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Partition Values Class 11 Commerce Maths 2 Chapter 1 Exercise 1.3 Answers Maharashtra Board

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

Std 11 Maths 2 Exercise 1.3 Solutions Commerce Maths

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

11th Commerce Maths Digest Pdf

• 11th Commerce Maths Exercise 1.1 Solutions
• 11th Commerce Maths Exercise 1.2 Solutions
• 11th Commerce Maths Exercise  1.3 Solutions
• 11th Commerce Maths Miscellaneous Exercise 1 Solutions

Correlation Class 11 Commerce Maths 2 Chapter 5 Exercise 5.1 Answers Maharashtra Board

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

Std 11 Maths 2 Exercise 5.1 Solutions Commerce Maths

Question 1.
Draw a scatter diagram for the data given below and interpret it.

Solution:

Since all the points lie in a band rising from left to right.
Therefore, there is a positive correlation between the values of X and Y respectively.

Question 2.
For the following data of marks of 7 students in Physics (x) and Mathematics (y), draw scatter diagram and state the type of correlation.

Solution:
We take marks in Physics on X-axis and marks in Mathematics on Y-axis and plot the points as below.

We get a band of points rising from left to right. This indicates the positive correlation between marks in Physics and marks in Mathematics.

Question 3.
Draw a scatter diagram for the data given below. Is there any correlation between Aptitude score and Grade points?

Solution:

The points are completely scattered i.e., no trend is observed.
∴ there is no correlation between Aptitude score (X) and Grade point (Y).

Question 4.
Find correlation coefficient between x andy series for the following data:
n = 15, \(\bar{x}\) = 25, \(\bar{y}\) = 18, σ_x = 3.01, σ_y = 3.03, \(\sum\left(x_{\mathrm{i}}-\bar{x}\right)\left(y_{\mathrm{i}}-\bar{y}\right)\) = 122
Solution:
Here, n = 15, \(\bar{x}\) = 25, \(\bar{y}\) = 18, σ_x = 3.01, σ_y = 3.03, \(\sum\left(x_{\mathrm{i}}-\bar{x}\right)\left(y_{\mathrm{i}}-\bar{y}\right)\) = 122

Question 5.
The correlation coefficient between two variables x and y are 0.48. The covariance is 36 and the variance of x is 16. Find the standard deviation of y.
Solution:
Given, r = 0.48, Cov(X, Y) = 36
Since \(\sigma_{X}^{2}\) = 16
∴ σ_x = 4

∴ the standard deviation of y is 18.75.

Question 6.
In the following data, one of the values of y is missing. Arithmetic means of x and y series are 6 and 8 respectively. (√2 = 1.4142)

(i) Estimate missing observation.
(ii) Calculate correlation coefficient.
Solution:
(i) Let X = x_i, Y = y_i and missing observation be ‘a’.
Given, \(\bar{x}\) = 6, \(\bar{y}\) = 8, n = 5
∴ 8 = \(\frac{35+a}{5}\)
∴ 40 = 35 + a
∴ a = 5
(ii) We construct the following table:

Question 7.
Find correlation coefficient from the following data. [Given: √3 = 1.732]

Solution:

Question 8.
The correlation coefficient between x and y is 0.3 and their covariance is 12. The variance of x is 9, find the standard deviation of y.
Solution:
Given, r = 0.3, Cov(X, Y) = 12,
\(\sigma_{X}^{2}\) = 9
∴ \(\sigma_{\mathrm{X}}\) = 3

∴ the standard deviation of y is 13.33.

Read More

• 11th Commerce Maths Exercise 5.1 Solutions
• 11th Commerce Maths Miscellaneous Exercise 5 Solutions

Locus and Straight Line Class 11 Commerce Maths 1 Chapter 5 Exercise 5.4 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 5 Locus and Straight Line Ex 5.4 Questions and Answers.

Std 11 Maths 1 Exercise 5.4 Solutions Commerce Maths

Question 1.
Find the slope, x-intercept, y-intercept of each of the following lines.
(a) 2x + 3y – 6 = 0
(b) x + 2y = 0
Solution:
(a) Given equation of the line is 2x + 3y – 6 = 0
Comparing this equation with ax + by + c = 0, we get
a = 2, b = 3, c = -6
∴ Slope of the line = \(\frac{-a}{b}=\frac{-2}{3}\)
x-intercept = \(\frac{-\mathrm{c}}{\mathrm{a}}=\frac{-(-6)}{2}\) = 3
y-intercept = \(\frac{-\mathrm{c}}{\mathrm{b}}=\frac{-(-6)}{3}\) = 2

(b) Given equation of the line is x + 2y = 0
Comparing this equation with ax + by + c = 0, we get
a = 1, b = 2, c = 0
∴ Slope of the line = \(\frac{-a}{b}=\frac{-1}{2}\)
x-intercept = \(\frac{-c}{a}=\frac{-0}{1}\) = 0
y-intercept = \(\frac{-\mathrm{c}}{\mathrm{b}}=\frac{-0}{2}\) = 0

Question 2.
Write each of the following equations in ax + by + c = 0 form.
(a) y = 2x – 4
(b) y = 4
(c) \(\frac{x}{2}+\frac{y}{4}=1\)
(d) \(\frac{x}{3}=\frac{y}{2}\)
Solution:
(a) y = 2x – 4
∴ 2x – y – 4 = 0 is the equation in ax + by + c = 0 form.

(b) y = 4
∴ 0x + 1y – 4 = 0 is the equation in ax + by + c = 0 form.

(c) \(\frac{x}{2}+\frac{y}{4}=1\)
∴ \(\frac{2 x+y}{4}=1\)
∴ 2x + y = 4
∴ 2x + y – 4 = 0 is the equation in ax + by + c = 0 form.

