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Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 5 Locus and Straight Line Ex 5.2 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 5 Locus and Straight Line Ex 5.2

Question 1.
Find the slope of each of the following lines which pass through the points:
(a) (2, -1), (4, 3)
(b) (-2, 3), (5, 7)
(c) (2, 3), (2, -1)
(d) (7, 1), (-3, 1)
Solution:
(a) Let A = (x₁, y₁) = (2, -1) and B = (x₂, y₂) = (4, 3).
Slope of line AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{3-(-1)}{4-2}\)
= \(\frac{4}{2}\)
= 2

(b) Let C = (x₁, y₁) = (-2, 3) and D = (x₂, y₂) = (5, 7)
Slope of line CD = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{7-3}{5-(-2)}\)
= \(\frac{4}{7}\)

(c) Let E = (2, 3) = (x₁, y₁) and F = (2, -1) = (x₂, y₂)
Since x₁ = x₂ = 2
∴ The slope of EF is not defined. ……[EF || y-axis]

(d) Let G = (7, 1) = (x₁, y₁) and H = (-3, 1) = (x₂, y₂) say.
Since y₁ = y₂
∴ The slope of GH = 0 …..[GH || x-axis]

Question 2.
If the X and Y-intercepts of line L are 2 and 3 respectively, then find the slope of line L.
Solution:
Given, x-intercept of line L is 2 and y-intercept of line L is 3
∴ the line L intersects X-axis at (2, 0) and Y-axis at (0, 3).
i.e. the line L passes through (2, 0) = (x₁, y₁) and (0, 3) = (x₂, y₂) say.
Slope of line L = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{3-0}{0-2}\)
= \(\frac{-3}{2}\)

Question 3.
Find the slope of the line whose inclination is 30°.
Solution:
Given, inclination (θ) = 30°
Slope of the line = tan θ = tan 30° = \(\frac{1}{\sqrt{3}}\)

Question 4.
Find the slope of the line whose inclination is 45°.
Solution:
Given, inclination (θ) = 45°
Slope of the line = tan θ = tan 45° = 1

Question 5.
A line makes intercepts 3 and 3 on the co-ordinate axes. Find the slope of the line.
Solution:
Given, x-intercept of line is 3 and y-intercept of line is 3
∴ The line intersects X-axis at (3, 0) and Y-axis at (0, 3).
i.e. the line passes through (3, 0) = (x₁, y₁) and (0, 3) = (x₂, y₂) say.
Slope of line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{3-0}{0-3}\)
= -1

Question 6.
Without using Pythagoras theorem, show that points A(4, 4), B(3, 5) and C(-1, -1) are the vertices of a right-angled triangle.
Solution:
Given, A(4, 4) = (x₁, y₁), B(3, 5) = (x₂, y₂), C(-1, -1) = (x₃, y₃)
Slope of AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{5-4}{3-4}=-1\)
Slope of BC = \(\frac{y_{3}-y_{2}}{x_{3}-x_{2}}=\frac{-1-5}{-1-3}=\frac{-6}{-4}=\frac{3}{2}\)
Slope of AC = \(\frac{y_{3}-y_{1}}{x_{3}-x_{1}}=\frac{-1-4}{-1-4}=\frac{-5}{-5}=1\)
Slope of AB × slope of AC = -1 × 1 = -1
∴ side AB ⊥ side AC
∴ ΔABC is a right angled triangle, right angled at A.
∴ The given points are the vertices of a right angled triangle.

Question 7.
Find the slope of the line which makes angle of 45° with the positive direction of the Y-axis measured clockwise.
Solution:

Since, the line makes an angle of 45° with positive direction of Y-axis in anticlockwise direction.
∴ Inclination of the line (θ) = (90° + 45°)
∴ Slope of the line = tan(90° + 45°)
= -cot 45° …….[tan(90 + θ°) = -cot θ]
= -1

Question 8.
Find the value of k for which the points P(k, -1), Q(2, 1), and R(4, 5) are collinear.
Solution:
Given, points P(k, -1), Q(2, 1), and R(4, 5) are collinear.
∴ Slope of PQ = Slope of QR
∴ \(\frac{1-(-1)}{2-k}=\frac{5-1}{4-2}\)
∴ \(\frac{2}{2-k}=\frac{4}{2}\)
∴ 1 = 2 – k
∴ k = 2 – 1 = 1

Check:
For collinear points P, Q, R,
Slope of PQ = Slope of QR = Slop of PR
For k = 1, if the given points are collinear, then our answer is correct.
P(1, -1), Q(2, 1) and R(4, 5)
Slope of PQ = \(\frac{1-(-1)}{2-1}=\frac{2}{1}=2\)
Slope of QR = \(\frac{5-1}{4-2}=\frac{4}{2}=2\)
Slope of PQ = Slope of QR
∴ The given points are collinear.
Thus, our answer is correct.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 5 Locus and Straight Line Ex 5.1 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 5 Locus and Straight Line Ex 5.1

Question 1.
If A(1, 3) and B(2, 1) are points, find the equation of the locus of point P such that PA = PB.
Solution:
Let P(x, y) be any point on the required locus.
Given, A(1, 3) and B(2, 1).
PA = PB
∴ PA² = PB²
∴ (x – 1)² + (y – 3)² = (x – 2)² + (y – 1)²
∴ x² – 2x + 1 + y² – 6y + 9 = x² – 4x + 4 + y² – 2y + 1
∴ -2x – 6y + 10 = -4x – 2y + 5
∴ 2x – 4y + 5 = 0
∴ The required equation of locus is 2x – 4y + 5 = 0.

Question 2.
A(-5, 2) and B(4, 1). Find the equation of the locus of point P, which is equidistant from A and B.
Solution:
Let P(x, y) be any point on the required locus.
P is equidistant from A(-5, 2) and B(4, 1).
∴ PA = PB
∴ PA² = PB²
∴ (x + 5)² + (y – 2)² = (x – 4)² + (y – 1)²
∴ x² + 10x + 25 + y² – 4y + 4 = x² – 8x + 16 + y² – 2y + 1
∴ 10x – 4y + 29 = -8x – 2y + 17
∴ 18x – 2y + 12 = 0
∴ 9x – y + 6 = 0
∴ The required equation of locus is 9x – y – 6 = 0

Question 3.
If A(2, 0) and B(0, 3) are two points, find the equation of the locus of point P such that AP = 2BP.
Solution:
Let P(x, y) be any point on the required locus.
Given, A(2, 0), B(0, 3) and AP = 2BP
∴ AP² = 4BP²
∴ (x – 2)² + (y – 0)² = 4[(x – 0)² + (y – 3)²]
∴ x² – 4x + 4 + y² = 4(x² + y² – 6y + 9)
∴ x² – 4x + 4 + y² = 4x² + 4y² – 24y + 36
∴ 3x² + 3y² + 4x – 24y + 32 = 0
∴ The required equation of locus is 3x² + 3y² + 4x – 24y + 32 = 0

Question 4.
If A(4, 1) and B(5, 4), find the equation of the locus of point P if PA² = 3PB².
Solution:
Let P(x, y) be any point on the required locus.
Given, A(4, 1), B(5, 4) and PA² = 3PB²
∴ (x – 4)² + (y – 1)² = 3[(x – 5)² + (y – 4)²]
∴ x² – 8x + 16 + y² – 2y + 1 = 3(x² – 10x + 25 + y² – 8y + 16)
∴ x² – 8x + y² – 2y + 17 = 3x² – 30x + 75 + 3y² – 24y + 48
∴ 2x² + 2y² – 22x – 22y + 106 = 0
∴ x² + y² – 11x – 11y + 53 = 0
∴ The required equation of locus is x² + y² – 11x – 11y + 53 = 0.

Question 5.
A(2, 4) and B(5, 8), find the equation of the locus of point P such that PA² – PB² = 13.
Solution:
Let P(x, y) be any point on the required locus.
Given, A(2, 4), B(5, 8) and PA² – PB² = 13
∴ [(x – 2)² + (y – 4)²] – [(x – 5)² + (y – 8)²] = 13
∴ (x² – 4x + 4 + y² – 8y + 16) – (x² – 10x + 25 + y² – 16y + 64) = 13
∴ 6x + 8y – 69 = 13
∴ 6x + 8y – 82 = 0
∴ 3x + 4y – 41 = 0
∴ The required equation of locus is 3x + 4y – 41 = 0

Question 6.
A(1, 6) and B(3, 5), find the equation of the locus of point P such that segment AB subtends a right angle at P. (∠APB = 90°)
Solution:
Let P(x. y) be any point on the required locus.
Given, A(1, 6) and B(3, 5), ∠APB = 90°
∴ ΔAPB is a right-angled triangle.

By Pythagoras theorem,
AP² + PB² = AB²
∴ [(x – 1)² + (y – 6)²] + [(x – 3)² + (y – 5)²] = (1 – 3)² + (6 – 5)²
∴ x² – 2x + 1 + y² – 12y + 36 + x² – 6x + 9 + y² – 10y + 25 = 4 + 1
∴ 2x² + 2y² – 8x – 22y + 66 = 0
∴ x² + y² – 4x – 11y + 33 = 0
∴ The required equation of locus is x² + y² – 4x – 11y + 33 = 0

Question 7.
If the origin is shifted to the point O'(2, 3), the axes remaining parallel to the original axes, find the new co-ordinates of the points (a) A(1, 3) (b) B(2, 5)
Solution:
Origin is shifted to (2, 3) = (h, k)
Let the new co-ordinates be (X, Y).
∴ x = X + h and y = Y + k
∴ x = X + 2 and y = Y + 3 …..(i)
(a) Given, A(x, y) = A(1, 3)
x = X + 2 and y = Y + 3 …..[From (i)]
∴ 1 = X + 2 and 3 = Y + 3
∴ X = -1 and Y = 0
∴ the new co-ordinates of point A are (-1, 0).

(b) Given, B(x, y) = B(2, 5)
x = X + 2 andy = Y + 3 ……[From (i)]
∴ 2 = X + 2 and 5 = Y + 3
∴ X = 0 and Y = 2
∴ the new co-ordinates of point B are (0, 2).

