) and P (F|E)
(ii) P (E|G) and P (G|E)
(iii) P ((E∪F)|G) and P (E∩F)|G)
Answer:

Question 12.
Assume that each born child is equally likely to be a girl. If a family has two children, what is the conditional probability that both are girls given that
(i) the youngest is a girl.
(ii) at least one is girl ?
Ans:
E : Both are girls {GG, BB, GB, BG}
F: Youngest is a girl
G: At least one is girl

Question 13.
An instructor has a question bank  consisting of 300 easy True/False questions, 200 difficult True/False questions, 500 easy multiple choice question and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question ?
Answer:
E : Easy question
F: Multiple choice question Given 300 – E (T|F)
500 – E multiple 200 – D (T|F)
400 – D (multiple)

Question 14.
Given that the two numbers appearing on throwing two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4
Answer:
E : Sum of the number is 4 [(1,3), (3,1), (2,2)]
F : The number appearing are different E∩F different number but sum is 4 throwing two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4′.
Answer:
E : Sum of the number is 4 [(1,3), (3,1), (2,2)]
F : The number appearing are different
E∩F : different number but sum is 4
\(P(F)=\frac{30}{36}\)
{except (1,1) , (2,2) , (3,3) ,(4,4) ,(,5,5) ,(6,6)}

Question 15.
Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the 1 die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’.
Answer:
{(3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (1,H), (1,T), (2,H), (2,T), (4,T), (4,H), (5,T), (5,H), (6,T),(6,H)}
E: coin shows a tail

∴ Event is called impossible event
In each of the Exercise 16 and 17 choose the correct answer:

Question 16.
If P(A) = \(\frac{1}{2}\),P(B) = 0, then P( A|B) is
(A) 0
(B) \(\frac{1}{2}\)
(C) not defined
(D) 1
Answer:

defined
Option “C”

Question 17.
If A and B are events such that P (A|B) = P (B|A), then
(A) A⊂B but A≠B
(B) A = B
(C) A∩B =φ
(D) P(A) = P(B)
Answer:

Option ‘D’

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Karnataka 2nd PUC Basic Maths Previous Year Question Paper March 2019

Time: 3 Hours 15 Min.
Max.Markss: 100

Instructions:

• Write SI Nos of question correctly
• Visually challenged students need to answer Question No 31(B) instead of map Question No 31(A) in Part D
• Answer the questions according to the instructions given for the questions

Part A

Answer all the Questions: (10 × 1 = 10)

Question 1.
If A = \(\left[ \begin{matrix} 1 & -2 \\ 3 & 4 \end{matrix} \right] \) then find 2A’.
Answer:
2A’ = \(\left[ \begin{matrix} 2 & 6 \\ -4 & 8 \end{matrix} \right]\)

Question 2.
In how many ways can 10 people be seated around a table?
Answer:

Question 3.
Symbolise the proposition:
“If Oxygen is a gas then gold is a compound”.
Answer:
p → 9

Question 4.
Find the triplicate ratio of 5 : 4.
Answer:
5³ : 4³ or 125 : 64

Question 5.
Find the income obtained by investing ₹ 3,600 in 5% stock at 90.
Answer:
Income = \(\frac{5 \times 3600}{90}\) = ₹200

Question 6.
Express sin 5A cos 3A as sum of difference of two trigonometric functions.
Answer:
\(\frac { 1 }{ 2 }\)[sin8A + sin 2A]

Question 7.
Find the centre of the circle x² + y² – 4xy – 5 = 0.
Answer:
\(\left(2,+\frac{1}{2}\right)\)

Question 8.
Evaluate : \(\lim _{x \rightarrow 3}\left(\frac{x^{2}-4 x}{x-2}\right)\)
Answer:
-3

Question 9.
Differentiate (5e^x – logx – 3√x) with respect to x.
Answer:
\(5 e^{x}-\frac{1}{x}-\frac{3}{2 \sqrt{x}}\)

Question 10.
Integrate (x² – \(\frac { 6 }{ x }\) + 5e^x) with respect to x.
Answer:
\(\frac{x^{3}}{3} 6 \log _{e} x+5 e^{x}+C\)

Part- B

Answer any ten of the following Questions. (10 × 2 = 20)

