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Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 5.3 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 5 Co-ordinate Geometry.

10th Standard Maths 2 Practice Set 5.3 Chapter 5 Co-ordinate Geometry Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Practice Set 5.3 Chapter 5 Co-ordinate Geometry Questions With Answers Maharashtra Board

Practice Set 5.3 Geometry Class 10 Question 1. Angles made by the line with the positive direction of X-axis are given. Find the slope of these lines.
i. 45°
ii. 60°
iii. 90°
Solution:
i. Angle made with the positive direction of
X-axis (θ) = 45°
Slope of the line (m) = tan θ
∴ m = tan 45° = 1
∴ The slope of the line is 1.

ii. Angle made with the positive direction of X-axis (θ) = 60°
Slope of the line (m) = tan θ
∴ m = tan 60° = \(\sqrt { 3 }\)
∴ The slope of the line is \(\sqrt { 3 }\).

iii. Angle made with the positive direction of
X-axis (θ) = 90°
Slope of the line (m) = tan θ
∴ m = tan 90°
But, the value of tan 90° is not defined.
∴ The slope of the line cannot be determined.

Practice Set 5.3 Geometry Question 2. Find the slopes of the lines passing through the given points.
i. A (2, 3), B (4, 7)
ii. P(-3, 1), Q (5, -2)
iii. C (5, -2), D (7, 3)
iv. L (-2, -3), M (-6, -8)
v. E (-4, -2), F (6, 3)
vi. T (0, -3), s (0,4)
Solution:
i. A (x₁, y₁) = A (2, 3) and B (x₂, y₂) = B (4, 7)
Here, x₁ = 2, x₂ = 4, y₁ = 3, y₂ = 7

∴ The slope of line AB is 2.

ii. P (x₁, y₁) = P (-3, 1) and Q (x₂, y₂) = Q (5, -2)
Here, x₁ = -3, x₂ = 5, y₁ = 1, y₂ = -2

∴ The slope of line PQ is \(\frac { -3 }{ 8 } \)

iii. C (x₁, y₁) = C (5, -2) and D (x₂, y₂) = D (7, 3)
Here, x₁ = 5, x₂ = 7, y₁ = -2, y₂ = 3

∴ The slope of line CD is \(\frac { 5 }{ 2 } \)

iv. L (x₁, y₁) = L (-2, -3) and M (x₂,y₂) = M (-6, -8)
Here, x₁ = -2, x₂ = – 6, y₁ = – 3, y₂ = – 8

∴ The slope of line LM is \(\frac { 5 }{ 4 } \)

v. E (x₁, y₁) = E (-4, -2) and F (x₂, y₂) = F (6, 3)
Here,x₁ = -4, x₂ = 6, y₁ = -2, y₂ = 3

∴ The slope of line EF is \(\frac { 1 }{ 2 } \).

vi. T (x₁, y₁) = T (0, -3) and S (x₂, y₂) = S (0, 4)
Here, x₁ = 0, x₂ = 0, y₁ = -3, y₂ = 4

∴ The slope of line TS cannot be determined.

5.3.5 Practice Question 3. Determine whether the following points are collinear.
i. A (-1, -1), B (0, 1), C (1, 3)
ii. D (- 2, -3), E (1, 0), F (2, 1)
iii. L (2, 5), M (3, 3), N (5, 1)
iv. P (2, -5), Q (1, -3), R (-2, 3)
v. R (1, -4), S (-2, 2), T (-3,4)
vi. A(-4,4),K[-2,\(\frac { 5 }{ 2 } \)], N (4,-2)
Solution:


∴ slope of line AB = slope of line BC
∴ line AB || line BC
Also, point B is common to both the lines.
∴ Both lines are the same.
∴ Points A, B and C are collinear.

∴ slope of line DE = slope of line EF
∴ line DE || line EF
Also, point E is common to both the lines.
∴ Both lines are the same.
∴ Points D, E and F are collinear.

∴ slope of line LM ≠ slope of line MN
∴ Points L, M and N are not collinear.

∴ slope of line PQ = slope of line QR
∴ line PQ || line QR
Also, point Q is common to both the lines.
∴ Both lines are the same.
∴ Points P, Q and R are collinear.

∴ slope of line RS = slope of line ST
∴ line RS || line ST
Also, point S is common to both the lines.
∴ Both lines are the same.
∴ Points R, S and T are collinear.

∴ slope of line AK = slope of line KN
∴ line AK || line KN
Also, point K is common to both the lines.
∴ Both lines are the same.
∴ Points A, K and N are collinear.

Practice Set 5.3 Geometry 9th Standard Question 4. If A (1, -1), B (0,4), C (-5,3) are vertices of a triangle, then find the slope of each side.
Solution:

∴ The slopes of the sides AB, BC and AC are -5, \(\frac { 1 }{ 5 } \) and \(\frac { -2 }{ 3 } \) respectively.

Geometry 5.3 Question 5. Show that A (-4, -7), B (-1, 2), C (8, 5) and D (5, -4) are the vertices of a parallelogram.
Proof:


∴ Slope of side AB = Slope of side CD … [From (i) and (iii)]
∴ side AB || side CD
Slope of side BC = Slope of side AD … [From (ii) and (iv)]
∴ side BC || side AD
Both the pairs of opposite sides of ꠸ABCD are parallel.
꠸ABCD is a parallelogram.
Points A(-4, -7), B(-1, 2), C(8, 5) and D(5, -4) are the vertices of a parallelogram.

Question 6.
Find k, if R (1, -1), S (-2, k) and slope of line RS is -2.
Solution:
R(x₁, y₁) = R (1, -1), S (x₂, y₂) = S (-2, k)
Here, x₁ = 1, x₂ = -2, y₁ = -1, y₂ = k

But, slope of line RS is -2. … [Given]
∴ -2 = \(\frac { k+1 }{ -3 } \)
∴ k + 1 = 6
∴ k = 6 – 1
∴ k = 5

5.3 Class 10 Question 7. Find k, if B (k, -5), C (1, 2) and slope of the line is 7.
Solution:
B(x₁, y₁) = B (k, -5), C (x₂, y₂) = C (1, 2)
Here, x₁ = k, x₂ = 1, y₁ = -5, y₂ = 2

But, slope of line BC is 7. …[Given]
∴ 7 = \(\frac { 7 }{ 1-k } \)
∴ 7(1 – k) = 7
∴ 1 – k = \(\frac { 7 }{ 7 } \)
∴ 1 – k = 1
∴ k = 0

Question 8.
Find k, if PQ || RS and P (2, 4), Q (3, 6), R (3,1), S (5, k).
Solution:

But, line PQ || line RS … [Given]
∴ Slope of line PQ = Slope of line RS
∴ 2 = \(\frac { k-1 }{ 2 } \)
∴ 4 = k – 1
∴ k = 4 + 1
∴ k = 5

Class 10 Maths Digest

• Practice Set 4.1 Class 10 Answers
• Practice Set 4.2 Class 10 Answers
• Problem Set 4 Class 10 Answers
• Practice Set 5.1 Class 10 Answers
• Practice Set 5.2 Class 10 Answers
• Practice Set 5.3 Class 10 Answers
• Problem Set 5 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 5.2 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 5 Co-ordinate Geometry.

10th Standard Maths 2 Practice Set 5.2 Chapter 5 Co-ordinate Geometry Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Practice Set 5.2 Chapter 5 Co-ordinate Geometry Questions With Answers Maharashtra Board

Question 1.
Find the co-ordinates of point P if P divides the line segment joining the points A (-1, 7) and B (4, -3) in the ratio 2:3.
Solution:
Let the co-ordinates of point P be (x, y) and A (x₁, y₁) B (x₂, y₂) be the given points.
Here, x₁ = -1, y₁ = 7, x₂ = 4, y₂ = -3, m = 2, n = 3
∴ By section formula,

∴ The co-ordinates of point P are (1,3).

Question 2.
In each of the following examples find the co-ordinates of point A which divides segment PQ in the ratio a : b.
i. P (-3, 7), Q (1, -4), a : b = 2 : 1
ii. P (-2, -5), Q (4, 3), a : b = 3 : 4
iii. P (2, 6), Q (-4, 1), a : b = 1 : 2
Solution:
Let the co-ordinates of point A be (x, y).
i. Let P (x₁, y₁), Q (x₂, y₂) be the given points.
Here, x₁ = -3, y₁ = 7, x₂ = 1, y₂ = -4, a = 2, b = 1
∴ By section formula,

∴ The co-ordinates of point A are (\(\frac { -1 }{ 3 } \),\(\frac { -1 }{ 3 } \)).

ii. Let P (x₁,y₁), Q (x₂, y₂) be the given points.
Here, x₁ = -2, y₁ = -5, x₂ = 4, y₂ = 3, a = 3, b = 4
By section formula,

∴ The co-ordinates of point A are (\(\frac { 4 }{ 7 } \),\(\frac { -11 }{ 7 } \))

iii. Let P (x₁, y₁), Q (x₂, y₂) be the given points.
Here,x₁ = 2,y₁ = 6, x₂ = -4, y₂ = 1, a = 1,b = 2
∴ By section formula,

∴ The co-ordinates of point A are (0,\(\frac { 13 }{ 3 } \))

Question 3.
Find the ratio in which point T (-1, 6) divides the line segment joining the points P (-3,10) and Q (6, -8).
Solution:
Let P (x₁, y₁), Q (x₂, y₂) and T (x, y) be the given points.
Here, x₁ = -3, y₁ = 10, x₂ = 6, y₂ = -8, x = -1, y = 6
∴ By section formula,

∴ Point T divides seg PQ in the ratio 2 : 7.

Question 4.
Point P is the centre of the circle and AB is a diameter. Find the co-ordinates of point B if co-ordinates of point A and P are (2, -3) and (-2,0) respectively.
Solution:
Let A (x₁, y₁), B (x₂, y₂) and P (x, y) be the given points.
Here, x₁ = 2, y₁ =-3,
x = -2, y = 0

Point P is the midpoint of seg AB.
∴ By midpoint formula,

∴ The co-ordinates of point B are (-6,3).

Question 5.
Find the ratio in which point P (k, 7) divides the segment joining A (8, 9) and B (1,2). Also find k.
Solution:
Let A (x₁, y₁), B (x₂, y₂) and P (x, y) be the given points.
Here, x₁ = 8, y₁ = 9, x₂ = 1, y₂ = 2, x = k, y = 7
∴ By section formula,

∴ Point P divides seg AB in the ratio 2 : 5, and the value of k is 6.

Question 6.
Find the co-ordinates of midpoint of the segment joining the points (22, 20) and (0,16).
Solution:
Let A (x₁, y₁) = A (22, 20),
B (x₂,y₂) = B (0, 16)
Let the co-ordinates of the midpoint be P (x,y).
∴ By midpoint formula,

The co-ordinates of the midpoint of the segment joining (22, 20) and (0, 16) are (11,18).

Question 7.
Find the centroids of the triangles whose vertices are given below.
i. (-7, 6), (2,-2), (8, 5)
ii. (3, -5), (4, 3), (11,-4)
iii. (4, 7), (8, 4), (7, 11)
Solution:
i. Let A (x₁, y₁) = A (-7, 6),
B (x₂, y₂) = B (2, -2),
C (x₃, y₃) = C(8, 5)
∴ By centroid formula,

∴ The co-ordinates of the centroid are (1,3).

ii. Let A (x₁ y₁) = A (3, -5),
B (x₂, y₂) = B (4, 3),
C(x₃, y₃) = C(11,-4)
∴ By centroid formula,

∴ The co-ordinates of the centroid are (6, -2).

iii. Let A (x₁, y₁) = A (4, 7),
B (x₂, y₂) = B (8,4),
C (x₃, y₃) = C(7,11)
∴ By centroid formula,

∴ The co-ordinates of the centroid are (\(\frac { 19 }{ 3 } \),\(\frac { 22 }{ 3 } \))

Question 8.
In ∆ABC, G (-4, -7) is the centroid. If A (-14, -19) and B (3, 5), then find the co-ordinates of C.
Solution:
G (x, y) = G (-4, -7),
A (x₁, y₁) = A (-14, -19),
B(x₂, y₂) = B(3,5)
Let the co-ordinates of point C be (x₃, y₃).
G is the centroid.
By centroid formula,

∴ The co-ordinates of point C are (-1, – 7).