Question 8.
If the origin is shifted to the point O'(1, 3), the axes remaining parallel to the original axes, find the old co-ordinates of the points (a) C(5, 4) (b) D(3, 3)
Solution:
Origin is shifted to (1, 3) = (h, k)
Let the new co-ordinates be (X, Y)
x = X + h and y = Y + k
∴ x = X + 1 and 7 = Y + 3 …..(i)
(a) Given, C(X, Y) = C(5, 4)
∴ x = X + 1 andy = Y + 3 …..[From(i)]
∴ x = 5 + 1 = 6 and y = 4 + 3 = 7
∴ the old co-ordinates of point C are (6, 7).

(b) Given, D(X, Y) = D(3, 3)
∴ x = X + 1 and y = Y + 3 …..[From (i)]
∴ x = 3 + 1 = 4 and y = 3 + 3 = 6
∴ the old co-ordinates of point D are (4, 6).

Question 9.
If the co-ordinates (5, 14) change to (8, 3) by the shift of origin, find the co-ordinates of the point, where the origin is shifted.
Solution:
Let the origin be shifted to (h, k).
Given, (x,y) = (5, 14), (X, Y) = (8, 3)
Since, x = X + h and y = Y + k
∴ 5 = 8 + h and 14 = 3 + k
∴ h = -3 and k = 11
∴ the co-ordinates of the point, where the origin is shifted are (-3, 11).

Question 10.
Obtain the new equations of the following loci if the origin is shifted to the point O'(2, 2), the direction of axes remaining the same:
(a) 3x – y + 2 = 0
(b) x² + y² – 3x = 7
(c) xy – 2x – 2y + 4 = 0
Solution:
Given, (h, k) = (2, 2)
Let (X, Y) be the new co-ordinates of the point (x, y).
∴ x = X + h and y = Y + k
∴ x = X + 2 and y = Y + 2
(a) Substituting the values of x and y in the equation 3x – y + 2 = 0, we get
3(X + 2) – (Y + 2) + 2 = 0
∴ 3X + 6 – Y – 2 + 2 = 0
∴ 3X – Y + 6 = 0, which is the new equation of locus.

(b) Substituting the values of x and y in the equation x² + y² – 3x = 7, we get
(X + 2)² + (Y + 2)² – 3(X + 2) = 7
∴ X² + 4X + 4 + Y² + 4Y + 4 – 3X – 6 = 7
∴ X² + Y² + X + 4Y – 5 = 0, which is the new equation of locus.

(c) Substituting the values of x and y in the equation xy – 2x – 2y + 4 = 0, we get
(X + 2) (Y + 2) – 2(X + 2) – 2(Y + 2) + 4 = 0
∴ XY + 2X + 2Y + 4 – 2X – 4 – 2Y – 4 + 4 = 0
∴ XY = 0, which is the new equation of locus.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 4 Sequences and Series Miscellaneous Exercise 4 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Miscellaneous Exercise 4

Question 1.
In a G.P., the fourth term is 48 and the eighth term is 768. Find the tenth term.
Solution:

Question 2.
For a G.P. a = \(\frac{4}{3}\) and t₇ = \(\frac{243}{1024}\), find the value of r.
Solution:

Question 3.
For a sequence, if t_n = \(\frac{5^{n-2}}{7^{n-3}}\), verify whether the sequence is a G.P. If it is a G.P., find its first term and the common ratio.
Solution:
The sequence (t_n) is a G.P., if \(\frac{5^{n-2}}{7^{n-3}}\) = constant, for all n ∈ N.

∴ the sequence is a G.P. with common ratio = \(\frac{5}{7}\)
∴ first term = t₁ = \(\frac{5^{1-2}}{7^{1-3}}=\frac{5^{-1}}{7^{-2}}=\frac{7^{2}}{5}=\frac{49}{5}\)

Question 4.
Find three numbers in G.P., such that their sum is 35 and their product is 1000.
Solution:
Let the three numbers in G.P. be \(\frac{a}{r}\), a, ar.
According to the first condition,


∴ the three numbers in G.P. are 20, 10, 5 or 5, 10, 20.

Question 5.
Find 4 numbers in G. P. such that the sum of the middle 2 numbers is \(\frac{10}{3}\) and their product is 1.
Solution:
Let the four numbers in G.P. be \(\frac{a}{r^{3}}, \frac{a}{r}, a r, a r^{3}\).
According to the second condition,
\(\frac{\mathrm{a}}{\mathrm{r}^{3}}\left(\frac{\mathrm{a}}{\mathrm{r}}\right)(\mathrm{ar})\left(\mathrm{ar}^{3}\right)=1\)
∴ a⁴ = 1
∴ a = 1
According to the first condition,

Question 6.
Find five numbers in G.P. such that their product is 243 and the sum of the second and fourth numbers is 10.
Solution:
Let the five numbers in G.P. be
\(\frac{a}{r^{2}}, \frac{a}{r}, a, a r, a r^{2}\)
According to the first condition,

Question 7.
For a sequence, S_n = 4(7ⁿ – 1), verify whether the sequence is a G.P.
Solution:

Question 8.
Find 2 + 22 + 222 + 2222 + …… upto n terms.
Solution:
S_n = 2 + 22 + 222 +….. upto n terms
= 2(1 + 11 + 111 +…… upto n terms)
= \(\frac{2}{9}\) (9 + 99 + 999 + … upto n terms)
= \(\frac{2}{9}\) [(10 – 1) + (100 – 1) + (1000 – 1) +…… upto n terms]
= \(\frac{2}{9}\) [(10 + 100 + 1000 + … upto n terms) – (1 + 1 + 1 + ….. n times)]
Since, 10, 100, 1000, …… n terms are in G.P.
with a = 10, r = \(\frac{100}{10}\) = 10

Question 9.
Find the nth term of the sequence 0.6, 0.66, 0.666, 0.6666,…..
Solution:
0.6, 0.66, 0.666, 0.6666, ……
∴ t₁ = 0.6
t₂ = 0.66 = 0.6 + 0.06
t₃ = 0.666 = 0.6 + 0.06 + 0.006
Hence, in general
t_n = 0.6 + 0.06 + 0.006 + …… upto n terms.
The terms are in G.P.with
a = 0.6, r = \(\frac{0.06}{0.6}\) = 0.1
∴ t_n = the sum of first n terms of the G.P.

Question 10.
Find \(\sum_{r=1}^{n}\left(5 r^{2}+4 r-3\right)\).
Solution:

Question 11.
Find \(\sum_{\mathbf{r}=1}^{\mathbf{n}} \mathbf{r}(\mathbf{r}-\mathbf{3})(\mathbf{r}-\mathbf{2})\).
Solution:

Question 12.
Find \(\sum_{r=1}^{n} \frac{1^{2}+2^{2}+3^{2}+\ldots+r^{2}}{2 r+1}\)
Solution:
We know that,

Question 13.
Find \(\sum_{r=1}^{n} \frac{1^{3}+2^{3}+3^{3}+\ldots+r^{3}}{(r+1)^{2}}\)
Solution:

Question 14.
Find 2 × 6 + 4 × 9 + 6 × 12 + …… upto n terms.
Solution:
2, 4, 6, … are in A.P.
∴ rth term = 2 + (r – 1)2 = 2r
6, 9, 12, … are in A.P.
∴ rth term = 6 + (r – 1) (3) = (3r + 3)
∴ 2 × 6 + 4 × 9 + 6 × 12 +…… upto n terms

= n(n + 1) (2n + 1 + 3)
= 2n(n + 1)(n + 2)

Question 15.
Find 12² + 13² + 14² + 15² + …… + 20².
Solution:
12² + 13² + 14² + 15² + …… + 20²
= (1² + 2² + 3² + 4² + ……. + 202) – (1² + 2² + 3² + 4² + …… + 11²)

= 2870 – 506
= 2364

Question 16.
Find (50² – 49²) + (48² – 47²) + (46² – 45²) + …… + (2² – 1²).
Solution:
(50² – 49²) + (48² – 47²) + (46² – 45²) + …… + (2² – 1²)
= (50² + 48² + 46² + …… + 2²) – (49² + 47² + 45² + …… + 1²)
= \(\sum_{r=1}^{25}(2 r)^{2}-\sum_{r=1}^{25}(2 r-1)^{2}\)

= 1300 – 25
= 1275

Question 17.
In a G.P., if t₂ = 7, t₄ = 1575, find r.
Solution:

Question 18.
Find k so that k – 1, k, k + 2 are consecutive terms of a G.P.
Solution:
Since k – 1, k, k + 2 are consecutive terms of a G.P.
∴ \(\frac{k}{k-1}=\frac{k+2}{k}\)
∴ k² = k² + k – 2
∴ k – 2 = 0
∴ k = 2

Question 19.
If pth, qth and rth terms of a G.P. are x, y, z respectively, find the value of \(x^{q-r} \cdot y^{r-p} \cdot z^{p-q}\).
Solution:
Let a be the first term and R be the common ratio of the G.P.
∴ t_n = \(\text { a. } R^{n-1}\)
∴ x = \(\text { a. } R^{p-1}\), y = \(\text { a. } R^{q-1}\), z = \(\text { a. } R^{r-1}\)

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 4 Sequences and Series Ex 4.5 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.5

Question 1.
Find the sum \(\sum_{r=1}^{n}(r+1)(2 r-1)\).
Solution:

Question 2.
Find \(\sum_{r=1}^{n}\left(3 r^{2}-2 r+1\right)\).
Solution:

Question 3.
Find \(\sum_{r=1}^{n} \frac{1+2+3+\ldots+r}{r}\).
Solution:

Question 4.
Find \(\sum_{r=1}^{n} \frac{1^{3}+2^{3}+\ldots+r^{3}}{r(r+1)}\).
Solution:
We know that,

Question 5.
Find the sum 5 × 7 + 7 × 9 + 9 × 11 + 11 × 13 + …… upto n terms.
Solution:
5 × 7 + 7 × 9 + 9 × 11 + 11 × 13 + ….. upto n terms
Now, 5, 7, 9, 11, … are in A.P.
rth term = 5 + (r – 1) (2) = 2r + 3
7, 9, 11,. … are in A.P.
rth term = 7 + (r – 1) (2) = 2r + 5
∴ 5 × 7 + 7 × 9 + 9 × 11 + 11 × 13 + …… upto n terms