(d) \(\frac{x}{3}=\frac{y}{2}\)
∴ 2x = 3y
∴ 2x – 3y + 0 = 0 is the equation in ax + by + c = 0 form.

Question 3.
Show that the lines x – 2y – 7 = 0 and 2x – 4y + 5 = 0 are parallel to each other.
Solution:
Let m₁ be the slope of the line x – 2y – 7 = 0.
∴ m₁ = \(\frac{-1}{-2}=\frac{1}{2}\)
Let m₂ be the slope of the line 2x – 4y + 5 = 0.
∴ m₂ = \(\frac{-2}{-4}=\frac{1}{2}\)
Since, m₁ = m₂
∴ The given lines are parallel to each other.

Question 4.
If the line 3x + 4y = p makes a triangle of area 24 square units with the co-ordinate axes, then find the value of p.
Solution:
Let the line 3x + 4y = p cuts the X and Y-axes at points A and B respectively.

3x + 4y = p
∴ \(\frac{3 x}{\mathrm{p}}+\frac{4 y}{\mathrm{p}}=1\)
∴ \(\frac{x}{\frac{p}{3}}+\frac{y}{\frac{p}{4}}=1\)
The equation is of the form \(\frac{x}{a}+\frac{y}{b}=1\), with a = \(\frac{p}{3}\) and b = \(\frac{p}{4}\)
∴ A = (a, 0) = (\(\frac{p}{3}\), 0) and B = (0, b) = (0, \(\frac{p}{4}\))
∴ OA = \(\frac{p}{3}\) and OB = \(\frac{p}{4}\)
Given, A(∆OAB) = 24 sq. units
∴ \(\frac{1}{2}\) × OA × OB = 24
∴ \(\frac{1}{2}\) × \(\frac{p}{3}\) × \(\frac{p}{4}\) = 24
∴ p² = 576
∴ p = ±24

Question 5.
Find the co-ordinates of the circumcentre of the triangle whose vertices are A(-2, 3), B(6, -1), C(4, 3).
Solution:
Here, A(-2, 3), B(6, -1), C(4, 3) are the vertices of ∆ABC.
Let F be the circumcentre of ∆ABC.
Let FD and FE be the perpendicular bisectors of the sides BC and AC respectively.

∴ D and E are the midpoints of side BC and AC respectively.
∴ D = \(\left(\frac{6+4}{2}, \frac{-1+3}{2}\right)\) = (5, 1)
and E = \(\left(\frac{-2+4}{2}, \frac{3+3}{2}\right)\) = (1, 3)
Now, slope of BC = \(\frac{-1-3}{6-4}\) = -2
∴ slope of FD = \(\frac{1}{2}\) …..[∵ FD ⊥ BC]
Since, FD passes through (5, 1) and has slope \(\frac{1}{2}\)
∴ Equation of FD is y – 1 = \(\frac{1}{2}\)(x – 5)
∴ 2(y – 1) = x – 5
∴ x – 2y – 3 = 0 ……(i)
Since, both the points A and C have same y co-ordinates i.e. 3
∴ the points A and C lie on the liney = 3.
Since, FE passes through E(1, 3).
∴ the equation of FE is x = 1. …….(ii)
To find co-ordinates of circumcentre, we have to solve equations (i) and (ii).
Substituting the value of x in (i), we get
1 – 2y – 3 = 0
∴ y = -1
∴ Co-ordinates of circumcentre F = (1, -1).

Question 6.
Find the equation of the line whose x-intercept is 3 and which is perpendicular to the line 3x – y + 23 = 0.
Solution:
Slope of the line 3x – y + 23 = 0 is 3.
∴ slope of the required line which is perpendicular to 3x – y + 23 = 0 is \(\frac{-1}{3}\).
Since, the x-intercept of the required line is 3.
∴ it passes through (3, 0).
∴ the equation of the required line is
y – 0 = \(\frac{-1}{3}\) (x – 3)
∴ 3y = -x – 3
∴ x – 3y = 3

Question 7.
Find the distance of the point A(-2, 3) from the line 12x – 5y – 13 = 0.
Solution:
Let p be the perpendicular distance of the point A(-2, 3) from the line 12x – 5y – 13 = 0
Here, a = 12, b = -5, c = -13, x₁ = -2, y₁ = 3

Question 8.
Find the distance between parallel lines 9x + 6y – 1 = 0 and 9x + 6y – 32 = 0.
Solution:
Equations of the given parallel lines are 9x + 6y – 7 = 0 and 9x + 6y – 32 = 0.
Here, a = 9, b = 6, C₁ = -7 and C₂ = -32
∴ Distance between the parallel lines

Question 9.
Find the equation of the line passing through the point of intersection of lines x + y – 2 = 0 and 2x – 3y + 4 = 0 and making intercept 3 on the X-axis.
Solution:
Given equations of lines are
x + y – 2 = 0 ……(i)
and 2x – 3y – 4 = 0 ……(ii)
Multiplying equation (i) by 3, we get
3x – 3y – 6 = 0 …..(iii)
Adding equation (ii) and (iii), we get
5x – 2 = 0
∴ x = \(\frac{2}{5}\)
Substituting x = \(\frac{2}{5}\) in equation (i), we get
\(\frac{2}{5}\) + y – 2 = 0
∴ y = 2 – \(\frac{2}{5}\) = \(\frac{8}{5}\)
∴ The required line passes through point (\(\frac{2}{5}\), \(\frac{8}{5}\)).
Also, the line makes intercept of 3 on X-axis
∴ it also passes through point (3, 0).
∴ required equation of line passing through points (\(\frac{2}{5}\), \(\frac{8}{5}\)) and (3, 0) is

∴ 13(5y – 8) = -8(5x – 2)
∴ 65y – 104 = -40x + 16
∴ 40x + 65y – 120 = 0
∴ 8x + 13y – 24 = 0 which is the equation of the required line.