Question 9.
A (h, -6), B (2, 3) and C (-6, k) are the co-ordinates of vertices of a triangle whose centroid is G (1,5). Find h and k.
Solution:
A(x₁,y₁) = A(h, -6),
B (x₂, y₂) = B(2, 3),
C (x₃, y₃) = C (-6, k)
∴ centroid G (x, y) = G (1, 5)
G is the centroid.
By centroid formula,

∴ 3 = h – 4
∴ h = 3 + 4
∴ h = 7

∴ 15 = -3 + k
∴ k = 15 + 3
∴ k = 18
∴ h = 7 and k = 18

Question 10.
Find the co-ordinates of the points of trisection of the line segment AB with A (2,7) and B (-4, -8).
Solution:
A (2, 7), B H,-8)
Suppose the points P and Q trisect seg AB.
∴ AP = PQ = QB

∴ Point P divides seg AB in the ratio 1:2.
∴ By section formula,

Co-ordinates of P are (0, 2).
Point Q is the midpoint of PB.
By midpoint formula,

Co-ordinates of Q are (-2, -3).
∴ The co-ordinates of the points of trisection seg AB are (0,2) and (-2, -3).

Question 11.
If A (-14, -10), B (6, -2) are given, find the co-ordinates of the points which divide segment AB into four equal parts.
Solution:
Let the points C, D and E divide seg AB in four equal parts.

Point D is the midpoint of seg AB.
∴ By midpoint formula,

∴ Co-ordinates of D are (-4, -6).
Point C is the midpoint of seg AD.
∴ By midpoint formula,

∴ Co-ordinates of C are (-9, -8).
Point E is the midpoint of seg DB.
∴ By midpoint formula,


∴ Co-ordinates of E are (1,-4).
∴ The co-ordinates of the points dividing seg AB in four equal parts are C(-9, -8), D(-4, -6) and E(1, – 4).

Question 12.
If A (20, 10), B (0, 20) are given, find the co-ordinates of the points which divide segment AB into five congruent parts.
Solution:
Suppose the points C, D, E and F divide seg AB in five congruent parts.
∴ AC = CD = DE = EF = FB

∴ co-ordinates of C are (16, 12).
E is the midpoint of seg CB.
By midpoint formula,

∴ co-ordinates of E are (8, 16).
D is the midpoint of seg CE.

∴ co-ordinates of F are (4, 18).
∴ The co-ordinates of the points dividing seg AB in five congruent parts are C (16, 12), D (12, 14), E (8, 16) and F (4, 18).

Maharashtra Board Class 10 Maths Chapter 5 Co-ordinate Geometry Intext Questions and Activities

Question 1.
A (15, 5), B (9, 20) and A-P-B. Find the ratio in which point P (11, 15) divides segment AB. Find the ratio using x and y co-ordinates. Write the conclusion. (Textbook pg. no. 113)
Solution:
Suppose point P (11,15) divides segment AB in the ratio m : n.
By section formula,

∴ Point P divides seg AB in the ratio 2 : 1.
The ratio obtained by using x and y co-ordinates is the same.

Question 2.
External division: (Textbook pg. no. 115)
Suppose point R divides seg PQ externally in the ratio 3:1.

Let the common multiple be k.
Let PR = 3k and QR = k
Now, PR = PQ + QR … [P – Q – R]
∴ 3k = PQ + k
∴ \(\frac { PQ }{ QR } \) = \(\frac { 2k }{ k } \) = \(\frac { 2 }{ 1 } \)
∴ Point Q divides seg PR in the ratio 2 : 1 internally.
Thus, we can find the co-ordinates of point R, when co-ordinates of points P and Q are given.

Class 10 Maths Digest

• Practice Set 4.1 Class 10 Answers
• Practice Set 4.2 Class 10 Answers
• Problem Set 4 Class 10 Answers
• Practice Set 5.1 Class 10 Answers
• Practice Set 5.2 Class 10 Answers
• Practice Set 5.3 Class 10 Answers
• Problem Set 5 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 5.1 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 5 Co-ordinate Geometry.

10th Standard Maths 2 Practice Set 5.1 Chapter 5 Co-ordinate Geometry Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Practice Set 5.1 Chapter 5 Co-ordinate Geometry Questions With Answers Maharashtra Board

Practice Set 5.1 Geometry Class 10 Question 1. Find the distance between each of the following pairs of points.
i. A (2, 3), B (4,1)
ii. P (-5, 7), Q (-1, 3)
iii. R (0, -3), S (0,\(\frac { 5 }{ 2 } \))
iv. L (5, -8), M (-7, -3)
v. T (-3, 6), R (9, -10)
vi. W(\(\frac { -7 }{ 2 } \),4), X(11, 4)
Solution:
i. Let A (x₁, y₁) and B (x₂, y₂) be the given points.
∴ x₁ = 2, y₁ = 3, x₂ = 4, y₂ = 1
By distance formula,

∴ d(A, B) = 2\(\sqrt { 2 }\) units
∴ The distance between the points A and B is 2\(\sqrt { 2 }\) units.

ii. Let P (x₁, y₁ ) and Q (x₂, y₂) be the given points.
∴ x₁ = -5, y₁ = 7, x₂ = -1, y₂ = 3
By distance formula,

∴ d(P, Q) = 4\(\sqrt { 2 }\) units
∴ The distance between the points P and Q is 4\(\sqrt { 2 }\) units.

iii. Let R (x₁, y₁) and S (x₂, y₂) be the given points.
∴ x₁ = 0, y₁ = -3, x₂ = 0, y₂ = \(\frac { 5 }{ 2 } \)
By distance formula,

∴ d(R, S) = \(\frac { 11 }{ 2 } \) units
∴ The distance between the points R and S is \(\frac { 11 }{ 2 } \) units.

iv. Let L (x₁, y₁) and M (x₂, y₂) be the given points.
∴ x₁ = 5, y₁ = -8, x₂ = -7, y₂ = -3
By distance formula,

∴ d(L, M) = 13 units
∴ The distance between the points L and M is 13 units.

v. Let T (x₁,y₁) and R (x₂, y₂) be the given points.
∴ x₁ = -3, y₁ = 6,x₂ = 9,y₂ = -10
By distance formula,

∴ d(T, R) = 20 units
∴ The distance between the points T and R 20 units.

vi. Let W (x₁, y₁) and X (x₂, y₂) be the given points.

∴ d(W, X) = \(\frac { 29 }{ 2 } \) units
∴ The distance between the points W and X is \(\frac { 29 }{ 2 } \) units.

Practice Set 5.1 Geometry 10th Question 2. Determine whether the points are collinear.
i. A (1, -3), B (2, -5), C (-4, 7)
ii. L (-2, 3), M (1, -3), N (5, 4)
iii. R (0, 3), D (2, 1), S (3, -1)
iv. P (-2, 3), Q (1, 2), R (4, 1)
Solution:
i. By distance formula,


∴ d(A, B) = \(\sqrt { 5 }\) …(i)
On adding (i) and (iii),
d(A, B) + d(A, C)= \(\sqrt { 5 }\) + 5\(\sqrt { 5 }\) = 6\(\sqrt { 5 }\)
∴ d(A, B) + d(A, C) = d(B, C) … [From (ii)]
∴ Points A, B and C are collinear.

ii. By distance formula,

On adding (i) and (iii),
d(L, M) + d(L, N) = 3\(\sqrt { 5 }\) + 5\(\sqrt { 2 }\) ≠ \(\sqrt { 65 }\)
∴ d(L, M) + d(L, N) ≠ d(M, N) … [From (ii)]
∴ Points L, M and N are not collinear.

iii. By distance formula,

On adding (i) and (ii),
∴ d(R, D) + d(D, S) = \(\sqrt { 8 }\) + \(\sqrt { 5 }\) ≠ 5
∴ d(R, D) + d(D, S) ≠ d(R, S) … [From (iii)]
∴ Points R, D and S are not collinear.

iv. By distance formula,

On adding (i) and (ii),
d(P, Q) + d(Q, R) = \(\sqrt { 10 }\) + \(\sqrt { 10 }\) = 2\(\sqrt { 10 }\)
∴ d(P, Q) + d(Q, R) = d(P, R) … [From (iii)]
∴ Points P, Q and R are collinear.

Coordinate Geometry Class 10 Practice Set 5.1 Question 3. Find the point on the X-axis which is equidistant from A (-3,4) and B (1, -4).
Solution:
Let point C be on the X-axis which is equidistant from points A and B.
Point C lies on X-axis.
∴ its y co-ordinate is 0.
Let C = (x, 0)
C is equidistant from points A and B.
∴ AC = BC

∴ (x + 3)² + (-4)² = (x- 1)² + 4²
∴ x² + 6x + 9 + 16 = x² – 2x + 1 + 16
∴ 8x = – 8
∴ x = – \(\frac { 8 }{ 8 } \) = -1
∴ The point on X-axis which is equidistant from points A and B is (-1,0).

10th Geometry Practice Set 5.1 Question 4. Verify that points P (-2, 2), Q (2, 2) and R (2, 7) are vertices of a right angled triangle.
Solution:
Distance between two points

Consider, PQ² + QR² = 42 + 52 = 16 + 25 = 41 … [From (i) and (ii)]
∴ PR² = PQ² + QR² … [From (iii)]
∴ ∆PQR is a right angled triangle. … [Converse of Pythagoras theorem]
∴ Points P, Q and R are the vertices of a right angled triangle.

Question 5.
Show that points P (2, -2), Q (7, 3), R (11, -1) and S (6, -6) are vertices of a parallelogram.
Proof:
Distance between two points

PQ = RS … [From (i) and (iii)]
QR = PS … [From (ii) and (iv)]
A quadrilateral is a parallelogram, if both the pairs of its opposite sides are congruent.
∴ □ PQRS is a parallelogram.
∴ Points P, Q, R and S are the vertices of a parallelogram.

Question 6.
Show that points A (-4, -7), B (-1, 2), C (8, 5) and D (5, -4) are vertices of rhombus ABCD.
Proof:
Distance between two points


∴ AB = BC = CD = AD …[From (i), (ii), (iii) and (iv)]
In a quadrilateral, if all the sides are equal, then it is a rhombus.
∴ □ ABCD is a rhombus.
∴ Points A, B, C and D are the vertices of rhombus ABCD.

Practice Set 5.1 Question 7. Find x if distance between points L (x, 7) and M (1,15) is 10.
Solution:
X₁ = x, y₁ = 7, x₂ = 1, y₂ = 15
By distance formula,

∴ 1 – x = ± 6
∴ 1 – x = 6 or l – x = -6
∴ x = – 5 or x = 7
∴ The value of x is – 5 or 7.

Geometry 5.1 Question 8. Show that the points A (1, 2), B (1, 6), C (1 + 2\(\sqrt { 3 }\), 4) are vertices of an equilateral triangle.
Proof:
Distance between two points

∴ AB = BC = AC … [From (i), (ii) and (iii)]
∴ ∆ABC is an equilateral triangle.
∴ Points A, B and C are the vertices of an equilateral triangle.