Question 11.
If A = \(\left[ \begin{matrix} 1 & 3 & -1 \\ -1 & 0 & 2 \end{matrix} \right] \) and B = \(\left[ \begin{matrix} 4 & -1 & 2 \\ 1 & 3 & -2 \end{matrix} \right] \)
Answer:
3B = \(\left[ \begin{matrix} 12 & -3 & 6 \\ -4 & -9 & 8 \end{matrix} \right] \)
∴ A = 3B = \(\left[ \begin{matrix} -11 & 6 & -7 \\ -4 & -9 & 8 \end{matrix} \right] \)

Question 12.
In how many ways the word “CARROM” be arranged such that the 2R’s are always together?
Answer:
₹ 2 as 1 unit
∴ No. of letters = 6 – 2 + 1 = 5

Question 13.
A box contains 8 red marbles 6 green marbles and 10 pink marbles. One marble in drawn at random from the box. What is the probability that the marble drawn is either red or green?
Answer:
Writing 8R + 6G + 10P = 24 OR No. of R & G = 14
∴ Probability = \(\frac{14}{24}=\frac{7}{12}\)

Question 14.
If the truth values of the propositions p,q,r are T,T,F respectively, then find the truth values of p → (q ^∧ r).
Answer:

Question 15.
What must be added to each term in the ratio 5 : 6 so that it becomes 8 : 9?
Answer:
Let x be added
\(\therefore \quad \frac{5+x}{6+x}=\frac{8}{9}\)
Solving to get x = 3

Question 16.
True discount on a bill was ₹ 900 and Banker’s gain was ₹ 27. What is the face value of the bill?
Answer:
Getting BD = ₹ 927
₹ F = Rs. 30,900

Question 17.
Find the value of cos 15°.
Answer:
cos 15° = cos (45° – 30°)
= cos 45°. cos30° + sin 45°. sin 30°
Getting answer = \(\frac{\sqrt{3}+1}{2 \sqrt{2}}\)

Question 18.
If tan A = \(\frac { 1 }{ 2 }\), tan B = \(\frac { 1 }{ 3 }\), then find tan (A + B).
Answer:

Simplifying to get answer = 1

Question 19.
Find the equation of parabola whose vertex is (0,0) and focus is (3,0).
Answer:
Identifying y² = 4ax
Getting a = 3 and hence y² = 12x

Question 20.

Answer:
Getting LHL = 0 and RHL = -2
Concluding f(x) is not continuous at x = 1.

Question 21.
If y = (a² – x²)¹⁰, find \(\frac{d y}{d x}\)
Answer:
\(\frac{d y}{d x}\) = 10(a² – x²)⁹(0 – 2x)
= – 20x(a² – x²)⁹

Question 22.
If the cost function of a firm is given by C(x) = x³ – 3x + 7. Find the average cost and marginal cost.
Answer:
AC = x² – 3 + \(\frac{7}{x}\)
MC = 3x² – 3

Question 23.
Evaluate : \(\int \frac{4 x+3}{2 x^{2}+3 x+5} d x\)
Answer:
Substituting 2x² + 3x + 5 = t
∴ (4x + 3)dx = dt

Question 24.
Evaluate : \(\int_{1}^{2} x e^{4} d x\)
Answer:
Writing \(\left[x e^{x}\right]_{1}^{2}-\int_{1}^{2} e^{x} \cdot 1 d x\)
Simplifying to get answer = e².

Part-C

Answer any ten of the following questions: (10 × 3 = 30)

Question 25.
If A = \(\left[ \begin{matrix} 2 & 3 \\ -4 & 1 \end{matrix} \right] \) and B = \(\left[ \begin{matrix} -1 & 5 \\ 6 & 2 \end{matrix} \right] \)
Answer:

Question 26.
Prove that: \(\left| \begin{matrix} 1+a & b & c \\ a & 1+b & c \\ a & b & 1+c \end{matrix} \right| \) = 1 + a + b + c
Answer:

Question 27.
From a class of 9 boys and 7 girls 12 students are to be chosen for a competition which includes atleast 6 boys and atleast 4 girls. In how many ways can this be done if a particular boys is always chosen?
Answer:
No. of ways
= ⁸C₅ × ⁷C₆ + ⁸C₆ × ⁷C₅ + ⁸C₇ × ⁷C₄
= 56 × 7 + 28 × 21 + 8 × 35 = 1260