Question 10.
D(-1, 8), E(4, -2), F(-5, -3) are midpoints of sides BC, CA and AB of ∆ABC. Find
(i) equations of sides of ∆ABC.
(ii) co-ordinates of the circumcentre of ∆ABC.
Solution:
(i) Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of ∆ABC.
Given, points D, E and F are midpoints of sides BC, CA and AB respectively of ∆ABC.

For x-coordinates:
Adding (i), (iii) and (v), we get
2x₁ + 2x₂ + 2x₃ = -4
∴ x₁ + x₂ + x₃ = -2 …..(vii)
Solving (i) and (vii), we get x₁ = 0
Solving (iii) and (vii), we get x₂ = -10
Solving (v) and (vii), we get x₃ = 8
For y-coordinates:
Adding (ii), (iv) and (vi), we get
2y₁ + 2y₂ + 2y₃ = 6
∴ y₁ + y₂ + y₃ = 3 …..(viii)
Solving (ii) and (viii), we get y₁ = -13
Solving (iv) and (viii), we get y₂ = 7
Solving (vi) and (viii), we get y₃ = 9
∴ Vertices of ∆ABC are A(0, -13), B(-10, 7), C(8, 9)

(ii) Here, A(0, -13), B(-10, 7), C(8, 9) are the vertices of ∆ABC.
Let F be the circumcentre of ∆ABC.
Let FD and FE be perpendicular bisectors of the sides BC and AC respectively.

∴ D and E are the midpoints of side BC and AC.
∴ D = \(\left(\frac{-10+8}{2} \cdot \frac{7+9}{2}\right)\) = (-1, 8)
and E = \(\left(\frac{0+8}{2}, \frac{-13+9}{2}\right)\) = (4, -2)
Now, slope of BC = \(\frac{7-9}{-10-8}=\frac{1}{9}\)
∴ slope of FD = -9 ……[∵ FD ⊥ BC]
Since, FD passes through (-1, 8) and has slope -9
∴ Equation of FD is y – 8 = -9(x + 1)
∴ y – 8 = -9x – 9
∴ y = -9x – 1 …..(i)
Also, slope of AC = \(\frac{-13-9}{0-8}=\frac{11}{4}\)
∴ Slope of FE = \(\frac{-4}{11}\) ….[∵ FE ⊥ AC]
Since, FE passes through (4, -2) and has slope \(\frac{-4}{11}\)
∴ Equation of FE is y + 2 = \(\frac{-4}{11}\)(x – 4)
∴ 11(y + 2) = -4(x – 4)
∴ 11y + 22 = -4x + 16
∴ 4x + 11y = -6 ….(ii)
To find co-ordinates of circumcentre, we have to solve equations (i) and (ii).
Substituting the value of y in (ii), we get
∴ 4x + 11(-9x – 1) = -6
∴ 4x – 99x – 11 = -6
∴ -95x = 5
∴ x = \(\frac{-1}{19}\)
Substituting the value of x in (i), we get
y = -9(\(\frac{-1}{19}\)) – 1 = \(\frac{-10}{19}\)
∴ Co-ordinates of circumcentre F = \(\left(\frac{-1}{19}, \frac{-10}{19}\right)\)

11th Commerce Maths Digest Pdf

• 11th Commerce Maths Exercise 5.1 Solutions
• 11th Commerce Maths Exercise 5.2 Solutions
• 11th Commerce Maths Exercise 5.3 Solutions
• 11th Commerce Maths Exercise 5.4 Solutions
• 11th Commerce Maths Miscellaneous Exercise 5 Solutions

Limits Class 11 Commerce Maths 1 Chapter 7 Exercise 7.2 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 7 Limits Ex 7.2 Questions and Answers.

Std 11 Maths 1 Exercise 7.2 Solutions Commerce Maths

I. Evaluate the following limits:

Question 1.
\(\lim _{z \rightarrow 2}\left[\frac{z^{2}-5 z+6}{z^{2}-4}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow-3}\left[\frac{x+3}{x^{2}+4 x+3}\right]\)
Solution:

Question 3.
\(\lim _{y \rightarrow 0}\left[\frac{5 y^{3}+8 y^{2}}{3 y^{4}-16 y^{2}}\right]\)
Solution:

Question 4.
\(\lim _{x \rightarrow-2}\left[\frac{-2 x-4}{x^{3}+2 x^{2}}\right]\)
Solution:

II. Evaluate the following limits:

Question 1.
\(\lim _{u \rightarrow 1}\left[\frac{u^{4}-1}{u^{3}-1}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow 3}\left[\frac{1}{x-3}-\frac{9 x}{x^{3}-27}\right]\)
Solution:

Question 3.
\(\lim _{x \rightarrow 2}\left[\frac{x^{3}-4 x^{2}+4 x}{x^{2}-1}\right]\)
Solution:

III. Evaluate the following limits:

Question 1.
\(\lim _{x \rightarrow-2}\left[\frac{x^{7}+x^{5}+160}{x^{3}+8}\right]\)
Solution:

Question 2.
\(\lim _{y \rightarrow \frac{1}{2}}\left[\frac{1-8 y^{3}}{y-4 y^{3}}\right]\)
Solution:

Question 3.
\(\lim _{v \rightarrow \sqrt{2}}\left[\frac{v^{2}+v \sqrt{2}-4}{v^{2}-3 v \sqrt{2}+4}\right]\)
Solution:

Question 4.
\(\lim _{x \rightarrow 3}\left[\frac{x^{2}+2 x-15}{x^{2}-5 x+6}\right]\)
Solution:

11th Commerce Maths Digest Pdf

• 11th Commerce Maths Exercise 7.1 Solutions
• 11th Commerce Maths Exercise 7.2 Solutions
• 11th Commerce Maths Exercise 7.3 Solutions
• 11th Commerce Maths Exercise 7.4 Solutions
• 11th Commerce Maths Miscellaneous Exercise 7 Solutions

Complex Numbers Class 11 Commerce Maths 1 Chapter 3 Exercise 3.2 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 3 Complex Numbers Ex 3.2 Questions and Answers.