Maharashtra Board Class 10 Maths Chapter 5 Coordinate Geometry Intext Questions and Activities

Question 1.
In the figure, seg AB || Y-axis and seg CB || X-axis. Co-ordinates of points A and C are given. To find AC, fill in the boxes given below. (Textbook pa. no. 102)

Solution:
In ∆ABC, ∠B = 900
∴ (AB)² + (BC)² = [(Ac)² …(i) … [Pythagoras theorem]
seg CB || X-axis
∴ y co-ordinate of B = 2
seg BA || Y-axis
∴ x co-ordinate of B = 2
∴ co-ordinate of B is (2, 2) = (x₁,y₁)
co-ordinate of A is (2, 3) = (x₂, Y₂)
Since, AB || to Y-axis,
d(A, B) = Y₂ – Y₁
d(A,B) = 3 – 2 = 1
co-ordinate of C is (-2,2) = (x₁,y₁)
co-ordinate of B is (2, 2) = (x₂, y₂)
Since, BC || to X-axis,
d(B, C) = x₂ – x₁
d(B,C) = 2 – -2 = 4
∴ AC² = 12 + 42 …[From (i)]
= 1 + 16 = 17
∴ AC = \(\sqrt { 17 }\) units …[Taking square root of both sides]

Class 10 Maths Digest

• Practice Set 4.1 Class 10 Answers
• Practice Set 4.2 Class 10 Answers
• Problem Set 4 Class 10 Answers
• Practice Set 5.1 Class 10 Answers
• Practice Set 5.2 Class 10 Answers
• Practice Set 5.3 Class 10 Answers
• Problem Set 5 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 1.2 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 1 Similarity.

10th Standard Maths 2 Practice Set 1.2 Chapter 1 Similarity Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Practice Set 1.2 Chapter 1 Similarity Questions With Answers Maharashtra Board

Question 1.
Given below are some triangles and lengths of line segments. Identify in which figures, ray PM is the bisector of ∠QPR.

Solution:
In ∆ PQR,
\(\frac { PQ }{ PR } \) = \(\frac { 7 }{ 3 } \) (i)
\(\frac { QM }{ RM } \) = \(\frac { 3.5 }{ 1.5 } \) = \(\frac { 35 }{ 15 } \) = \(\frac { 7 }{ 3 } \) (ii)
∴ \(\frac { PQ }{ PR } \) = \(\frac { QM }{ RM } \) [From (i) and (ii)]
∴ Ray PM is the bisector of ∠QPR. [Converse of angle bisector theorem]

ii. In ∆PQR,
\(\frac { PQ }{ PR } \) = \(\frac { 10 }{ 7 } \) (i)
\(\frac { QM }{ RM } \) = \(\frac { 8 }{ 6 } \) = \(\frac { 4 }{ 3 } \) (ii)
∴ \(\frac { PQ }{ PR } \) ≠ \(\frac { QM }{ RM } \) [From (i) and (ii)]
∴ Ray PM is not the bisector of ∠QPR

iii. In ∆PQR,
\(\frac { PQ }{ PR } \) = \(\frac { 9 }{ 10 } \) (i)
\(\frac { QM }{ RM } \) = \(\frac { 3.6 }{ 4 } \) = \(\frac { 36 }{ 40 } \) = \(\frac { 9 }{ 10 } \) (ii)
∴ \(\frac { PQ }{ PR } \) = \(\frac { QM }{ RM } \) [From (i) and (ii)]
∴ Ray PM is the bisector of ∠QPR [Converse of angle bisector theorem]

Question 2.
In ∆PQR PM = 15, PQ = 25, PR = 20, NR = 8. State whether line NM is parallel to side RQ. Give reason.

Solution:
PN + NR = PR [P – N – R]
∴ PN + 8 = 20
∴ PN = 20 – 8 = 12
Also, PM + MQ = PQ [P – M – Q]
∴ 15 + MQ = 25

∴ line NM || side RQ [Converse of basic proportionality theorem]

Question 3.
In ∆MNP, NQ is a bisector of ∠N. If MN = 5, PN = 7, MQ = 2.5, then find QP.

Solution:
In ∆MNP, NQ is the bisector of ∠N. [Given]
∴\(\frac { PN }{ MN } \) = \(\frac { QP }{ MQ } \) [Property of angle bisector of a triangle]
∴\(\frac { 7 }{ 5 } \) = \(\frac { QP }{ 2.5 } \)
∴ QP = \(\frac { 7\times 2.5 }{ 5 } \)
∴ QP = 3.5 units

Question 4.
Measures of some angles in the figure are given. Prove that \(\frac { AP }{ PB } \) = \(\frac { AQ }{ QC } \)

Solution:
Proof
∠APQ = ∠ABC = 60° [Given]
∴ ∠APQ ≅ ∠ABC
∴ side PQ || side BC (i) [Corresponding angles test]
In ∆ABC,
sidePQ || sideBC [From (i)]
∴\(\frac { AP }{ PB } \) = \(\frac { AQ }{ QC } \) [Basic proportionality theorem]

Question 5.
In trapezium ABCD, side AB || side PQ || side DC, AP = 15, PD = 12, QC = 14, find BQ.

Solution:
side AB || side PQ || side DC [Given]
∴\(\frac { AP }{ PD } \) = \(\frac { BQ }{ QC } \) [Property of three parallel lines and their transversals]
∴\(\frac { 15 }{ 12 } \) = \(\frac { BQ }{ 14 } \)
∴ BQ = \(\frac { 15\times 14 }{ 12 } \)
∴ BQ = 17.5 units

Question 6.
Find QP using given information in the figure.

Solution:
In ∆MNP, seg NQ bisects ∠N. [Given]
∴\(\frac { PN }{ MN } \) = \(\frac { QP }{ MQ } \) [Property of angle bisector of a triangle]
∴\(\frac { 40 }{ 25 } \) = \(\frac { QP }{ 14 } \)
∴ QP = \(\frac { 40\times 14 }{ 25 } \)
∴ QP = 22.4 units

Question 7.
In the adjoining figure, if AB || CD || FE, then find x and AE.

Solution:
line AB || line CD || line FE [Given]
∴\(\frac { BD }{ DF } \) = \(\frac { AC }{ CE } \) [Property of three parallel lines and their transversals]
∴\(\frac { 8 }{ 4 } \) = \(\frac { 12 }{ X } \)
∴ X = \(\frac { 12\times 4 }{ 8 } \)
∴ X = 6 units
Now, AE AC + CE [A – C – E]
= 12 + x
= 12 + 6
= 18 units
∴ x = 6 units and AE = 18 units

Question 8.
In ∆LMN, ray MT bisects ∠LMN. If LM = 6, MN = 10, TN = 8, then find LT.

Solution:
In ∆LMN, ray MT bisects ∠LMN. [Given]
∴\(\frac { LM }{ MN } \) = \(\frac { LT }{ TN } \) [Property of angle bisector of a triangle]
∴\(\frac { 6 }{ 10 } \) = \(\frac { LT }{ 8 } \)
∴ LT = \(\frac { 6\times 8 }{ 10 } \)
∴ LT = 4.8 units

Question 9.
In ∆ABC,seg BD bisects ∠ABC. If AB = x,BC x+ 5, AD = x – 2, DC = x + 2, then find the value of x.

Solution:
In ∆ABC, seg BD bisects ∠ABC. [Given]
∴\(\frac { AB }{ BC } \) = \(\frac { AD }{ CD } \) [Property of angle bisector of a triangle]
∴\(\frac { x }{ x+5 } \) = \(\frac { x-2 }{ x+2 } \)
∴ x(x + 2) = (x – 2)(x + 5)
∴ x2 + 2x = x2 + 5x – 2x – 10
∴ 2x = 3x – 10
∴ 10 = 3x – 2x
∴ x = 10

Question 10.
In the adjoining figure, X is any point in the interior of triangle. Point X is joined to vertices of triangle. Seg PQ || seg DE, seg QR || seg EF. Fill in the blanks to prove that, seg PR || seg DF.

Solution:

Question 11.
In ∆ABC, ray BD bisects ∠ABC and ray CE bisects ∠ACB. If seg AB = seg AC, then prove that ED || BC.

Solution:
In ∆ABC, ray BD bisects ∠ABC. [Given]
∴\(\frac { AB }{ BC } \) = \(\frac { AE }{ EB } \) (i) [Property of angle bisector of a triangle]
Also, in ∆ABC, ray CE bisects ∠ACB. [Given]
∴\(\frac { AC }{ BC } \) = \(\frac { AE }{ EB } \) (ii) [Property of angle bisector of a triangle]
But, seg AB = seg AC (iii) [Given]
∴\(\frac { AB }{ BC } \) = \(\frac { AE }{ EB } \) (iv) [From (ii) and (iii)]
∴\(\frac { AD }{ DC } \) = \(\frac { AE }{ EB } \) [From (i) and (iv)]
∴ ED || BC [Converse of basic proportionality theorem]

Question 1.
i. Draw a ∆ABC.
ii. Bisect ∠B and name the point of intersection of AC and the angle bisector as D.
iii. Measure the sides.

iv. Find ratios \(\frac { AB }{ BC } \) and \(\frac { AD }{ DC } \)
v. You will find that both the ratios are almost equal.
vi. Bisect remaining angles of the triangle and find the ratios as above. Verify that the ratios are equal. (Textbook pg. no. 8)
Solution:

Note: Students should bisect the remaining angles and verify that the ratios are equal.

Question 2.
Write another proof of the above theorem (property of an angle bisector of a triangle). Use the following properties and write the proof.
i. The areas of two triangles of equal height are proportional to their bases.
ii. Every point on the bisector of an angle is equidistant from the sides of the angle. (Textbook pg. no. 9)

Given: In ∆CAB, ray AD bisects ∠A.
To prove: \(\frac { AB }{ AC } \) = \(\frac { BD }{ DC } \)
Construction: Draw seg DM ⊥ seg AB A – M – B and seg DN ⊥ seg AC, A – N – C.
Solution:
Proof:
In ∆ABC,
Point D is on angle bisector of ∠A. [Given]
∴DM = DN [Every point on the bisector of an angle is equidistant from the sides of the angle]
\(\frac{A(\Delta A B D)}{A(\Delta A C D)}=\frac{A B \times D M}{A C \times D N}\) [Ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights]
∴ \(\frac{A(\Delta A B D)}{A(\Delta A C D)}=\frac{A B}{A C}\) (ii) [From (i)]
Also, ∆ABD and ∆ACD have equal height.
∴ \(\frac{\mathrm{A}(\Delta \mathrm{ABD})}{\mathrm{A}(\Delta \mathrm{ACD})}=\frac{\mathrm{BD}}{\mathrm{CD}}\) (iii) [Triangles having equal height]
∴\(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{BD}}{\mathrm{DC}}\) [From (ii) and (iii)]

Question 3.
i. Draw three parallel lines.
ii. Label them as l, m, n.
iii. Draw transversals t₁ and t₂.
iv. AB and BC are intercepts on transversal t₁.
v. PQ and QR are intercepts on transversal t₂.
vi. Find ratios \(\frac { AB }{ BC } \) and \(\frac { PQ }{ QR } \). You will find that they are almost equal. Verify that they are equal.(Textbook pg, no. 10)
Solution:

(Students should draw figures similar to the ones given and verify the properties.)

Question 4.
In the adjoining figure, AB || CD || EF. If AC = 5.4, CE = 9, BD = 7.5, then find DF.(Textbook pg, no. 12)

Solution:

Question 5.
In ∆ABC, ray BD bisects ∠ABC. A – D – C, side DE || side BC, A – E – B, then prove that \(\frac { AB }{ BC } \) = \(\frac { AE }{ EB } \) (Textbook pg, no. 13)

Solution:

Class 10 Maths Digest

• Practice Set 1.1 Class 10 Answers
• Practice Set 1.2 Class 10 Answers
• Practice Set 1.3 Class 10 Answers
• Practice Set 1.4 Class 10 Answers
• Problem Set 1 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 1.1 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 1 Similarity.

10th Standard Maths 2 Practice Set 1.1 Chapter 1 Similarity Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Practice Set 1.1 Chapter 1 Similarity Questions With Answers Maharashtra Board

Question 1.
Base of a triangle is 9 and height is 5. Base of another triangle is 10 and height is 6. Find the ratio of areas of these triangles.
Solution:
Let the base, height and area of the first triangle be b₁, h₁, and A₁ respectively.
Let the base, height and area of the second triangle be b₂, h₂ and A₂ respectively.

[Since Ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights]
∴ The ratio of areas of the triangles is 3:4.