Question 28.
A couple appears in an interview for two vacancies in the same post. The probability of husband getting selected is \(\frac { 1 }{ 7 }\) and the probability of wife getting selected is \(\frac { 1 }{ 5 }\) . What is the probability that
a. Both of them will be selected?
b. Only one of them will be selected?
c. None of them will be selected?
Answer:
Writing:
P(husband not selected) = \(\frac { 6 }{ 7 }\)  and
P(Wife not selected) = \(\frac { 4 }{ 5 }\)
(OR) Getting:
a. P(Both slected) = \(\frac { 1 }{ 35 }\)
b. P(Only one selected) = \(\frac { 10 }{ 35 }\) = \(\frac { 2 }{ 7 }\)
C. P(None selected) = \(\frac { 24 }{ 35 }\)

Question 29.
In a fort, there was ration for 560 soldiers that would last the soldiers for 70 days after 20 days, 60 soldiers left the fort. For how many days the remaining ration can support the remaining soldiers?
Answer:
Writing:

Writing:
560 : 500 = x : 50
⇒ x = 56 days

Question 30.
For 512.50 due 6 months at 15% p.a. Find the true present value and discounted cash value.
Answer:
P = \(\frac{F}{1+t r}\) = ₹ 7476.74

OR
Getting P = ₹ 476.74
Getting BD = F.tr = ₹ 38.43

Question 31.
Sanjana invests 7 3240 in a stock at 108 and sells when price falls to 104. How much stock at 130 can Sanjana now by?
Answer:
Case 1: (Buying) stock at 108
Worth of stock purchased = \(\frac{100 \times 3240}{108}\)
= ₹ 3000

Case 2: Stock sold at 104

Case 3: Stock sold at 130

Question 32.
Sanju goes to a shop to buy a bicycle quoted at ₹ 2,000. The rate of sales tax is 12% on it. He asks the shopkeeper for a rebate on the price of the bicycle to such an extent that he has to pay ₹ 2016 inclusive of sales tax. Find the rebate percentage on the price of the bicycle.
Answer:
SP = 2016, ST = 12% Let MP = X
SP = MP + 12% MP
∴ 2016 = x + 0.12%
⇒ x = 1800
⇒ Rebate = 2000 – 1800 = ₹ 200
Rebate \(\%=\frac{200}{2000} \times 100=10 \%\)

Question 33.
Find the equation of parabola when vertex is (0,0) and axis is y-axis and passes through the point(-1,-3)
Answer:
Identifying x² = 4ay [or x² = – 4ay]
Substituting x = -1, y = -3

Question 34.
Differentiate y = \(\frac{e^{x}-1}{e^{x}+1}\) with respect to x.
Answer:
\(\frac{d y}{d x}=\frac{\left(e^{x}+1\right) e^{x}-\left(e^{x}-1\right)\left(e^{x}\right)}{\left(e^{x}+1\right)^{2}}\)
Substituting x = -1, y=-3.

Question 35.
Find the maximum and minimum value of y = x³ – 9x² + 15 x – 1.
Answer:
For max or min.
\(\frac{d y}{d x}\) = 0 ⇒ 3x² – 18x + 5 = 0
Getting x = 1, x = 5
\(\frac{d^{2} y}{d x^{2}}\) 6x – 18
At x = 1, \(\frac{d^{2} y}{d x^{2}}\) = -12
∴ y has max at x = 1
⇒ y_min = 6
At x = 5, \(\frac{d^{2} y}{d x^{2}}\) = 12
∴ y has min at x = 5
⇒ y_min = -26

Question 36.
A square plate is expanding uniformly, the side is increasing at the rate of 5cm / sec what is the rate at which the area and its perimeter is increasing when the side is 20 cm long?
Answer:
Area, A = l² ⇒ \(\frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}\) = 2l × \(\frac{\mathrm{d} l}{\mathrm{dt}}\)
= 2 × 20 × 5
\(\frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}\) = 200cm² / sec.
P = 41 ⇒ \(\frac{\mathrm{d} \mathrm{P}}{\mathrm{dt}}\) = 4 \(\frac{\mathrm{d} l}{\mathrm{dt}}\) 4 × 5= 20cm / sec