Question 6.
Find the sum 2² + 4² + 6² + 8² + …… upto n terms.
Solution:
2² + 4² + 6² + 8² + …… upto n terms
= (2 × 1)² + (2 × 2)² + (2 × 3)² + (2 × 4)² + ……

Question 7.
Find (70² – 69²) + (68² – 67²) + (66² – 65²) + ……. + (2² – 1²)
Solution:
Let S = (70² – 69²) + (68² – 67²) + …… +(2² – 1²)
∴ S = (2² – 1²) + (4² – 3²) + …… + (70² – 69²)
Here, 2, 4, 6,…, 70 is an A.P. with rth term = 2r
and 1, 3, 5,….., 69 in A.P. with rth term = 2r – 1

Question 8.
Find the sum 1 × 3 × 5 + 3 × 5 × 7 + 5 × 7 × 9 + …… + (2n – 1) (2n + 1) (2n + 3)
Solution:
1 × 3 × 5 + 3 × 5 × 7 + 5 × 7 × 9 + …… + (2n – 1) (2n + 1) (2n + 3)
Now, 1, 3, 5, 7, … are in A.P. with a = 1 and d = 2.
∴ rth term = 1 + (r – 1)2 = 2r – 1
3, 5, 7, 9, … are in A.P. with a = 3 and d = 2
∴ rth term = 3 + (r – 1)2 = 2r + 1
and 5, 7, 9, 11, … are in A.P. with a = 5 and d = 2
∴ rth term = 5 + (r – 1)2 = 2r + 3
∴ 1 × 3 × 5 + 3 × 5 × 7 + 5 × 7 × 9 + …… upto n terms

= n(n + 1)[2n(n + 1) + 4n + 2 – 1] – 3n
= n(n + l)(2n² + 6n + 1) – 3n
= n(2n³ + 8n² + 7n + 1 – 3)
= n(2n³ + 8n² + 7n – 2)

Question 9.
Find n, if \(\frac{1 \times 2+2 \times 3+3 \times 4+4 \times 5+\ldots+\text { upto } n \text { terms }}{1+2+3+4+\ldots+\text { upto } n \text { terms }}\) = \(\frac{100}{3}\)
Solution:

Question 10.
If S₁, S₂, and S₃ are the sums of first n natural numbers, their squares, and their cubes respectively, then show that:
9\(S_{2}^{2}\) = S₃(1 + 8S₁).
Solution:

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 machine 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 ₹)

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

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 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).
Answer:
(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.
Answer:
(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.
Answer:
(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) \(\frac{235}{3}\)
Answer:
(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
Answer:
(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)
Answer:
(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
Answer:
(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
Answer:
(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 = \(\frac{1}{2}\)
Answer:
(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)
Answer:
(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)
Answer:
(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
Answer:
(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)
Answer:
(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)
Answer:
(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)
Answer:
(c) (3, 4)

(II) Fill in the blanks.

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

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

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

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

Question 5.
The constraint that a factory has to employ more women (y) than men (x) is given by _________
Answer:
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.
Answer:
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 _________
Answer:
x ≥ 2y

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

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

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

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

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 1 Partition Values Miscellaneous Exercise 1 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 1 Partition Values Miscellaneous Exercise 1

Question 1.
The data gives the number of accidents per day on a railway track. Compute Q₂, P₁₇, and D₇.
4, 2, 3, 5, 6, 3, 4, 1, 2, 3, 2, 3, 4, 3, 2
Solution:
The given data can be arranged in ascending order as follows:
1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 5, 6
Here, n = 15
Q₂ = value of 2\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of 2\(\left(\frac{15+1}{4}\right)^{\text {th }}\) observation
= value of (2 × 4)th observation
= value of 8th observation
∴ Q₂ = 3
P₁₇ = value of 17\(\left(\frac{n+1}{100}\right)^{\text {th }}\) observation
= value of 17\(\left(\frac{15+1}{100}\right)^{\text {th }}\) observation
= value of (17 × 0.16)th observation
= value of (2.72)th observation
= value of 2nd observation + 0.72 (value of 3rd observation – value of 2nd observation)
= 2 + 0.72 (2 – 2)
∴ P₁₇ = 2
D₇ = value of 7\(\left(\frac{n+1}{10}\right)^{\text {th }}\) observation
= value of 7\(\left(\frac{15+1}{10}\right)^{\text {th }}\) observation
= value of (7 × 1.6)th observation
= value of (11.2)th observation
= value of 11th observation + 0.2(value of 12th observation – value of 11th observation)
= 4 + 0. 2(4 – 4)
∴ D₇ = 4

Question 2.
The distribution of daily sales of shoes (size-wise) for 100 days from a certain shop is as follows:

Compute Q₁, D₂, and P₉₅.
Solution:
By arranging the given data in ascending order, we construct the less than cumulative frequency table as given below:

Here, n = 100
Q₁ = value of \(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of \(\left(\frac{100+1}{4}\right)^{\text {th }}\) observation
= value of (25.25)th observation
Cumulative frequency which is just greater than (or equal) to 25.25 is 27.
∴ Q₁ = 3
D₂ = value of 2\(\left(\frac{n+1}{10}\right)^{\text {th }}\) observation
= value of 2\(\left(\frac{100+1}{10}\right)^{\text {th }}\) observation
= value of (2 × 10.1)th observation
= value of (20.2)th observation
Cumulative frequency which is just greater than (or equal) to 20.2 is 27.
∴ D₂ = 3
P₉₅ = value of 95\(\left(\frac{n+1}{100}\right)^{\text {th }}\) observation
= value of 95\(\left(\frac{100+1}{100}\right)^{\text {th }}\) observation
= value of (95 × 1.01)th observation
= value of (95.95)th observation
The cumulative frequency which is just greater than (or equal) to 95.95 is 100.
∴ P₉₅ = 8

Question 3.
Ten students appeared for a test in Mathematics and Statistics and they obtained the marks as follows:

If the median will be the criteria, in which subject, the level of knowledge of the students is higher?
Solution:
Marks in Mathematics can be arranged in ascending order as follows:
23, 23, 25, 25, 32, 35, 36, 37, 38, 42
Here, n = 10
∴ Median = value of \(\left(\frac{n+1}{2}\right)^{\text {th }}\) observation
Median = value of \(\left(\frac{10+1}{2}\right)^{\text {th }}\) observation
= value of (5.5)th observation
= value of 5th observation + 0.5(value of 6th observation – value of 5th observation)
= 32 + 0.5 (35 – 32)
= 32 + 0.5(3)
= 32 + 1.5
= 33.5
Marks in Statistics can be arranged in ascending order as follows:
22, 23, 26, 28, 29, 32, 34, 36, 45, 50
Here, n = 10
∴ Median = value of \(\left(\frac{n+1}{2}\right)^{\text {th }}\) observation
= value of \(\left(\frac{10+1}{2}\right)^{\text {th }}\) observation
= value of (5.5)th observation
= value of 5th observation + 0.5(value of 6th observation – value of 5th observation)
= 29 + 0.5(32 – 29)
= 29 + 0.5(3)
= 29 + 1.5
= 30.5
∴ Median marks for Mathematics = 33.5 and
Median marks for Statistics = 30.5
∴ The level of knowledge in Mathematics is higher than that of Statistics.

Question 4.
In the frequency distribution of families given below, the number of families corresponding to expenditure group 2000 – 4000 is missing from the table. However, the value of the 25th percentile is 2880. Find the missing frequency.

Solution:
Let x be the missing frequency of expenditure group 2000 – 4000.
We construct the less than cumulative frequency table as given below:

Here, N = 75 + x
Given, P₂₅ = 2880
∴ P₂₅ lies in the class 2000 – 4000.
∴ L = 2000, h = 2000, f = x, c.f. = 14
∴ P₂₅ = L + \(\frac{h}{f}\left(\frac{25 \mathrm{~N}}{100}-\text { c.f. }\right)\)
∴ 2880 = 2000 + \(\frac{2000}{x}\left(\frac{75+x}{4}-14\right)\)
∴ 2880 – 2000 = \(\frac{2000}{x}\left(\frac{75+x-56}{4}\right)\)
∴ 880x = 500(x + 19)
∴ 880x = 500x + 9500
∴ 880x – 500x = 9500
∴ 380x = 9500
∴ x = 25
∴ 25 is the missing frequency of the expenditure group 2000 – 4000.

Question 5.
Calculate Q₁, D₆, and P₁₅ for the following data:

Solution:
Since the difference between any two consecutive mid values is 50, the width of each class interval is 50.
∴ the class intervals will be 0 – 50, 50 – 100, etc.
We construct the less than cumulative frequency table as given below:

Here, N = 500
Q₁ class = class containing \(\left(\frac{\mathrm{N}}{4}\right)^{\text {th }}\) observation
∴ \(\frac{\mathrm{N}}{4}=\frac{500}{4}\) = 125
Cumulative frequency which is just greater than (or equal) to 125 is 160.
Q₁ lies in the class 100 – 150.
∴ L = 100, h = 50, f = 80, c.f. = 80
∴ Q₁ = L + \(\left(\frac{\mathrm{N}}{4}-\text { c.f. }\right)\)
= 100 + \(\frac{50}{80}\)(125 – 80)
= 100 + \(\frac{5}{8}\)(45)
= 100 + 28.125
= 128.125
D₆ class = class containing \(\left(\frac{6 \mathrm{~N}}{10}\right)^{\text {th }}\) observation
∴ \(\frac{6 \mathrm{~N}}{10}=\frac{6 \times 500}{10}\) = 300
Cumulative frequency which is just greater than (or equal) to 300 is 410.
∴ D₆ lies in the class 200 – 250.
∴ L = 200, h = 50, f = 150, c.f. = 260
∴ D₆ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{6 \mathrm{~N}}{10}-\text { c.f. }\right)\)
= 200 + \(\frac{50}{150}\)(300 – 260)
= 200 + \(\frac{1}{3}\)(40)
= 200 + 13.33
= 213.33
P₁₅ class = class containing \(\left(\frac{15 \mathrm{~N}}{100}\right)^{\text {th }}\) observation
∴ \(\frac{15 \mathrm{~N}}{100}=\frac{15 \times 500}{100}\) = 75
Cumulative frequency which is just greater than (or equal) to 75 is 80.
∴ P15 lies in the class 50 – 100.
∴ L = 50, h = 50, f = 70, c.f. = 10
∴ P₁₅ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{15 \mathrm{~N}}{100}-\text { c.f. }\right)\)
= 50 + \(\frac{50}{70}\) (75 – 10)
= 50 + \(\frac{5}{7}\) (65)
= 50 + \(\frac{325}{7}\)
= 50 + 46.4286
= 96.4286
∴ Q₁ = 128.125, D₆ = 213.33, P₁₅ = 96.4286