Question 2.
In the adjoining figure, BC ± AB, AD _L AB, BC = 4, AD = 8, then find \(\frac{A(\Delta A B C)}{A(\Delta A D B)}\)

Solution:
∆ABC and ∆ADB have same base AB.

[Since Triangles having equal base]

Question 3.
In the adjoining figure, seg PS ± seg RQ, seg QT ± seg PR. If RQ = 6, PS = 6 and PR = 12, then find QT.

Solution:
In ∆PQR, PR is the base and QT is the corresponding height.
Also, RQ is the base and PS is the corresponding height.
\(\frac{A(\Delta P Q R)}{A(\Delta P Q R)}=\frac{P R \times Q T}{R Q \times P S}\) [Ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights]
∴ \(\frac{1}{1}=\frac{P R \times Q T}{R Q \times P S}\)
∴ PR × QT = RQ × PS
∴ 12 × QT = 6 × 6
∴ QT = \(\frac { 36 }{ 12 } \)
∴ QT = 3 units

Question 4.
In the adjoining figure, AP ⊥ BC, AD || BC, then find A(∆ABC) : A(∆BCD).

Solution:
Draw DQ ⊥ BC, B-C-Q.

AD || BC [Given]
∴ AP = DQ   (i)  [Perpendicular distance between two parallel lines is the same]
∆ABC and ∆BCD have same base BC.

Question 5.
In the adjoining figure, PQ ⊥ BC, AD ⊥ BC, then find following ratios.

Solution:
i. ∆PQB and tPBC have same height PQ.

ii. ∆PBC and ∆ABC have same base BC.

iii. ∆ABC and ∆ADC have same height AD.

Question 1.
Find \(\frac{A(\Delta A B C)}{A(\Delta A P Q)}\)

Solution:
In ∆ABC, BC is the base and AR is the height.
In ∆APQ, PQ is the base and AR is the height.

Class 10 Maths Digest

• Practice Set 1.1 Class 10 Answers
• Practice Set 1.2 Class 10 Answers
• Practice Set 1.3 Class 10 Answers
• Practice Set 1.4 Class 10 Answers
• Problem Set 1 Class 10 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 1.1 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 1 Basic Concepts in Geometry.

9th Standard Maths 2 Practice Set 1.1 Chapter 1 Basic Concepts in Geometry Textbook Answers Maharashtra Board

Class 9 Maths Part 2 Practice Set 1.1 Chapter 1 Basic Concepts in Geometry Questions With Answers Maharashtra Board

Question 1.
Find the distances with the help of the number line given below.

i. d(B, E)
ii. d (J, J)
iii. d(P, C)
iv. d(J, H)
v. d(K, O)
vi. d(O, E)
vii. d(P, J)
viii. d(Q, B)
Solution:
i. Co-ordinate of the point B is 2.
Co-ordinate of the point E is 5.
Since, 5 > 2
∴ d(B, E) = 5 – 2
∴ d(B, E) = 3

ii. Co-ordinate of the point J is -2.
Co-ordinate of the point A is 1.
Since, 1 > -2
∴ d(J, A) = 1 – (-2)
= 1 + 2
∴ d(J, A) = 3

iii. Co-ordinate of the point P is -4.
Co-ordinate of the point C is 3.
Since, 3 > -4
∴ d(P,C) = 3 – (-4)
= 3 + 4
∴ d(P,C) = 7

iv. Co-ordinate of the point J is -2.
Co-ordinate of the point H is -1.
Since, -1 > -2
∴ d(J,H) = – 1 – (-2)
= -1 + 2
∴ d(J,H) = 1

v. Co-ordinate of the point K is -3.
Co-ordinate of the point O is 0.
Since,0 > -3
∴ d(K, O) = 0 – (-3)
= 0 + 3
∴ d(K, O) = 3

vi. Co-ordinate of the point O is 0.
∴ Co-ordinate of the point E is 5.
Since, 5 > 0
∴ d(O, E) = 5 – 0
∴ d(O, E) = 5

vii. Co-ordinate of the point P is -4.
Co-ordinate of the point J is -2.
Since -2 > -4
∴ d(P, J) = -2 – (-4)
= – 2+ 4
∴ d(P, J) = 2

viii. Co-ordinate of the point Q is -5.
Co-ordinate of the point B is 2.
Since,2 > -5
∴ d(Q,B) = 2 – (-5)
= 2 + 5
∴ d(Q, B) = 7

Question 2.
If the co-ordinate of A is x and that of B is . y, find d(A, B).
i. x = 1, y = 7
ii. x = 6, y = -2
iii. x = -3, y = 7
iv. x = -4, y = -5
v. x = -3, y = -6
vi. x = 4, y = -8
Solution:
i. Co-ordinate of point A is x = 1.
Co-ordinate of point B is y = 7
Since, 7 > 1
∴ d(A, B) = 7 – 1
∴ d(A, B) = 6

ii. Co-ordinate of point A is x = 6.
Co-ordinate of point B is y = -2.
Since, 6 > -2
∴ d(A, B) = 6 – ( -2) = 6 + 2
∴ d(A, B) = 8

iii. Co-ordinate of point A is x = -3.
Co-ordinate of point B is y = 7.
Since, 7 > -3
∴ d(A, B) = 7 – (-3) = 7 + 3
∴ d(A, B) = 10

iv. Co-ordinate of point A is x = -4.
Co-ordinate of point B is y = -5.
Since, -4 > -5
∴ d(A, B) = -4 – (-5)
= -4 + 5
∴ d(A, B) = 1

v. Co-ordinate of point A is x =-3.
Co-ordinate of point B is y = -6.
Since, -3 > -6
∴ d(A, B) = -3 – (-6)
= -3 + 6
∴ d(A, B) = 3

vi. Co-ordinate of point A is x = 4.
Co-ordinate of point B is y = -8.
Since, 4 > -8
∴ d(A, B) = 4 – (-8)
= 4 + 8
∴d(A, B) = 12

Question 3.
From the information given below, find which of the point is between the other two. If the points are not collinear, state so.
i. d(P, R) = 7, d(P, Q) = 10, d(Q, R) = 3
ii. d(R, S) = 8, d(S, T) = 6, d(R, T) = 4
iii. d(A, B) = 16, d(C, A) = 9, d(B, C) = 7
iv. d(L, M) =11, d(M, N) = 12, d(N, L) = 8
v. d(X, Y) = 15, d(Y, Z) = 7, d(X, Z) = 8
vi. d(D, E) = 5, d(E, F) = 8, d(D, F) = 6
Solution:
i. Given, d(P, R) = 7, d(P, Q) = 10, d(Q, R) = 3
d(P, Q) = 10 …(i)
d(P, R) + d(Q, R) = 7 + 3 = 10 .. .(ii)
∴ d(P, Q) = d(P, R) + d(Q, R) …[From (i) and (ii)]
∴ Point R is between the points P and Q
i. e., P – R – Q or Q – R – P.
∴ Points P, R, Q are collinear.

ii. Given, d(R, S) = 8, d(S, T) = 6, d(R, T) = 4
d(R, S) = 8 …(i)
d(S, T) + d(R, T) = 6 + 4 = 10 …(h)
∴ d(R, S) ≠ d(S, T) + d(R, T) … [From (i) and (ii)]
∴ The given points are not collinear.

iii. Given, d(A, B) = 16, d(C, A) = 9, d(B, C) = 7
d(A, B) = 16 …(i)
d(C, A) + d(B, C) = 9 + 7 = 16 …(ii)
∴ d(A, B) = d(C, A) + d(B, C) …[From(i) and (ii)]
∴ Point C is between the points A and B.
i. e., A – C – B or B – C – A.
∴ Points A, C, B are collinear

iv. Given, d(L, M) = 11, d(M, N) = 12, d(N, L) = 8
d(M, N) = 12 …(i)
d(L, M) + d(N, L) = 11 + 8 = 19 …(ii)
∴d(M, N) + d(L, M) + d(N, L) … [From (i) and (ii)]
∴ The given points are not collinear.

v. Given, d(X, Y) = 15, d(Y, Z) = 7, d(X, Z) = 8
d(X, Y) = 15 …(i)
d(X,Z) + d(Y, Z) = 8 + 7= 15 …(ii)
∴ d(X, Y) = d(X, Z) + d(Y, Z) …[From (i) and (ii)]
∴ Point Z is between the points X and Y
i. e.,X – Z – Y or Y – Z – X.
∴ Points X, Z, Y are collinear.

vi. Given, d(D, E) = 5, d(E, F) = 8, d(D, F) = 6
d(E, F) = 8 …(i)
d(D, E) + d(D, F) = 5 + 6 = 11 …(ii)
∴ d(E, F) ≠ d(D, E) + d(D, F) … [From (i) and (ii)]
∴ The given points are not collinear.

Question 4.
On a number line, points A, B and C are such that d(A, C) = 10, d(C, B) = 8. Find d(A, B) considering all possibilities.
Solution:
Given, d(A, C) = 10, d(C, B) = 8.

Case I: Points A, B, C are such that, A – B – C.

∴ d(A, C) = d(A, B) + d(B, C)
∴ 10 = d(A, B) + 8
∴ d(A, B) = 10 – 8
∴ d(A, B) = 2

Case II: Points A, B, C are such that, A – C – B.

∴ d(A, B) = d(A, C) + d(C, B)
= 10 + 8
∴ d(A, B) = 18

Case III: Points A, B, C are such that, B – A – C.

From the diagram,
d (A, C) > d(B, C)
Which is not possible
∴ Point A is not between B and C.
∴ d(A, B) = 2 or d(A, B) = 18.

Question 5.
Points X, Y, Z are collinear such that d(X, Y) = 17, d(Y, Z) = 8, find d(X, Z).
Solution:
Given,d(X, Y) = 17, d(Y, Z) = 8
Case I: Points X, Y, Z are such that, X – Y – Z.

∴ d(X, Z) = d(X, Y) + d(Y, Z)
= 17 + 8
∴ d(X, Z) = 25

Case II: Points X, Y, Z are such that, X – Z – Y.

∴ d(X,Y) = d(X,Z) + d(Z,Y)
∴ 17 = d(X, Z) + 8
∴ d(X, Z) = 17 – 8
∴ d(X, Z) = 9

Case III: Points X, Y, Z are such that, Z – X – Y.

From the diagram,
d(X, Y) > d (Y, Z)
Which is not possible
∴ Point X is not between Z and Y.
∴ d(X, Z) = 25 or d(X, Z) = 9.

Question 6.
Sketch proper figure and write the answers of the following questions. [2 Marks each]
i. If A – B – C and l(AC) = 11,
l(BC) = 6.5, then l(AB) = ?
ii. If R – S – T and l(ST) = 3.7,
l(RS) = 2.5, then l(RT) = ?
iii. If X – Y – Z and l(XZ) = 3√7,
l(XY) = √7, then l(YZ) = ?
Solution:
i. Given, l(AC) =11, l(BC) = 6.5

l(AC) = l(AB) + l(BC) … [A – B – C]
∴ 11= l(AB) + 6.5
∴ l(AB) = 11 – 6.5
∴ l(AB) = 4.5

ii. Given, l(ST) = 3.7, l(RS) = 2.5

l(RT) = l(RS) + l(ST) … [R – S – T]
= 2.5 + 3.7
∴ (RT) = 6.2

iii. l(XZ) = 3√7 , l(XY) = √7,

l(XZ) = l(X Y) + l(YZ) … [X – Y – Z]
∴ 3 √7 ⇒ √7 + l(YZ)
∴ l(YZ)= 3√7 – √7
∴ l(YZ) = 2 √7

Question 7.
Which figure is formed by three non-collinear points?
Solution:
Three non-collinear points form a triangle.