Std 11 Maths 1 Exercise 3.2 Solutions Commerce Maths

Question 1.
Find the square root of the following complex numbers:
(i) -8 – 6i
Solution:
Let \(\sqrt{-8-6 i}\) = a + bi, where a, b ∈ R
Squaring on both sides, we get
-8 – 6i = (a + bi)²
-8 – 6i = a² + b²i² + 2abi
-8 – 6i = (a² – b²) + 2abi …..[∵ i² = -1]
Equating real and imaginary parts, we get

(ii) 7 + 24i
Solution:
Let \(\sqrt{7+24 i}\) = a + bi, where a, b ∈ R
Squaring on both sides, we get
7 + 24i = (a + bi)²
7 + 24i = a² + b²i² + 2abi
7 + 24i = (a² – b²) + 2abi …..[∵ i² = -1]
Equating real and imaginary parts, we get

(iii) 1 + 4√3i
Solution:
Let \(\sqrt{1+4 \sqrt{3} i}\) = a + bi, where a, b ∈ R
Squaring on both sides, we get
1 + 4√3i = (a + bi)²
1 + 4√3i = a² + b²i² + 2abi
1 +4√3i = (a² – b²) + 2abi ……[∵ i² = -1]
Equating real and imaginary parts, we get

(iv) 3 + 2√10i
Solution:
Let \(\sqrt{3+2 \sqrt{10}} i\) = a + bi, where a, b ∈ R
Squaring on both sides, we get
3 + 2√10i = (a + bi)²
3 + 2√10i = a² + b²i² + 2abi
3 + 2√10i = (a² – b²) + 2abi …..[∵ i² = -1]
Equating real and imaginary parts, we get
a² – b² = 3 and 2ab = 2√10
a² – b² = 3 and b = \(\frac{\sqrt{10}}{a}\)

(v) 2(1 – √3i)
Solution:
Let \(\sqrt{2(1-\sqrt{3} i)}\) = a + bi, where a, b ∈ R
Squaring on both sides, we get
2(1 – √3i) = (a + bi)²
2(1 – √3i) = a² + b²i² + 2abi
2 – 2√3i = (a² – b²) + 2abi …..[∵ i² = -1]
Equating real and imaginary parts, we get

Question 2.
Solve the following quadratic equations.
(i) 8x² + 2x + 1 = 0
Solution:
Given equation is 8x² + 2x + 1 = 0
Comparing with ax² + bx + c = 0, we get
a = 8, b = 2, c = 1
Discriminant = b² – 4ac
= (2)² – 4 × 8 × 1
= 4 – 32
= -28 < 0
So, the given equation has complex roots.
These roots are given by

∴ the roots of the given equation are \(\frac{-1+\sqrt{7} \mathrm{i}}{8}\) and \(\frac{-1-\sqrt{7} \mathrm{i}}{8}\)

(ii) 2x² – √3x + 1 = 0
Solution:
Given equation is 2x² – √3x + 1 = 0
Comparing with ax² + bx + c = 0, we get
a = 2, b = -√3, c = 1
Discriminant = b² – 4ac
= (-√3)² – 4 × 2 × 1
= 3 – 8
= -5 < 0
So, the given equation has complex roots.
These roots are given by

∴ the roots of the given equation are \(\frac{\sqrt{3}+\sqrt{5} i}{4}\) and \(\frac{\sqrt{3}-\sqrt{5} i}{4}\)

(iii) 3x² – 7x + 5 = 0
Solution:
Given equation is 3x² – 7x + 5 = 0
Comparing with ax² + bx + c = 0, we get
a = 3, b = -7, c = 5
Discriminant = b² – 4ac
= (-7)² – 4 × 3 × 5
= 49 – 60
= -11 < 0
So, the given equation has complex roots.
These roots are given by

∴ the roots of the given equation are \(\frac{7+\sqrt{11} i}{6}\) and \(\frac{7-\sqrt{11} i}{6}\)

(iv) x² – 4x + 13 = 0
Solution:
Given equation is x² – 4x + 13 = 0
Comparing with ax² + bx + c = 0, we get
a = 1, b = -4, c = 13
Discriminant = b² – 4ac
= (-4)² – 4 × 1 × 13
= 16 – 52
= -36 < 0
So, the given equation has complex roots.
These roots are given by

∴ the roots of the given equation are 2 + 3i and 2 – 3i.

Question 3.
Solve the following quadratic equations.
(i) x² + 3ix + 10 = 0
Solution:
Given equation is x² + 3ix + 10 = 0
Comparing with ax² + bx + c = 0, we get
a = 1, b = 3i, c = 10
Discriminant = b² – 4ac
= (3i)² – 4 × 1 × 10
= 9i² – 40
= -9 – 40 …..[∵ i² = -1]
= -49
So, the given equation has complex roots.
These roots are given by

∴ x = 2i or x = -5i
∴ the roots of the given equation are 2i and -5i.
Check:
If x = 2i and x = -5i satisfy the given equation, then our answer is correct.
L.H.S. = x² + 3ix + 10
= (2i)² + 3i(2i) + 10i
= 4i² + 6i² + 10
= 10i² + 10
= -10 + 10 ……[∵ i² = -1]
= 0
= R.H.S.
L.H.S. = x² + 3ix + 10
= (-5i)² + 3i(-5i) + 10
= 25i² – 15i² + 10
= 10i² + 10
= -10 + 10 …..[∵ i² = -1]
= 0
= R.H.S.
Thus, our answer is correct.

