## Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 8 Probability Distributions Miscellaneous Exercise 8 Questions and Answers.

## Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8

(I) Choose the correct alternative.

Question 1.
F(x) is c.d.f. of discreter r.v. X whose p.m.f. is given by P(x) = $$k\left(\begin{array}{l} 4 \\ x \end{array}\right)$$, for x = 0, 1, 2, 3, 4 & P(x) = 0 otherwise then F(5) = __________
(a) $$\frac{1}{16}$$
(b) $$\frac{1}{8}$$
(c) $$\frac{1}{4}$$
(d) 1
(d) 1

Question 2.
F(x) is c.d.f. of discrete r.v. X whose distribution is

then F(-3) = __________
(a) 0
(b) 1
(c) 0.2
(d) 0.15
(a) 0

Question 3.
X : number obtained on uppermost face when a fair die is thrown then E(X) = __________
(a) 3.0
(b) 3.5
(c) 4.0
(d) 4.5
(b) 3.5

Question 4.
If p.m.f. of r.v. X is given below.

then Var(X) = __________
(a) p2
(b) q2
(c) pq
(d) 2pq
(d) 2pq

Question 5.
The expected value of the sum of two numbers obtained when two fair dice are rolled is __________
(a) 5
(b) 6
(c) 7
(d) 8
(c) 7

Question 6.
Given p.d.f. of a continuous r.v. X as
f(x) = $$\frac{x^{2}}{3}$$ for -1 < x < 2
= 0 otherwise then F(1) =
(a) $$\frac{1}{9}$$
(b) $$\frac{2}{9}$$
(c) $$\frac{3}{9}$$
(d) $$\frac{4}{9}$$
(b) $$\frac{2}{9}$$

Question 7.
X is r.v. with p.d.f.
f(x) = $$\frac{k}{\sqrt{x}}$$, 0 < x < 4
= 0 otherwise then E(X) = __________
(a) $$\frac{1}{3}$$
(b) $$\frac{4}{3}$$
(c) $$\frac{2}{3}$$
(d) 1
(b) $$\frac{4}{3}$$

Question 8.
If X follows B(20, $$\frac{1}{10}$$) then E(X) = __________
(a) 2
(b) 5
(c) 4
(d) 3
(a) 2

Question 9.
If E(X) = m and Var(X) = m then X follows __________
(a) Binomial distribution
(b) Possion distribution
(c) Normal distribution
(d) none of the above
(b) Possion distribution

Question 10.
If E(X) > Var(X) then X follows __________
(a) Binomial distribution
(b) Possion distribution
(c) Normal distribution
(d) none of the above
(a) Binomial distribution

(II) Fill in the blanks.

Question 1.
The values of discrete r.v. are generally obtained by __________
counting

Question 2.
The values of continuous r.v. are generally obtained by __________
measurement

Question 3.
If X is dicrete random variable takes the values x1, x2, x3, …… xn then $$\sum_{i=1}^{n} p\left(x_{i}\right)$$ = __________
1

Question 4.
If f(x) is distribution function of discrete r.v. X with p.m.f. p(x) = $$\frac{x-1}{3}$$ for x = 1, 2, 3, and p(x) = 0 otherwise then F(4) = __________
1

Question 5.
If f(x) is distribution function of discrete r.v. X with p.m.f. p(x) = $$k\left(\begin{array}{l} 4 \\ x \end{array}\right)$$ for x = 0, 1, 2, 3, 4, and p(x) = 0 otherwise then F(-1) = __________
0

Question 6.
E(X) is considered to be __________ of the probability distribution of X.
centre of gravity

Question 7.
If X is continuous r.v. and f(xi) = P(X ≤ xi) = $$\int_{-\infty}^{x_{i}} f(x) d x$$ then f(x) is called __________
Cumulative Distribution Function

Question 8.
In Binomial distribution probability of success ________ from trial to trial.
remains constant/independent

Question 9.
In Binomial distribution, if n is very large and probability success of p is very small such that np = m (constant) then ________ distribution is applied.
Possion

(III) State whether each of the following is True or False.

Question 1.
If P(X = x) = $$k\left(\begin{array}{l} 4 \\ x \end{array}\right)$$ for x = 0, 1, 2, 3, 4, then F(5) = $$\frac{1}{4}$$ when f(x) is c.d.f.
False

Question 2.

If F(x) is c.d.f. of discrete r.v. X then F(-3) = 0.
True

Question 3.
X is the number obtained on the uppermost face when a die is thrown the E(X) = 3.5.
True

Question 4.
If p.m.f. of discrete r.v.X is

then E(X) = 2p.
True

Question 5.
The p.m.f. of a r.v. X is p(x) = $$\frac{2 x}{n(n+1)}$$, x = 1, 2,……n
= 0 otherwise,
Then E(X) = $$\frac{2 n+1}{3}$$
True

Question 6.
If f(x) = kx (1 – x) for 0 < x < 1
= 0 otherwise then k = 12
False

Question 7.
If X ~ B(n, p) and n = 6 and P(X = 4) = P(X = 2) then p = $$\frac{1}{2}$$.
True

Question 8.
If r.v. X assumes values 1, 2, 3,………, n with equal probabilities then E(X) = $$\frac{(n+1)}{2}$$
True

Question 9.
If r.v. X assumes the values 1, 2, 3,………, 9 with equal probabilities, E(X) = 5.
True

(IV) Solve the following problems.

Part – I

Question 1.
Identify the random variable as discrete or continuous in each of the following. Identify its range if it is discrete.
(i) An economist is interested in knowing the number of unemployed graduates in the town with a population of 1 lakh.
Solution:
X = No. of unemployed graduates in a town.
∵ The population of the town is 1 lakh
∴ X takes finite values
∴ X is a Discrete Random Variable
∴ Range of = {0, 1, 2, 4, …. 1,00,000}

(ii) Amount of syrup prescribed by a physician.
Solution:
X : Amount of syrup prescribed.
∴ X Takes infinite values
∴ X is a Continuous Random Variable.

(iii) A person on a high protein diet is interested in the weight gained in a week.
Solution:
X : Gain in weight in a week.
X takes infinite values
∴ X is a Continuous Random Variable.

(iv) Twelve of 20 white rats available for an experiment are male. A scientist randomly selects 5 rats and counts the number of female rats among them.
Solution:
X : No. of female rats selected
X takes finite values.
∴ X is a Discrete Random Variable.
Range of X = {0, 1, 2, 3, 4, 5}

(v) A highway safety group is interested in the speed (km/hrs) of a car at a checkpoint.
Solution:
X : Speed of car in km/hr
X takes infinite values
∴ X is a Continuous Random Variable.

Question 2.
The probability distribution of a discrete r.v. X is as follows.

(i) Determine the value of k.
(ii) Find P(X ≤ 4), P(2 < X < 4), P(X ≥ 3).
Solution:
(i) Assuming that the given distribution is a p.m.f. of X
∴ Each P(X = x) ≥ 0 for x = 1, 2, 3, 4, 5, 6
k ≥ 0
ΣP(X = x) = 1 and
k + 2k + 3k + 4k + 5k + 6k = 1
∴ 21k = 1 ∴ k = $$\frac{1}{21}$$

(ii) P(X ≤ 4) = 1 – P(X > 4)
= 1 – [P(X = 5) + P(X = 6)]
= 1 – $\frac{5}{21}+\frac{6}{21}$
= 1 – $$\frac{11}{21}$$
= $$\frac{10}{21}$$
P(2 < X < 6) = p(3) + p(4) + p(5)
= 3k + 4k + 5k
= $$\frac{3}{21}+\frac{4}{21}+\frac{5}{21}$$
= $$\frac{12}{21}$$
= $$\frac{4}{7}$$

(iii) P(X ≥ 3) = p(3) + p(4) + p(5) + p(6)
= 3k + 4k + 5k + 6k

Question 3.
Following is the probability distribution of an r.v. X.

Find the probability that
(i) X is positive.
(ii) X is non-negative.
(iii) X is odd.
(iv) X is even.
Solution:
(i) P(X is positive)
P(X = 0) = p(1) + p(2) + p(3)
= 0.25 + 0.15 + 0.10
= 0.50

(ii) P(X is non-negative)
P(X ≥ 0) = p(0) + p(1) + p(2) + p(3)
= 0.20 + 0.25 + 0.15 + 0.10
= 0.70

(iii) P(X is odd)
P(X = -3, -1, 1, 3)
= p(- 3) +p(-1) + p(1) + p(3)
= 0.05 + 0.15 + 0.25 + 0.10
= 0.55

(iv) P(X is even)
= 1 – P(X is odd)
= 1 – 0.55
= 0.45

Question 4.
The p.m.f of a r.v. X is given by
$$P(X=x)= \begin{cases}\left(\begin{array}{l} 5 \\ x \end{array}\right) \frac{1}{2^{5}}, & x=0,1,2,3,4,5 . \\ 0 & \text { otherwise }\end{cases}$$
Show that P(X ≤ 2) = P(X ≥ 3).
Solution:
For x = 0, 1, 2, 3, 4, 5

Question 5.
In the following probability distribution of an r.v. X

Find a and obtain the c.d.f. of X.
Solution:
Given distribution is p.m.f. of r.v. X
ΣP(X = x) = 1
∴ p(1) + p(2) + p(3) + p(4) + p(5) = 1

Question 6.
A fair coin is tossed 4 times. Let X denote the number of heads obtained. Identify the probability distribution of X and state the formula for p.m.f. of X.
Solution:
A fair coin is tossed 4 times
∴ Sample space contains 16 outcomes
Let X = Number of heads obtained
∴ X takes the values x = 0, 1, 2, 3, 4.
∴ The number of heads obtained in a toss is an even

Question 7.
Find the probability of the number of successes in two tosses of a die, where success is defined as (i) number greater than 4 (ii) six appearing in at least one toss.
Solution:
S : A die is tossed two times
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(S) = 36
(i) X : No. is greater than 4
Range of X = {0, 1, 2}

(ii) X : Six appears on aleast one die.
Range of X = {0, 1, 2}

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

Determine (i) k, (ii) P(X < 3), (iii) P(X > 6), (iv) P(0 < X < 3).
Solution:
(i) It is a p.m.f. of r.v. X
Σp(x) = 1
p(1) + p(2) + p(3) + p(4) + p(5) + p(6) + p(7) = 1
k + 2k + 2k + 3k + k2 + 2k2 + 7k2 + k = 1
9k + 10k2 = 1
10k2 + 9k – 1 = 0
10k2 +10k – k – 1 = 0
∴ 10k(k + 1) – 1(k + 1) = 0
∴ (10k – 1) (k + 1) = 0
∴ 10k – 1 = 0r k + 1 = 0
∴ k = $$\frac{1}{10}$$ or k = -1
k = -1 is not accepted, p(x) ≥ 0, ∀ x ∈ R
∴ k = $$\frac{1}{10}$$

(ii) P(X < 3) = p(1) + p(2)
= k + 2k
= 3k
= 3 × $$\frac{1}{10}$$
= $$\frac{3}{10}$$

(iii) P(X > 6) = p(7)
= 7k2 + k
= $$7\left(\frac{1}{10}\right)^{2}+\frac{1}{10}$$
= $$\frac{7}{100}+\frac{1}{10}$$
= $$\frac{17}{100}$$

(iv) P(0 < X < 3) = p(1) + p(2)
= k + 2k
= 3k
= 3 × $$\frac{1}{10}$$
= $$\frac{3}{10}$$

Question 9.
The following is the c.d.f. of a r.v. X.