Class 9 Maths Digest

• Practice Set 1.1 Class 9 Answers
• Practice Set 1.2 Class 9 Answers
• Practice Set 1.3 Class 9 Answers
• Problem Set 1 Class 9 Answers
• Practice Set 2.1 Class 9 Answers
• Practice Set 2.2 Class 9 Answers
• Problem Set 2 Class 9 Answers

Multiplication and Division Class 5 Problem Set 14 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 4 Multiplication and Division

Question 1.
Multiply the following:

(1) 327 × 92
Solution:
3 2 7
x
9 2
_____
6 5 4
+
2 9 4 3 0
Answer:
3 0 0 8 4

(2) 807 × 126
Solution:
8 0 7
x
1 2 6
______
4 8 4 2
+
1 6 1 4 0
+
8 0 7 0 0
Answer:
1 0 1 6 8 2

(3) 567 × 890
Solution:
5 6 7
x
8 9 0
______
0 0 0
+
5 1 0 3 0
+
4 5 3 6 0 0
Answer:
5 0 4 6 3 0

(4) 4317 × 824
Solution:
4 3 1 7
8 2 4
1 7 2 6 8
+ 8 6 3 4 0
3 4 5 3 6 0 0
Answer:
3 5 5 7 2 0 8

(5) 6092 × 203
Solution:
6 0 9 2
x
2 0 3
______
1 8 2 7 6
+
0 0 0 0 0
+
1 2 1 8 4 0 0
Answer:
1 2 3 6 6 7 6

(6) 1177 × 99
Solution:
1 1 7 7
x
9 9
1 0 5 9 3
+
1 0 5 9 3 0
Answer:
1 1 6 5 2 3

(7) 456 × 187
Solution:
4 5 6
x
1 8 7
3 1 9 2
+
3 6 4 8 0
+
4 5 6 0 0
Answer:
8 5 2 7 2

(8) 6543 × 79
Solution:
6 5 4 3
x
7 9
5 8 8 8 7
+
4 5 8 0 1 0
Answer:
5 1 6 8 9 7

(9) 2306 × 832
Solution:
2 3 0 6
x
8 3 2
______
4 6 1 2
+
6 9 1 8 0
+
1 8 4 4 8 0 0
______
Answer:
1 9 1 8 5 9 2

(10) 6429 × 509
Solution:
6 4 2 9
x
5 0 9
______
5 7 8 6 1
+
0 0 0 0 0
+
3 2 1 4 5 0 0
Answer:
3 2 7 2 3 6 1

(11) 4,321 × 678
Solution:
4 3 2 1
x
6 7 8
_____
3 4 5 6 8
3 0 2 4 7 0
2 5 9 2 6 0 0
_______
Answer:
2 9 2 9 6 3 8

(12) 20,304 × 87
Solution:
2 0 3 0 4
x
8 7
1 4 2 1 2 8
+
1 6 2 4 3 2 0
_________
Answer:
1 7 6 6 4 4 8

Question 2.
As part of the ‘Avoid Plastic’ campaign, each of 745 students made 25 paper bags. What was the total number of paper bags made ?
Solution:
7 4 5 Number of students
x
2 5 bags made by each
___________
3 7 2 5
+
1 4 9 0 0
__________
1 8 6 2 5
__________

Answer:
18,625 bags made

Question 3.
In a plantation, saplings of 215 medicinal trees have been planted in each of the 132 rows of trees. How many saplings are there in the plantation altogether ?
Solution:
2 1 5 Saplings in each now
x
1 3 2 Number of rows
__________
4 3 0
+
6 4 5 0
+
2 1 5 0 0
__________
2 8 3 8 0
__________

Answer:
Altogether there is 28,380 saplings.

Question 4.
One computer costs 27,540 rupees. How much will 18 such computers cost ?

Solution:
2 7 5 4 0 Cost of 1 computer
x
1 8 No. of computers
__________
2 2 0 3 2 0
+
2 7 5 4 0 0
___________
4 9 5 7 2 0
__________

Answer:
4,95,720 rupees cost of 18 computers.

Question 5.
Under the ‘Inspire Awards’ scheme, 5000 rupees per student were granted for the purchase of science project materials. If 154 students in a certain taluka were covered under the scheme, find the total amount granted to that taluka.

Solution:
₹ 5 0 0 0 Granted per student
x 1 5 4 Number of students
__________
2 0 0 0 0
+
2 5 0 0 0 0
+
5 0 0 0 0 0
______________
₹ 7 7 0 0 0 0
______________

Answer:
7,70,0000 granted totally

Question 6.
If a certain two-wheeler costs 53,670 rupees, how much will 35 such two-wheelers cost ?
Solution:
5 3 7 6 0 Cost of 1 two-wheeler
x 3 5 No. of two-wheelers
___________
2 6 8 8 0 0
+
1 6 1 2 8 0 0
_____________
1 8 8 1 6 0 0
______________

Answer:
18,81,600 is the total cost of 35- two-wheelers

Question 7.
One hour has 3,600 seconds. How many seconds do 365 hours have ?
Solution:
3 6 0 0 Seconds of 1 hour
x
3 6 5 No. of hours
_________
1 8 0 0 0
+
2 1 6 0 0 0
+
1 0 8 0 0 0 0
______________
1 3 1 4 0 0 0
______________

Answer:
13,14,000 seconds for 365 hours.

Question 8.
Frame a multiplication word problem with the numbers 5473 and 627 and solve it.

Solution:
Cost of one mobile is 5,473. What is the cost of such 627 mobiles?
5 4 7 3 Cost of 1 mobile
x 6 2 7 Number of mobiles
__________
3 8 3 1 1
+
1 0 9 4 6 0
+
3 2 8 3 8 0 0
_____________
3 4 3 1 5 7 1
_____________

Answer:
34,31,571 cost for 627 mobiles.

Question 9.
Find the product of the biggest three-digit number and the biggest four-digit number.

Solution:
9 9 9 9 Biggest four digit no.
x 9 9 9 Biggest three-digit no.
_________
8 9 9 9 1
+
8 9 9 9 1 0
+
8 9 9 9 1 0 0
_____________
9 9 8 9 0 0 1
_____________

Answer:
99,89,001

Question 10.
One traveller incurs a cost of 7,650 rupees for a certain journey. What will be the cost for 26 such travellers?
Solution:
7 6 5 0 Cost of one traveller
x 2 6 No. of travellers
______
4 5 9 0 0
+
1 5 3 0 0 0
_____________
1 9 8 9 0 0
_____________

Answer:
1,98,900 cost of 26 travellers.

Pairing off objects from two groups in different ways

(1) Ajay wants to travel light. So he took with him three shirts – one red, one green and one blue and two pairs of trousers – one white and one black. How many different ways does he have of pairing off a shirt with trousers?

Writing ‘S’ for shirt and ‘T’ for trousers, the possible different pairs are :

(2) Suresh has three balls of different colours marked A, B and C and three bats marked P, Q and R. He wishes to take only one bat and one ball to the playground. In how many ways can he pair off a ball and a bat to take with him?

How many different pairs have been shown here?

(3) The three friends, Sanju, John and Ali went to the fair. A shop there, had five different types of hats. Each of the boys had photos taken of himself, wearing every type of hat, in turn. Find how many photographs were taken in all.

How many different pairs were formed ? That is, how many photos were taken ?

Take two collections, each containing the given number of objects. Make as many different pairs as possible, taking one object from each collection every time. Thus, complete the table below.

What does this table tell us ?
The number of different pairs formed by pairing off objects from two groups is equal to the product of the number of objects in the two groups.

Division
Teacher : You have learnt some things about division. For example, we know that division means making equal parts of a given number, or, subtracting a number repeatedly from a given number. What else do you know ?
Shubha : We know that we get two divisions from one multiplication. From 9 × 4 = 36, we get the divisions 36 ÷ 4 = 9 and 36 ÷ 9 = 4.
Teacher : Very good! Right now, there’s nothing new to learn about division. Only the number of digits in the dividend and the divisor will grow. Tell me what is 354 ÷ 6 ?
Sarang : 354 = 300 + 54. 300 divided by six is 50. And 54 ÷ 6 = 9. Hence the quotient is 50 + 9 = 59.
Teacher : Right! Now let’s learn, step by step, how to divide a four-digit number by a one-digit number. So now, divide 4925 by 7 and tell me the quotient and the remainder.
Shubha : We cannot divide 4 thousands by 7 into whole thousands. Now, 4 Th = 40 H. So let us instead take the 40 hundreds together with 9 hundreds and divide 49 hundreds. 49 ÷ 7 = 7. So, everyone gets 7 hundreds. Now, we cannot divide 2T equally among 7 people. So we must write 0 in the tens place in the quotient. Then on dividing 25 by seven, we get quotient 3 and the remainder is 4. Thus, the answer is quotient 703, remainder 4.

Teacher : Very good ! Now divide 7439 by 9.
Sarang : It’s difficult to do this mentally. I’ll write it down on paper. The quotient is 826 and the remainder, 5.
Teacher : We use the same method to divide a four-digit number by a two-digit number. If necessary, we can prepare the table of the divisor before we start.

Study the solved examples shown below.
Example (1)

Quotient 170, Remainder 4

Example (2)

Quotient 305, Remainder 23

Example (3) Divide. 9842 ÷ 45

We can prepare the 45 times table to do this division.
But when the divisor is a big number, we can solve the example by first guessing what the quotient will be. Let us see how to do that.
We have 0 in the thousands place in the quotient.
Now, to guess the quotient when dividing 98 by 45, look at the first digits in both – the dividend and the divisor. These are 9 and 4, respectively.
Dividing 9 by 4, we will get 2 in the quotient. Let us see if 2 times 45 can be subtracted from 98. 45 × 2 = 90. 90 < 98. So, we write 2 in the hundreds place in the quotient.
Next, dividing 84 by 45 we can easily see that as 90 > 84, we have to write 1 in the tens place in the quotient.
Now, we have to divide 392 by 45. As 3 < 4, let us look at 39, the number formed by the first 2 digits, to guess the next digit in the quotient.
4 × 9 = 36 and 36 < 39. Let us check if the next digit can be 9. 45 × 9 = 405 and 405 > 392. Therefore, 9 cannot be the next digit in the quotient.
Let us check for 8. 45 × 8 = 360. 360 < 392. So, we write 8 in the units place of the quotient.
We subtract 8 × 45 from 392 and complete the division.
The quotient is 218 and the remainder, 32.

Example (4)
If 35 kilograms of wheat cost 910 rupees, what is the rate of wheat per kg?

Weight of wheat in kg × rate of wheat per kg = cost of wheat Hence, 35 × rate per kg = 910
Therefore, when we divide 910 by 35, we will get the per kg rate of wheat.
The rate per kilogram of wheat is 26 rupees.

Multiplication and Division Problem Set 14 Additional Important Questions and Answers

Multiply the following:

(1) 2132 x 231
Solution:
2 1 3 2
x
2 3 1
2 1 3 2
+
6 3 9 6 0
+
4 2 6 4 0 0
____________
Answer:
4 9 2 4 9 2

(2) 1863 x 432
Solution:
1 8 6 3
x
4 3 2
3 7 2 6
+
5 5 8 9 0
+
7 4 5 2 0 0
___________
Answer:
8 0 4 8 1 6

Solve the following word problems:

(1) A factory manufactures 34,796 pairs of socks in one hour. How many pairs will the factory manufacture in one day?
Solution:
3 4 7 9 6 Pairs of socks
x 2 4 No. of hours
______
1 3 9 1 8 4
+
6 9 5 9. 2 0
_____________
8 3 5 1 0 4
_____________

Answer:
8,35,104 pairs of socks manufactured in one day

(2) There are 375 toffees in a box. How many toffees will be there in 632 such boxes?
Solution:
3 7 5 No. of toffees
x 6 3 2 No. of boxes
_______
7 5 0
+
1 1 2 5 0
+
2 2 5 0 0 0
___________
2 3 7 0 0 0
_____________

Answer:
There will be 2,37,000 toffees.

(3) There are 144 articles in a gross. How many articles are there in 2174 gross?
Solution:
2 1 7 4 No. of gross
x 1 4 4 Articles in I gross
______
8 6 9 6
+
8 6 9 6 0
+
2 1 7 4 0 0
_____________
3 1 3 0 5 6
_____________

Answer:
There are 3,13,056 articles in 2174 gross.