Question 6.
Daily income for a group of 100 workers are given below:

P₃₀ for this group is ₹ 110. Calculate the missing frequencies.
Solution:
Let a and b be the missing frequencies of class 50 – 100 and class 200 – 250 respectively.
We construct the less than cumulative frequency table as given below:

Here, N = 62 + a + b
Since, N = 100
∴ 62 + a + b = 100
∴ a + b = 38 ……(i)
Given, P30 = 110
∴ P30 lies in the class 100 – 150.
∴ L = 100, h = 50, f = 25, c.f. = 7 + a
\(\frac{30 \mathrm{~N}}{100}=\frac{30 \times 100}{100}\) = 30
∴ P₃₀ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{30 \mathrm{~N}}{100}-\text { c.f. }\right)\)
∴ 110 = 100 + \(\frac{50}{25}\) [30 – (7 + a)]
∴ 110 – 100 = 2(30 – 7 – a)
∴ 10 = 2(23 – a)
∴ 5 = 23 – a
∴ a = 23 – 5
∴ a = 18
Substituting the value of a in equation (i), we get
18 + b = 38
∴ b = 38 – 18
∴ b = 20
∴ 18 and 20 are the missing frequencies of the class 50 – 100 and class 200 – 250 respectively.

Question 7.
The distribution of a sample of students appearing for a C.A. examination is:

Help C.A. institute to decide cut-off marks for qualifying for an examination when 3% of students pass the examination.
Solution:
To decide cut-off marks for qualifying for an examination when 3% of students pass, we have to find P97.
We construct the less than cumulative frequency table as given below:

Here, N = 1100
P97 class = class containing \(\left(\frac{97 \mathrm{~N}}{100}\right)^{\text {th }}\) observation
∴ \(\frac{97 \mathrm{~N}}{100}=\frac{97 \times 1100}{100}\) = 1067
Cumulative frequency which is just greater than (or equal) to 1067 is 1100.
∴ P₉₇ lies in the class 500 – 600.
∴ L = 500, h = 100, f = 130, c.f. = 970
∴ P₉₇ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{97 \mathrm{~N}}{100}-\text { c.f. }\right)\)
= 500 + \(\frac{100}{130}\)(1067 – 970)
= 500 + \(\frac{10}{13}\) (97)
= 500 + 74.62
= 574.62 ~ 575
∴ the cut off marks for qualifying an examination is 575.

Question 8.
Determine graphically the value of median, D₃, and P₃₅ for the data given below:

Solution:
To draw an ogive curve, we construct the less than cumulative frequency table as given below:

The points to be plotted for less than ogive are (15, 8), (20, 22), (25, 30), (30, 55), (35, 70), (40, 84), (45, 90).

N = 90
For median, consider \(\frac{\mathrm{N}}{2}=\frac{90}{2}\) = 45
For D3, consider \(\frac{3 \mathrm{~N}}{10}=\frac{3 \times 90}{10}\) = 27
For P35, consider \(\frac{35 \mathrm{~N}}{100}=\frac{35 \times 90}{100}\) = 31.5
∴ We take the values 45, 27 and 31.5 on the Y-axis and draw lines from these points parallel to X-axis.
From the points where they intersect the less than ogive, we draw perpendicular on the X-axis.
Foot of the perpendicular represent the values of median, D3 and P35 respectively.
∴ Median ~ 29, D₃ ~ 23.5, P₃₅ ~ 26

Question 9.
The I.Q. test of 500 students of a college is as follows:

Find graphically the number of students whose I.Q. is more than 55 graphically.
Solution:
To draw an ogive curve, we construct the less than cumulative frequency table as given below:

The points to be plotted for less than ogive are (30, 41), (40, 93), (50, 157), (60, 337), (70, 404), (80, 449), (90, 489), (100, 500)

To find the number of students whose I.Q. is more than 55, we consider the value 55 on the X-axis.
From this point, we draw a line that is parallel to Y-axis.
From the point this line intersects the less than ogive, we draw a perpendicular on the Y-axis.
The foot of perpendicular gives the number of students whose I.Q. is less than 55.
∴ The foot of perpendicular ~ 244
∴ No. of students whose I.Q. is less than 55 ~ 244
∴ No. of Students whose I.Q. is more than 55 = 500 – 244 = 256

Question 10.
Draw an ogive for the following distribution. Determine the median graphically and verify your result by a mathematical formula.

Solution:
To draw an ogive curve, we construct the less than cumulative frequency table as given below:

The points to be plotted for less than ogive are (150, 2), (155, 7), (160, 16), (165, 31), (170, 47), (175, 54), (180, 59) and (185, 60).

N = 60
∴ \(\frac{\mathrm{N}}{2}=\frac{60}{2}\) = 30
∴ We take the value 30 on the Y-axis and from this point, we draw a line parallel to X-axis.
From the point where this line intersects the less than ogive, we draw a perpendicular on X-axis.
The foot perpendicular gives the value of the median.
∴ Median ~ 164.67
Now, let us calculate the median from the mathematical formula.
∴ \(\frac{\mathrm{N}}{2}\) = 30
The median lies in the class interval 160 – 165.
∴ L = 160, h = 5, f = 15, c.f. = 16
Median = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{\mathrm{N}}{2}-\mathrm{c} . \mathrm{f} .\right)\)
= 160 + \(\frac{5}{15}\) (30 – 16)
= 160+ \(\frac{1}{3}\) × 14
= 160 + 4.67
= 164.67

Question 11.
In a group of 25 students, 7 students failed and 6 students got distinction and the marks of the remaining 12 students are 61, 36, 44, 59, 52, 56, 41, 37, 39, 38, 41, 64. Find the median marks of the whole group.
Solution:
n = 25
Median = \(\frac{\mathrm{n}+1}{2}=\frac{25+1}{2}\) = 13th observation
We have been stated that 7 students failed (assuming passing marks on 35) and 6 students got distinction (assuming distinction as 70+), and the marks of the remaining 12 students (who will be situated between the two groups mentioned above, if arranged in ascending order), we have,
F, F, F, F, F, F, F, 36, 37, 38, 39, 41, 41, 44, 52, 56, 59, 61, 64, D, D, D, D, D, D
∴ median = 13th observation = 41.

Question 12.
The median weight of a group of 79 students is found to be 55 kg. 6 more students are added to this group whose weights are 50, 51, 52, 59.5, 60, 61 kg. What will be the value of the median of the combined group if the lowest and the highest weights were 53 kg and 59 kg respectively?
Solution:
n = 79
Median = 55kg
Lowest observation = 53 kg
Flighest observation = 59 kg
6 new students are added to the group having weights in Kg as follows:
50, 51, 52, 59.5, 60, 61
From the above, we see that of the 6 new students, 3 have weights which are below the lowest weight of the earlier group and 3 have weights which are above the highest weight of the earlier group.
∴ the median remains the same
∴ median = 55 kg.

Question 13.
The median of the following incomplete table is 92. Find the missing frequencies:

Solution:
Let a and b be the missing frequencies of class 50 – 70 and class 110 – 130 respectively.
We construct the less than cumulative frequency table as given below:

Here, N = 54 + a + b
Since, N = 80
∴ 54 + a + b = 80
∴ a + b = 26 …..(i)
Given, Median = Q₂ = 92
∴ Q₂ lies in the class 90 – 110.
∴ L = 90, h = 20, f = 20, c.f. = 24 + a
\(\frac{2 \mathrm{~N}}{4}=\frac{2 \times 80}{4}\) = 40
∴ Q₂ = L + \(\frac{h}{f}\left(\frac{2 N}{4}-\text { c.f. }\right)\)
∴ 92 = 90 + \(\frac{20}{20}\) [40 – (24 + a)
∴ 92 – 90 = 40 – 24 – a
∴ 2 = 16 – a
∴ a = 14
Substituting the value of a in equation (i), we get
14 + b = 26
∴ b = 26 – 14 = 12
∴ 14 and 12 are the missing frequencies of the class 50 – 70 and class 110 – 130 respectively.