Find the probability distribution of X and P(-1 ≤ X ≤ 2).
Solution:

P(-1 ≤ X ≤ 2) = p(-1) + p(0) + p(1) + p(2)
= 0.2 + 0.15 + 0.10 + 0.10
= 0.55

Question 10.
Find the expected value and variance of the r.v. X if its probability distribution is as follows.
(i)

Solution:

(ii)

Solution:
E(X) = Σx . p(x)

(iii)

Solution:

(iv)

Solution:

= 1.25
S.D. of X = σx = √Var(X)
= √1.25
= 1.118

Question 11.
A player tosses two coins. He wins ₹ 10 if 2 heads appear, ₹ 5 if 1 head appears, and ₹ 2 if no head appears. Find the expected value and variance of the winning amount.
Solution:
S : Two fair coin are tossed
S = {HH, HT, TT, TH}
n(S) = 4
∴ Range of X = {0, 1, 2}
∴ Let Y = amount received corresponds to values of X

Expected winning amount
E(Y) = Σpy = $$\frac{22}{4}$$ = ₹ 5.5
V(Y) = Σpy2 – (Σpy)2
= $$\frac{154}{4}$$ – (5.5)2
= 38.5 – 30.25
= ₹ 8.25

Question 12.
Let the p.m.f. of the r.v. X be
$$p(x)= \begin{cases}\frac{3-x}{10} & \text { for } x=-1,0,1,2 \\ 0 & \text { otherwise }\end{cases}$$
Calculate E(X) and Var(X).
Solution:

Question 13.
Suppose error involved in making a certain measurement is a continuous r.v. X with p.d.f.
$$f(x)= \begin{cases}k\left(4-x^{2}\right) & \text { for }-2 \leq x \leq 2 \\ 0 & \text { otherwise }\end{cases}$$
Compute (i) P(X > 0), (ii) P(-1 < X < 1), (iii) P(X < -0.5 or X > 0.5)
Solution:
We know that

Question 14.
The p.d.f. of the r.v. X is given by
$$f(x)= \begin{cases}\frac{1}{2 a} & \text { for } 0<x<2 a \\ 0 & \text { otherwise }\end{cases}$$
Show that P(X < $$\frac{a}{2}$$) = P(X > $$\frac{3a}{2}$$)
Solution:

Question 15.
Determine k if
$$f(x)= \begin{cases}k e^{-\theta x} & \text { for } 0 \leq x<\infty, \theta>0 \\ 0 & \text { otherwise }\end{cases}$$
is the p.d.f. of the r.v. X. Also find P(X > $$\frac{1}{\theta}$$). Find M if P(0 < X < M) = $$\frac{1}{2}$$
Solution:
We know that

Question 16.
The p.d.f. of the r.v. X is given by
$$f_{x}(x)=\left\{\begin{array}{l} \frac{k}{\sqrt{x}}, 0<x<4 \\ 0, \text { otherwise } \end{array}\right.$$
Determine k, c.d.f. of X and hence find P(X ≤ 2) and P(X ≥ 1).
Solution:
We know that

Question 17.
Let X denote the reaction temperature (in °C) of a certain chemical process. Let X be a continuous r.v. with p.d.f.
$$f(x)= \begin{cases}\frac{1}{10}, & -5 \leq x \leq 5 \\ 0, & \text { otherwise }\end{cases}$$
Compute P(X < 0).
Solution:
Given p.d.f. is f(x) = $$\frac{1}{10}$$, for -5 ≤ x ≤ 5
Let its c.d.f. F(x) be given by

Part – II

Question 1.
Let X ~ B(10, 0.2). Find (i) P(X = 1) (ii) P(X ≥ 1) (iii) P(X ≤ 8)
Solution:
X ~ B(10, 0.2)
n = 10, p = 0.2
∴ q = 1 – p = 1 – 0.2 = 0.8
(i) P(X = 1) = 10C1 (0.2)1 (0.8)9 = 0.2684

(ii) P(X ≥ 1) = 1 – P(X < 1)
= 1 – P(X = 0)
= 1 – 10C0 (0.2)0 (0.8)10
= 1 – 0.1074
= 0.8926

(iii) P(X ≤ 8) = 1 – P(x > 1)
= 1 – [p(9) + p(10)]
= 1 – [10C9 (0.2)9 (0.8)1 + 10C10 (0.2)10]
= 1 – 0.00000041984
= 0.9999

Question 2.
Let X ~ B(n, p) (i) If n = 10 and E(X) = 5, find p and Var(X), (ii) If E(X) = 5 and Var(X) = 2.5, find n and p.
Solution:
X ~ B(n, p)
(i) n = 10, E(X) = 5
∴ np = 5
∴ 10p = 5
∴ p = $$\frac{1}{2}$$
∴ q = 1 – p = 1 – $$\frac{1}{2}$$ = $$\frac{1}{2}$$
V(X) = npq
= 10 × $$\frac{1}{2}$$ × $$\frac{1}{2}$$
= 2.5

(ii) E(X) = 5, V(X) = 2.5
∴ np = 5, ∴ npq = 2.5
∴ 5q = 2.5
∴ q = $$\frac{2.5}{5}$$ = 0.5, p = 1 – 0.5 = 0.5
But np = 5
∴ n(0.5) = 5
∴ n = 10

Question 3.
If a fair coin is tossed 4 times, find the probability that it shows (i) 3 heads, (ii) head in the first 2 tosses, and tail in the last 2 tosses.
Solution:
n : No. of times a coin is tossed
∴ n = 4
P : Probability of getting heads

Question 4.
The probability that a bomb will hit the target is 0.8. Find the probability that, out of 5 bombs, exactly 2 will miss the target.
Solution:
X : No. of bombs miss the target
p : Probability that bomb miss the target
∴ q = 0.8
∴ p = 1 – q = 1 – 0.8 = 0.2
n = No. of bombs = 5
∴ X ~ B(5, 0.2)
∴ p(x) = nCx px qn-x
P(X = 2) = 5C2 (0.2)2 (0.8)5-2
= 10 × 0.04 × (0.8)3
= 10 × 0.04 × 0.512
= 0.4 × 0.512
= 0.2048

Question 5.
The probability that a lamp in the classroom will burn is 0.3. 3 lamps are fitted in the classroom. The classroom is unusable if the number of lamps burning in it is less than 2. Find the probability that the classroom can not be used on a random occasion.
Solution:
X : No. of lamps not burning
p : Probability that the lamp is not burning
∴ q = 0.3
∴ p = 1 – q = 1 – 0.3 = 0.7
n = No. of lamps fitted = 3
∴ X ~ B(3, 0.7)
∴ p(x) = nCx px qn-x
P(classroom cannot be used)
P(X < 2) = p(0) + p(1)
= 3C0 (0.7)0 (0.3)3-0 + 3C1 (0.7)1 (0.3)3-1
= 1 × 1 × (0.3)3 + 3 × 0.7 × (0.3)2
= (0.3)2 [0.3 + 3 × 0.7]
= 0.09 [0.3 + 2.1]
= 0.09 [2.4]
= 0.216

Question 6.
A large chain retailer purchases an electric device from the manufacturer. The manufacturer indicates that the defective rate of the device is 10%. The inspector of the retailer randomly selects 4 items from a shipment. Find the probability that the inspector finds at most one defective item in the 4 selected items.
Solution:
X : No. of defective items
n : No. of items selected = 4
p : Probability of getting defective items
∴ p = 0.1
∴ q = 1 – p = 1 – 0.1 = 0.9
P(At most one defective item)
P(X ≤ 1) = p(0) + p(1)
= 4C0 (0.1)0 (0.9)4-0 + 4C1 (0.1)1 (0.9)4-1
= 1 × 1 × (0.9)4 + 4 × 0.1 × (0.9)3
= (0.9)3 [0.9 + 4 × 0.1]
= (0.9)3 × [0.9 + 0.4]
= 0.729 × 1.3
= 0.9477

Question 7.
The probability that a component will survive a check test is 0.6. Find the probability that exactly 2 of the next 4 components tested survive.
Solution:
p = 0.6, q = 1 – 0.6 = 0.4, n = 4
x = 2
∴ p(x) = nCx px qn-x
P(X = 2) = 4C2 (0.6)2 (0.4)2 = 0.3456

Question 8.
An examination consists of 5 multiple choice questions, in each of which the candidate has to decide which one of 4 suggested answers is correct. A completely unprepared student guesses each answer randomly. Find the probability that this student gets 4 or more correct answers.
Solution:
n : No. of multiple-choice questions
∴ n = 5
X : No. of correct answers
p : Probability of getting correct answer
∵ There are 4 options out of which one is correct
∴ p = $$\frac{1}{4}$$
∴ q = 1 – p = 1 – $$\frac{1}{4}$$ = $$\frac{3}{4}$$
∵ X ~ B(5, $$\frac{1}{4}$$)
∴ p(x) = nCx px qn-x
P(X ≥ 4) = p(4) + p(5)

Question 9.
The probability that a machine will produce all bolts in a production run with in the specification is 0.9. A sample of 3 machines is taken at random. Calculate the probability that all machines will produce all bolts in a production run within the specification.
Solution:
n : No. of samples selected
∴ n = 3
X : No. of bolts produce by machines
p : Probability of getting bolts
∴ p = 0.9
∴ q = 1 – p = 1 – 0.9 = 0.1
∴ X ~ B(3, 0.9)
∴ p(x) = nCx px qn-x
P(Machine will produce all bolts)
P(X = 3) = 3C3 (0.9)3 (0.1)3-3
= 1 × (0.9)3 × (0.1)0
= 1 × (0.9)3 × 1
= (0.9)3
= 0.729

Question 10.
A computer installation has 3 terminals. The probability that anyone terminal requires attention during a week is 0.1, independent of other terminals. Find the probabilities that (i) 0 (ii) 1 terminal requires attention during a week.
Solution:
n : No. of terminals
∴ n = 3
X : No. of terminals need attention
p : Probability of getting terminals need attention
∴ p = 0.1
∴ q = 1 – p = 1 – 0.1 = 0.9
∵ X ~ B(3, 0.1)
∴ p(x) = nCx px qn-x
(i) P(No attention)
∴ P(X = 0) = 3C0 × (0.1)0 (0.9)3-1
= 1 × 1 × (0.9)3
= 0.729

(ii) P(One terminal need attention)
∴ P(X = 1) = 3C1 (0.1)1 (0.9)3-1
= 3 × 0.1 × (0.9)2
= 0.3 × 0.81
= 0.243

Question 11.
In a large school, 80% of the students like mathematics. A visitor asks each of 4 students, selected at random, whether they like mathematics, (i) Calculate the probabilities of obtaining an answer yes from all of the selected students, (ii) Find the probability that the visitor obtains the answer yes from at least 3 students.
Solution:
X : No. of students like mathematics
p: Probability that students like mathematics
∴ p = 0.8
∴ q = 1 – p = 1 – 0.8 = 0.2
n : No. of students selected
∴ n = 4
∵ X ~ B(4, 0.8)
∴ p(x) = nCx px qn-x
(i) P(All students like mathematics)
∴ P(X = 4) = 4C4 (0.8)4 (0.2)4-4
= 1 × (0.8)4 × (0.2)0
= 1 × (0.8)4 × 1
= 0.4096

(ii) P(Atleast 3 students like mathematics)
∴ P(X ≥ 3) = p(3) + p(4)
= 4C3 (0.8)3 (0.2)4-3 + 0.4096
= 4 × (0.8)3 (0.2)1 + 0.4096
= 0.8 × (0.8)3 + 0.4096
= (0.8)4 × 0.4096
= 0.4096 + 0.4096
= 0.8192

Question 12.
It is observed that it rains on 10 days out of 30 days. Find the probability that
(i) it rains on exactly 3 days of a week.
(ii) it rains at most 2 days a week.
Solution:
X : No. of days it rains in a week
p : Probability that it rains
∴ p = $$\frac{10}{30}=\frac{1}{3}$$
∴ q = 1 – p = 1 – $$\frac{1}{3}$$ = $$\frac{2}{3}$$
n : No. of days in a week
∴ n = 7
∴ X ~ B(7, $$\frac{1}{3}$$)
(i) P(Rains on Exactly 3 days of a week)

(ii) P(Rains on at most 2 days of a week)
∴ P(X ≤ 2) = p(0) + p(1) + p(2)

Question 13.
If X follows Poisson distribution such that P(X = 1) = 0.4 and P(X = 2) = 0.2, find variance of X.
Solution:
X : Follows Possion Distribution

∴ m = 1
∴ Mean = m = Variance of X = 1

Question 14.
If X has Poisson distribution with parameter m, such that
$$\frac{P(X=x+1)}{P(X=x)}=\frac{m}{x+1}$$
find probabilities P(X = 1) and P(X = 2), when X follows Poisson distribution with m = 2 and P(X = 0) = 0.1353.
Solution:
Given that the random variable X follows the Poisson distribution with parameter m = 2
i.e. X ~ P(2)
Its p.m.f. is satisfying the given equation.
$$\frac{P(X=x+1)}{P(X=x)}=\frac{m}{x+1}$$
When x = 0,
$$\frac{\mathrm{P}(\mathrm{X}=1)}{\mathrm{P}(\mathrm{X}=0)}=\frac{2}{0+1}$$
P(X = 1) = 2P(X = 0)
= 2(0.1353)
= 0.2706
When x = 1,
$$\frac{\mathrm{P}(\mathrm{X}=2)}{\mathrm{P}(\mathrm{X}=1)}=\frac{2}{1+1}$$
P(X = 2) = P(X = 1) = 0.2706

## Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.4

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 8 Probability Distributions Ex 8.4 Questions and Answers.

## Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.4

Question 1.
If X has Poisson distribution with m = 1, then find P(X ≤ 1) given e-1 = 0.3678.
Solution:
∵ m = 1
∵ X follows Poisson Distribution

= e-m × 1 + e-m × 1
= e-1 + e-1
= 2 × e-1
= 2 × 0.3678
= 0.7356

Question 2.
If X ~ P($$\frac{1}{2}$$), then find P(X = 3) given e-0.5 = 0.6065.
Solution:

Question 3.
If X has Poisson distribution with parameter m and P(X = 2) = P(X = 3), then find P(X ≥ 2). Use e-3 = 0.0497
Solution:
∵ X follows Poisson Distribution

Question 4.
The number of complaints which a bank manager receives per day follows a Poisson distribution with parameter m = 4. Find the probability that the manager receives (i) only two complaints on a given day, (ii) at most two complaints on a given day. Use e-4 = 0.0183.
Solution:
∵ m = 1
∵ X ~ P(m = 4)
∴ p(x) = $$\frac{e^{-m} \cdot m^{x}}{x !}$$
X = No. of complaints recieved
(i) P(Only two complaints on a given day)

(ii) P(Atmost two complaints on a given day)
P(X ≤ 2) = p(0) + p(1) + p(2)
= $$\frac{e^{-4} \times 4^{0}}{0 !}+\frac{e^{-4} \times 4^{1}}{1 !}$$ + 0.1464
= e-4 + e-4 × 4 + 0.1464
= e-4 [1 + 4] + 0.1464
= 0.0183 × 5 + 0.1464
= 0.0915 + 0.1464
= 0.2379

Question 5.
A car firm has 2 cars, which are hired out day by day. The number of cars hired on a day follows a Poisson distribution with a mean of 1.5. Find the probability that
(i) no car is used on a given day.
(ii) some demand is refused on a given day, given e-1.5 = 0.2231.
Solution:
Let X = No. of demands for a car on any day
∴ No. of cars hired
n = 2
m = 1.5
∵ X ~ P(m = 1.5)

Question 6.
Defects on plywood sheets occur at random with an average of one defect per 50 sq. ft. Find the probability that such a sheet has (i) no defect, (ii) at least one defect. Use e-1 = 0.3678.
Solution:
∵ X = No. of defects on a plywood sheet
∵ m = -1
∵ X ~ P(m = -1)
∴ p(x) = $$\frac{e^{-m} \cdot m^{x}}{x !}$$
(i) P(No defect)
P(X = 0) = $$\frac{e^{-1} \times 1^{0}}{0 !}$$
= e-1
= 0.3678

(ii) P(At least one defect)
P(X ≥ 1) = 1 – P(X < 1)
= 1 – p(0)
= 1 – 0.3678
= 0.6322

Question 7.
It is known that, in a certain area of a large city, the average number of rats per bungalow is five. Assuming that the number of rats follows Poisson distribution, find the probability that a randomly selected bungalow has
(i) exactly 5 rats
(ii) more than 5 rats
(iii) between 5 and 7 rats, inclusive. Given e-5 = 0.0067.
Solution:
X = No. of rats
∵ m = 5
∴ X ~ P(m = 5)
∴ p(x) = $$\frac{e^{-m} \cdot m^{x}}{x !}$$
(i) P(Exactly five rats)

(ii) P(More than five rats)
P(X > 5) = 1 – P(X ≤ 5)

(iii) P(between 5 and 7 rats, inclusive)
P(5 ≤ x ≤ 7) = p(5) + p(6) + p(7)

= 0.0067 × 3125 × 0.02
= 0.0067 × 62.5
= 0.42

## Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.3

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 8 Probability Distributions Ex 8.3 Questions and Answers.

## Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.3

Question 1.
A die is thrown 4 times. If ‘getting an odd number’ is a success, find the probability of (i) 2 successes (ii) at least 3 successes (iii) at most 2 successes.
Solution:
X: Getting an odd no.
p: Probability of getting an odd no.
A die is thrown 4 times
∴ n = 4
∵ p = $$\frac{3}{6}=\frac{1}{2}$$
∴ q = 1 – p = 1 – $$\frac{1}{2}$$ = $$\frac{1}{2}$$
∵ X ~ B(3, $$\frac{1}{2}$$)
∴ p(x) = $${ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}$$
(i) P(Two Successes)

(ii) P(Atleast 3 Successes)

(iii) P(Atmost 2 Successes)

Question 2.
A pair of dice is thrown 3 times. If getting a doublet is considered a success, find the probability of two successes.
Solution:
n: No. of times die is thrown = 3
X: No. of doublets
p: Probability of getting doublets
Getting a doublet means, same no. is obtained on 2 throws of a die
There are 36 outcomes
No. of doublets are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)

Question 3.
There are 10% defective items in a large bulk of items. What is the probability that a sample of 4 items will include not more than one defective item?
Solution:
n: No of sample items = 4
X: No of defective items
p: Probability of getting defective items
∴ p = 0.1
∴ q = 1 – p = 1 – 0.1 = 0.9
X ~ B(4, 0.1)
∴ p(x) = $${ }^{n} \mathrm{C}_{x} p^{x} \mathrm{q}^{n-x}$$
P(Not include more than 1 defective)
P(X ≤ 1) = p(0) + p(1)
= 4C0 (0.1)0 (0.9)4 + 4C1 (0.1)1 (0.9)4-1
= 1 × 1 × (0.9)4 + 4 × 0.1 × (0.9)3
= (0.9)3 [0.9 + 0.4]
= (0.9)3 × 1.3
= 0.977

Question 4.
Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Find the probability that (i) all the five cards are spades, (ii) only 3 cards are spades, (iii) none is a spade.
Solution:
Number of cards drawn
∴ n = 5
p: Probability of getting spade card

(i) P(All five cards are spades)

(ii) P(Only 3 cards are spades)

Question 5.
The probability that a bulb produced by a factory will use fuse after 200 days of use is 0.2. Let X denote the number of bulbs (out of 5) that fuse after 200 days of use. Find the probability of (i) X = 0, (ii) X ≤ 1, (iii) X > 1, (iv) X ≥ 1.
Solution:
X : No. of bulbs fuse after 200 days of use
p : Probability of getting fuse bulbs
No. of bulbs in a sample
∴ n = 5
∴ p = 0.2
∴ q = 1 – p = 1 – 0.2 = 0.8
∵ X ~ B(5, 0.2)
∴ p(x) = $${ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}$$
(i) P(X = 0) = 5C0 (0.2)0 (0.8)5-0
= 1 × 1 × (0.8)5
= (0.8)5

(ii) P(X ≤ 1) = p(0) + p(1)
= 5C0 (0.2)0 (0.8)5-0 + 5C1 (0.2)1 (0.8)5-1
= 1 × 1 × (0.8)5 + 5 × 0.2 × (0.8)4
= (0.8)4 [0.8 + 1]
= 1.8 × (0.8)4

(iii) P(X > 1) = 1 – [p(0) + p(1)]
= 1 – 1.8 × (0.8)4

(iv) P(X ≥ 1) = 1 – p(0)
= 1 – (0.8)5

Question 6.
10 balls are marked with digits 0 to 9. If four balls are selected with replacement. What is the probability that none is marked 0?
Solution:
X : No. of balls drawn marked with the digit 0
n : No. of balls drawn
∴ n = 4
p : Probability of balls marked with 0.
∴ p = $$\frac{1}{10}$$
∴ q = 1 – p = 1 – $$\frac{1}{10}$$ = $$\frac{9}{10}$$
p(x) = $${ }^{n} C_{x} p^{x} q^{n-x}$$
P(None of the ball is marked with digit 0)

Question 7.
In a multiple-choice test with three possible answers for each of the five questions, what is the probability of a candidate getting four or more correct answers by random choice?
Solution:
n: No. of Questions
∴ n = 5
X: No. of correct answers by guessing
p: Probability of getting correct answers

Question 8.
Find the probability of throwing at most 2 sixes in 6 throws of a single die.
Solution:
X : No. of sixes in 6 throws
n : No. of times dice thrown
∴ n = 6
p : Probability of getting six
∴ p = $$\frac{1}{6}$$
∴ q = 1 – p = 1 – $$\frac{1}{6}$$ = $$\frac{5}{6}$$
∵ X ~ B(6, $$\frac{1}{6}$$)
∴ p(x) = $${ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}$$
P(At most 2 sixes)
P(X ≤ 2) = p(0) + p(1) + p(2)

Question 9.
Given that X ~ B(n, p),
(i) if n = 10 and p = 0.4, find E(X) and Var(X).
(ii) if p = 0.6 and E(X) = 6, find n and Var(X).
(iii) if n = 25, E(X) = 10, find p and Var(X).
(iv) if n = 10, E(X) = 8, find Var(X).
Solution:
∵ X ~ B (n, p), E(X) = np, V(X) = npq, q = 1 – p
(i) E(X) = np = 10 × 0.4 = 4
∵ q = 1 – p = 1 – 0.4 = 0.6
V(X) = npq = 10 × 0.4 × 0.6 = 2.4

(ii) ∵ p = 0.6
∴ q = 1 – p = 1 – 0.6 = 0.4
E(X) = np
∴ 6 = n × 0.6
∴ n = 10
∴ V(X) = npq = 10 × 0.6 × 0.4 = 2.4

(iii) E(X) = np
∴ 10 = 25 × p
∴ p = 0.4
∴ q = 1, p = 1 – 0.4 = 0.6
∴ S.D.(X) = √V(X)
= $$\sqrt{n p q}$$
= $$\sqrt{25 \times 0.4 \times 0.6}$$
= √6
= 2.4494

(iv) ∵ E(X) = np
∴ 8 = 10p
∴ p = 0.8
∴ q = 1 – p = 1 – 0.8 = 0.2
∵ V(X) = npq = 10 × 0.8 × 0.2 = 1.6

## Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 8 Probability Distributions Ex 8.2 Questions and Answers.

## Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2

Question 1.
Check whether each of the following is p.d.f.
(i) $$f(x)= \begin{cases}x & \text { for } 0 \leq x \leq 1 \\ 2-x & \text { for } 1<x \leq 2\end{cases}$$
Solution:
Given function is
f(x) = x, 0 ≤ x ≤ 1
Each f(x) ≥ 0, as x ≥ 0.

∴ The given function is a p.d.f. of x.

(ii) f(x) = 2 for 0 < x < 1
Solution:
Given function is
f(x) = 2 for 0 < x < 1 Each f(x) > 0,

∴ The given function is not a p.d.f.

Question 2.
The following is the p.d.f. of a r.v. X.
$$f(x)= \begin{cases}\frac{x}{8} & \text { for } 0<x<4 \\ 0 & \text { otherwise }\end{cases}$$
Find (i) P(X < 1.5), (ii) P(1 < X < 2), (iii) P(X > 2)
Solution:

Question 3.
It is felt that error in measurement of reaction temperature (in Celsius) in an experiment is a continuous r.v. with p.d.f.
$$f(x)= \begin{cases}\frac{x^{3}}{64} & \text { for } 0 \leq x \leq 4 \\ 0 & \text { otherwise }\end{cases}$$
(i) Verify whether f(x) is a p.d.f.
(ii) Find P(0 < X ≤ 1).
(iii) Find the probability that X is between 1 and 3.
Solution:
(i) f(x) is p.d.f. of r.v. X if
(a) f(x) ≥ 0, ∀ x ∈ R
(b) $$\int_{0}^{4} f(x) d x$$ = 1

Question 4.
Find k, if the following function represents the p.d.f. of a r.v. X.
(i) $$f(x)= \begin{cases}k x & \text { for } 0<x<2 \\ 0 & \text { otherwise }\end{cases}$$
Also find P[$$\frac{1}{4}$$ < X < $$\frac{1}{2}$$]
Solution:

(ii) $$f(x)= \begin{cases}k x(1-x) & \text { for } 0<x<1 \\ 0 & \text { otherwise }\end{cases}$$
Also find (a) P[$$\frac{1}{4}$$ < X < $$\frac{1}{2}$$], (b) P[X < $$\frac{1}{2}$$]
Solution:
We know that

Question 5.
Let X be the amount of time for which a book is taken out of the library by a randomly selected student and suppose that X has p.d.f.
$$f(x)= \begin{cases}0.5 x & \text { for } 0 \leq x \leq 2 \\ 0 & \text { otherwise }\end{cases}$$
Calculate (i) P(X ≤ 1), (ii) P(0.5 ≤ X ≤ 1.5), (iii) P(X ≥ 1.5).
Solution:
Given p.d.f. of X is f(x) = 0.5x for 0 ≤ x ≤ 2
∴ Its c.d.f. F(x) is given by

(i) P(X < 1) = F(1)
= 0.25(1)2
= 0.25

(ii) P(0.5 < X < 1.5) = F(1.5) – F(0.5)
= 0.25(1.5)2 – 0.25(0.5)2
= 0.25[2.25 – 0.25]
= 0.25(2)
= 0.5

(iii) P(X ≥ 1.5) = 1 – P(X ≤ 1.5)
= 1 – F(1.5)
= 1 – 0.25(1.5)2
= 1 – 0.25(2.25)
= 1 – 0.5625
= 0.4375

Question 6.
Suppose X is the waiting time (in minutes) for a bus and its p.d.f. is given by
$$f(x)=\left\{\begin{array}{cl} \frac{1}{5} & \text { for } 0 \leq x \leq 5 \\ 0 & \text { otherwise } \end{array}\right.$$
Find the probability that (i) waiting time is between 1 and 3 minutes, (ii) waiting time is more than 4 minutes.
Solution:
p.d.f. of r.v. X is given by
f(x) = $$\frac{1}{5}$$ for 0 ≤ x ≤ 5
This is a constant function.
(i) Probability that waiting time X is between 1 and 3 minutes

(ii) Probability that waiting time X is more than 4 minutes

Question 7.
Suppose error involved in making a certain measurement is a continuous r.v. X with p.d.f.
$$f(x)= \begin{cases}k\left(4-x^{2}\right) & \text { for }-2 \leq x \leq 2 \\ 0 & \text { otherwise }\end{cases}$$
Compute (i) P(X > 0), (ii) P(-1 < X < 1), (iii) P(X < -0.5 or X > 0.5)
Solution:
Since given f(x) is a p.d.f. of r.v. X
Since -2 ≤ x ≤ 2
∴ x2 ≤ 4
∴ 4 – x2 ≥ 0
∴ k(4 – x2) ≥ 0
∴ k ≥ 0 [∵ f(x) ≥ 0]

Question 8.
Following is the p.d.f. of a continuous r.v. X.
$$f(x)= \begin{cases}\frac{x}{8} & \text { for } 0<x<4 \\ 0 & \text { otherwise }\end{cases}$$
(i) Find an expression for the c.d.f. of X.
(ii) Find F(x) at x = 0.5, 1.7, and 5.
Solution:
The p.d.f. of a continuous r.v. X is
$$f(x)= \begin{cases}\frac{x}{8} & \text { for } 0<x<4 \\ 0 & \text { otherwise }\end{cases}$$
(i) c.d.f. of continuous r.v. X is given by

(ii) F(0.5) = $$\frac{(0.5)^{2}}{16}=\frac{0.25}{16}=\frac{1}{64}$$ = 0.015
F(1.7) = $$\frac{(1.7)^{2}}{16}=\frac{2.89}{16}$$ = 0.18
For any of x greater than or equal to 4, F(x) = 1
∴ F(5) = 1

Question 9.
The p.d.f. of a continuous r.v. X is
$$f(x)=\left\{\begin{array}{cl} \frac{3 x^{2}}{8} & \text { for } 0<x<2 \\ 0 & \text { otherwise } \end{array}\right.$$
Determine the c.d.f. of X and hence find (i) P(X < 1), (ii) P(X < -2), (iii) P(X > 0), (iv) P(1 < X < 2).
Solution:
The p.d.f. of a continuous r.v. X is

Question 10.
If a r.v. X has p.d.f.
$$f(x)= \begin{cases}\frac{c}{x} & \text { for } 1<x<3, c>0 \\ 0 & \text { otherwise }\end{cases}$$
Find c, E(X) and V(X). Also find f(x).
Solution:
The p.d.f. of r.v. X is
f(x) = $$\frac{c}{x}$$, 1 < x < 3, c > 0
For p.d.f. of X, we have

## Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 8 Probability Distributions Ex 8.1 Questions and Answers.

## Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1

Question 1.
Let X represent the difference between a number of heads and the number of tails obtained when a coin is tossed 6 times. What are the possible values of X?
Solution:
∵ A coin is tossed 6 times
S = {6H and 0T, 5H and 1T, 4H and 2T, 3H and 3T, 2H and 4T, 1H and 5T, 0H and 6T}
X: Difference between no. of heads and no. of tails.
X = 6 – 0 = 6
X = 5 – 1 = 4
X = 4 – 2 = 2
X = 3 – 3 = 0
X = 2 – 4 = -2
X = 1 – 5 = -4
X = 0 – 6 = -6
X = {-6, -4, -2, 0, 2, 4, 6}

Question 2.
An urn contains 5 red and 2 black balls. Two balls are drawn at random. X denotes the number of black balls drawn. What are the possible values of X?
Solution:
S : Two bolts are drawn from the Urn
S = {RR, RB, BR, BB}
X : No. of black balls
∴ X = {0, 1, 2}

Question 3.
Determine whether each of the following is a probability distribution. Give reasons for your answer.
(i)

Solution:
Here, P(X = x) ≥ 0, ∀ x and
$$\sum_{x=0}^{2} \mathrm{P}(\mathrm{X}=x)$$ = p(0) + p(1) + p(2)
= 0.4 + 0.4 + 0.2
= 1
∴ The function is a p.m.f.

(ii)

Solution:
Here, p(3) = -0.1 < 0
∴ P(X = x) ≯ 0, ∀ x
∴ The function is not a p.m.f.

(iii)

Solution:
Here, P(X = x) ≥ 0, ∀ x and
$$\sum_{x=0}^{2} \mathrm{P}(\mathrm{X}=x)$$ = p(0) + p(1) + p(2)
= 0.1 + 0.6 + 0.3
= 1
∴ The function is a p.m.f.

(iv)

Solution:
Here, P(Z = z) ≥ 0, ∀ z and
$$\sum_{x=-1}^{3} \mathrm{P}(\mathrm{Z}=z)$$ = p(-1) + p(0) + p(1) + p(2) + p(3)
= 0.05 + 0 + 0.4 + 0.2 + 0.3
= 0.95
≠ 1
∴ The function is not a p.m.f.

(v)

Solution:
Here, P(Y = y) ≥ 0, ∀ y and
$$\sum_{x=-1}^{2} \mathrm{P}(\mathrm{Y}=y)$$ = p(-1) + p(0) + p(1)
= 0.1 + 0.6 + 0.2
= 0.9
≠ 1
∴ The function is not a p.m.f.

(vi)

Solution:
Here, P(X = x) ≥ 0, ∀ x and
$$\sum_{x=0}^{0} \mathrm{P}(X=x)$$ = p(-2) + p(-1) + p(0)
= 0.3 + 0.4 + 0.2
= 0.9
≠ 1
∴ The function is not a p.m.f.

Question 4.
Find the probability distribution of
(i) number of heads in two tosses of a coin,
(ii) number of trails in three tosses of a coin,
(iii) number of heads in four tosses of a coin.
Solution:
(i) S: Coin is tossed two times
S = {HH, HT, TH, TT}
n(S) = 4
Range of X = {0, 1, 2}
p.m.f. Table

(ii) S: 3 coin are tossed
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
Range of X = {0, 1, 2, 3}
p.m.f. Table

(iii) S: Four coin are tossed
S = {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}
n(S) = 16
Range of X = {0, 1, 2, 3, 4}
p.m.f. Table

Question 5.
Find the probability distribution of the number of successes in two tosses of a die if successes are defined as getting a number greater than 4.
Solution:
S = A die is tossed 2 times
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(S) = 36
X = No. getting greater than 4
Range of X = {0, 1, 2}
p(0) = $$\frac{16}{36}=\frac{4}{9}$$
p(1) = $$\frac{16}{36}=\frac{4}{9}$$
p(2) = $$\frac{4}{36}=\frac{1}{9}$$

Question 6.
A sample of 4 bulbs is drawn at random with replacement from a lot of 30 bulbs which includes 6 defective bulbs. Find the probability distribution of the number of defective bulbs.
Solution:
Total no. of bulbs = 30
No. of defective bulbs = 6
A sample of 4 bulbs are drawn from 30 bulbs.
∴ n(S) = $${ }^{30} \mathrm{C}_{4}$$
∴ No. of non-defective bulbs = 24
Let X = No. of defective bulbs drawn in sample of 4 bulbs.

Question 7.
A coin is biased so that the head is 3 times as likely to occur as the tail. Find the probability distribution of a number of tails in two tosses.
Solution:
Here, the head is 3 times as likely to occur as the tail.
i.e., If 4 times coin is tossed, 3 times there will be a head and 1 time there will be the tail.
∴ p(H) = $$\frac{3}{4}$$ and p(T) = $$\frac{1}{4}$$
Let X : No. of tails in two tosses.
And coin is tossed twice.
∴ X = {0, 1, 2}
For X = 0,
= p(H) × p(H)
= $$\frac{3}{4} \times \frac{3}{4}$$
= $$\frac{9}{16}$$
For X = 1,
p(1) = p(HT or TH)
= p(HT) + p(TH)
= p(H) × p(T) + p(T) × p(H)
= $$\frac{3}{4} \times \frac{1}{4}+\frac{1}{4} \times \frac{3}{4}$$
= $$\frac{6}{16}$$
For X = 2,
p(2) = p(both tails)
= p(T) × p(T)
= $$\frac{1}{4} \times \frac{1}{4}$$
= $$\frac{1}{16}$$
The probability distribution of the number of tails in two tosses is

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

Determine (i) k, (ii) P(X < 3), (iii) P(0 < X < 3), (iv) P(X > 4).
Solution:
(i) It is a p.m.f. of r.v. X
∴ Σp(x) = 1
∴ p(1) + p(2) + p(3) + p(4) + p(5) + p(6) + p(7) = 1
∴ k + 2k + 2k + 3k + k2 + 2k2 + (7k2 + k) = 1
∴ 10k2 + 9k = 1
∴ 10k2 + 9k – 1 = 0
∴ 10k2 + 10k – k – 1 = 0
∴ 10k(k + 1) – (k + 1) = 0
∴ (10k – 1)(k + 1) = 0
∴ 10k – 1 = 0 or k + 1 = 0
∴ k = $$\frac{1}{10}$$ or k = -1
but k = -1 is not accepted
∴ k = $$\frac{1}{2}$$ is accepted

(ii) P(X < 3) = p(1) + p(2)
= k + 2k
= 3k
= 3 × $$\frac{1}{10}$$
= $$\frac{3}{10}$$

(iii) P(0 < X < 3) = p(1) + p(2)
= k + 2k
= 3k
= 3 × $$\frac{1}{10}$$
= $$\frac{3}{10}$$

(iv) P(X > 4) = p(5) + p(6) + p(7)
= k2 + 2k2 + (7k2 + k)
= 10k2 + k
= $$10\left(\frac{1}{10}\right)^{2}+\frac{1}{10}$$
= $$\frac{2}{10}$$
= $$\frac{1}{5}$$

Question 9.
Find expected value and variance of X using the following p.m.f.

Solution:

E(X) = Σxp = -0.05
V(X) = Σx2p – (Σxp)2
= 2.25 – (-0.05)2
= 2.25 – 0.0025
= 2.2475

Question 10.
Find expected value and variance of X, the number on the uppermost face of a fair die.
Solution:
S : A fair die is thrown
S = {1, 2, 3, 4, 5, 6}
n(S) = 6
X: No obtained on uppermost face of die
Range of X = {1, 2, 3, 4, 5, 6}

E(X) = Σxp = $$\frac{21}{6}=\frac{7}{2}$$ = 3.5
V(X) = Σx2p – (Σxp)2
= $$\frac{91}{6}$$ – (3.5)2
= 15.17 – 12.25
= 2.92

Question 11.
Find the mean of the number of heads in three tosses of a fair coin.
Solution:
S : A coin is tossed 3 times
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
Range of X = {0, 1, 2, 3}

∴ Mean = E(X) = Σxp = $$\frac{12}{8}=\frac{3}{2}$$ = 1.5

Question 12.
Two dice are thrown simultaneously. If X denotes the number of sixes, find the expectation of X.
Solution:
S : Two dice are thrown
S = {(1, 1), (1, 2), (1, 3), ……, (6, 6)}
n(S) = 36
Range of X = {0, 1, 2}
First 6 positive integers are 1, 2, 3, 4, 5, 6
X = Larger two numbers selected
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(S) = 36

∴ E(X) = Σxp = $$\frac{12}{36}=\frac{1}{3}$$

Question 13.
Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers. Find E(X).
Solution:
First 6 positive integers are 1, 2, 3, 4, 5, 6
X : The larger of the selected two numbers
S = {(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}
n(S) = 30

E(X) = Σxp = $$\frac{140}{30}=\frac{14}{3}$$ = 4.67

Question 14.
Let X denote the sum of the numbers obtained when two fair dice are rolled. Find the variance of X.
Solution:
S : Two fair dice are rolled
S = {(1, 1), (1, 2), (1, 4), ……, (6, 6)}
n(S) = 36
X : Sum of the two numbers.
Range of X = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

V(X) = Σx2p – (Σxp)2
= $$\frac{1952}{36}-\left(\frac{252}{36}\right)^{2}$$
= 54.22 – (7)2
= 5.22
SD(X) = √V(X) = √5.22 = 2.28

Question 15.
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. If X denotes the age of a randomly selected student, find the probability distribution of X. Find the mean and variance of X.
Solution:

Question 16.
70% of the member’s favour and 30% oppose a proposal in a meeting. The random variable X takes the value 0 if a member opposes the proposal and the value 1 if a member is in favour. Find E(X) and V(X).
Solution:

E(X) = Σxp = 0.7
V(X) = Σx2p – (Σxp)2
= 0.7 – (0.7)2
= 0.7 – 0.49
= 0.21

## Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 Questions and Answers.

## Maharashtra State Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7

(I) Choose the correct alternative.