(ii) 2x² + 3ix + 2 = 0
Solution:
Given equation is 2x² + 3ix + 2 = 0
Comparing with ax² + bx + c = 0, we get
a = 2, b = 3i, c = 2
Discriminant = b² – 4ac
= (3i)² – 4 × 2 × 2
= 9i² – 16
= -9 – 16
= -25 < 0
So, the given equation has complex roots.
These roots are given by

∴ the roots of the given equation are \(\frac{1}{2}\)i and -2i.

(iii) x² + 4ix – 4 = 0
Solution:
Given equation is x² + 4ix – 4 = 0
Comparing with ax² + bx + c = 0, we get
a = 1, b = 4i, c = -4
Discriminant = b² – 4ac
= (4i)² – 4 × 1 × -4
= 16i² + 16
= -16 + 16 …..[∵ i² = -1]
= 0
So, the given equation has equal roots.
These roots are given by

∴ the roots of the given equation are -2i and -2i.

(iv) ix² – 4x – 4i = 0
Solution:
ix² – 4x – 4i = 0
Multiplying throughout by i, we get
i²x² – 4ix – 4i² = 0
∴ -x² – 4ix + 4 = 0 ……[∵ i² = -1]
∴ x² + 4ix – 4 = 0
Comparing with ax² + bx + c = 0, we get
a = 1, b = 4i, c = -4
Discriminant = b² – 4ac
= (4i)² – 4 × 1 × -4
= 16i² + 16
= -16 + 16 …..[∵ i² = -1]
= 0
So, the given equation has equal roots.
These roots are given by

∴ the roots of the given equation are -2i and -2i.

Question 4.
Solve the following quadratic equations.
(i) x² – (2 + i) x – (1 – 7i) = 0
Solution:
Given equation is x² – (2 + i)x – (1 – 7i) = 0
Comparing with ax² + bx + c = 0, we get
a = 1, b = -(2 + i), c = -(1 – 7i)
Discriminant = b² – 4ac
= [-(2 + i)]² – 4 × 1 × -(1 – 7i)
= 4 + 4i + i² + 4 – 28i
= 4 + 4i – 1 + 4 – 28i …….[∵ i² = -1]
= 7 – 24i
So, the given equation has complex roots.
These roots are given by

(ii) x² – (3√2 + 2i) x + 6√2i = 0
Solution:
Given equation is x² – (3√2 + 2i) x + 6√2i = 0
Comparing with ax² + bx + c = 0, we get
a = 1, b = -(3√2 + 2i), c = 6√2i
Discriminant = b² – 4ac
= [-(3√2 + 2i)]² – 4 × 1 × 6√2i
= 18 + 12√2i + 4i² – 24√2i
= 18 – 12√2i – 4 …..[∵ i² = -1]
= 14 – 12√2i
So, the given equation has complex roots.
These roots are given by

(iii) x² – (5 – i) x + (18 + i) = 0
Solution:
Given equation is x² – (5 – i)x + (18 + i) = 0
Comparing with ax² + bx + c = 0, we get
a = 1, b = -(5 – i), c = 18 + i
Discriminant = b² – 4ac
= [-(5 – i)]² – 4 × 1 × (18 + i)
= 25 – 10i + i² – 72 – 4i
= 25 – 10i – 1 – 72 – 4i …..[∵ i² = -1]
= -48 – 14i
So, the given equation has complex roots.
These roots are given by

(iv) (2 + i) x² – (5 – i) x + 2(1 – i) = 0
Solution:
Given equation is
(2 + i) x² – (5 – i) x + 2(1 – i) = 0
Comparing with ax² + bx + c = 0, we get
a = 2 + i, b = -(5 – i), c = 2(1 – i)
Discriminant = b² – 4ac
= [-(5 – i)]² – 4 × (2 + i) × 2(1 – i)
= 25 – 10i + i² – 8(2 + i)(1 – i)
= 25 – 10i + i² – 8(2 – 2i + i – i²)
= 25 – 10i – 1 – 8(2 – i + 1) …..[∵ i² = -1]
= 25 – 10i – 1 – 16 + 8i – 8
= -2i
So, the given equation has complex roots.
These roots are given by

11th Commerce Maths Digest Pdf 

• 11th Commerce Maths Exercise 3.1 Solutions
• 11th Commerce Maths Exercise 3.2 Solutions
• 11th Commerce Maths Exercise 3.3 Solutions
• 11th Commerce Maths Miscellaneous Exercise 3 Solutions

Limits Class 11 Commerce Maths 1 Chapter 7 Exercise 7.1 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 7 Limits Ex 7.1 Questions and Answers.

Std 11 Maths 1 Exercise 7.1 Solutions Commerce Maths

I. Evaluate the following limits:

Question 1.
\(\lim _{x \rightarrow 3}\left[\frac{\sqrt{x+6}}{x}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow 2}\left[\frac{x^{-3}-2^{-3}}{x-2}\right]\)
Solution:

Question 3.
\(\lim _{x \rightarrow 5}\left[\frac{x^{3}-125}{x^{5}-3125}\right]\)
Solution:

Question 4.
If \(\lim _{x \rightarrow 1}\left[\frac{x^{4}-1}{x-1}\right]=\lim _{x \rightarrow a}\left[\frac{x^{3}-a^{3}}{x-a}\right]\), find all possible values of a.
Solution:

II. Evaluate the following limits:

Question 1.
\(\lim _{x \rightarrow 7}\left[\frac{(\sqrt[3]{x}-\sqrt[3]{7})(\sqrt[3]{x}+\sqrt[3]{7})}{x-7}\right]\)
Solution:

Question 2.
If \(\lim _{x \rightarrow 5}\left[\frac{x^{k}-5^{k}}{x-5}\right]=500\), find all possible values of k.
Solution:

Question 3.
\(\lim _{x \rightarrow 0}\left[\frac{(1-x)^{8}-1}{(1-x)^{2}-1}\right]\)
Solution:

III. Evaluate the following limits:

Question 1.
\(\lim _{x \rightarrow 0}\left[\frac{\sqrt[3]{1+x}-\sqrt{1+x}}{x}\right]\)
Solution:

Question 2.
\(\lim _{y \rightarrow 1}\left[\frac{2 y-2}{\sqrt[3]{7+y}-2}\right]\)
Solution:

Question 3.
\(\lim _{z \rightarrow a}\left[\frac{(z+2)^{\frac{3}{2}}-(a+2)^{\frac{3}{2}}}{z-a}\right]\)
Solution:

Question 4.
\(\lim _{x \rightarrow 5}\left[\frac{x^{3}-125}{x^{2}-25}\right]\)
Solution:

11th Commerce Maths Digest Pdf

• 11th Commerce Maths Exercise 7.1 Solutions
• 11th Commerce Maths Exercise 7.2 Solutions
• 11th Commerce Maths Exercise 7.3 Solutions
• 11th Commerce Maths Exercise 7.4 Solutions
• 11th Commerce Maths Miscellaneous Exercise 7 Solutions

Differentiation Class 11 Commerce Maths 1 Chapter 9 Exercise 9.1 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 9 Differentiation Ex 9.1 Questions and Answers.

Std 11 Maths 1 Exercise 9.1 Solutions Commerce Maths

I. Find the derivatives of the following functions w.r.t. x.

Question 1.
x¹²
Solution:
Let y = x¹²
Differentiating w.r.t. x, we get

Question 2.
x^-9
Solution:
Let y = x^-9
Differentiating w.r.t. x, we get

Question 3.
\(x^{\frac{3}{2}}\)
Solution:
Let y = \(x^{\frac{3}{2}}\)
Differentiating w.r.t. x, we get

Question 4.
7x√x
Solution:

Question 5.
3⁵
Solution:
Let y = 3⁵
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}=\frac{d}{d x} 3^{5}=0\) …..[3⁵ is a constant]

II. Differentiate the following w.r.t. x.

Question 1.
x⁵ + 3x⁴
Solution:
Let y = x⁵ + 3x⁴
Differentiating w.r.t. x, we get

Question 2.
x√x + log x – e^x
Solution:
Let y = x√x + log x – e^x
= \(x^{\frac{3}{2}}+\log x-e^{x}\)
Differentiating w.r.t. x, we get

Question 3.
\(x^{\frac{5}{2}}+5 x^{\frac{7}{5}}\)
Solution:
Let y = \(x^{\frac{5}{2}}+5 x^{\frac{7}{5}}\)
Differentiating w.r.t. x, we get

Question 4.
\(\frac{2}{7} x^{\frac{7}{2}}+\frac{5}{2} x^{\frac{2}{5}}\)
Solution:
Let y = \(\frac{2}{7} x^{\frac{7}{2}}+\frac{5}{2} x^{\frac{2}{5}}\)
Differentiating w.r.t. x, we get

Question 5.
\(\sqrt{x}\left(x^{2}+1\right)^{2}\)
Solution:
Let y = \(\sqrt{x}\left(x^{2}+1\right)^{2}\)

III. Differentiate the following w.r.t. x.

Question 1.
x³ log x
Solution:
Let y = x³ log x
Differentiating w.r.t. x, we get

Question 2.
\(x^{\frac{5}{2}} e^{x}\)
Solution:
Let y = \(x^{\frac{5}{2}} e^{x}\)
Differentiating w.r.t. x, we get

Question 3.
e^x log x
Solution:
Let y = e^x log x
Differentiating w.r.t. x, we get

Question 4.
x³ . 3^x
Solution:
Let y = x³ . 3^x
Differentiating w.r.t. x, we get

IV. Find the derivatives of the following w.r.t. x.

Question 1.
\(\frac{x^{2}+a^{2}}{x^{2}-a^{2}}\)
Solution:

Question 2.
\(\frac{3 x^{2}+5}{2 x^{2}-4}\)
Solution:

Question 3.
\(\frac{\log x}{x^{3}-5}\)
Solution:

Question 4.
\(\frac{3 e^{x}-2}{3 e^{x}+2}\)
Solution:

Question 5.
\(\frac{x \mathrm{e}^{x}}{x+\mathrm{e}^{x}}\)
Solution:

V. Find the derivatives of the following functions by the first principle:

Question 1.
3x² + 4
Solution:
Let f(x) = 3x² + 4
∴ f(x + h) = 3(x + h)² + 4
= 3(x² + 2xh + h²) + 4
= 3x² + 6xh + 3h² + 4
By first principle, we get

Question 2.
x√x
Solution:
Let f(x) = x√x
∴ f(x + h) = \((x+h)^{\frac{3}{2}}\)
By first principle, we get

Question 3.
\(\frac{1}{2 x+3}\)
Solution:
Let f(x) = \(\frac{1}{2 x+3}\)
∴ f(x + h) = \(\frac{1}{2(x+\mathrm{h})+3}=\frac{1}{2 x+2 \mathrm{~h}+3}\)
By first principle, we get

Question 4.
\(\frac{x-1}{2 x+7}\)
Solution:
Let f(x) = \(\frac{x-1}{2 x+7}\)
∴ f(x + h) = \(\frac{x+\mathrm{h}-1}{2(x+\mathrm{h})+7}=\frac{x+\mathrm{h}-1}{2 x+2 \mathrm{~h}+7}\)
By first principle, we get

11th Commerce Maths Digest Pdf

• 11th Commerce Maths Exercise 9.1 Solutions
• 11th Commerce Maths Exercise 9.2 Solutions
• 11th Commerce Maths Miscellaneous Exercise 9 Solutions

Locus and Straight Line Class 11 Commerce Maths 1 Chapter 5 Exercise 5.3 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 5 Locus and Straight Line Ex 5.3 Questions and Answers.