Question 37.
Integrate x² sin x with respect to x.
Answer:
∫x sin xdx = -x² cosx – ∫ -cosx.2xdx
= -x² cosx + 2[xsinx – ∫sinx.dx]
= -x² cosx + 2xsinx + 2cosx + C

Question 38.
Evaluate: ∫(2x + 3)(x² + 3x + 5)^3/2 dx.
Answer:
Substituting x² – 3x + 5 = t to get (2x + 3) dx = dt
∴ Given integral becomes ∫t^3/2dt
Integrating to get \(\frac{t^{52}}{52}+C=\frac{2}{5}\left(x^{2}+3 x+5\right)^{52}+C\)

Part – D

Answer any six questions: (6 × 5 = 30)

Question 39.
Find the middle term in the expansion of \(\left(\frac{2 x^{2}}{3}-\frac{3}{2 x}\right)^{10}\)
Answer:
Identifying middle term as T₆
T_r+1 = ¹⁰C_r\(\left(\frac{2 x^{2}}{3}\right)^{10-r}\left(\frac{-3}{2 x}\right)^{r}\)
Substituting r= 5 and simplifying to get T₆ = -252x⁵.

Question 40.
Resolve \(\frac{x-1}{x(x+2)(x+4)}\) into partial fractions.
Answer:

Getting Getting A = \(-\frac { 1 }{ 8 }\) B = \(\frac { 3 }{ 4 }\) C = \(\frac { -5 }{ 8 }\)
Conclusion

Question 41.
Verify whether the proposition (~p^q)^~r is a tautology or a contradiction or neither.
Answer:

Question 42.
Walking 4 kmph a student reaches his college 5 minutes late and if he walks at 5 kmph he reaches \(2 \frac{1}{2}\) minutes early. What is the distance from his house to the college?
Answer:
Let required distance = x kms
At 4 kmph, time taken = \(\frac { x }{ 4 }\)
At 5 kmph, time taken = \(\frac { x }{ 5 }\)
Given \(\frac{x}{4}=\frac{5}{60}=\frac{x}{5}+\frac{2.5}{60} s\)
Getting x = 2.5 km

Question 43.
A motor company Ltd. Has observed that a 90° learning effect applies to all labour related costs whenever a new product is taken up for production the anticipated production to 320 units. For the coming year. The at ₹ 15/hour. Calculate the total labour hours and labour cost to manufacture 320 units.
Answer:
10 units = 1 lot

∴ Total hours = 18,895.68
Total labour cost = 18,956.68 × 15 = ₹ 2,83,435.20

Question 44.
Solve the following LPP graphically:
Maximize Z = 60x + 15y
Subjected to the constraints: x + y ≤ 50
3x + y ≤ 90
And x, y ≥ 0
Answer:

Question 45.
If A + B + C = 180°, then prove that sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C.
Answer:
sin 2A + sin 2B + sin 2C
\(=2 \sin \left(\frac{2 \mathrm{A}+2 \mathrm{B}}{2}\right) \cdot \cos \left(\frac{2 \mathrm{A}-2 \mathrm{B}}{2}\right)+2 \sin \mathrm{C} \cos C\)
= 2 sin C. cos (A – B) + 2 sin C. cos C
= 2 sin C (cos(A – B) + cos C]
= 2 sin C[cos (A – B) + cos {180 – (A + B)}]
= 2 sin C[cos (A – B) – cos (A + B)]
= 2 sin C[-2 sin A . sin(- B)]
= 4 sin A. sin B.sin C

Question 46.
Find the equation of the circle passing through the points (5,3), (1,5) and (3,-1).
Answer:
Getting 3 equations as
10g + 6f + C = – 34
2g + 10f + C = -26
6g – 2f + C = – 10
Solving to get g = -, f = -2, C = -2
Final equation: x² + y² – 4x – 4y – 2 = 0

Question 47.
If y = log\((x+\sqrt{x^{2}+1})\) show that (x² + 1)y₂ + xy₁ = 0.
Answer:

Question 48.
Find the area bounded by the parabola y² = 4x and the line x = y.
Answer:
y² = 4x … (1)
x = y …. (2)
(1) & (2) ⇒ x = 0, x = 4
Required area
\(=\int_{0}^{4}[f(x)-g(x)] d x\)
\(=\int_{0}^{4}[2 \sqrt{x}-x] d x=\left[2 \times \frac{2}{3} x^{3 / 2}-\frac{x^{2}}{2}\right]_{0}^{4}\)
\(=\frac{32}{3}-\frac{16}{2}=\frac{8}{3}\) sq.units

Part – E

Answer any one of the following questions: (1 × 10 = 10)

Question 49.
(a). A sales person Samanth has the following record of sales. He is paid a commission at fixed rate per unit but varying rates for products P, Q and R.