Question 1.
In sequencing, an optimal path is one that minimizes ___________
(a) Elapsed time
(b) Idle time
(c) Both (a) and (b)
(c) Both (a) and (b)

Question 2.
If job A to D have processing times as 5, 6, 8, 4 on first machine and 4, 7, 9, 10 on second machine then the optimal sequence is:
(a) CDAB
(b) DBCA
(c) BCDA
(d) ABCD
(b) DBCA

Question 3.
The objective of sequence problem is
(a) to find the order in which jobs are to be made
(b) to find the time required for the completing all the job on hand
(c) to find the sequence in which jobs on hand are to be processed to minimize the total time required for processing the jobs
(d) to maximize the cost
(c) to find the sequence in which jobs on hand are to be processed to minimize the total time required for processing the jobs

Question 4.
If there are n jobs and m machines, then there will be ___________ sequences of doing the jobs.
(a) mn
(b) m(n!)
(c) nm
(d) (n!)m
(d) (n!)m

Question 5.
The Assignment Problem is solved by
(a) Simple method
(b) Hungarian method
(c) Vector method
(d) Graphical method
(b) Hungarian method

Question 6.
In solving 2 machine and n jobs sequencing problem, the following assumption is wrong
(a) No passing is allowed
(b) Processing times are known
(c) Handling times is negligible
(d) The time of passing depends on the order of machining
(d) The time of passing depends on the order of machining

Question 7.
To use the Hungarian method, a profit maximization assignments problem requires
(a) Converting all profit to opportunity losses
(b) A dummy person or job
(c) Matrix expansion
(d) Finding the maximum number of lines to cover all the zeros in the reduced matrix
(a) Converting all profits to opportunity losses

Question 8.
Using the Hungarian method the optimal assignment obtained for the following assignment problem to minimize the total cost is:

(a) 1 – C, 2 – B, 3 – D, 4 – A
(b) 1 – B, 2 – C, 3 – A, 4 – D
(c) 1 – A, 2 – B, 3 – C, 4 – D
(d) 1 – D, 2 – A, 3 – B, 4 – C
(a) 1 – C, 2 – B, 3 – D, 4 – A

Question 9.
The assignment problem is said to be unbalanced if
(a) Number of rows is greater than the number of columns
(b) Number of rows is lesser than number of columns
(c) Number of rows is equal to the number of columns
(d) Both (a) and (b)
(d) Both (a) and (b)

Question 10.
The assignment problem is said to be balanced if
(a) Number of rows is greater than the number of columns
(b) Number of rows is lesser than number of columns
(c) Number of rows is equal to the number of columns
(d) If the entry of rows is zero
(c) Number of rows is equal to number of columns

Question 11.
The assignment problem is said to be balanced if it is a
(a) Square matrix
(b) Rectangular matrix
(c) Unit matrix
(d) Triangular matrix
(a) Square matrix

Question 12.
In an assignment problem if the number of rows is greater than the number of columns then
(c) Row with cost 1 is added
(d) Column with cost 1 is added

Question 13.
In a 3 machine and 5 jobs problem, the least of processing times on machines A, B, and C are 5, 1 and 3 hours and the highest processing times are 9, 5 and 7 respectively, then it can be converted to a 2 machine problem if the order of the machines is:
(a) B – A – C
(b) A – B – C
(c) C – B – A
(d) Any order
(b) A – B – C

Question 14.
The objective of an assignment problem is to assign
(a) Number of jobs to equal number of persons at maximum cost
(b) Number of jobs to equal number of persons at minimum cost
(c) Only the maximize cost
(d) Only to minimize cost
(b) Number of jobs to equal number of persons at minimum cost

(II) Fill in the blanks.

Question 1.
An assignment problem is said to be unbalanced when ___________
the number of rows is not equal to the number of columns

Question 2.
When the number of rows is equal to the Number of columns then the problem is said to be ___________ assignment problem.
balanced

Question 3.
For solving assignment problem the matrix should be a ___________
square matrix

Question 4.
If the given matrix is not a ___________ matrix, the assignment problem is called an unbalanced problem.
square

Question 5.
A dummy row(s) or column(s) with the cost elements as ___________ the matrix of an unbalanced assignment problem as a square matrix.
zero

Question 6.
The time interval between starting the first job and completing the last, job including the idle time (if any) in a particular order by the given set of machines is called ___________
Total elapsed time

Question 7.
The time for which a machine j does not have a job to process to the start of job i is called ___________
Idle time

Question 8.
The maximization assignment problem is transformed to minimization problem by subtracting each entry in the table from the ___________ value in the table.
maximum

Question 9.
When the assignment problem has more than one solution, then it is ___________ optimal solution.
multiple

Question 10.
The time required for printing four books A, B, C, and D is 5, 8, 10, and 7 hours. While its data entry requires 7, 4, 3, and 6 hrs respectively. The sequence that minimizes total elapsed time is ___________
A – D – B – C

(III) State whether each of the following is True or False.

Question 1.
One machine – one job is not an assumption in solving sequencing problems.
False

Question 2.
If there are two least processing times for machine A and machine B, priority is given for the processing time which has the lowest time of the adjacent machine.
True

Question 3.
To convert the assignment problem into a maximization problem, the smallest element in the matrix is deducted from all other elements.
False

Question 4.
The Hungarian method operates on the principle of matrix reduction, whereby the cost table is reduced to a set of opportunity costs.
True

Question 5.
In a sequencing problem, the processing times are dependent on the order of processing the jobs on machines.
False

Question 6.
The optimal assignment is made in the Hungarian method to cells in the reduced matrix that contain a Zero.
True

Question 7.
Using the Hungarian method, the optimal solution to an assignment problem is fund when the minimum number of lines required to cover the zero cells in the reduced matrix equals the number of people.
True

Question 8.
In an assignment problem, if a number of columns are greater than the number of rows, then a dummy column is added.
False

Question 9.
The purpose of a dummy row or column in an assignment problem is to obtain a balance between a total number of activities and a total number of resources.
True

Question 10.
One of the assumptions made while sequencing n jobs on 2 machines is: two jobs must be loaded at a time on any machine.
False

(IV) Solve the following problems.

Part – I

Question 1.
A plant manager has four subordinates, and four tasks to be performed. The subordinates differ in efficiency and the tasks differ in their intrinsic difficulty. This estimate of the times each man would take to perform each task is given in the effectiveness matrix below.

How should the tasks be allocated, one to a man, as to minimize the total man-hours?
Solution:
The hr matrix is given by

Subtracting row minimum from all values in that row we get

Subtracting column minimum from all values in that column we get

The minimum no. of lines covering ail the zeros (4) is equal to the order of the matrix (4)
∴ The assignment is possible.

The assignment is
A → I, B → III, C → II, D → IV
For the minimum hrs. take the corresponding value from the hr matrix.
Minimum hrs = 7 + 3 + 18 + 9 = 37 hrs

Question 2.
A dairy plant has five milk tankers, I, II, III, IV & V. These milk tankers are to be used on five delivery routes A, B, C, D & E. The distances (in kms) between the dairy plant and the delivery routes are given in the following distance matrix.

How should the milk tankers be assigned to the chilling centre so as to minimize the distance travelled?
Solution:
The distance matrix is given by

Subtracting row minimum from all values in that row we get

Subtracting column minimum from each value in that column we get

The number of lines covering all the zeros (3) is less than the order of the matrix (5) so the assignment is not possible. The modification is required.
The minimum uncovered value (15) is subtracted from uncovered values and added to the values at the intersection. The numbers on the lines remain the same. We get

The minimum lines covering all the zeros (4) are less than the order of the matrix (5) so the assignment is not possible. The modification is required the minimum uncovered value (5) is subtracted from uncovered values and added to the values at the intersection. The numbers on the lines remain the same we get

The minimum number of lines covering all the zeros (5) is equal to the order of the matrix (5) So assignment is possible.
The assignment is
A → II, B → III, C → V, D → I, E → IV
Total minimum distance is = 120 + 120 + 175 + 40 + 70 = 525 kms.

Question 3.
Solve the following assignment problem to maximize sales:

Solution:
As it is a maximization problem so we need to convert it into a minimization problem.
Subtracting all the values from the maximum value (19) we get

Also, it is an unbalanced problem so we need to add a dummy row (E) with all values zero, we get

Subtracting row minimum from all values in that row we get

Subtracting column minimum from all values in that column we get the same matrix

The minimum number of lines covering all the zero (4) is less than the order of the matrix (5) So assignment is not possible. The modification is required. The minimum uncovered value (2) is subtracted from the uncovered values and added to the values at the intersection. The values on the lines remain the same. We get

The minimum number of lines covering all the zeros (4) is less than the order of the matrix (5) so the assignment is not possible. The modification is required. The minimum uncovered value (1) is subtracted from the uncovered value and added to the values at the intersection. The values on the lines remain the same. We get

The minimum number of lines covering all the zeros (5) is equal to the order of the matrix (5) so the assignment is possible.

The assignment is
A → V, B → II, C → IV, D → III, E → I
No salesman goes to I as E is a dummy row.
For the maximum value take the corresponding values from the original matrix.
We get Maximum value = 15 + 19 + 14 + 17 + 0 = 65 units

Question 4.
The estimated sales (tons) per month in four different cities by five different managers are given below:

Find out the assignment of managers to cities in order to maximize sales.
Solution:
This is a maximizing problem. To convert it into minimizing problem subtract all the values of the matrix from the maximum (largest) value (39) we get

Also as it is an unbalanced problem so we have to add a dummy column (T) with all the values as zero. We get

Subtracting row minimum from all values in that row we get the same matrix
Subtracting column minimum from all values in that column we get

The minimum number of lines covering all the zeros (4) is less than the order of the matrix (5) so assignments are not possible. The modification is required. The minimum uncovered value (1) is subtracted from the uncovered values and added to the values at the intersection. The values on the lines remain the same. We get

The minimum number of lines covering all the zeros (5) is equal to the order of the matrix (5) so the assignment is possible.

So I → S, II → T, III → Q, IV → P, V → R.
As T is dummy manager II is not given any city.
To find the maximum sales we take the corresponding value from the original matrix
Total maximum sales = 35 + 39 + 36 + 35 = 145 tons

Question 5.
Consider the problem of assigning five operators to five machines. The assignment costs are given in the following table.

Operator A cannot be assigned to machine 3 and operator C cannot be assigned to machine 4. Find the optimal assignment schedule.
Solution:
This is a restricted assignment problem, so we assign a very high cost (oo) to the prohibited cells we get

Subtracting row minimum from all values in that row we get.

Subtracting column minimum from all values in that column we get

As the minimum number of lines covering all the zeros (4) is equal to the order of the matrix (5) so the assignment is not possible. The modification is required. The minimum uncovered value (2) is subtracted from all the uncovered values and added to the values at the intersection. The values on the lines remain the same. We get

As the minimum number of lines covering all the zeros (5) is equal to the order of the matrix, assignment is the possible

So A → 4, B → 3, C → 2, D → 1, E → 5
For the minimum cost take the corresponding values from the cost matrix we get
Total minimum cost = 3 + 3 + 4 + 3 + 7 = 20 units

Question 6.
A chartered accountant’s firm has accepted five new cases. The estimated number of days required by each of their five employees for each case are given below, where-means that the particular employee can not be assigned the particular case. Determine the optimal assignment of cases of the employees so that the total number of days required to complete these five cases will be minimum. Also, find the minimum number of days.

Solution:
This is a restricted assignment problem so we assign a very high cost (∞) to all the prohibited cells. The day matrix becomes

Subtracting row minimum from all values in that row we get

Subtracting column minimum from all values in that column we get

The minimum number of lines covering all the zeros (4) is less than the order of the matrix (5) so the assignment is not possible, The modification is required. The minimum uncovered value (1) is subtracted from all the uncovered values and added to the values at the intersection. The values on the lines remain the same, we get

The minimum number of lines covering all the zeros (5) is equal to the order of the matrix (5) so the assignment is possible.

So E1 → I, E2 → IV, E3 → II, E4 → V, E5 → III
To find the minimum number of days we take the corresponding values from the day matrix.
Total minimum number of days = 6 + 6 + 6 + 6 + 3 = 27 days

Part – II

Question 1.
A readymade garments manufacture has to process 7 items through two stages of production, namely cutting and sewing. The time taken in hours for each of these items in different stages are given below:

Find the sequence in which these items are to be processed through these stages so as to minimize the total processing time. Also, find the idle time of each machine.
Solution:
Let A = cutting and B = sewing. So we have

Observe min {A, B} = 2 for item 1 for B.

The problem reduces to

Now min {A, B} = 3 for item 3 for A

The problem reduces to

New min {A , B} = 4 for item 4 for A.

The problem reduces to

Now min(A, B} = 5 for item 6 for B

The problem reduces to

Now min {A, B} = 6 for item 5 for A and item 2 for B

Now only 7 is left
∴ The optimal sequence is

Worktable

Total elapsed time = 46 hrs
Idle time for A (cutting) = 46 – 44 = 2 hrs
Idle time for B (Sewing) = 4 hrs

Question 2.
Five jobs must pass through a lathe and a surface grinder, in that order. The processing times in hours are shown below. Determine the optimal sequence of the jobs. Also, find the idle time of each machine.

Solution:
Let A = lathe and B = surface grinder. We have

Observe min {A, B} = 1 for job II for A

The problem reduces to

Now min {A, B} = 2 for job IV for A

The problem reduces to

Now min {A, B} = 3 for job I for B

The problem reduces to

Now min {A, B} = 5 for jobs III and V for A
∴ We have two options

or

We take the first one.
Worktable

Total elapsed time = 21 hrs
Idle time for A (lathe) = 21 – 17 = 4 hrs
Idle time for B (surface grinder) = 3 hrs

Question 3.
Find the sequence that minimizes the total elapsed time to complete the following jobs. Each job is processed in order AB.

Determine the sequence for the jobs so as to minimize the processing time. Find the total elapsed time and the idle time for both machines.
Solution:
Observe min {A, B} = 3 for job VII on B.

The problem reduces to

Now min {A, B} = 4 for job IV on B.

The problem reduces to

Now min {A, B} = 5 for job III & V on A. we have two options

or

We take the first one
The problem reduces to

Now min {A, B} = 5 for job II on A

The problem reduces to

Now min {A, B} = 7 for a job I on B and for job VI on A
∴ The optional sequence is

Worktable

Total elapsed time = 55 units
Idle time for A = 55 – 52 = 3 units
Idle time for B = 9 units.

Question 4.
A toy manufacturing company has five types of toys. Each toy has to go through three machines A, B, C in the order ABC. The time required in hours for each process is given in the following table.