Question 14.
A company produces tables which are packed in batches of 100. An analysis of the defective tubes in different batches has received the following information:

estimate the number of defective tubes in the central batch.
Solution:
To find the number of defective tubes in the central batch, we have to find Q₂.
Since the given data is not continuous, we have to convert it into a continuous form by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of every class interval.
∴ the class intervals will be Less than 4.5, 4.5 – 9.5, etc.
We construct the less than cumulative frequency table as given below:

Here, N = 251
Q₂ class = class containing \(\left(\frac{2 \mathrm{~N}}{4}\right)^{\mathrm{th}}\) observation
∴ \(\frac{2 \mathrm{~N}}{4}=\frac{2 \times 251}{4}\) = 125.5
Cumulative frequency which is just greater than (or equal to) 125.5 is 180.
∴ Q₂ lies in the class 9.5 – 14.5.
∴ L = 9.5, h = 5, f = 84, c.f. = 96
∴ Q₂ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{2 \mathrm{~N}}{4}-\text { c.f. }\right)\)
= 9.5 + \(\frac{5}{84}\) (125.5 – 96)
= 9.5 + \(\frac{5}{84}\) × 29.5
= 9.5 + \(\frac{147.5}{84}\)
= 9.5 + 1.76
= 11.26

Question 15.
In a college, there are 500 students in junior college, 5% score less than 25 marks, 68 scores from 26 to 30 marks, 30% score from 31 to 35 marks, 70 scores from 36 to 40 marks, 20% score from 41 to 45 marks and the rest score 46 and above marks. What are the median marks?
Solution:
Given data can be written in tabulated form as follows:

Since the given data is not continuous, we have to convert it into the continuous form by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of every class interval.
∴ the class intervals will be Less than 25.5, 25.5 – 30.5, etc.
We construct the less than cumulative frequency table as given below:

Here, N = 500
Q₂ class = class containing \(\left(\frac{2 \mathrm{~N}}{4}\right)^{\text {th }}\) observation
∴ \(\frac{2 \mathrm{~N}}{4}=\frac{2 \times 500}{4}\) = 250
Cumulative frequency which is just greater than (or equal to) 250 is 313.
∴ Q₂ lies in the class 35.5 – 40.5.
∴ L = 35.5, h = 5, f = 70, c.f. = 243
∴ Median = Q₂ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{2 \mathrm{~N}}{4}-\text { c.f. }\right)\)
= 35.5 + \(\frac{5}{70}\) (250 – 243)
= 35.5 + \(\frac{1}{14}\) (7)
= 35.5 + 0.5
= 36

Question 16.
Draw a cumulative frequency curve more than typical for the following data and hence locate Q₁ and Q₃. Also, find the number of workers with daily wages
(i) Between ₹ 170 and ₹ 260
(ii) less than ₹ 260

Solution:
For more than ogive points to be plotted are (100, 200), (150, 188), (200, 160), (250, 124), (300, 74), (350, 49), (400, 31), (450, 15), (500, 5)

Here, N = 200
For Q₁, \(\frac{\mathrm{N}}{4}=\frac{200}{4}\) = 4
For Q₃, \(\frac{3 \mathrm{~N}}{4}=\frac{3 \times 200}{4}\) = 150
We take the points having Y co-ordinates 50 and 150 on Y-axis.
From these points, we draw lines which are parallel to X-axis.
From the points of intersection of these lines with the curve, we draw perpendicular on X-axis.
X-Co-ordinates of these points gives the values of Q₁ and Q₃.
Since X-axis has daily wages more than and not less than the given amounts.
∴ Q₁ = Q₃ and Q₃ = Q₁
∴ Q₂ ~ 215 , Q₃ ~ 348

(i) To find the number of workers with daily wages between ₹ 170 and ₹ 260,
Take the values 170 and 260 on X-axis. From these points, we draw lines parallel to Y-axis.
From the point where they intersect the more than ogive, we draw perpendiculars on Y-axis.
The points where they intersect the Y-axis gives the values 178 and 114.
∴ Number of workers having daily wages between ₹ 170 and ₹ 260 = 178 – 114 = 64

(ii) To find the number of workers having daily wages less than ₹ 260, we consider the value 260 on the X-axis.
From this point, we draw a line that is parallel to Y-axis.
From the point where the line intersects the more than ogive, we draw a perpendicular on the Y-axis.
The foot of perpendicular gives the number of workers having daily wages of more than 260.
The foot of perpendicular ~ 114
∴ No. of workers whose daily wages are more than ₹ 260 ~ 114
∴ No. of workers whose daily wages are less than ₹ 260 = 200 – 114 = 86

Question 17.
Draw ogive of both the types for the following frequency distribution and hence find the median.

Solution:

For less than given points to be plotted are (10, 5), (20, 10), (30, 18), (40, 30), (50, 46), (60, 61), (70, 71), (80, 79), (90, 84), (100, 86)
For more than given points to be plotted are (0, 86), (10, 81), (20, 76), (30, 68), (40, 56), (50, 40), (60, 25), (70, 15), (80, 7), (90, 2)

From the point of intersection of two ogives. We draw a perpendicular on X-axis.
The point where it meets the X-axis gives the value of the median.

Question 18.
Find Q1, D6 and P78 for the following data:

Solution:
Since the given data is not in the form of a continuous frequency distribution, we have to convert it into that form by subtracting 0.025 from the lower limit and adding 0.025 to the upper limit of each class interval.
∴ the class intervals will be 7.975 – 8.975, 8.975 – 9.975, etc.
We construct the less than cumulative frequency table as given below:

Here, N = 50
Q₁ class = class containing \(\left(\frac{\mathrm{N}}{4}\right)^{\text {th }}\) observation
∴ \(\frac{N}{4}=\frac{50}{4}\) = 12.5
Cumulative frequency which is just greater than (or equal) to 12.5 is 15.
∴ Q₁ lies in the class 8.975 – 9.975.
∴ L = 8.975, h = 1, f = 10, c.f. = 5
Q₁ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{\mathrm{N}}{4}-\mathrm{c} . \mathrm{f} .\right)\)
= 8.975 + \(\frac{1}{10}\) (12.5 – 5)
= 8.975 + 0.1(7.5)
= 8.975 + 0.75
= 9.725
D₆ class = class containing \(\left(\frac{6 \mathrm{~N}}{10}\right)^{\text {th }}\) observation
∴ \(\frac{6 \mathrm{~N}}{10}=\frac{6 \times 50}{10}\) = 30
Cumulative frequency which is just greater than (or equal) to 30 is 35.
∴ D6 lies in the class 9.975 – 10.975.
∴ L = 9.975, h = 1, f = 20, c.f. = 15
D₆ = L + \(\frac{h}{f}\left(\frac{6 N}{10}-\text { c.f. }\right)\)
= 9.975 + \(\frac{1}{20}\) (30 – 15)
= 9.975 + 0.05(15)
= 9.975 + 0.75
= 10.725
P₇₈ class = class containing \(\left(\frac{78 \mathrm{~N}}{100}\right)^{\text {th }}\) observation
\(\frac{78 \mathrm{~N}}{100}=\frac{78 \times 50}{100}\) = 39
Cumulative frequency which is just greater than (or equal) to 39 is 45.
∴ P₇₈ lies in the class 10.975 – 11.975.
∴ L = 10.975, h = 1, f = 10, c.f. = 35
∴ P₇₈ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{78 \mathrm{~N}}{100}-\text { c.f. }\right)\)
= 10.975 + \(\frac{1}{10}\) (39 – 35)
= 10.975 + 0.1(4)
= 10.975 + 0.4
= 11.375

Question 19.

For the above data, find all quartiles and number of persons weighing between 57 kg and 72 kg.
Solution:
We construct the less than cumulative frequency table as given below:

Here, N = 111
Q₁ class = class containing \(\left(\frac{\mathrm{N}}{4}\right)^{\text {th }}\) observation
∴ \(\frac{N}{4}=\frac{111}{4}\) = 27.75
Cumulative frequency which is just greater than (or equal) to 27.75 is 39.
∴ Q₁ lies in the class 50 – 55.
∴ Q₁ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{\mathrm{N}}{4}-\text { c.f. }\right)\)
= 50 + \(\frac{5}{20}\) (27.75 – 19)
= 50 + \(\frac{1}{4}\) × 8.75
= 50 + 2.1875
= 52.1875
Q₂ class = class containing \(\left(\frac{2 \mathrm{~N}}{4}\right)^{\mathrm{th}}\) observation
∴ \(\frac{2 N}{4}=\frac{2 \times 111}{4}\) = 55.5
Cumulative frequency which is just greater than (or equal) to 55.5 is 69.
∴ Q₂ lies in the class 55 – 60.
∴ L = 55, h = 5, f = 30, c.f. = 39
∴ Q₂ = L + \(\frac{h}{f}\left(\frac{2 N}{4}-\text { c.f. }\right)\)
= 55 + \(\frac{5}{30}\) (55.5 – 39)
= 55 + \(\frac{1}{6}\) × 16.5
= 55 + 2.75
= 57.75
Q₃ class = class containing \(\left(\frac{3 \mathrm{~N}}{4}\right)^{\mathrm{th}}\) observation
∴ \(\frac{3 \mathrm{~N}}{4}=\frac{3 \times 111}{4}\) = 83.25
Cumulative frequency which is just greater than (or equal) to 83.25 is 89.
∴ Q₃ lies in the class 60 – 65.
∴ L = 60, h = 5, f = 20, c.f. = 69
∴ Q₃ = L + \(\frac{h}{f}\left(\frac{3 N}{4}-c . f .\right)\)
= 60 + \(\frac{5}{20}\) (83.25 – 69)
= 60 + \(\frac{1}{4}\) × 14.25
= 60 + 3.5625
= 63.5625
In order to find the number of persons between 57 kg and 72 kg,
We need to find x in P_x, where P_x = 57 kg and y in P_y, where P_y = 72 kg
Then (y – x) would be the % of persons weighing between 57 kg and 72 kg
P_x = 57
∴ L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{x \times \mathrm{N}}{100}-\mathrm{c} . \mathrm{f} .\right)\) = 57
∴ 55 + \(\frac{5}{30}\) (1.11x – 39) = 57
∴ \(\frac{1}{6}\) (1.11x – 39) = 2
∴ 1.11x – 39 = 12
∴ 1.11x = 51
∴ x = 45.95
∴ P_y = 72
∴ L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{y \times \mathrm{N}}{100}-\mathrm{c} . \mathrm{f} .\right)\) = 72
∴ 70 + \(\frac{5}{8}\) (1.11y – 99) = 72
∴ 0.625(1.11y – 99) = 2
∴ 1.11y – 99 = 3.2
∴ 1.11y = 102.2
∴ y = 92.07
∴ % of people weighing between 57 kg and 72 kg = 92.07 – 45.95 = 46.12 %
∴ No. of people weighing between 57 kg and 72 kg = 111 × 46.12% = 51.1932 ~ 51

Question 20.
For the following data showing weights of 100 employees, find the maximum weight of the lightest 25% of employees.