Class 5 Maths Solution Maharashtra Board

• Multiplication and Division Problem Set 14 Class 5 Maths Solutions
• Multiplication and Division Problem Set 15 Class 5 Maths Solutions
• Multiplication and Division Problem Set 16 Class 5 Maths Solutions

Balbharti Maharashtra State Board Class 10 English Solutions Unit 1.3 On Wings of Courage Notes, Textbook Exercise Important Questions and Answers.

Class 10 English Chapter 1.3 Question Answer Maharashtra Board

On Wings of Courage Poem 10th Std Question Answer

Question 1.
The ranks of officers in Indian Army, Navy and Air Force are jumbled up. Discuss with your group and put them in the appropriate boxes.

Commander, Brigadier, Wing-Commander, Vice-Admiral, Squadron-Leader, Major, Colonel, Field Marshal, Air Marshal, Admiral of Fleet, Lieutenant-General, Flying Officer, Commodore, Rear Admiral, Air-Commodore.

ARMY	NAVY	AIR FORCE

Answer:

Army	Navy	Air Force
Brigadier,	Commander,	Wing-
Major, Colonel,	Vice-Admiral,	Commander,
Field Marshal,	Admiral	Squadron-
Lieutenant-	of Fleet,	Leader, Air
General	Commodore,	Marshal, Flying
	Rear Admiral	Officer, Air-Commodore

Question 2.
Homophones/ Homographs
(A) Make sentences to bring out the difference between-
(1) (a) wear ……………………………………..
(b) ware ……………………………………..
(2) (a) here ……………………………………..
(b) hear ……………………………………..
(3) (a) there ……………………………………..
(b) their ……………………………………..
(4) (a) cell ……………………………………..
(b) sell ……………………………………..
Answer:
(1) (a) wear: The little girl wanted to wear a pink, frilly dress.
(b) ware: The silver ware laid out on the King’s table was exquisite.

(2) (a) here: “You must sit here,” said the man to his guest.
(b) hear: The children could hear the sound of the planes quite clearly.

(3) (a) there: “I had kept my bag there,” said the woman to the policeman.
(b) their: The girls picked up their bags and went home.

(4) (a) cell: The prisoner sat in the dark cell without talking.
(b) sell: The hawker wanted to sell all his wares before evening.

(B) Write what the underlined Homographs in the following sentences mean.
(1) (a) A bear is an omnivorous animal. ……………………………………..
(b) She could not bear the injustice. ……………………………………..
(2) (a) A bat is the only bird which is a mammal. ……………………………………..
(b) His bat broke as it struck the ball. ……………………………………..
(3) (a) He had to pay a fine for breaking the traffic signal. ……………………………………..
(b) Use a fine cloth for the baby’s clothes. ……………………………………..
(4) (a) We enjoyed a lot at the temple fair. ……………………………………..
(b) She has a fair complexion. ……………………………………..
Answer:
(1) (a) A bear is an omnivorous animal.
bear – a large, heavy animal
(b) She could not bear the injustice,
bear – to tolerate

(2) (a) A bat is the only bird which is a mammal.
bat – a mammal that flies
(b) His bat broke as it struck the ball.
bat – a wooden implement used for hitting the ball in many games.

(3) (a) He had to pay a fine for breaking the traffic signal.
fine – penalty
(b) Use a fine cloth for the baby’s clothes,
fine – delicate, soft

(4) (a) We enjoyed a lot at the temple fair.
fair – a gathering of stalls and amusements for public entertainment
(b) She has a fair complexion, fair – light, not dark

Maharashtra Board Class 10 English Kumarbharati Unit 1.3 Questions and Answers

Question 1.
Read the text and fill in the flow chart of the promotions received by Arjan Singh.

Answer:

Question 2.
With the help of facts given in the text prepare a Fact file of Air Marshal Arjan Singh.
(a) Date of Birth
(b) Place of Birth
(c) Education
(d) First Assignments
(e) Important posts held
(a) In Air Force
(b) After retirement
(f) Awards
(g) Most outstanding contribution in IAF
(h) Retirement
Answer:
(a) Date of birth: April 15, 1919
(b) Place of birth: Lyalpur
(c) Education: at Montgomery; Empire Pilot Training Course at RAF (Cranwell)
(d) First Assignment: to fly Westland Wapiti biplanes in the North-Western Frontier Province as a member of the No. 1 RIAF Squadron
(e) Important posts held:
(1) In Air Force: Member of No. 1. RIAF, Flying Officer, Squadron Leader, Wing Commander, Group Captain, Air Commodore, Air Officer Commanding, Air Vice Marshal, Air Officer Commanding-in-Chief, Deputy Chief of Air Staff, Vice Chief of Air Staff, Chief of Air Staff, Air Chief Marshal.
(2) After retirement: Ambassador to Switzerland Lieutenant Governor of Delhi
(f) Awards: Distinguished Flying Cross (1944); Padma Vibhushan
(g) Most outstanding contribution in IAF: Transforming the IAF into one of the most potent air forces globally and the fourth biggest in the world.
(h) Retirement: in August 1969.

Question 3.
Fill in the web.

Answer:
(1) Singh had successfully led a young IAF during the 1965 Indo-Pak war.
(2) Singh played a major role in transforming the IAF into one of the most potent air forces globally and the fourth biggest in the world.
(3) Singh was honoured with the rank of Marshal on the Republic Day in 2002.
(4) Singh’s contribution was most outstanding during the 1965 Indo-Pak war.

Question 4.
Say what actions preceded the following promotions of Arjan Singh in his career in the IAF.
(a) Selected for Empire Pilot training course at RAF
(b) Promoted to Squadron Leader
(c) Leader of a flypast of over 100 aircraft at Red Fort, Delhi
(d) Awarded Padma Vibhushan
(e) First Air Chief Marshal of Indian Air Force
Answer:
(a) The authorities selected Singh for the Empire Pilot training course.
(b) He flew against the tribal forces and moved back to No. 1 Squadron as a Flying Officer to fly the Hawker Hurricane.
(c) On 15th August 1947, Arjan Singh achieved the unique honour of leading a fly-past of over a hundred IAF aircraft over the Red Fort in Delhi.
(d) He was awarded the Padma Vibhushan for his astute leadership of the Air Force and for inspiring the IAF to victory in the 1965 Indo-Pak war.
(e) He was a source of inspiration to all the personnel of the Armed Forces through the years.

Question 5.
Replace the underlined words/phrases with the appropriate ones, to retain the proper meaning.
(be the epitome of, gear up, a brief stint, play a major role, in recognition of, take over reins)
(a) He contributed notably in bringing up the school.
(b) Our school cricket team got ready for the final match against P. Q. R. High School.
(c) After a short period of working as a lecturer, Ravi took up an important post in a multi-national company.
(d) Our class monitor is a perfect symbol of duty and discipline.
(e) Accepting the great value of his research; they awarded him with a Ph.D. (degree)
(f) After the murder of King Duncan, Macbeth took over the control of Scotland.
Answer:
(a) He played a major role in bringing up the school.
(b) Our school cricket team geared up for the final match against P.Q.R.High School.
(c) After a brief stint as a lecturer, Ravi took up an important post in a multinational company.
(d) Our class monitor is the epitome of duty and discipline.
(e) In recognition of his research, they awarded him with a Ph.D. (degree)
(f) After the murder of King Duncan, Macbeth took over the reins of Scotland.

Question 6.
Build the word wall with the words related to ‘Military’.

Answer:

Question 7.
(A) State the different meanings of the following pairs of Homophones and make sentences of your own with each of them.

Word	Meaning	Sentence
(a) led lead(b) role roll(c) air heir(d) feat feet(e) reign rein rain	………………………….. ………………………………………………………… ……………………………………………………… ………………………………………………………. ………………………………………………………. …………………………..	………………………….. ………………………………………………………. ………………………………………………………. ………………………………………………………. ………………………………………………………. …………………………..

Answer:

Word	Meaning	Sentence
(a) led	past participle of lead (to guide or conduct)	The captain led his team to safety.
lead	graphite used as part of a pencil	Do you have a lead pencil?
(b) role	a part (in a play, film, etc.)	Marie got the leading role in the new movie.
roll	move in a particular direction by turning over and over	The boy wanted to roll in the mud while playing.
(c) air	the invisible gaseous substance surrounding the earth	There Is a lot of humidity in the air during the monsoon.
heir	successor or inheritor	The family did not know who the heir to the property was.
(d) feat	a great achievement	Climbing Mt. Everest is a feat.
feet	a unit of measurement	The girl saw to her shock that the lion was only a few feet away.
(e) reign	rule as king or queen	Queen Elizabeth’s reign has been a long one.
rein	a restraining influence	The new manager kept a tight rein on her employees.
rain	water that falls In drops from clouds in the sky	Children love to play in the rain.

(B) The following Homographs have the same spelling and pronunciation but can have different meanings. Make sentences of your own to show the difference.

Answer:
(a) firm: (i) My neighbour recently Joined an electronics firm as Sales Executive.
(ii) Many people feel that they must be firm with their children when they are growing.

(b) train: (i) The train left from platform 2 at seven p.m. sharp.
(ii) You must always train your pets to obey you.

(c) type: (i) The man asked his secretary to type the letter immediately.
(ii) Cows eat only a particular type of grass.

(d) post: (i) My aunt quit her job because she felt that the post was not suitable for her.
(ii) The little boy ran to the post office to post the letter to Santa Claus.

(e) current : (i) The minister was disturbed when he read about the current situation of unrest In the country.
(ii) It is a difficult task to row against the current in a river.

Question 8.
Glance through the text and prepare notes from the information that you get. Take only relevant points. Don’t use sentences. Arrange the points in the same order. You may use symbols or short forms. Present the points sequentially. Use highlighting techniques.
Answer:
Air Force Marshal Arjart Singh—Icon of India’s Military History

1. Date of Birth: 15 April, 1919
2. Qualifications: Empire Pilot Training Course at RAF (Cranwell)
3. Responsibilities:

• first assignment to fly Westland Wapiti biplanes in No.l RIAF Squadron
• brief stint in No.2 RIAF Squadron; moved back to No. 1 RIAF Squadron as Flying Officer
• overall commander of ‘Shiksha’
• led the IAF during the 1965 Indo-Pak war
• led a squadron against the Japanese during the Arakan Campaign; assisted the advance of Allied Forces to Yangoon
• led a fly-past on August 15, 1947
• commanded Ambala in the rank of Group Captain; took over as AOC of an operational command
• took over reins of the IAF
• ambassador to Switzerland; Lieutenant Governor of Delhi

(4) Achievements:

• selected for the Empire Pilot Training Course at RAF (Cranwell) in 1938, at age 19
• promoted to the rank of Squadron Leader in 1944
• led a fly-past over the Red Fort on August 15, 1947
• promoted to the rank of Wing Commander; promoted to the rank of Air Commodore in 1949
• longest tenure as AOC (1949-1952 and 19571961)
• appointed as Deputy Chief of Air Staff at the end of the 1962 war; appointed as Vice Chief of Air Staff in 1963
• rank of Air Marshal in August 1964; took over reins of IAF
• successfully led the IAF in 1965 Indo-Pak war
• promoted as Air Vice Marshal; appointed as AOC-in-C of an operational command
• first Air Chief to keep his flying currency till his CAS rank; has flown more than 60 different types of * aircraft
• first and only Air Chief Marshal of the IAF

(5) Awards:

– Distinguished Flying Cross (1944)
– Padma Vibhushan

(6) After retirement: Ambassador to Switzerland; Lieutenant Governor of Delhi
(Students can put these points attractively in boxes and use highlighting techniques.)