Find the rate of commission payable on P, Q and R per unit sold
Answer:
Let rates of commission per unit of sales be
x,y and z for P, Q and Rrespectively
Writing 9x + 10y + 2z = 780
15x + 5y + 4z = 900
6x + 10y + 3z = 820

Note: Proportionate marks should be given to any other appropriate method.

(b). Find the value of (12.)⁵5 using Binomial theorem upto 4 places of decimals.
Answer:
(1.2)⁵5 = (1 + 0.2)⁵5
= 1⁵5 + ⁵C₁(0.2) + ⁵C₂(0.2)² + ⁵C₃(0.2)³ + ⁵C₄(0.2)⁴ + (0.2)⁵
= 1 + 5(0.2) + 10 (0.04) + 10(0.008) + 5(00004) + (0.00032)
= 2.4883

Question 50.
(a). If’n’ is a rational number and ‘a’ is a non zero real number, then prove that \(\lim _{x \rightarrow-}\left(\frac{x^{n}-a^{n}}{x-a}\right)=n a^{n-1}\)
Answer:
Proof for n positive integer
Proof for n negative integer
Proof for n rational number

(b). A person standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is 60°. When he returns 80 mts from the bank he finds the angle to be 30°. Find the height of the tree and the width of the river.
Answer:

tan 60° = \(\frac { h }{ x }\) ∴ √3 = \(\frac { h }{ x }\) …. (1)
tan 30° = \(\frac{h}{x+80}\) ∴ \(\frac{1}{\sqrt{3}}=\frac{h}{x+80}\) …. (2)
Solving (1) & (2) to get h = 40√3m
And x = 40 m

Students can Download Maths Chapter 12 Linear Programming Miscellaneous Exercise Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Maths Question Bank Chapter 12 Linear Programming Miscellaneous Exercise

Question 1.
Refer to Example 9. How many packets of each food should be used to maximise the amount of vitamin A in the diet? What is the maximum amount of vitamin A in the diet?
Answer:

Amount of	P(x)	Q (y)
Calcium	12	3
Iron	4	20
Chlosterol	6	4
Vitamin A	6	3

Z = 6x + 3y subject to the constraints
(i) 12x + 3y ≥240 ⇒ 4x + y ≥ 80
(ii) 4x + 20y ≥ 460 ⇒x + 5y ≥115
(iii) 6x + 4y ≤ 300 ⇒ 3x + 2y≤150
(iv) x ≥0 ; y≥0

ABC is the solution region
A (15,20) Z = 6 x 15+ 3 x 20= 150
B(40,15) Z = 6 x 40+ 3 x 15 = 285
C (2,72) Z = 6 x 2 + 3 x 72 = 228

Vitamin A is maximised at B (40,15) to 285 if P contains 40 units and Q containts 15 units.

Question 2.
A farmer mixes two brands P and Q of cattle feed. Brand P, costing ? 250 per bag, contains 3 units of nutritional element A, 2.5 units of element B and 2 units of element C. Brand Q costing ?200 per bag contains 1.5 units of nutritional element A, 11.25 units of element B, and 3 units of element C. The minimum requirements of nutrients A, B and C are 18 units, 45 units and 24 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag? What is the minimum cost of the mixture per bag?
Answer:


ABCD is the unbounded solution region
A (3,6)     Z = ₹ 1950
B (9,2)     Z = ₹ 2650
C (0,12)    Z = ₹ 2400
D (18,0) Z = ₹ 4500
cost is minimum at A (3,6)
Z = ₹ 1950 with 3 packets of P and 6 packets of Q.