Solution:
We construct the less than cumulative frequency table as given below:

Here, N = 100
Q₁ class = class containing \(\left(\frac{N}{4}\right)^{\text {th }}\) observation
∴ \(\frac{\mathrm{N}}{4}=\frac{100}{4}\) = 25
Cumulative frequency which is just greater than (or equal) to 25 is 29.
∴ Q₁ lies in the class 55 – 60.
∴ L = 55, h = 5, f = 15, c.f. = 14
∴ Q₁ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{\mathrm{N}}{4}-\mathrm{c} . \mathrm{f} .\right)\)
= 55 + \(\frac{5}{15}\) (25 – 14)
= 55 + \(\frac{1}{3}\) × 11
= 55 + 3.67
= 58.67
∴ Maximum weight of the lightest 25% of employees is 58.67 kg.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 1 Partition Values Ex 1.3 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.3

Question 1.
The following table gives the frequency distribution of marks of 100 students in an examination.

Determine D₆, Q₁, and P₈₅ graphically.
Solution:
To draw an ogive curve, we construct the less than cumulative frequency table as given below:

The points to be plotted for less than ogive are (20, 9), (25, 21), (30, 44), (35, 75), (40, 85), (45, 93), (50, 100).

Here, N = 100
For D₆, \(\frac{6 \mathrm{~N}}{10}=\frac{6 \times 100}{10}\) = 60
For Q₁, \(\frac{\mathrm{N}}{4}=\frac{100}{4}\) = 25
For P₈₅, \(\frac{85 \mathrm{~N}}{100}=\frac{85 \times 100}{100}\) = 85
∴ We take the points having Y co-ordinates 60, 25 and 85 on Y-axis.
From these points, we draw lines parallel to X-axis.
From the points where these lines intersect the curve, we draw perpendiculars on X-axis.
X co-ordinates of these points give the values of D₆, Q₁ and P₈₅.
∴ D₆ = 32.5, Q₁ = 26, P₈₅ = 40

Question 2.
The following table gives the distribution of daily wages of 500 families in a certain city.

Draw a ‘less than’ ogive for the above data. Determine the median income and obtain the limits of income of central 50% of the families.
Solution:
To draw an ogive curve, we construct the less than cumulative frequency table as given below:

The points to be plotted for less than ogive are (100, 50), (200, 200), (300, 380), (400, 430), (500, 470), (600, 490) and (700, 500).

Here, N = 500
For Q₁, \(\frac{\mathrm{N}}{4}=\frac{500}{4}\) = 125
For Q₂, \(\frac{\mathrm{N}}{2}=\frac{500}{2}\) = 250
For Q₃, \(\frac{3 \mathrm{~N}}{4}=\frac{3 \times 500}{4}\) = 375
∴ We take the points having Y co-ordinates 125, 250 and 375 on Y-axis.
From these points we draw lines parallel to X-axis.
From the points where these lines intersect the curve, we draw perpendiculars on X-axis.
X-Co-ordinates of these points give the values of Q₁, Q₂ and Q₃.
∴ Q₁ ~ 150, Q₂ ~ 228, Q₃ ~ 297
∴ Median = 228
50% families lie between Q₁ and Q₃
∴ Limits of income of central 50% families are from ₹ 150 to ₹ 297

Question 3.
From the following distribution, determine the median graphically.

Solution:
To draw an ogive curve, we construct the less than and more than cumulative frequency table as given below:

The points to be plotted for less than ogive are (400, 50), (500, 121), (600, 310), (700, 415), (800, 475), (900, 513) and (1000, 520) and that for more than ogive are (300, 520), (400, 470), (500, 399), (600, 210), (700, 105), (800, 45), (900, 7).

From the point of intersection of two ogives, we draw a perpendicular on X-axis.
The point where it meets the X-axis gives the value of the median.
∴ Median ~ 574

Question 4.
The following frequency distribution shows the profit (in ₹) of shops in a particular area of the city.

Find graphically
(i) the Unfits of middle 40% shops.
(ii) the number of shops having a profit of fewer than 35,000 rupees.
Solution:
To draw an ogive curve, we construct a less than cumulative frequency table as given below:

Points to be plotted are (10, 12), (20, 30), (30, 57), (40, 77), (50, 94), (60, 100).

The Middle 40% value lies in between P₃₀ and P₇₀.
N = 100
For P₃₀ = \(\frac{30 \mathrm{~N}}{100}=\frac{30 \times 100}{100}\) = 30
For P₇₀ = \(\frac{70 \mathrm{~N}}{100}=\frac{70 \times 100}{100}\) = 70
∴ We take the points having Y co-ordinates 30 and 70 on Y-axis. From these points we draw lines parallel to X-axis.
From the points where these lines intersect the curve, we draw perpendiculars on X-axis.
X-Co-ordinates of these points give the values of P₃₀ and P₇₀.
∴ P₃₀ ~ 20, P₇₀ ~ 36
Limits of middle 40% shops lie between ₹ 20,000 to ₹ 36,000
To find the number of shops having a profit of less than ₹ 35,000, we take the value 35 on the X-axis.
From this point, we draw a line parallel to Y-axis, and from the point where it intersects the less than ogive we draw a perpendicular on Y-axis. It intersects the Y-axis at approximately 67.
∴ No. of shops having profit less than ₹ 35,000 is 67.

Question 5.
The following is the frequency distribution of overtime (per week) performed by various workers from a certain company. Determine the values of D₂, Q₂, and P₆₁ graphically.

Solution:
To draw an ogive curve, we construct a less than cumulative frequency table as given below:

Points to be plotted are (8, 4), (12, 12), (16, 28), (20, 46), (24, 66) and (28, 80)
Here, N = 80

For D₂, we have to consider \(\frac{2 \mathrm{~N}}{10}=\frac{2 \times 80}{10}\) = 16
For Q₂, we have to consider \(\frac{\mathrm{N}}{2}=\frac{80}{2}\) = 40
and for P₆₁, we have to consider \(\frac{61 \mathrm{~N}}{100}=\frac{61 \times 80}{100}\) = 48.8
∴ We consider the values 16, 40 and 48.8 on the Y-axis.
From these points, we draw the lines which are parallel to the X-axis.
From the points where they intersect the less than ogive, we draw perpendiculars to X-axis.
The values at the foot of perpendiculars represent the values of D₂, Q₂, and P₆₁ respectively.
∴ D₂ ~ 13, Q₂ ~ 19, P₆₁ ~ 20.5

Question 6.
Draw ogive for the following data and hence find the values of D₁, Q₁, and P₄₀.

Solution:
N = 100
To draw the less than ogive we have to plot the points (10, 4), (20, 6), (30, 24), (40, 46), (50, 67), (60, 86), (70, 96), (80, 99), (90, 100).

For D₁, we have to consider \(\frac{\mathrm{N}}{10}=\frac{100}{10}\) = 10
For Q₁, we have to consider \(\frac{\mathrm{N}}{4}=\frac{100}{4}\) = 25
For P₄₀, we have to consider \(\frac{40 \mathrm{~N}}{100}=\frac{40 \times 100}{100}\) = 40
∴ We consider the values 10, 25 and 40 on the Y-axis. From these points we draw lines parallel to X-axis.
From the points where they intersect the less than ogive, we draw perpendiculars on the X-axis.
The values at the foot of perpendicular represent the values of D₁, Q₁ and P₄₀ respectively.
∴ D₁ ~ 22, Q₁ ~ 30.5, P₄₀ ~ 37

Question 7.
The following table shows the age distribution of heads of the families in a certain country. Determine the third, fifth, and eighth decile of the distribution graphically.

Solution:
To draw an ogive curve, we construct a less than cumulative frequency table as given below:

Points to be plotted are (35, 46), (45, 131), (55, 195), (65, 270), (75, 360), (85, 400).

N = 400
For D₃, we have to consider \(\frac{3 \mathrm{~N}}{10}=\frac{3 \times 400}{10}\) = 120
For D₅, we have to consider \(\frac{5 \mathrm{~N}}{10}=\frac{5 \times 400}{10}\) = 200
For D₈, we have to consider \(\frac{8 \mathrm{~N}}{10}=\frac{8 \times 400}{10}\) = 320
∴ We consider the values 120, 200 and 320 on Y-axis. From these points we draw the lines parallel to X-axis.
From the points where they intersect the less than ogive, we draw perpendiculars on the X-axis.
The foot of perpendicular represent the values of D₃, D₅ and D₈.
∴ D₃ ~ 44, D₅ ~ 55.5 and D₈ ~ 70

Question 8.
The following table gives the distribution of females in an Indian village. Determine the median age graphically.

Solution:
To draw an ogive curve, we construct the less than cumulative frequency table as given below:


Points to be plotted are (10, 175), (20, 275), (30, 343), (40, 391), (50, 416), (60, 466), (70, 489), (80, 497), (90, 499), (100, 500).

N = 500
For median we have to consider \(\frac{\mathrm{N}}{2}=\frac{500}{2}\) = 250
∴ We consider the value 250 on Y-axis. From this point, we draw a line parallel to X-axis.
From the point it intersects the less than ogive, we draw a perpendicular to X-axis.
The foot perpendicular represents the value of the median.
∴ Median ~ 17.5

Question 9.
Draw ogive for the following distribution and hence find graphically the limits of the weight of middle 50% fishes.

Solution:
Since the given data is not continuous, we have to convert it into the continuous form by subtracting 5 from the lower limit and adding 5 to the upper limit of every class interval.
To draw an ogive curve, we construct the less than cumulative frequency table as given below:

Points to be plotted are (895, 8), (995, 24), (1095, 44), (1195, 69), (1295, 109), (1395, 115), (1495, 120).

N = 120
For Q₁ and Q₃ we have to consider
\(\frac{\mathrm{N}}{4}=\frac{120}{4}\) = 30
\(\frac{3 \mathrm{~N}}{4}=\frac{3 \times 120}{4}\) = 90
For finding Q₁ and Q₃ we consider the values 30 and 90 on the Y-axis.
From these points, we draw the lines which are parallel to X-axis.
From the points where these lines intersect the less than ogive, we draw perpendicular on X-axis.
The feet of perpendiculars represent the values Q₁ and Q₂.
∴ Q₁ ~ 1025 and Q₃ ~ 1248
∴ the limits of the weight of the middle 50% of fishes lie between 1025 to 1248.