Std 11 Maths 1 Exercise 5.3 Solutions Commerce Maths

Question 1.
Write the equation of the line:
(a) parallel to the X-axis and at a distance of 5 units from it and above it.
(b) parallel to the Y-axis and at a distance of 5 units from it and to the left of it.
(c) parallel to the X-axis and at a distance of 4 units from the point (-2, 3).
Solution:
(a) Equation of a line parallel to the X-axis is y = k.
Since the line is at a distance of 5 units above the X-axis.
∴ k = 5
∴ the equation of the required line is y = 5.

(b) Equation of a line parallel to the Y-axis is x = h.
Since the line is at a distance of 5 units to the left of the Y-axis.
∴ h = -5
∴ the equation of the required line is x = -5.

(c) Equation of a line parallel to the X-axis is of the form y = k (k > 0 or k < 0).
Since, the line is at a distance of 4 units from the point (-2, 3).
∴ k = 3 + 4 = 7 or k = 3 – 4 = -1
∴ the equation of the required line is y = 7 or y = -1.

Question 2.
Obtain the equation of the line:
(a) parallel to the X-axis and making an intercept of 3 units on the Y-axis.
(b) parallel to the Y-axis and making an intercept of 4 units on the X-axis.
Solution:
(a) Equation of a line parallel to X-axis with y-intercept ‘k’ is y = k.
Here, y-intercept = 3
∴ the equation of the required line is y = 3.

(b) Equation of a line parallel to Y-axis with x-intercept ‘h’ is x = h.
Here, x-intercept = 4
∴ the equation of the required line is x = 4.

Question 3.
Obtain the equation of the line containing the point:
(a) A(2, -3) and parallel to the Y-axis.
(b) B(4, -3) and parallel to the X-axis.
Solution:
(a) Equation of a line parallel to the Y-axis is of the form x = h.
Since, the line passes through A(2, -3).
∴ h = 2
∴ the equation of the required line is x = 2.

(b) Equation of a line parallel to the X-axis is of the form y = k.
Since, the line passes through B(4, -3)
∴ k = -3
∴ the equation of the required line is y = -3.

Question 4.
Find the equation of the line passing through the points A(2, 0) and B(3, 4).
Solution:
The required line passes through the points A(2, 0) = (x₁, y₁) and B(3, 4) = (x₂, y₂) say.
Equation of the line in two-point form is
\(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
∴ the equation of the required line is
∴ \(\frac{y-0}{4-0}=\frac{x-2}{3-2}\)
∴ \(\frac{y}{4}=\frac{x-2}{1}\)
∴ y = 4(x – 2)
∴ y = 4x – 8
∴ 4x – y – 8 = 0

Check:
If the points A(2, 0) and B(3, 4) satisfy 4x – y – 8 = 0, then our answer is correct.
For point A(2, 0),
L.H.S. = 4x – y – 8
= 4(2) – 0 – 8
= 8 – 8
= 0
= R.H.S.
For point B(3, 4),
L.H.S. = 4x – y – 8
= 4(3) – 4 – 8
= 12 – 12
= 0
= R.H.S.
Thus, our answer is correct.

Question 5.
Line y = mx + c passes through the points A(2, 1) and B(3, 2). Determine m and c.
Solution:
Given, A(2, 1) and B(3, 2).
Equation of a line in two-point form is
\(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
∴ the equation of the passing through A and B line is
∴ \(\frac{y-1}{2-1}=\frac{x-2}{3-2}\)
∴ \(\frac{y-1}{1}=\frac{x-2}{1}\)
∴ y – 1 = x – 2
∴ y = x – 1
Comparing this equation with y = mx + c, we get
m = 1 and c = -1

Alternate method:
Points A(2, 1) and B(3, 2) lie on the line y = mx + c.
∴ They must satisfy the equation.
∴ 2m + c = 1 ……..(i)
and 3m + c = 2 ……(ii)
equation (ii) – equation (i) gives m = 1
Substituting m = 1 in (i), we get
2(1) – c = 1
∴ c = 1 – 2 = -1

Question 6.
The vertices of a triangle are A(3, 4), B(2, 0), and C(-1, 6). Find the equations of
(a) side BC
(b) the median AD
(c) the midpoints of sides AB and BC.
Solution:
Vertices of ∆ABC are A(3, 4), B(2, 0) and C(-1, 6).
(a) Equation of a line in two-point form is
\(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
∴ the equation of the side BC is
\(\frac{y-0}{6-0}=\frac{x-2}{-1-2}\) ……[B = (x₁, y₁) = (2, 0), C = (x₂, y₂) = (-1, 6)]
∴ \(\frac{y}{6}=\frac{x-2}{-3}\)
∴ y = -2(x – 2)
∴ 2x + y – 4 = 0

(b) Let D be the midpoint of side BC.
Then, AD is the median through A.
∴ D = \(\left(\frac{2-1}{2}, \frac{0+6}{2}\right)=\left(\frac{1}{2}, 3\right)\)
The median AD passes through the points A(3, 4) and D(\(\frac{1}{2}\), 3)

∴ the equation of the median AD is
\(\frac{y-4}{3-4}=\frac{x-3}{\frac{1}{2}-3}\)
∴ \(\frac{y-4}{-1}=\frac{x-3}{-\frac{5}{2}}\)
∴ \(\frac{1}{2}\) (y – 4) = x – 3
∴ 5y – 20 = 2x – 6
∴ 2x – 5y + 14 = 0