Solve the problem for minimizing the total elapsed time.
Solution:
Min A = 12, Max B = 12
As min A ≥ max B.
The problem can be converted into two machine problems.
Let G and H be two fictitious machines such that G = A + B and H = B + C, We get

Now min {G, H} = 16 for type 3 on G

The problem reduces to

Min (G, H} = 18 for type 1, 4 & 5 on H
We have more than one option, we take

Now only type 2 is left.
∴ The optional sequence is

Worktable

Total elapsed time = 102 hours
Idle time for A = 102 – 84 = 18 hours
Idle time for B = 54 + (102 – 94) = 62 hours
Idle time for C = 38 hours

Question 5.
A foreman wants to process 4 different jobs on three machines: a shaping machine, a drilling machine, and a tapping, the sequence of operations being shaping-drilling-tapping. Decide the optimal sequence for the four jobs to minimize the total elapsed time. Also, find the total elapsed time and the idle time for every machine.

Solution:
The time matrix is

Min A = 8, Max B = 8, as min A ≥ max B.
The problem can be converted into a two-machine problem.
Let G and H be two fictitious machines such that
G = A + B and H = B + C we get

Observe min (G, H} = 12 for job 2 on H

The problem reduces to

Now min {G, H} = 14 for job 3 on G and job 4 on H

Now only job 1 is left.
∴ The optimal sequence is

Worktable

Total elapsed time = 74 min
Idle time for A (shapping) = 74 – 62 = 12 min
Idle time for B (Drilling) = 47 + (74 – 70) = 51 min
Idle time for C (trapping) = 31 min

## Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Ex 7.2

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 7 Assignment Problem and Sequencing Ex 7.2 Questions and Answers.

## Maharashtra State Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Ex 7.2

Question 1.
A machine operator has to perform two operations, turning and threading on 6 different jobs. The time required to perform these operations (in minutes) for each job is known. Determine the order in which the jobs should be processed in order to minimize the total time required to complete all the jobs. Also, find the total processing time and idle times for turning and threading operations.

Solution:
Let turning to be A and threading be B.

∴ Observe Min{A, B} = 1 for job 6 on B.

Then the problem reduces to

∴ Now Min {A, B} = 2 for job 4 on A

Then the problem reduce to

Now Min {A, B} = 3 for job 1 on A and job 5 on B

Then the problem reduces to

Now Min {A, B) = 5 for job 3 on A

Only job 2 is left so the optimal sequence is

Worktable is given by

Total elapsed time = 43 minutes
Idle time for A (turning) = 43 – 42 = 1 min
Idle time for B (threshing) = 2 + 4 = 6 min

Question 2.
A company has three jobs on hand, Each of these must be processed through two departments, in the AB where
Department A: Press shop and
Department B: Finishing
The table below gives the number of days required by each job each department

Find the sequence in which the three jobs should be processed so as to take minimum time to finish all the three jobs. Also find idle time for both the departments.
Solution:
Observe Min {A, B} = 3 for job II on B.

Then the problem is reduced to

Now Min {A, B} = 4 for job III at B

Now only job I in left
∴ the optimal sequence is given by

The work table is

Total elapsed time = 23 days
Idle time for A = 23 – 19 = 4 days
Idle time for B = 8 days

Question 3.
An insurance company receives three types of policy application bundles daily from its head office for data entry and filing. The time (in minutes) required for each type for these two operations is given in the following table:

Find the sequence that minimizes the total time required to complete the entire task. Also, find the total elapsed time and idle times for each operation.
Solution:
Let Data entry be A and filing be B. So

Observe min {A, B} = 90 for policy 1 at A

Then the problem reduces to

Observe min {A, B} = 100 for policy 3 at B

Now only policy 2 is left
∴ The optimal sequence is

Worktable

So Total elapsed time = 490 min
Idle time for A (data entry) = 490 – 390 = 100 min
Idle time for B (filing) = 140 min.

Question 4.
There are five jobs, each of which must go through two machines in the order XY. Processing times (in hours) are given below. Determine the sequence for the jobs that will minimize the total elapsed time. Also, find the total elapse time and idle time for each machine.

Solution:
Observe min {x, y} = 2 for job B on x

The problem reduces to

Now min [x, y] = 4 for job A on x

The problem reduces to

Now min [x, y] = 6 for job D on x

The problem reduces to

Now min [x, y] = 8 for job E on y

Now only job C in left
∴ The optimal sequence is

Worktable

Total elapsed time = 60 hrs
Idle time for X = 60 – 56 = 4 hrs
Idle time for Y = 6 hrs

Question 5.
Find the sequence that minimizes the total elapsed time to complete the following jobs in the order AB. Find the total elapsed time and idle times for both machines.

Solution:
Observe min {A, B} = 5 for job VI for B and job VII for A

The problem reduces to

Now min {A, B] = 7 for job I on A

The problem reduces to

Now min {A, B] = 10 for job IV on A and B so we have two options.

Or

we take the 1st one.
The problem reduces to

Now min {A, B} = 14 for job V on A and job II and III for job B.
∴ We have

Or

We take the optimal sequence as.
VII – I – IV – V – III – II – VI
Worktable

Total elapsed time = 91 units
Idle time for A = 91 – 86 = 5 units
Idle time for B = 13 units

Question 6.
Find the optimal sequence that minimizes the total time required to complete the following jobs in the order ABC. The processing times are given in hrs.

Solution:
(i) Min A = 5, Max B = 5
As min A ≥ max B.
The problem can be converted into two machine problems.
Let G and H be two fictitious machines such that G = A + B and H = B + C we get

Now min {G, H} = 7 for job III & V for G and job I for H
∴ We have two options

Or

We take the first one
The problem reduces to

Min {G, H} = 9 for job IV on H

The problem reduces to

Now min {G, H} = 10 for job II for G and job VII for H

Now job VI is left
∴ The optimal sequence is

The work table is

Total elapsed time = 61 hrs
Idle time for A = 61 – 54 = 7 hrs
Idle time for B = 35 + [61 – 58] = 38 hrs
Idle time for C = 15 hrs

(ii) Min A = 5, Max B = 5
Min A ≥ Max B.
The problem can be converted into two machine problems.
Let G and H be two fictitious machines such that G = A + B and H = B + C we get

Now min {G, H} = 5 for job 1 for H.

The problem reduces to

Now min {G, H} = 8 for job 2 for G and job H also job 5 for G
∴ We have two options

Or

We take the first one
The problem reduces to

Now min {G, H} = 9 for job 3 for H

Now only job 4 is left
∴ The optimal sequence is

Worktable

Total elapsed time = 40 hrs
Idle time for A = 40 – 32 = 8 hrs
Idle time for B = 19 + [40 – 34] = 25 hrs
Idle time for C = 12 hrs

Question 7.
A publisher produces 5 books on Mathematics. The books have to go through composing, printing, and binding was done by 3 machines P, Q, E. The time schedule for the entire task in the proper unit is as follows.

Determine the optimum time required to finish the entire task.
Solution:
Min R = 6, Max Q = 6
As min R ≥ max Q.
The problem can be converted into a two-machine problem.
Let G and H be two fictitious machines such that G = P + Q and H = Q + R we get

Min {G, H} = 9 for books A, D, E for G.
∴ We have more than one option we take

The problem reduces to

Min {G, H} = 8 for book C on H

Now only B is left. So the optimal sequence is

Worktable

Total elapsed time = 51 units
Idle time for P = 51 – 32 = 19 units
Idle time for Q = 14 + [51 – 34] = 31 units
Idle time for R = 9 units

## Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Ex 7.1

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 7 Assignment Problem and Sequencing Ex 7.1 Questions and Answers.

## Maharashtra State Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Ex 7.1

Question 1.
A job production unit has four jobs A, B, C, D which can be manufactured on each of the four machines P, Q, R, and S. The processing cost of each job for each machine is given in the following table:

Find the optimal assignment to minimize the total processing cost.
Solution:
The cost matrix is given by

Subtracting row minimum from all the elements in that row we get

Subtracting column minimum from all the elements in that column we get the same matrix.
As all the rows and columns have single zeros the allotment can be done as follows.

As per the table, the job allotments are
P → II, Q → IV, R → I, S → III
The total minimum cost = 25 + 21 + 19 + 34 = ₹ 99

Question 2.
Five wagons are available at stations 1, 2, 3, 4, and 5. These are required at 5 stations I, II, III, IV, and V. The mileage between various stations are given in the table below. How should the wagons be transported so as to minimize the mileage covered?

Solution:
The mileage matrix is given by

Subtracting row minimum from all elements in that row we get

Subtracting column minimum from all elements in that column we get

Draw minimum lines covering all the zeros

The number of lines covering all the zeros (3) is less than the order of the matrix (5). Hence an assignment is not possible. The modification is required. The minimum uncovered value 1 is subtracted from uncovered values and added to the values at the intersection. The numbers on the lines remain the same we get

Drawing a minimum number of lines covering all the zeros.
No. of lines covering all the zeros (4) is less than the order of the matrix (5).
Hence assignment is not possible.
Again modification is required. The minimum uncovered value 3 is subtracted from the uncovered values and added to the values at the intersection.
The numbers on the lines remain the same we get

No. of lines covering all the zeros (5) are equal to the order of the matrix so the assignment is possible.

According to the table the assignment is
1 → I, 2 → II, 3 → IV, 4 → II, 5 → V
Total minimum mileage = 10 + 6 + 4 + 9 + 10 = 39 units

Question 3.
Five different machines can do any of the five required jobs, with different profits five required jobs, with different profits resulting from each assignment as shown below:

Find the optimal assignment schedule.
Solution:
This profit matrix has to be reduced to cost matrix by subtracting all the values of the matrix from the largest value (62) we get

Subtracting row minimum value from all the elements in that column we get

Subtracting column minimum from all the elements in that column we get

Drawing minimum lines covering all zeros we get

No. of lines (4) is less than the order of the matrix (5). Hence assignment is not possible. The modification is required. The minimum uncovered value (4) is subtracted from the uncovered value and added to the values at the intersection. The values on the lines remain the same, we get

No. of lines (5) are equal to the order of the matrix (5). So assignments are possible
1 → C, 2 → E, 3 → A, 4 → D, 5 → B
For the minimum profit look at the corresponding in the profit matrix given.
Maximum profit = 40 + 36 + 40 + 36 + 62 = 214 units

Question 4.
Four new machines M1, M2, M3, and M4 are to be installed in a machine shop. There are five vacant places A, B, C, D, and E available. Because of limited space, machine M2 cannot be placed at C and M1 cannot be placed at A. The cost matrix is given below.

Find the optimal assignment schedule.
Solution:
This is a restricted assignment so we assign a very high cost ‘∞’ to the prohibited all.
Also as it is an unbalanced problem we add a dummy row M5 with all values as ‘0’, we get

Subtracting row minimum from all the elements in that row, we get

Subtracting column minimum from all the elements in that column we get the same matrix.

As minimum no. of lines covering all zeros (5) is equal to the order of the matrix, Assignment is possible

The assignments are given by
M1 → A, M2 → B, M3 → E, M4 → D, M5 → C
As M5 is dummy no machine is installed at C
For minimum cost taking the corresponding values in the cost matrix we get
Minimum cost = 4 + 4 + 2 + 2 = 12 units

Question 5.
A company has a team of four salesmen and there is four districts where the company wants to start its business. After taking into account the capabilities of salesmen and the nature of districts, the company estimates that the profit per day in rupees for each salesman in each district is as below:

Find the assignments of a salesman to various districts which will yield maximum profit.
Solution:
The profit matrix has to be reduced to the cost matrix. Subtracting all the values from the maximum value (16) we get

Subtracting row minimum from all values in that row we get

Subtracting column minimum from each column we get the same matrix

As minimum no. of lines covering all zeros (4) is equal to the order of the matrix (4) Assignment is possible

∴ A → 1, B → 3, C → 2, D → 4
For maximum profit, we take the corresponding values in the profit matrix. We get
Maximum profit = 16 + 15 + 15 + 15 = ₹ 61

Question 6.
In the modification of a plant layout of a factory four new machines M1, M2, M3, and M4 are to be installed in a machine shop. There are five vacant places A, B, C, D, and E available. Because of limited space, machine M2 can not be placed at C and M3 can not be placed at A the cost of locating a machine at a place (in hundred rupees) is as follows.

Find the optimal assignment schedule.
Solution:
This is an unbalanced problem so we add a dummy row M5 with all values as ‘0’.
Also, this is on restricted assignment problem. So we assign a very high-cost W to the prohibited cells we have

Subtracting row minimum from all values in that row we get

Subtracting column minimum from all values in that column we get the same matrix

As minimum no. of lines covering all zeros (5) is equal to the order of the matrix (5) assignment is possible.

The assignment is
M1 → A, M2 → B, M3 → E, M4 → D, M5 → C
As M5 is dummy, no machine is installed at C.
The minimum cost is found by taking the corresponding values in the cost matrix
Minimum cost = 9 + 9 + 7 + 7 + 0 = 32 (in hundred ₹)

## Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Linear Programming Miscellaneous Exercise 6

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 6 Linear Programming Miscellaneous Exercise 6 Questions and Answers.

## Maharashtra State Board 12th Commerce Maths Solutions Chapter 6 Linear Programming Miscellaneous Exercise 6

(I) Choose the correct alternative.