Question 10.
Find graphically the values of D₃ and P₆₅ for the data given below:

Solution:
Since the given data is not continuous, we have to convert it into a continuous form by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of every class interval.
To draw an ogive curve, we construct the less than cumulative frequency table as given below:

Points to be plotted are (69.5, 20), (79.5, 60), (89.5, 110), (99.5, 160), (109.5, 180), (119.5, 190), (129.5, 200).

N = 200
For D₃, \(\frac{3 \mathrm{N}}{10}=\frac{3 \times 200}{10}\) = 60
For P₆₅, \(\frac{65 \mathrm{N}}{100}=\frac{65 \times 200}{100}\) = 130
∴ We take the values 60 and 130 on the Y-axis.
From these points we draw lines parallel to X-axis and from the points where these lines intersect less than ogive, we draw perpendiculars on X-axis.
The foot of perpendiculars represents the median of the values, D₃ and P₆₅.
∴ D₃ = 79.5, P₆₅ = 93.5

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Determinants Ex 6.3 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 6 Determinants Ex 6.3

Question 1.
Solve the following equations using Cramer’s Rule.
(i) x + 2y – z = 5, 2x – y + z = 1, 3x + 3y = 8
Solution:
Given equations are
x + 2y – z = 5
2x – y + z = 1
3x + 3y = 8 i.e. 3x + 3y + 0z = 8
∴ D = \(\left|\begin{array}{ccc}
1 & 2 & -1 \\
2 & -1 & 1 \\
3 & 3 & 0
\end{array}\right|\)
= 1(0 – 3) – 2(0 – 3) – 1(6 + 3)
= -3 + 6 – 9
= -6
D_x = \(\left|\begin{array}{ccc}
5 & 2 & -1 \\
1 & -1 & 1 \\
8 & 3 & 0
\end{array}\right|\)
= 5(0 – 3) – 2(0 – 8) + (-1)(3 + 8)
= -15 + 16 – 11
= -10
D_y = \(\left|\begin{array}{ccc}
1 & 5 & -1 \\
2 & 1 & 1 \\
3 & 8 & 0
\end{array}\right|\)
= 1(0 – 8) – 5(0 – 3) + 1(16 – 3)
= -8 + 15 – 13
= -6
D_z = \(\left|\begin{array}{ccc}
1 & 2 & 5 \\
2 & -1 & 1 \\
3 & 3 & 8
\end{array}\right|\)
= 1(-8 – 3) – 2(16 – 3) + 5(6 + 3)
= -11 – 26 + 45
= 8
By Cramer’s Rule,

x = \(\frac{5}{3}\), y = 1 and z = \(\frac{-4}{3}\) are the solutions of the given equations.

Check:
We can check if our answer is right or wrong.
In order to do so, substitute the values of x, y and z in the given equations.
x = \(\frac{5}{3}\), y = 1 and z = \(\frac{-4}{3}\) satisfy the given equations.
If either one of the equations is not satisfied, then our answer is wrong.
If x = \(\frac{5}{3}\), y = 1 and z = \(\frac{-4}{3}\) are the solutions of the given equations.
L.H.S. = x + 2y – z
= \(\frac{5}{3}+2-\frac{4}{3}\)
= \(\frac{7}{3}\)
≠ R.H.S.
L.H.S. = 2x – y + z
= \(\frac{10}{3}-1+\frac{4}{3}\)
= \(\frac{11}{3}\)
≠ R.H.S.
L.H.S. = 3x + 3y
= 5 + 3
= 8
= R.H.S.

(ii) 2x – y + 6z = 10, 3x + 4y – 5z = 11, 8x – 7y – 9z = 12
Solution:
Given equations are
2x – y + 6z = 10
3x + 4y – 5z = 11
8x – 7y – 9z = 12
∴ D = \(\left|\begin{array}{ccc}
2 & -1 & 6 \\
3 & 4 & -5 \\
8 & -7 & -9
\end{array}\right|\)
= 2(-36 – 35) – (-1)(-27 + 40) + 6(-21 – 32)
= -142 + 13 – 318
= -447
D_x = \(\left|\begin{array}{ccc}
10 & -1 & 6 \\
11 & 4 & -5 \\
12 & -7 & -9
\end{array}\right|\)
= 10(-36 – 35) – (-1)(-99 + 60) + 6(-77 – 48)
= -710 – 39 – 750
= -1499
D_y = \(\left|\begin{array}{ccc}
2 & 10 & 6 \\
3 & 11 & -5 \\
8 & 12 & -9
\end{array}\right|\)
= 2(-99 + 60) – 10(-27 + 40) + 6(36 – 88)
= -78 – 130 – 312
= -520
D_z = \(\left|\begin{array}{ccc}
2 & -1 & 10 \\
3 & 4 & 11 \\
8 & -7 & 12
\end{array}\right|\)
= 2(48 + 77) – (-1)(36 – 88) + 10(-21 – 32)
= 250 – 52 – 530
= -332
By Cramer’s Rule,

∴ x = \(\frac{1499}{447}\), y = \(\frac{520}{447}\) and z = \(\frac{332}{447}\) are the solutions of the given equations.

(iii) 11x – y – z = 31, x – 6y + 2z = -26, x + 2y – 7z = -24
Solution:
Given equations are
11x – y – z = 31
x – 6y + 2z = -26
x + 2y – 7z = -24
D = \(\left|\begin{array}{ccc}
11 & -1 & -1 \\
1 & -6 & 2 \\
1 & 2 & -7
\end{array}\right|\)
= 11(42 – 4) – (-1)(-7 – 2) + (-1)(2 + 6)
= 418 – 9 – 8
= 401
D_x = \(\left|\begin{array}{ccc}
31 & -1 & -1 \\
-26 & -6 & 2 \\
-24 & 2 & -7
\end{array}\right|\)
= 31(42 – 4) – (-1)(182 + 48) + (-1)(-52 – 144)
= 1178 + 230 + 196
= 1604
D_y = \(\left|\begin{array}{ccc}
11 & 31 & -1 \\
1 & -26 & 2 \\
1 & -24 & -7
\end{array}\right|\)
= 11(182 + 48) – 31(-7 – 2) + (-1)(-24 + 26)
= 2530 + 279 – 2
= 2807
D_z = \(\left|\begin{array}{ccc}
11 & -1 & 31 \\
1 & -6 & -26 \\
1 & 2 & -24
\end{array}\right|\)
= 11(144 + 52) – (-1)(-24 + 26) + 31(2 + 6)
= 2156 + 2 + 248
= 2406
By Cramer’s Rule,

∴ x = 4, y = 7 and z = 6 are the solutions of the given equations.

(iv) \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=-2\), \(\frac{1}{x}-\frac{2}{y}+\frac{1}{z}=3\), \(\frac{2}{x}-\frac{1}{y}+\frac{3}{z}=-1\)
Solution:
Let \(\frac{1}{x}\) = p, \(\frac{1}{y}\) = q, \(\frac{1}{z}\) = r
The given equations become
p + q + r = -2
p – 2q + r = 3
2p – q + 3r = -1
D = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & -2 & 1 \\
2 & -1 & 3
\end{array}\right|\)
= 1(-6 + 1) – 1(3 – 2) + 1(-1 + 4)
= -5 – 1 + 3
= -3
D_p = \(\left|\begin{array}{rrr}
-2 & 1 & 1 \\
3 & -2 & 1 \\
-1 & -1 & 3
\end{array}\right|\)
= -2(-6 + 1) – 1(9 + 1) + 1(-3 – 2)
= 10 – 10 – 5
= -5
D_q = \(\left|\begin{array}{ccc}
1 & -2 & 1 \\
1 & 3 & 1 \\
2 & -1 & 3
\end{array}\right|\)
= 1(9 + 1) + 2(3 – 2) + 1(-1 – 6)
= 10 + 2 – 7
= 5
D_r = \(\left|\begin{array}{rrr}
1 & 1 & -2 \\
1 & -2 & 3 \\
2 & -1 & -1
\end{array}\right|\)
= 1(2 + 3) – 1(-1 – 6) – 2(-1 + 4)
= 5 + 7 – 6
= 6
By Cramer’s Rule,

∴ x = \(\frac{3}{5}\), y = \(\frac{-3}{5}\), z = \(\frac{-1}{2}\) are the solutions of the given equations.

(v) \(\frac{2}{x}-\frac{1}{y}+\frac{3}{z}=4, \frac{1}{x}-\frac{1}{y}+\frac{1}{z}=2, \frac{3}{x}+\frac{1}{y}-\frac{1}{z}=2\)
Solution:
Let \(\frac{1}{x}\) = p, \(\frac{1}{y}\) = q, \(\frac{1}{z}\) = r
The given equations become
2p – q – 3r = 4
p – q + r = 2
3p + q – r = 2
D = \(\left|\begin{array}{ccc}
2 & -1 & 3 \\
1 & -1 & 1 \\
3 & 1 & -1
\end{array}\right|\)
= 2(1 – 1) – (-1)(-1 – 3) + 3(1 + 3)
= 0 – 4 + 12
= 8
D_p = \(\left|\begin{array}{ccc}
4 & -1 & 3 \\
2 & -1 & 1 \\
2 & 1 & -1
\end{array}\right|\)
= 4(1 – 1) – (-1)(-2 – 2) + 3(2 + 2)
= 0 – 4 + 12
= 8
D_q = \(\left|\begin{array}{ccc}
2 & 4 & 3 \\
1 & 2 & 1 \\
3 & 2 & -1
\end{array}\right|\)
= 2(-2 – 2) – 4(-1 – 3) + 3(2 – 6)
= -8 + 16 – 12
= -4
D_r = \(\left|\begin{array}{ccc}
2 & -1 & 4 \\
1 & -1 & 2 \\
3 & 1 & 2
\end{array}\right|\)
= 2(-2 – 2) – (-1)(2 – 6) + 4(1 + 3)
= -8 – 4 + 16
= 4
By Cramer’s Rule,

∴ x = 1, y = -2 and z = 2 are the solutions of the given equations.