(c) Let D and E be the midpoints of side AB and side BC respectively.
∴ D = \(\left(\frac{3+2}{2}, \frac{4+0}{2}\right)=\left(\frac{5}{2}, 2\right)\) and
E = \(\left(\frac{2-1}{2}, \frac{0+6}{2}\right)=\left(\frac{1}{2}, 3\right)\)

the equation of the line DE is
\(\frac{y-2}{3-2}=\frac{x-\frac{5}{2}}{\frac{1}{2}-\frac{5}{2}}\)
∴ \(\frac{y-2}{1}=\frac{2 x-5}{-4}\)
∴ -4(y – 2) = 2x – 5
∴ -4y + 8 = 2x – 5
∴ 2x + 4y – 13 = 0

Question 7.
Find the x and y-intercepts of the following lines:
(a) \(\frac{x}{3}+\frac{y}{2}=1\)
(b) \(\frac{3 x}{2}+\frac{2 y}{3}=1\)
(c) 2x – 3y + 12 = 0
Solution:
(a) Given equation of the line is \(\frac{x}{3}+\frac{y}{2}=1\)
This is of the form \(\frac{x}{a}+\frac{y}{b}=1\),
where x-intercept = a, y-intercept = b
∴ x-intercept = 3, y-intercept = 2

(b) Given equation of the line is \(\frac{3 x}{2}+\frac{2 y}{3}=1\)
∴ \(\frac{x}{\left(\frac{2}{3}\right)}+\frac{y}{\left(\frac{3}{2}\right)}=1\)
This is of the form \(\frac{x}{a}+\frac{y}{b}=1\),
where x-intercept = a, y-intercept = b
∴ x-intercept = \(\frac{2}{3}\) and y-intercept = \(\frac{3}{2}\)

(c) Given equation of the line is 2x – 3y + 12 = 0
∴ 2x – 3y = -12
∴ \(\frac{2 x}{(-12)}-\frac{3 y}{(-12)}=1\)
∴ \(\frac{x}{-6}+\frac{y}{4}=1\)
This is of the form \(\frac{x}{a}+\frac{y}{b}=1\),
where x-intercept = a, y-intercept = b
∴ x-intercept = -6 and y-intercept = 4

Question 8.
Find the equations of a line containing the point A(3, 4) and make equal intercepts on the co-ordinate axes.
Solution:
Let the equation of the line be
\(\frac{x}{a}+\frac{y}{b}=1\) …..(i)
Since, the required line make equal intercepts on the co-ordinate axes.
∴ a = b
∴ (i) reduces to x + y = a …..(ii)
Since the line passes through A(3, 4).
∴ 3 + 4 = a
i.e. a = 7
Substituting a = 7 in (ii) to get
x + y = 7

Question 9.
Find the equations of the altitudes of the triangle whose vertices are A(2, 5), B(6, -1) and C(-4, -3).
Solution:

A(2, 5), B(6, -1), C(-4, -3) are the vertices of ∆ABC.
Let AD, BE and CF be the altitudes through the vertices A, B and C respectively of ∆ABC.
Slope of BC = \(\frac{-3-(-1)}{-4-6}=\frac{-2}{-10}=\frac{1}{5}\)
∴ slope of AD = -5 ……..[∵ AD ⊥ BC]
Since, altitude AD passes through (2, 5) and has slope -5.
∴ the equation of the altitude AD is
y – 5 = -5(x – 2)
∴ y – 5 = – 5x + 10
∴ 5x + y – 15 = 0
Now, slope of AC = \(\frac{-3-5}{-4-2}=\frac{-8}{-6}=\frac{4}{3}\)
∴ slope of BE = \(\frac{-3}{4}\) …..[∵ BE ⊥ AC]
Since, altitude BE passes through (6, -1) and has slope \(\frac{-3}{4}\).
∴ the equation of the altitude BE is
y – (-1) = \(\frac{-3}{4}\)(x – 6)
∴ 4(y + 1) = -3(x – 6)
∴ 3x + 4y – 14 = 0
Also, slope of AB = \(\frac{-1-5}{6-2}=\frac{-6}{4}=\frac{-3}{2}\)
∴ slope of CF = \(\frac{2}{3}\) ………[∵ CF ⊥ AB]
Since, altitude CF passes through (-4, -3) and has slope \(\frac{2}{3}\).
∴ the equation of the altitude CF is
y – (-3) = \(\frac{2}{3}\) [x – (-4)]
∴ 3(y + 3) = 2(x + 4)
∴ 2x – 3y – 1 = 0

11th Commerce Maths Digest Pdf

• 11th Commerce Maths Exercise 5.1 Solutions
• 11th Commerce Maths Exercise 5.2 Solutions
• 11th Commerce Maths Exercise 5.3 Solutions
• 11th Commerce Maths Exercise 5.4 Solutions
• 11th Commerce Maths Miscellaneous Exercise 5 Solutions

Sequences and Series Class 11 Commerce Maths 1 Chapter 4 Exercise 4.5 Answers Maharashtra Board

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

Std 11 Maths 1 Exercise 4.5 Solutions Commerce Maths

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:

11th Commerce Maths Digest Pdf

• 11th Commerce Maths Exercise 4.1 Solutions
• 11th Commerce Maths Exercise 4.2 Solutions
• 11th Commerce Maths Exercise 4.3 Solutions
• 11th Commerce Maths Exercise 4.4 Solutions
• 11th Commerce Maths Exercise 4.5 Solutions
• 11th Commerce Maths Miscellaneous Exercise 4 Solutions

Locus and Straight Line Class 11 Commerce Maths 1 Chapter 5 Exercise 5.2 Answers Maharashtra Board

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

Std 11 Maths 1 Exercise 5.2 Solutions Commerce Maths

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.

11th Commerce Maths Digest Pdf

• 11th Commerce Maths Exercise 5.1 Solutions
• 11th Commerce Maths Exercise 5.2 Solutions
• 11th Commerce Maths Exercise 5.3 Solutions
• 11th Commerce Maths Exercise 5.4 Solutions
• 11th Commerce Maths Miscellaneous Exercise 5 Solutions