Question 1.
The value of the objective function is maximized under linear constraints.
(a) at the centre of feasible region
(b) at (0, 0)
(c) at any vertex of feasible region.
(d) The vertex which is at maximum distance from (0, 0).
(a) at the centre of feasible region

Question 2.
Which of the following is correct?
(a) Every LPP has on optional solution
(b) Every LPP has unique optional solution
(c) If LPP has two optional solutions then it has infinitely many solutions
(d) The set of all feasible solutions LPP may not be a convex set.
(c) If LPP has two optional solutions then it has infinitely many solutions

Question 3.
Objective function of LPP is
(a) a constraint
(b) a function to be maximized or minimized
(c) a relation between the decision variables
(d) a feasible region.
(b) a function to be maximized or minimized

Question 4.
The maximum value of z = 5x + 3y. subject to the constraints 3x + 5y = 15; 5x + 2y ≤ 10, x, y ≥ 0 is
(a) 235
(b) $$\frac{235}{9}$$
(c) $$\frac{235}{19}$$
(d) $$\frac{235}{3}$$
(c) $$\frac{235}{19}$$

Question 5.
The maximum value of z = 10x + 6y. subject to the constraints 3x + y ≤ 12, 2x + 5y ≤ 34, x ≥ 0, y ≥ 0 is.
(a) 56
(b) 65
(c) 55
(d) 66
(a) 56

Question 6.
The point at which the maximum value of z = x + y subject to the constraint x + 2y ≤ 70, 2x + y ≤ 15, x ≥ 0, y ≥ 0 is
(a) (36, 25)
(b) (20, 35)
(c) (35, 20)
(d) (40, 15)
(d) (40, 15)

Question 7.
Of all the points of the feasible region the optimal value of z is obtained at a point
(a) Inside the feasible region
(b) at the boundary of the feasible region
(c) at vertex of feasible region
(d) on x -axis
(c) at vertex of feasible region

Question 8.
Feasible region; the set of points which satisfy
(a) The objective function
(b) All of the given function
(c) Some of the given constraints
(d) Only non-negative constraints
(b) All of the given function

Question 9.
Solution of LPP to minimize z = 2x + 3y subjected to x ≥ 0, y ≥ 0, 1 ≤ x + 2y ≤ 10 is
(a) x = 0, y = $$\frac{1}{2}$$
(b) x = $$\frac{1}{2}$$, y = 0
(c) x = 1, y = -2
(d) x = y = $$\frac{1}{2}$$
(a) x = 0, y = $$\frac{1}{2}$$

Question 10.
The corner points of the feasible region given by the inequation x + y ≤ 4, 2x + y ≤ 7, x ≥ 0, y ≥ 0, are
(a) (0, 0), (4, 0), (3, 1), (0, 4)
(b) (0, 0), ($$\frac{7}{2}$$, 0), (3, 1), (0, 4)
(c) (0, 0), ($$\frac{7}{2}$$, 0), (3, 1), (5, 7)
(d) (6, 0), (4, 0), (3, 1), (0, 7)
(b) (0, 0), ($$\frac{7}{2}$$, 0), (3, 1), (0, 4)

Question 11.
The corner point of the feasible region are (0, 0), (2, 0), ($$\frac{12}{7}$$, $$\frac{3}{7}$$) and (0, 1) then the point of maximum z = 6.5x + y = 13
(a) (0, 0)
(b) (2, 0)
(c) ($$\frac{11}{7}$$, $$\frac{3}{7}$$)
(d) (0, 1)
(b) (2, 0)

Question 12.
If the corner points of the feasible region are (0, 0), (3, 0), (2, 1) and (0, $$\frac{7}{3}$$) the maximum value of z = 4x + 5y is
(a) 12
(b) 13
(c) $$\frac{35}{2}$$
(d) 0
(b) 13

Question 13.
If the comer points of the feasible region are (0, 10), (2, 2), and (4, 0) then the point of minimum z = 3x + 2y is.
(a) (2, 2)
(b) (0, 10)
(c) (4, 0)
(d) (2, 4)
(a) (2, 2)

Question 14.
The half plane represented by 3x + 2y ≤ 0 contains the point.
(a) (1, $$\frac{5}{2}$$)
(b) (2, 1)
(c) (0, 0)
(d) (5, 1)
(c) (0, 0)

Question 15.
The half plane represented by 4x + 3y ≥ 14 contains the point
(a) (0, 0)
(b) (2, 2)
(c) (3, 4)
(d) (1, 1)
(c) (3, 4)

(II) Fill in the blanks.

Question 1.
Graphical solution set of the in equations x ≥ 0, y ≥ 0 is in _________ quadrant.
First

Question 2.
The region represented by the in equations x ≥ 0, y ≥ 0 lines in _________ quadrants.
First

Question 3.
The optimal value of the objective function is attained at the _________ points of feasible region.
End

Question 4.
The region represented by the inequality y ≤ 0 lies in _________ quadrants
Third and Fourth

Question 5.
The constraint that a factory has to employ more women (y) than men (x) is given by _________
y > x

Question 6.
A garage employs eight men to work in its showroom and repair shop. The constants that there must be not least 3 men in showroom and repair shop. The constrains that there must be at least 3 men in showroom and at least 2 men in repair shop are _________ and _________ respectively.
x ≥ 3 and y ≥ 2

Question 7.
A train carries at least twice as many first class passengers (y) as second class passangers (x). The constraint is given by _________
x ≥ 2y

Question 8.
A dishwashing machine hold up to 40 pieces of large crockery (x) this constraint is given by _________
x ≤ 40

(III) State whether each of the following is True or False.

Question 1.
The region represented by the inequalities x ≥ 0, y ≥ 0 lies in first quadrant.
True

Question 2.
The region represented by the inqualities x ≤ 0, y ≤ 0 lies in first quadrant.
False

Question 3.
The optimum value of the objective function of LPP occurs at the center of the feasible region.
False

Question 4.
Graphical solution set of x ≤ 0, y ≥ 0 in xy system lies in second quadrant.
True

Question 5.
Saina wants to invest at most ₹ 24000 in bonds and fixed deposits. Mathematically this constraints is written as x + y ≤ 24000 where x is investment in bond and y is in fixed deposits.
True

Question 6.
The point (1, 2) is not a vertex of the feasible region bounded by 2x + 3y ≤ 6, 5x + 3y ≤ 15, x ≥ 0, y ≥ 0.
True

Question 7.
The feasible solution of LPP belongs to only quadrant I. The Feasible region of graph x + y ≤ 1 and 2x + 2y ≥ 6 exists.
True

(IV) Solve the following problems.

Question 1.
Maximize z = 5x1 + 6x2, Subject to 2x1 + 3x2 ≤ 18, 2x1 + x2 ≤ 12, x ≥ 0, y ≥ 0
Solution:

OAED is the feasible region O(0, 0) A(0, 6) D (6, 0)
and E is the intersection of 2x1 + 3x2 = 18 and 2x1 + x2 = 12
For E, Solving, 2x1 + 3x2 = 18 …….(i)
2x1 + x2 = 12 ……(ii)
We get x1 = 4.5, x2 = 3
∴ E = (4.5, 3)

∴ Maximum value of z = 40.5 at E(4.5, 3)

Question 2.
Minimize z = 4x + 2y, Subject to 3x + y ≥ 27, x + y ≥ 21, x ≥ 0, y ≥ 0
Solution:

AED is the feasible region A(0, 27), D(21, 0)
and E is the point of intersection of 3x + y = 27 and x + y = 21
For E, Solving 3x + y = 27 ………(i)
x + y = 21 …….(ii)
We get x = 3, y = 18
∴ E(3, 18)

∴ Minimum value of z = 48 at (3, 18)

Question 3.
Maximize z = 6x + 10y, subject to 3x + 5y ≤ 10, 5x + 3y ≤ 15, x ≥ 0, y ≥ 0
Solution:

OAED is the feasible region;
O(0, 0), A (0, 2) D (3, 0) and E is the point of intersection of 3x + 5y = 10 and 5x + 3y = 15
For E, Solving 3x + 5y = 10
5x + 3y = 15
We get, x = $$\frac{45}{16}$$, y = $$\frac{5}{16}$$
∴ E($$\frac{45}{16}$$, $$\frac{5}{16}$$)

Since the maximum value of z = 20 at two points i.e, at A(0, 2) and E($$\frac{45}{16}$$, $$\frac{5}{16}$$).
z is maximum at all points on segment AE.
Hence it has infinite number of solutions.

Question 4.
Minimize z = 2x + 3y, Subject to x – y ≤ 1, x + y ≥ 3, x ≥ 0, y ≥ 0
Solution:

Shaded portion CE is the feasible region
Where C = (0, 3) and E is the point of intersection of x – y = 1 and x + y = 3
For E, Solving x – y = 1 …….(i)
x + y = 3 ………(ii)
We get, x = 2, y = 1
∴ E(2, 1)

∴ Minimum value z = 7 at E(2, 1)

Question 5.
Maximize z = 4x1 + 3x2, Subject to 3x1 + x2 ≤ 15, 3x1 + 4x2 ≤ 24, x ≥ 0, y ≥ 0
Solution:

OCEB is the feasible region where O(0, 0), C(0, 6), B(5, 0)
E is the point of intersection of 3x1 + x2 = 15 and 3x1 + 4x2 = 24
For E, Solving 3x1 + x2 = 15 ……..(i)
3x1 + 4x2 = 24 …….(ii)
We get, x1 = 4, x2 = 3
∴ E(4, 3)

∴ Maximum value of z = 25 at (4, 3)

Question 6.
Maximize z = 60x + 50y, Subject to x + 2y ≤ 40, 3x + 2y ≤ 60, x ≥ 0, y ≥ 0
Solution:

OAED is the feasible region O(0, 0), A(0, 20), D(20, 0)
E is x + 2y = 40 and 3x + 2y = 60
For E, Solving x + 2y = 40
3x + 2y = 60
We get, x = 10, y = 15
∴ E = (10, 15)

∴ Maximum value of z = 1350 at E(10, 15).

Question 7.
Minimize z = 4x + 2y, Subject to 3x + y ≥ 27, x + y ≥ 21, x + 2y ≥ 30, x ≥ 0, y ≥ 0
Solution:

AGHF is the feasible region where A(0, 27) F(30, 0)
G is the point of intersection of 3x + y = 27 and x + y = 21
H is the point of intersection of x + y = 21 and x + 2y = 30
For G, Solving 3x + y = 27 …….(i)
x + y = 21 ………(ii)
We get, x = 3, y = 18
∴ G(3, 8)
For H, Solving x + y = 21 …….(i)
x + 2y = 30 ……..(ii)
We get, x = 12, y = 9
∴ G = (12, 9)

∴ Maximum value of z = 48 at G(3, 18)

Question 8.
A carpenter makes chairs and table profit are ₹ 140 per chair and ₹ 210 per table Both products are processed on three machines, Assembling, Finishing and Polishing the time required for each product in hours and availability of each machine is given by the following table.

Formulate and solve the following Linear programming problem using graphical method.
Solution:
Let z be the profit which can be made by selling x chair and y table.
∴ x ≥ 0, y ≥ 0
Total profit = 140x + 210y
According to the table, the constraints can be written as
3x + 3y ≤ 36
5x + 2y ≤ 50
2x + 6y ≤ 60
∴ The given LPP can be formulated as.
Maximize z = 140x + 210y
Subject to 3x + 3y ≤ 36, 5x + 2y ≤ 50, 2x + 6y ≤ 60, x ≥ 0, y ≥ 0.

OEGH is the feasible region
O(0, 0), D(10.4, 0), E(0, 10)
For G, Solving 3x + 3y = 36 ……..(i)
2x + 6y = 60 …….(ii)
∴ G = (3, 9)
For H, Solving 5x + 2y = 52 ………(i)
3x + 3y = 36 ……..(ii)
We get, x = $$\frac{28}{3}$$, y = $$\frac{8}{3}$$
∴ H($$\frac{28}{3}$$, $$\frac{8}{3}$$)

∴ z in maximum at (3, 9) and maximum profit = ₹ 2310.

Question 9.
A company manufactures bicycles and tricycles, each of which must be processed through two machines A and B Maximum availability of machines A and B are respectively 120 and 180 hours. Manufacturing a bicycle requires 6 hours on machine A and 3 hours on machine B manufacturing a tricycle requires 4 hours on machine A and 10 hours on machine B. If profits are ₹ 180 for a bicycle and ₹ 220 on a tricycle, determine the number of bicycles and tricycles that should be manufactured in order to maximize the profit.
Solution:
Let x number of bicycle and y number of tricycle has to be manufactured and to be sold to get the profit (z)
∴ x ≥ 0, y ≥ 0
Total profit = 180x + 220y.
The given LPP can be tabulated as follows:

∴ The given LPP can be formulated as
Maximize z = 180x + 220y
Subject to 6x + 4y ≤ 120, 3x + 10y ≤ 180, x ≥ 0, y ≥ 0

OCEB is the feasible region where O(0, 0) C(0, 18) B(20, 0)
E is the point of intersection of 6x + 4y = 120 and 3x + 10y = 180
For E, Solving 6x + 4y = 120 ……..(i)
3x + 10y = 180 …….(ii)
We get, x = 10, y = 15
∴ E(10, 15)

∴ Maximum value of z is 5100 at E(10, 15)
Hence 10 bicycles and 15 tricycles should be produced to get maximum profit.

Question 10.
A factory produced two types of chemicals A and B. The following table gives the units of ingredients P and Q (per kg) of chemicals A and B as well as minimum requirements of P and Q and also cost per kg. Chemicals A and B:

Find the number of units of chemicals A and B should be produced sp as to minimize the cost.
Solution:
Let x be the no. of units of chemicals, A produced and y be the no. of units of chemical B produced.
Total cost is 4x + 6y
The LPP is. Minimise z = 4x + 6y
Subject to x + 2y ≥ 80, 3x + y ≥ 75, x ≥ 0, y ≥ 0.