Question 2.
An amount of ₹ 5,000 is invested in three plans at rates 6%, 7% and 8% per annum respectively. The total annual income from these investments is ₹ 350. If the total annual income from first two investments is ₹ 70 more than the income from the third, find the amount invested in each plan by using Cramer’s Rule.
Solution:
Let the amount of each investment be ₹ x, ₹ y and ₹ z.
According to the given conditions,
x + y + z = 5000
6%x + 7%y + 8%z = 350
∴ \(\frac{6}{100} x+\frac{7}{100} y-\frac{8}{100} z=350\)
∴ 6x + 7y + 8z = 35000
6%x + 7%y = 8%z + 70
∴ \(\frac{6}{100} x+\frac{7}{100} y=\frac{8}{100} z+70\)
∴ 6x + 7y = 8z + 7000
∴ 6x + 7y – 8z = 7000



∴ Amounts of investments are ₹ 1750, ₹ 1500, and ₹ 1750.

Check:
First condition:
1750 + 1500 + 1750 = 5000
Second condition:
6% of 1750 + 7% of 1500 + 8% of 1750
= 105 + 105 + 140
= 350
Third condition:
Combined income = 105 + 105
= 210
= 140 + 70
Thus, all the conditions are satisfied.

Question 3.
Show that the following equations are consistent.
2x + 3y + 4 = 0, x + 2y + 3 = 0, 3x + 4y + 5 = 0
Solution:
Given equations are
2x + 3y + 4 = 0
x + 2y + 3 = 0
3x + 4y + 5 = 0
∴ \(\left|\begin{array}{lll}
2 & 3 & 4 \\
1 & 2 & 3 \\
3 & 4 & 5
\end{array}\right|\)
= 2(10 – 12) – 3(5 – 9) + 4(4 – 6)
= 2(-2) – 3(-4) + 4(-2)
= -4 + 12 – 8
= 0
∴ The given equations are consistent.

Question 4.
Find k, if the following equations are consistent.
(i) x + 3y + 2 = 0, 2x + 4y – k = 0, x – 2y – 3k = 0
Solution:
Given equations are
x + 3y + 2 = 0
2x + 4y – k = 0
x – 2y – 3k = 0
Since, these equations are consistent.
∴ \(\left|\begin{array}{ccc}
1 & 3 & 2 \\
2 & 4 & -k \\
1 & -2 & -3 k
\end{array}\right|=0\)
∴ 1(-12k – 2k) – 3(-6k + k) + 2(-4 – 4) = 0
∴ -14k + 15k – 16 = 0
∴ k – 16 = 0
∴ k = 16
Check:
If the value of k satisfies the condition for the given equations to be consistent, then our answer is correct.
Substitute k = 16 in the given equation.
\(\left|\begin{array}{ccc}
1 & 3 & 2 \\
2 & 4 & -16 \\
1 & -2 & -48
\end{array}\right|\)
= 1(-192 – 32) – 3(-96 + 16) + 2(-4 – 4)
= 0
Thus, our answer is correct.

(ii) (k – 2)x + (k – 1)y = 17, (k – 1)x + (k – 2)y = 18, x + y = 5
Solution:
Given equations are
(k – 2)x + (k – 1)y = 17
(k – 1)x + (k – 2)y = 18
x + y = 5
Since, these equations are consistent.
∴ \(\left|\begin{array}{ccc}
k-2 & k-1 & -17 \\
k-1 & k-2 & -18 \\
1 & 1 & -5
\end{array}\right|=0\)
Applying R₁ → R₁ – R₂, we get
\(\left|\begin{array}{ccc}
-1 & 1 & 1 \\
k-1 & k-2 & -18 \\
1 & 1 & -5
\end{array}\right|=0\)
∴ -1(-5k + 10 + 18) – 1(-5k + 5 + 18) + 1(k – 1 – k + 2) = 0
∴ -1(-5k – 28) – 1(- 5k + 23) + 1(1) = 0
∴ 5k – 28 + 5k – 23 – 1 = 0
∴ 10k – 50 = 0
∴ k = 5

Question 5.
Find the area of the triangle whose vertices are:
(i) (4, 5), (0, 7), (-1, 1)
Solution:
Here, A(x₁, y₁) ≡ A(4, 5), B(x₂, y₂) ≡ B(0, 7), C(x₃, y₃) ≡ C(-1, 1)
Area of a triangle = \(\frac{1}{2}\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
∴ A(ΔABC) = \(\frac{1}{2}\left|\begin{array}{ccc}
4 & 5 & 1 \\
0 & 7 & 1 \\
-1 & 1 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [4(7 – 1) – 5(0 + 1) + 1(0 + 7)]
= \(\frac{1}{2}\) (24 – 5 + 7)
= 13 sq.units.

(ii) (3, 2), (-1, 5), (-2, -3)
Solution:
Here, A(x₁, y₁) ≡ A(3, 2), B(x₂, y₂) = B(-1, 5), C(x₃, y₃) ≡ C(-2, -3)
Area of a triangle = \(\frac{1}{2}\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
∴ A(ΔABC) = \(\frac{1}{2}\left|\begin{array}{ccc}
3 & 2 & 1 \\
-1 & 5 & 1 \\
-2 & -3 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [3(5 + 3) – 2(-1 + 2) + 1(3 + 10)]
= \(\frac{1}{2}\) (24 – 2 + 13)
= \(\frac{35}{2}\) sq. units

(iii) (0, 5), (0, -5), (5, 0)
Solution:
Here, A(x₁, y₁) ≡ A(0, 5), B(x₂, y₂) ≡ B(0, -5), C(x₃, y₃) ≡ C(5,0)
Area of a triangle = \(\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
∴ A(ΔABC) = \(\frac{1}{2}\left|\begin{array}{ccc}
0 & 5 & 1 \\
0 & -5 & 1 \\
5 & 0 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [0(-5 – 0) – 5(0 – 5) + 1(0 + 25)]
= \(\frac{1}{2}\) (0 + 25 + 25)
= \(\frac{50}{2}\)
= 25 sq.units

Question 6.
Find the value of k, if the area of the triangle with vertices at A(k, 3), B(-5, 7), C(-1, 4) is 4 square units.
Solution:
Here, A(x₁, y₁) ≡ A(k, 3), B(x₂, y₂) ≡ B(-5, 7), C(x₃, y₃) ≡ C(-1, 4)
A(ΔABC) = 4 sq.units
Area of a triangle = \(\frac{1}{2}\left|\begin{array}{lll}
x_{b} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
∴ \(\frac{1}{2}\left|\begin{array}{ccc}
k & 3 & 1 \\
-5 & 7 & 1 \\
-1 & 4 & 1
\end{array}\right|\) = ±4
∴ k(7 – 4) – 3(-5 + 1) + 1(-20 + 7) = ±8
∴ 3k + 12 – 13 = ±8
∴ 3k – 1 = ±8
∴ 3k – 1 = 8 or 3k – 1 = -8
∴ 3k = 9 or 3k = -7
∴ k = 3 or k = \(\frac{-7}{3}\)

Check:
For k = 3,

Thus, our answer is correct.

Question 7.
Find the area of the quadrilateral whose vertices are A(-3, 1), B(-2, -2), C(4, 1), D(2, 3).
Solution:
A(-3, 1), B(-2, -2), C(4, 1), D(2, 3)

A(ABCD) = A(ΔABC) + A(ΔACD)
= \(\frac{21}{2}\) + 7
= \(\frac{35}{2}\) sq.units.

Question 8.
By using determinant, show that the following points are collinear.
P(5, 0), Q(10, -3), R(-5, 6)
Solution:
Here, P(x₁, y₁) ≡ P(5, 0), Q(x₂, y₂) ≡ Q(10, -3), R(x₃, y₃) ≡ R(-5, 6)
If A(ΔPQR) = 0, then the points P, Q, R are collinear.
∴ A(ΔPQR) = \(\frac{1}{2}\left|\begin{array}{ccc}
5 & 0 & 1 \\
10 & -3 & 1 \\
-5 & 6 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [5(-3 – 6) – 0(10 + 5) + 1(60 – 15)]
= \(\frac{1}{2}\) (-45 + 0 + 45)
= 0
∴ A(ΔPQR) = 0
∴ Points P, Q and R are collinear.

Question 9.
The sum of three numbers is 15. If the second number is subtracted from the sum of first and third numbers, then we get 5. When the third number is subtracted from the sum of twice the first number and the second number, we get 4. Find the three numbers.
Solution:
Let the three numbers be x, y and z.
According to the given conditions,
x + y + z = 15
x + z – y = 5 i.e. x – y + z = 5
2x + y – z = 4
D = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & -1 & 1 \\
2 & 1 & -1
\end{array}\right|\)
= 1(1 – 1) – 1 (-1 – 2) + 1(1 + 2)
= 1(0) – 1(-3) + 1(3)
= 0 + 3 + 3
= 6 ≠ 0
D_x = \(\left|\begin{array}{ccc}
15 & 1 & 1 \\
5 & -1 & 1 \\
4 & 1 & -1
\end{array}\right|\)
= 15(1 – 1) – 1(-5 – 4) + 1(5 + 4)
= 15(0) – 1(-9) + 1(9)
= 0 + 9 + 9
= 18
D_y = \(\left|\begin{array}{ccc}
1 & 15 & 1 \\
1 & 5 & 1 \\
2 & 4 & -1
\end{array}\right|\)
= 1(-5 – 4) – 15(-1 – 2) + 1(4 – 10)
= 1(-9) – 15(-3) + 1(-6)
= -9 + 45 – 6
= 30
D_z = \(\left|\begin{array}{ccc}
1 & 1 & 15 \\
1 & -1 & 5 \\
2 & 1 & 4
\end{array}\right|\)
= 1(-4 – 5) – 1(4 – 10) + 15(1 + 2)
= 1(-9) – 1(-6) + 15(3)
= -9 + 6 + 45
= 42
By Cramer’s Rule,
x = \(\frac{D_{x}}{D}=\frac{18}{6}\) = 3
y = \(\frac{D_{y}}{D}=\frac{30}{6}\) = 5
z = \(\frac{D_{z}}{D}=\frac{42}{6}\) = 7
∴ The three numbers are 3, 5 and 7.