Question 3.
A dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg food is given below:

Food	Vitamin A	Vitamin B	Vitamin C
X	1	2	3
Y	2	2	1

One kg of food X costs ₹ 16 and one kg of food Y costs ₹ 20. Find the least cost of the mixture which will produce the required diet?
Answer:

Mininize
Z = 16x +20y Subject to the constraints
(i) x + 2y ≥ 10
(ii) 2x + 2y ≥ 12 ⇒ x + y ≥ 6
(iii) 3x + y ≥ 8
(iv) x ≥ 0; y ≥ 0

shaded region is the un bounded solution region
A (0,8)                     Z = ₹ 160
B (1,5)                     Z = ₹ 116
C (2,4)                     Z = ₹ 112
D (10,0)                  Z = ₹ 160
The cost is minimised  at C (2,4) Z = ₹ 112 2 kg of X and 4 kg of Y.

Question 4.
A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given below:

Types of Toys	Machines
	I	II          III
A	12	18         6
B	6	0          9

Each machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is ₹ 7.50 and that on each toy of type B is ₹ 5, show that 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit.
Answer:

Machines
Toys	I	II	III	Profit
A (x)	12	18	6	₹ 7.50
B(y)	6	0	9	₹ 5

\(Z=\frac{15}{2} x+5 y\) subject to the constraints
(i) 12x + 6y ≤ 360 ⇒ 2x+y ≤  60
(ii) 18x ≤ 360 ⇒ x ≤ 20
(iii) 6x + 9y ≤ 360 ⇒ 2x +3y ≤ 120
(iv) x ≥ 0 , y ≥ 0

ABCDE is the solution region
A (0, 0) Z = 0
B (20, 0) Z = ₹ 150
C (20, 20) Z = ₹ 250
D (15, 30) Z = ₹ 262.5                                                                                      ,
E (0, 40) Z = ₹ 200                                                                                  ‘
At D (15, 30) Z is maximised to ₹ 262. 5
Hence proved.

Question 5.
An aeroplane can carry a maximum of 200 passengers. A profit of t 1000 is made on-each executive class ticket and a profit of ₹ 600 is made on each economy class ticket. The airline reserves at least 20 seats for executive class. However, at least 4 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximise the profit for the airline. What is the maximum profit?
Answer:

Z = 1000 x + 600 y
subject to the constreint
(i) x + y ≤ 200
(ii) x ≥ 20
(iii) y ≥ 80
(iv) x ≥ 0; y ≥ 0

A (20,180) Z= ₹ 128000
B(120,80) Z= ₹ 168000
C (20,80) Z = ₹ 68000
Z is maximised at B (120, 80) ₹ 168000
120 executive at B 80 economic class tickets

Question 6.
Two godowns A and B have grain capacity of 100 quintals and 50 quintals respectively. They supply to 3 ration shops, D, E and F whose requirements are 60, 50 and 40 quintals respectively. The cost of transportation per quintal from the godowns to the shops are given in the following table: How should the supplies be transported in order that the transportation cost is minimum? What is the minimum cost?

Transportation Cost  Per quintal (in ₹)
From/To	A	B
D	6	4
E	3	2
F	2.50	3

Answer:

(i) 60 – x ≥ 0 ⇒ x ≤ 60
(ii) 50 – y ≥0 ⇒ y ≤ 50
(iii) 100 – (x + y) ≥ 0 ⇒ x + y ≤ 100
(iv) x + y – 60 ≥ 0 ⇒ x + y ≥ 60

ABCD is the solution region
A (10, 50) Z = ₹ 510
B(50, 50) Z = ₹ 610
C (60, 40) Z = ₹ 620
D (60, 0) Z = ₹ 560
Cost is minimised to ₹ 510 at A (10,50)

Question 7.
An oil company has two depots A and B with capacities of 7000 L and 4000 L respectively. The company is to supply oil to three petrol pumps, D, E and F whose requirements are 4500L, 3000L and 3500L respectively. The distances (in km) between the depots and the petrol pumps is given in the following table:

Distance in (Km)
From/To	A	B
D	7	3
E	6	4
F	3	2

Assuming that the transportation cost of 10 litres of oil is Re 1 per km, how should the delivery be scheduled in order that the transportation cost is minimum? What is the minimum cost?
Answer:

(ii) 3000 – y ≥ 0 = y ≤ 3000
(iii) x + y – 3500 ≥ 0 = x+y ≥ 3500  x ≥ 0
(iv) 7000 – (x + y) ≥ 0= x + y ≥ 3500 y ≥ 0

ABCD is the the solution region
A (500, 3000)  Z = ₹ 4400
B (3500,0)       Z = ₹ 5000
C (4500,0)      Z = ₹  5300
D (4500, 2500) Z = ₹ 5550
E (4000,3000)  Z = ₹ 5450
Z is minimised at (500,3000)
i.e. ₹ 4400
From A to D 500 Ltrs
From A to E 3000 Ltrs
From A to F 3500 Ltrs
From B to D 4000 Ltrs.