The shaded region BEC in the feasible region B(80, 0) C(0, 75)
and E is the point of intersection of x + 2y = 80 and 3x + y = 75
For E, Solving x + 2y = 80 …….(i)
3x + y = 75 …….(ii)
We get, x = 14, y = 33
∴ E(14, 33)

∴ z is minimum at E(14, 33) and the minimum value of z = 254.
Hence 14 units of chemical A and 33 units of chemical B are produced to get a minimum cost of ₹ 254.

Question 11.
A company produces mixers and food processors. Profit on selling one mixer and one food processor is ₹ 2,000/- and ₹ 3,000/- respectively. Both the products are processed through three Machines A, B, C. The time required in hours by each product and total time available in hours per week on each machine are as follows:

How many mixes and food processors should be produced to maximize profit?
Solution:
Let x be the no. of mixers produced and y be the no.of food processors produced.
The profit is 2000x + 3000y
The LPP is Maximize 2 = 2000x + 3000y
Subject to 3x + 3y ≤ 36, 5x + 2y ≤ 50, 2x + 6y ≤ 60 x, y ≥ 0.
Let 3x + 3y = 36,
i.e. x + y = 12
x = 0, y = 12, (0, 12)
y = 0, x = 12, (12, 0)
5x + 2y = 50
x = 0, y = 25, (0, 25)
y = 0, x = 10, (10, 0)
Let 2x + 6y = 60, x + 3y = 30
x = 0, y = 10 (0, 10)
y = 0, x = 30 (30, 0)

The Shaded region OABCD is the feasible region.
O(0, 0) A (10, 0) D(0, 10)
B is the intersection of x + y = 12 and 5x + 2y = 50
Solving we get x = 8.67, y = 3.33
∴ B(8.67, 3.33)
C is the intersection of x + y = 12 and x + 3y = 30
Solving we get x = 3, y = 9
∴ C(3, 9)

∴ Maximum value of z is 330000 at C(3, 9)
Hence 3 mixers and 9 food processors should be produced to get a maximum profit of ₹ 33,000.

Question 12.
A chemical company produces a chemical containing three basic elements A, B, C so that it has at least 16 liters of A, 24 liters of B, and 18 liters of C. This chemical is made by mixing two compounds I and II. Each unit of compound I has 4 liters of A, 12 liters of B, 2 liters of C. Each unit of compound II has 2 liters of A, 2 liters of B and 6 liters of C. The cost per unit of compound is ₹ 800/- and that of compound II is ₹ 640/- Formulate the problem as L.P.P and solve it to minimize the cost.
Solution:
Let x be the no. of units of compound I used and y be the no of units of compound II used.
The data can be tabulated as

The LPP is minimize z = 800x + 640y
Subject to 4x + 2y ≥ 16, 12x + 2y ≥ 24, 2x + 6y ≥ 18, x, y ≥ 0.
Let 4x + 2 y = 16,
2x + y = 8
x = 0, y = 8, (0, 8)
y = 0, x = 4, (4, 0)
12x + 2y = 24,
6x + y = 12
x = 0, y = 12, (0, 12)
y = 0, x = 2, (2, 0)
2x + 6y = 18
x + 3y = 9
x = 0, y = 3, (0, 3)
y = 0, x = 9, (9, 0)

The Shaded region ABCD is the feasible region.
A(0, 12) D(9, 0)
B is the intersection of 6x + y = 12 and 2x + y = 8
Solving we get x = 1, y = 6
∴ B(1, 6)
C is the intersection of 2x + y = 8 and x + 3y = 9
Solving we get x = 3, y = 2
∴ C(3, 2)

∴ The minimum value of z = 3680 at (3, 2)
Hence 3 unit of compound I and 2 units of compound II should be used to get the minimum cost of ₹ 3680.

Question 13.
A person who makes two types of gift items A and B requires the services of a cutter and a finisher. Gift item A requires 4 hours of the cutter’s time and 2 hours of the finisher’s time. B required 2 hours of the cutter’s time and 4 hours of finisher‘s time. The cutter and finisher have 208 hours and 152 hours available times respectively every month. The profit of one gift item of type A is ₹ 75/- and on gift item B is ₹ 125/-. Assuming that the person can sell all the gift items produced, determine how many gift items of each type should he make every month to obtain the best returns?
Solution:
Let x be the no. of gift A produced and y be the no. of gift B produced.
The data can be tabulated as

The LLP is maximize z = 75x + 125y
Subject to 4x + 2y ≤ 208, 2x + 4y ≤ 152, x, y ≥ 0
Let 4x + 2y = 208, 2x + 4y = 152,
2x + y = 104, x + 2y = 76,
x = 0, y = 104; (0, 104), x = 0, y = 38; (0, 38)
y = 0, x = 52; (52, 0), y = 0, x = 76; (76, 0)

The shaded region OABC is the feasible region.
O(0,0) A (52, 0) C(0, 38)
B is the intersection of 2x + y = 104 and x + 2y = 76
Solving we get x = 44, y = 16
∴ B(44, 16)

∴ Maximum value of z = 5300 at B(44, 16)
Hence he should produce 44 gifts of type A & 16 of type B to get a maximum profit of ₹ 5300.

Question 14.
A firm manufactures two products A and B on which profit is earned per unit ₹ 3/- and ₹ 4/- respectively. Each product is processed on two machines M1 and M2. Product A requires one minute of processing time on Mx and two minutes of processing time on M2. B requires one minute of processing time on M1 and one minute processing time on M2 Machine M1 is available for use for 450 minutes while M2 is available for 600 minutes during any working day. Final the number of units of products A and B to be manufactured to get the maximum profit.
Solution:
Let x denote the number of units of product A and y denote the number of units of product B.
The total profit in ₹ 3x + 4y,
The given statements can be tabulated as

The constraints are x + y ≤ 450, 2x + y ≤ 600, x ≥ 0, y ≥ 0
∴ The LPP can be formulated as
Maximize z = 3x + 4y
Subject to x + y ≤ 450, 2x + y ≤ 600, x ≥ 0, y ≥ 0.

OAED in the feasible region O(0, 0) A(0, 450) D(300, 0)
E is the intersection of x + y = 450 and 2x + y = 600
For E, Solving x + y = 450 ……..(i)
2x + y = 600 ……..(ii)
Solving x = 150, y = 300
∴ E(150, 300)

∴ Maximum value of z = 1800 at (0, 450)

Question 15.
A firm manufacturing two types of electrical items A and B, can make a profit of ₹ 20/- per unit of A and ₹ 30/- per unit of B. Both A and B make use of two essential components a motor and a transformer. Each unit of A requires 3 motors and 2 transformers and each unit of B required 2 motors and 4 transformers. The total supply of components per month is restricted to 210 motors and 300 transformers. How many units of A and B should they manufacture per month to maximize profit? How much is the maximum profit?
Solution:
Let x be the no. of electrical item A and y be the no. of electrical items to be manufactured per month to maximize the profit.
Total profit is ₹ 20x + 30y
The given condition can be tabulated as.

The given LPP can be formulated as
Maximize z = 20x + 30y
Subject of 3x + 2y ≤ 210, 2x + 4y ≤ 300, x ≥ 0, y ≥ 0.

BOCE is the feasible region
B(70, 0) O(0, 0) C(0, 75)
E is the intersection of 3x + 2y = 210 and 2x + 4y = 300
For E, Solving 3x + 2y = 210 ……….(i)
2x + 4y = 300 …….(ii)
We get x = 30, y = 60
∴ E(30, 60)

∴ Maximum value of z = ₹ 2400 at (30, 60)
Hence 30 units of A and 60 units of B should be manufactured per month to get the maximum profit of ₹ 2400

## Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Linear Programming Ex 6.2

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 6 Linear Programming Ex 6.2 Questions and Answers.

## Maharashtra State Board 12th Commerce Maths Solutions Chapter 6 Linear Programming Ex 6.2

Solve the following LPP by graphical method.

Question 1.
Maximize z = 11x + 8y, Subject to x ≤ 4 ,y ≤ 6 x + y ≤ 6, x ≥ 0, y ≥ 0.
Solution:

The feasible solution is AOBE
Where A(4, 0) O(0, 0) B(0, 6)
E is the point of intersection of x + y = 6 and x = 4.
∴ 4 + y = 6
∴ y = 2
∴ E = (4, 2)

∴ z is maximum at (4, 12) and the maximum value of z = 60

Question 2.
Maximize z = 4x + 6y, Subject to 3x + 2y ≤ 12, x + y ≥ 4 x, y ≥ 0.
Solution:

From figure, ABC is the feasible region
Where A(0, 6) B(4, 0) C(0, 4)

Maximum value of z = 36 at A(0, 6)

Question 3.
Maximize z = 7x + 11y, Subject to 3x + 5y ≤ 26, 5x + 3y ≤ 30, x ≥ 0, y ≥ 0.
Solution:

∴ AODE is the feasible region where
A(0, 5.2) O(0, 0) D(6, 0) and E is the intersection of 3x + 5y = 26 and 5x + 3y = 30
For E,
Solving 3x + 5y = 26 ……(i)
5x + 3y = 30 ……(ii)
We get, x = 4.5, y = 2.5
∴ E = (4.5, 2.5)

∴ Maximum value of z = 59 at E(4.5, 2.5)

Question 4.
Maximize z = 10x + 25y, Subject to 0 ≤ x ≤ 3, 0≤ y ≤ 3, x + y ≤ 5.
Solution:
The constraints can be written as, x ≤ 3, x ≥ 0, y ≥ 0, y ≤ 3, x + y ≤ 5

ABCDE is the feasible region where A(3, 0) B(0, 0) and C(0, 3) D is the intersection of y = 3 and x + 5y = 5 and E is the intersection of x = 3 and x + 7 = 5
For D,
Solving y = 3 ………(i)
x + y = 5 ……..(ii)
We get x = 2, y = 3
∴ D = (2, 3)
For E,
Solving x = 3 …….(i)
x + y = 5 ……….(ii)
We get x = 3, y = 2
∴ E = (3, 2)

∴ Maximum value of z = 90 at D(2, 3)

Question 5.
Maximize z = 3x + 5y, Subject to x + 4y ≤ 24, 3x + y ≤ 21, x + y ≤ 9, x ≥ 0, y ≥ 0.
Solution:

OAGHD is the feasible region where O(0, 0), A(0, 6), D(7, 0) G is the intersecting point of x + 4y = 24 and x + y = 9
H is the intersecting points of 3x + y = 21 and x + y = 9.
For G, Solving x + 4y = 24 …….(i)
x + y = 9 ………(ii)
We get, x = 4, y = 5
∴ G (4, 5)
For H, Solving x + y = 9 ………(i)
3x + y = 21 ……..(ii)
We get x = 6, y = 3
∴ H(6, 3)

∴ Maximum value of z = 37 at the point G(4, 5)

Question 6.
Minimize z = 7x + y Subject to 5x + y ≥ 5, x + y ≥ 3, x ≥ 0, y ≥ 0
Solution:

AED is the feasible region where A(0, 5) D(3, 0) and E is the point of intersection of 5x + y = 5 and x + y = 3.
For E, Solving 5x + y = 5 ………(i)
x + y = 3 ……..(ii)
We get, x = $$\frac{1}{2}$$, y = $$\frac{5}{2}$$
∴ E($$\frac{1}{2}$$, $$\frac{5}{2}$$)

∴ Minimum value of z = 5 at A(0, 5)

Question 7.
Minimize z = 8x + 10y, Subject to 2x + y ≥ 7, 2x + 3y ≥ 15 ,y ≥ 2, x ≥ 0, y ≥ 0
Solution:

AEG is the feasible solution where A(0, 7)
F is the point of intersection of 2x + y = 7 and 2x + 3y = 15
G is the point of intersection of y = 2 and 2x + 3y = 15
For F, Solving 2x + y = 7 ……..(i)
2x + 3y = 15 ……..(ii)
We get x = $$\frac{3}{2}$$, y = 4
∴ F = ($$\frac{3}{2}$$, 4)
For G, Solving 2x + 3y = 15 ……..(i)
y = 2 …….(ii)
We get x = 4.5, y = 2
∴ G = (4.5, 2)

∴ Minimum value of z = 52 at F($$\frac{3}{2}$$, 4)

Question 8.
Minimize z = 6x + 2y, Subject to x + 2y ≥ 3, x + 4y ≥ 4, 3x + y ≥ 3, x ≥ 0, y ≥ 0.
Solution:

DGHB is the feasible region where D(0, 3), B(4, 0)
G is the point of intersection of 3x + y = 3 and x + 2y = 3 and H is the point of intersection of x + 2y = 3 and x + 4y = 4
For G, Solving 3x + y = 3 ……(i)
x + 2y = 3 ………(ii)
W e get x = $$\frac{3}{5}$$, y = $$\frac{6}{5}$$
∴ G($$\frac{3}{5}$$, $$\frac{6}{5}$$)
For H, Solving x + 2y = 3 ……..(i)
x + 4y = 4 ………(ii)
We get, x = 2, y = $$\frac{1}{2}$$
∴ H(2, $$\frac{1}{2}$$)

∴ Minimum value of z = 22.5 at H(2, $$\frac{1}{2}$$)