Question 9.
Develop a story suitable to the conclusion/end given below. Suggest a suitable title.
………………………………………………….. (Title)
…………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………..
………………………………………………………….. and so, with tears of joy and pride, the 10 year old Sanyogita More received the National Bravery Award from the Prime Minister.
Answer:
A WONDERFUL ACT OF BRAVERY
It was the 26th of July in Mumbai. It was raining cats and dogs. Ten-year-old Sanyogita More stood at the door of her hut. The street was flooded with water. Sanyogita was frightened. Her parents had not returned from work and she was all alone.

Suddenly, she saw two little boys, Rohan and Sohan, come out from the neighbouring hut to play in the water. As Sanyogita watched, there came a sudden gush of water and the boys were dragged towards an open manhole, which had been marked with a pole. They caught hold of the pole, but the pole began to tilt. It would soon fall—and the boys would go down the manhole!

Sanyogita ran as fast as she could towards the boys. Pulling a rope from a nearby door, she looped it around a large stone. She held onto the rope and extended her hand towards the boys. “Catch my hand, Sohan, Rohan,” she shouted. “Catch! Catch soon!”

The boys were in a panic but they did as they were told. Sohan held Rohan’s leg, Rohan held Sanyogita’s hand, and Sanyogita held onto the rope.

“Help! Help! she shouted, knowing that if the rope broke or the stone was dislodged, they would all go into the manhole.

She stood there shivering, her arms numb, for nearly 15 minutes before help arrived. Sanyogita collapsed after the incident. The news of her brave deed spread far and wide, and reached the ears of 1 the Prime Minister, who decided to honour her with an award. And so, with tears of joy and pride, the 10- I year-old Sanyogita More received the National Bravery ‘ Award from the Prime Minister.

Question 10.
You wish to join any one of the Indian Armed Forces. Fill in the following application form.
To
The Advertiser
N/AF Recruitment Service
Purangaon – 456 789

Affix recent
passport size
photograph

Application For Recruitment
Rect notice No 1234

1. Post applied for
2. Name and surname of Candidate (in Block letters)
3. Father’s Name ………………………………… Mother’s Name …………………………………
4. Date of Birth
5. Contact details :
Tel. No. (Res) ………………….. . Mobile No.
Email ID ………………….. .
6. Permanent Address :
House No./Street/Village ………………….. .
Post Office ………………….. .
District ………………….. State ………………….. .
Pincode ………………….. .
7. Educational Qualifications :

Serial Number	Qualification	Name of School/College	Name of Board/University	Percentage obtained

8. Whether registered at any employment exchange Yes/ No ………………….. (If yes, mention registration number and the name of the Employment Exchange.)

9. Outstanding achievements in extra-curricular activities/ sports/ games, etc.
………………………………………………………………………………………………………………………………………….
………………………………………………………………………………………………………………………………………….

10. Why you wish to join Armed Forces. …………………………………………………………………
………………………………………………………………………………………………………………………………………….
………………………………………………………………………………………………………………………………………….

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• The Thief’s Story Question Answers
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Balbharti Maharashtra State Board Class 10 History Solutions Chapter 5 Mass Media and History Notes, Textbook Exercise Important Questions and Answers.

Std 10 History Chapter 5 Question Answer Mass Media and History Maharashtra Board

Class 10 History Chapter 5 Mass Media and History Question Answer Maharashtra Board

History Class 10 Chapter 5 Question Answer Maharashtra Board

Mass Media And History Class 10 Question 1.
(A) Choose the correct option from the given options and complete the statement.

(1) The first English newspaper in India was started by ………………………….. .
(a) James Augustus Hickey
(b) John Marshall
(c) Allen Hume
Answer:
(a) James Augustus Hickey

(2) Television is an ………………………….. medium.
(a) visual
(b) audio
(c) audio-visual
Answer:
(c) audio-visual

(B) Identify and write the wrong pair in the following set.
(1) ‘Prabhakar’ – Acharya P.K. Atre
(2) ‘Darpan’ – Balshastri Jambhekar
(3) ‘Deenbandhu’ – Krishnarao Bhalekar
(4) ‘Kesari’ – Bal Gangadhar Tilak
Answer:
(1) Wrong Pair: ‘Prabhakar’ – Acharya P.K. Atre

Mass Media And History Class 10 Question 2.
Write short notes :
(1) The role of newspaper in the Indian struggle for independence
Answer:
Newspapers played an important role in the Indian independence struggle. It is as follows

• Newspapers served as an important medium to create awareness during those times.
• They described greatness of Indian culture and history to gather support of masses for the freedom movement.
• They supported social, political and religious movements and opposed imperialism.
• They discussed various social and political issues.
• The ideas of social reformers and various organisations active in independence struggle reached people through newspapers.

(2) Why do we need mass media?
Answer:
Mass media includes print and electronic and various new media.

• It facilitated free flow of information to all strata of the society and brought the world closer.
• Editorials, various columns and supplements are essential parts of newspapers.
• Readers also get a platform to voice their opinions. In fact, newspapers can help to make democracy stronger.
• Akashrani broadcasted various programmes of the government as well entertainment.
• Awareness creating programmes. It fulfill the need of the government to connect with people.
• Television is an Audio-Visual medium which has made it possible to cross the inherent limitations of newspapers.
• Radio to show the actual visuals of an event to people.
• Mass Media is very important as it plays an important role to strengthen democracy.

(3) Mass Media and professional opportunities.
Answer:
There are many professional opportunities available in printed, electronic and digital media.

• Writers, columnists, editors are required to write articles, columns and editorials in news-papers.
• Newspapers also require reporters to gather news and technicians to work in the press.
• There is requirement of actors and technicians in electronic media.
• Artists are required to present programmes on television, in the same way news presenters, anchors are required.
• If the articles, columns and programmes are based on history, an expert in history is required.

Mass Media And History Class 10 Question Answer Question 3.
Explain the following statements with reasons.
(1) Any information received through mass media needs to be reviewed critically.
Answer:

• Information provided in the media may not represent the exact truth. We need to scan it carefully.
• We have to understand idealistic and investigative motives of newspapers, government policies and prevailing social conditions behiid the newspiece.
• The information received through Mass Media might be prejudiced or give a one-sided idea.
• ‘Stern’, a German weekly magazine, purchased and published a number of so called handwritten diaries of Hitler.
• It then sold them to a number of publication companies.
• However, later it was proved that those diaries were forged. Hence it is essential to verily the information received through Mass Media.

(2) Knowledge of history is essential for newspaper articles.
Answer:

1. In order to unfold the background of an event in the news, we have to resort to history.
2. Some columns are based on historical events. These columns provide historical information about economical, social and political events in the past.
3. Newspapers publish supplements in addition to the regular edition or special issues to mark the completion of 50 or 100 years of an event. On such occasions, one has to review history of that particular event.
4. Even while writing columns like what happened in history on this day it is necessary to know past event. Hence, the knowledge of history is essential for writings of such type.

(3) Television is the most popular medium.
Answer:

• Television being an audio-visual medium brings us into contact with events in an exciting and clarifying way.
• It crossed the inherent limitations of newspapers and radio to show actual visuals.
• It becomes possible for people to watch all the national and international events sitting at home.
• Social problems, discussion on education and economics and political events are viewed by people.
• In 1991, Indian government granted permission to private, national and international channels to telecast in India.
• Television became a treasure house of entertainment.

Therefore, the television is the most popular medium.

Class 10 History Chapter 5 Questions And Answers Ssc Board Question 4.
Read the following extract and answer the questions.
Radio: ‘Indian Broadcasting Company’ (IBC), a private radio company was the first one to broadcast daily programmes. Later the same company was taken over by the British Government and named as, ‘Indian State Broadcasting Service (ISBS). On 8th June 1936 it was renamed, as ‘All India Radio (AIR)’.

After Independence, AIR became an integral part of the Ministry of Information and Broadcasting (India). Initially, it broadcasted Governmental programmes and schemes. It was named as ‘Akashvani’ on the suggestion of the famous poet Pandit Narendra Sharma. Akashvani broadcasts various entertainment, awareness creating and literary programmes. It also broadcasts special programmes for farmers, workers, the youth and women. The ‘Vividh Bharati’ programmes are broadcasted in 24 regional languages as well as 146 dialects of Indian languages. Lately, various new channels like ‘Radio Mirchi’ are providing radio services.

(1) Akashavani (AIR) is an integral part of which ministry?
Answer:

1. Indian Broadcasting Company, a private radio company was taken over by the British Government in 1927 and named ‘Indian State Broadcasting Service (ISBS)’. On 8th June 1936, it was renamed as ‘All India Radio (AIR)’.
2. AIR became integral part of the Ministry of Information and Broadcasting after independence. It was renamed Akashvani on the suggestion of Pandit Narendra Sharma.
3. Initially it used to broadcast Government’s programmes and schemes. Later it started broadcasting various entertainment, awareness creating and literary programmes.
4. Akashvani started ‘Vividh Bharati1 programmes. It broadcasts special programmes for farmers, workers, the youth and women.
5. Vividh Bharati Programmes are broadcast in 24 regional languages and 146 dialects.

(2) What was the new name of IBC?
Answer:
Indian Broadcasting Company (IBC) was taken over by the British Government. It was named as the Indian State Broadcasting Services. (ISBS). On 8th June 1936, it was renamed as ‘All India Radio’ (AIR).

(3) In how many regional languages and local dialects are ‘Vividh Bharati’ programmes broadcasted?
Answer:
People get access to news through social media like Twitter, Instagram, Facebook, YouTube and from web news portals, web channels. This information is available in English and many other languages.

(4) How AIR was named ‘Akashavani’?
Answer:
AIR was named as Akashvani on the suggestion of the famous poet Pandit Narendra Sharma.

Question 5.
Complete the following concept chart.

	Newspapers	Radio	Television
Beginning/Background
Nature of information/programmes
Functions

Answer:

	Newspapers	Radio	Television
Background/Beginning	James Augustus Hickey started Calcutta General Advertiser or Bengal Gazette on 29th January, 1780. It was the first newspaper in English.	A private radio station was started known as Indian Broadcasting Company.	First Doordarshan centre was started in Delhi.
Information Programmes	News, articles, columns, opinions of the people, editorials, advertisements etc. are published.	Along with entertainment programmes, have literary, informative programmes on farmers, women and educative values.	Events around the world, movies, music, information about environmental and historical places, sports are shown either live or recorded.
Functions	(1) Report daily news (2) Public awareness and mass education. (3) Provide information and strengthen democracy. (4) Oppose injustice and give publicity to developmental work.	(1) Provide news from different sectors. (2) Entertain through music, dramas, songs, etc. (3) Present social problems and educate the masses about it. (4) Conduct discussions on various issues ranging from the environment to culture.	(1) Telecast daily events and entertain. (2) Educate the masses. (3) Publicise programmes which are for social benefit. (4) Bring about social awakening opposing evil traditions and practices.

Project
Write a review of a historical serial that you have watched.

Memory Map

Question 6.
Complete the sentences by choosing a correct option:
(a) ………………….. is the first newspaper in Marathi.
(a) Deenbandhu
(b) Prabhakar
(c) Darpan
(d) Kesari
Answer:
(c) Darpan

(b) 6th January is celebrated as ………………….. day in Maharashtra.
(a) Periodical Day
(b) Newspaper Day
(c) Printing Day
(d) Journalist Day
Answer:
(d) Journalist Day

(c) The letters ‘Shatpatre1 published in Prabhakar were written by ………………….. .
(a) Lokmanya Tilak
(b) Lokhitvadi
(c) Mahatma Gandhi
(d) Justice Ranade
Answer:
(b) Lokhitvadi.