Question 8.
A fruit grower can use two types of fertilizer in his garden, brand P and brand Q. The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least 240 kg of phosphoric acid, at least 270 kg of potash and at most 310 kg of chlorine. If the grower wants to minimise the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added in the garden?

kg per bag
	Brand P	Brand Q
Nitrogen	3	3.5
Phosphoric acid	1	2
Potash	3	1.5
Chlorine	1.5	2

Answer:

	P(x)	Q (y)
N2	3	3.5
Phosphoric	1	2
Potash	3	1.5
Chlorine	1.5	2

\(\text { To minimise } \mathrm{N}_{2} \mathrm{Z}=3 \mathrm{x}+\frac{7}{2} \mathrm{y}\)

∴ Amount of nitrogen is minimised equal to 470 kg when 40 bags of P and 100 bags of Q are mixed.

Question 9.
Refer to Question 8. If the grower wants to maximise the amount of nitrogen added to the garden, how many bags of each brand should be added? What is the maximum amount of nitrogen added?
Answer:
The amount of N₂ is maximum equal to 595kg when 140 bags of P and 50 bags of Q are mixed.

Question 10.
A toy company manufactures two types of dolls, A and B. Market tests and available resources have indicated that the combined production level should not exceed 1200 dolls per week and the demand for dolls of type B is at most half of that for dolls of type A. Further, the production level of dolls of type A can exceed three times the production of dolls of other type by at most 600 units. If the company makes profit of ₹ 12 and ₹ 16 per doll respectively on dolls A and B, how many of each should be produced weekly in order to maximise the profit?
Answer:
x be the no. of dolls A
y be the no. of dolls B
Maximize Z = 12x + 16 y subject to the constraints
(i) x + y < 1200
(ii) \(y \leq \frac{x}{2} \Rightarrow x-2 y \geq 0\)
(iii) x ≤ 3y + 600 ⇒ x -3y < 600
(iv) x ≥ 0; y ≥0

ABCD is the solution region
A (0,0)       Z = 0
B (600,0)   Z = ₹ 7200
C(1050,150) Z = ₹15000
D (800,400) Z = ₹ 16000
∴ Profit is maximum equal to ₹ 16000 when 800 balls A and 400 balls B are manufactured if sold.

2nd PUC Maths Linear Programming Miscellaneous Exercise Additional Question and Answers

Question 1.
One kind of cake required 300g of flour and 15g of fat, another kind requires 150gm of flour and 30gm of fat. Find the maximum number of cakes which can be made from 7.5kg of flour and 600gm of fat. (CBSE 2010)
Answer:
Let the 1st type of cake x and
2nd type of cake be y
∴ maximize z = x + y subject to
300x + 150 y ≤ 7500, 2x + y ≤ 50
15 x + 30 y ≤ 600 x + 2y ≤ 40
x ≤ 0, y ≤0

comer point are (0,0), (25,0) (20, 10), (0, 20)
At 0, Z = 0
A, Z = 25
B, Z = 30 – Maximum
C, C = 20
Maximum at B (20, 10)
∴ 20 pieces of 1 st type and
10 pieces of 2nd type cake

Question 2.
A merchant plans to sell two type of computer and desktop model and a portable model that will cost ₹ 25,000/ and ₹ 40,000. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computer should stock to get maximum profit if he does not want to invest more than 70 lakhs and his profit on the desktop models is ₹ 4500/ and on the portable model is ?5000/ (CBSE 2011)
Answer:
No. of desktop computer be x and no. of portable be y
Maximize
∴ Z = 4500 x + 5000 y
subject to x + y ≤ 250
25000 x + 4000 y ≤ 70,00,000
or 5x + 8y ≤ 1400, x ≥ 0, y
The corner points are.

0 (0,0), B (250,0), C (200,50), D (0,175)
At 0, Z = 0
At C  Z = 1150000
B Z= 1125000
D Z = 875000
Maximum of C (200, 50)
∴ No. of desktop model = 200
No. of portable = 50