(d) The honour of printing illustrations for the first time in an Indian newspaper goes to ………………….. .
(a) Dnyanoday
(b) Darpan
(c) Prabhakar
(d) Kesari
Answer:
(a) Dnyanoday

(e) Deenbandhu was started by ………………….. who was a close associate of Mahatma Phule.
(a) Dr. Babasaheb Ambedkar
(b) Lokmanya Tilak %
(c) Narayan Meghaji Lokhande
(d) Krishnarao Bhalekar
Answer:
(d) Krishnarao Bhalekar

(f) ………………….. newspaper was started by Agarkar and Lokmanya Tilak.
(a) Deenbandhu and Induprakash
(b) Darpan and Prabhakar
(c) Dnyanoday and .Digdarshan
(d) Kesari and Maratha
Answer:
(d) Kesari and Maratha

(g) Balshastri Jambhekar started ………………….. the first monthly magazine in Marathi.
(a) Digdarshan
(b) Prabhakar
(c) Darpan
(d) Dnyanoday
Answer:
(a) Digdarshan

(h) ………………….. was acknowledged as the fourth pillar of democracy.
(a) Representatives
(b) Periodicals
(c) Newspaper
(d) Books
Answer:
(c) Newspaper.

(i) The first English news bulletin was broadcast on 23rd July, 1927 from the …………………… radio station.
(a) Kolkata
(b) Madras
(c) Mumbai
(d) Dblhi
Answer:
(c) Mumbai

(j) Dr. Rajendra Prasad, the first President of India inaugurated the …………………… Doordarshan centre.
(a) Mumbai
(b) Bangalore
(c) Lucknow
(d) Delhi
Answer:
(d) Delhi

(k) Newspapers published special supplements or a special issue to commemorate occasions like completion of seventy-five years of …………………… in 2017.
(a) Khilafat Movement
(b) Non Co-operation Movement
(c) Civil Disobedience Movement
(d) Quit India Movement
Answer:
(d) Quit India Movement

(l) ……………………, a German weekly magazine, had purchased a number of so called handwritten diaries that were later proved forged.
(a) Time Magazine
(b) Statesman
(c) Stern
(d) Reuters
Answer:
(c) Stern

(m) Akashvani has preserved recordings of all speeches delivered by the …………………… on 15th August.
(a) President
(b) Wee President
(c) Prime Minister
(d) Army General
Answer:
(c) Prime Minister

(n) Akashwani comes under the Ministry of …………………… of the Indian Government.
(a) Social welfare
(b) Human Resource and Development
(c) Information and Broadcasting
(d) Education Technology
Answer:
(c) Information and Broadcasting.

Question 7.
Identify the wrong pair in the following and write it:

Newspaper	Editor
(1)  Prabhakar (2)  Darpan (3)  Deenbandhu (4)  Kesari	(a)   Acharya R K. Atre (b)   Balshastri Jambhekar (c)  Krishnarao Bhalekar (d)   Bal Gangadhar Tilak

Answer:
Wrong pair: Prabhakar – Acharya R K. Atre

(2)

Newspaper	Issues
(1)  Prabhakar (2)  Induprakash (3)  Deenbandhu (4)  Kesari	(a)   History of French revolution (b)  Advocated widow re­marriage (c)  Information on Telegraph (d)   Voiced social and political problems.

Answer:
Wrong pair Deenbandhu – Information on Telegraph

(3)

Newspaper/Magazine/Book	Editor
(1) Pragati	(a) Tryambak Shankar Shejwalkar
(2) Digdarshan	(b) Narendra Sharma
(3) Deenbandhu	(c) Krishnarao Bhalekar
(4) Discovery of India	(d) Pandit Nehru

Answer:
Wrong pair: Digdarshan – Narendra Sharma

Question 8.
Do as directed:
(a) Complete the graphical description

Answer:

(2)

Answer:

(b) Show the progress of Indian television Time-line:

Answer:

Question 9.
Explain the following concepts:

(a) Electronic or Digital Journalism or Web Journalism.
Answer:

1. In the modem times, the computer and internet have become indispensable parts of printing and publishing process. Computer technology has led to the widespread practice of digital journalism.
2. Websites run by newspapers are basically extensions of newspapers themselves. Modern periodicals are part of electronic or digital journalism.
3. People get access to news through social media like Twitter, Instagram, Facebook, YouTube and from web news portals, web channels. This information is available in English and many other languages.
4. Journalists working in this area today have to have many more skills than in the past when writing was the only requirement. Information available on these mediums should be reviewed critically and used with utmost care.

(b) E-newspapers
Answer:

1. In recent times, e-newspapers have got prominent place in mass media.
2. E-newspaper is not exactly like the printed one. In e-newspapers, news comes in sequence and not based on the nature and the importance of the news, like in printed newspaper e.g.. Front page news. Headline or Last page news.
3. The news which we want to read has to be clicked and then it appears on the screen in detail.
4. There is space provided for opinion of readers. In 1992, the first edition of the e-newspaper was published by ‘Chicago Tribune1.
5. At present, almost all newspapers are available as e-newspapers and people can read them anytime, anywhere using the Internet or computer, tab, laptop or mobile.
6. In recent times many newspapers have introduced e-newspapers. The e-newspapers are being received well by the readers.
7. Learn to read e-newspapers with the help of your teachers.

Question 10.
Write short notes:
(a) Bengal Gazette:
Answer:

• Bengal Gazette is the first newspaper which was started in India.
• It was started by James Augustus Hickey, an Irish national.
• It was first published on 29th January, .1780. It was also called “Calcutta General Advertiser’.
• Bengal Gazette laid the foundation of newspaper in India.

(b) News printed in ‘Darpan’:
Answer:
The ‘Darpan newspaper started by Balshastri Jambhekar printed all types of news like political, economic, social and cultural. Some of them are mentioned below:

• The Accounts of Expenditure from the Three Administrative Divisions of the East India Company.
• The Danger of Russian Attack on the Nation.
• Appointment of a Committee for Cleanliness of the City.
• Remarriage of Hindu Widows.
• The Inception of Theatre at Calcutta.
• Achievements of Raja Ram Mohan Roy in England. All these reports published in the paper throw light on various situations/events of those days.

(c) Television:
Answer:

• The first President of India, Dr. Rajendra Prasad inaugurated Delhi Doordarshan Centre.
• Mumbai Doordarshan started to telecast its programmes on 2nd October, 1972.
• Colour television started on 15th August, 1982. The Indian government granted permission to private, national and international channels in 1991 to telecast in India.

Question 11.
Explain the following sentences with reason:
(a) Newspaper is an important medium of education and information.
Answer:

• Newspapers report events which are interesting to the public. But the importance of newspaper stretches far beyond a passing humari interest in events.
• It covers a miscellany of topical issues. News would involve matters of higher importance like war, global warming, education, national elections or trivial issues such as scandals, gossips and debates on minor controversies.
• Newspapers have contributed significantly to the spread of literacy and the concept of human rights and democratic freedoms.
• They are integral to the development of democracy. In fact, they can help in making the democracy stronger.
• Newspapers not only report the events but continue to shape opinions in the global village.

(b) 6th January is observed as ‘Patrakar Din’ or ‘Journalist Day’ in Maharashtra.
Answer:

• Balshastri Jambhekar started the first newspaper in Marathi on 6th January, 1832 in Mumbai.
• He is referred to as the ‘First Editor’ as he was the first editor.
• He laid the foundation of Marathi newspaper by starting Darpan. As 6th January is his birth date, it is observed as ‘Patrkar Din’ or ‘Journalists’ Day’ in Maharashtra.

(c) Television and history are closely related.
Answer:

1. Television plays a major role in developing interest in history. While producing shows and serials based on history and mythology, it is essential to have an accurate knowledge of history and know the minute details.
2. ‘Bharat Ek Khoj’, Raja Shivchhatrapati, Ramayana, Mahabharata are among the few popular serials based on history and mythology. While producing these serials.
3. It was essential to know the prevalent social conditions, outfits, lifestyle, weaponry, lingual expressions of the people. Historians who had knowledge on these subjects are required.
4. While making programmes, based on sportsmen, literature, war, historical events, forts and animal life, it is important to give history of their development in that particular period.
5. While conducting discussions on television on topics like social problems, education, economics, health, it is important to give references from the past.

This shows that the knowledge of history is required in the making many of programmes on Television. Hence Television and history are closely related.

Question 12.
Answer the following question in 25-30 words:
(a) Explain the objectives of newspapers.
Answer:
The main objectives of newspapers are as follows:

• Newspapers provide various local, national and international news to the people and inform them about daily events.
• They narrate political, economic, cultural and social history of the country.
• Newspapers fulfill their role as the fourth column of democracy by creating public awareness and becoming a medium of mass education.
• They even condemn the anti-social elements in the society and support the weaker section.

(b) How is history helpful in the planning of the Akashvani programmes ?
Answer:
Akashvani broadcasts all types of programmes from celebration of independence day to entertainment programmes. In planning these programmes, the knowledge of history is essential.

1. Akashvani invites historians as experts for discussions while presenting programmes on various occasions such as the anniversaries of births and deaths of national leaders, anniversaries of historical events; speeches of all Prime Ministers/Presidents.
2. Programmes like ‘On This Day in History’ is a daily programme which highlights importance of that day and date in history.
3. Information has to be verified by historians before it reaches the people. Lectures on the contributions of various national leaders need to be supported by historical information. In the following ways history is helpful in the planning of Akashvani programmes.

(c) How were the message conveyed to the people in olden days?
Answer:
The following were a few means used to convey messages to the people in olden days:

• A town crier would run on the streets beating drums and crying out important news according to the orders of the king.
• So, the news would spread among people by word of mouth.
• Inscriptions with royal decrees were placed at public places.

Question 13.
Read the following passages and answer the questions:”
(a) Which programmes are broadcasted by Akashvani?
Answer:

• Initially, Akashvani broadcasted government programmes and schemes.
• Later it broadcasted various entertainment and literary programmes.
• Akashvani presents various programmes for creating awareness.
• Special programmes are also broadcasted for farmers, workers, youth and women.

(a) On which book is the serial ‘Bharat Ek Khoj’ based on?
Answer:
The Serial Bharat Ek Khoj is based on ‘Discovery of India’, a book written by Pandit Jawaharlal Nehru.

(b) Who directed the serial ‘Bharat Ek Khoj’?
Answer:
The serial was directed by Shyam Benegal.

(c) Which factors/aspects of Indian history are depicted in ‘Bharat Ek Khoj’? OR Why was ‘Bharat Ek Khoj’ a serial telecasted by Doordarshan admired in all parts of India?
Answer:
The television serial ‘Bharat Ek Khoj’ presented social, political and economic life from ancient to the modem period in India.

1. It portrayed various aspects of Indian history like Harappan civilisation, Vedic history and the interpretation of epics like Mahabharata and Ramayana.
2. It used the technique of dramatisation effectively to recreate the Mauryan period and show the impact of Turk-Afghan invasions.
3. The Mughal period and their contributions which have long-lasting effect on social and cultural fabric of India is shown. The rise of Bhakti movement, role of Chhatrapati Shivaji Maharaj in getting swarajya is portrayed.
4. The last episodes (finale) of the serial narrate social movements and India’s freedom struggle in modem period.

Thus, the serial effectively portrayed the journey of India from Harappan civilisation to the modern period and therefore was admired in all parts of India.

Question 14.
Answer the following questions in detail:
(a) What were the different means of communication known around the world before the advent of newspaper?
Answer:
The following means of communication were used to convey news before the advent of newspaper:

1. Inscriptions with royal decrees placed at public places was a custom in Egypt. Emperor Ashoka followed the sam method to reach out to his subjects.
2. In the Roman Empire, roytil decrees were written on papers and those were distribute’d in all regions. It also contained information of various events taking place in the nation and its capital.
3. During, the reign of’ Julius Caesar ‘Acta Diurna’, meaning acts of everyday used to be placed at public places in Rome.
4. In the 7th century C.E., in China, royal dictates were distributed among people at public places.
5. In England handouts were distributed occasionally among people at public places giving information about war or any important events.
6. Travellers arriving from different faraway places would add spice to stories from those places and narrate the same to local people. The ambassadors of a king posted at various places would send back important news to the royal court.

(b) Write information on Periodicals based on its types.
Answer:
Magazines and journals which are published at regular intervals are known as Periodicals.
Types:

• They are categorised as weekly, biweekly, monthly, bimonthly, quarterly, six monthly and annual.
• There are some chronicles which are published at no fixed time.

Classification: Periodicals can be classified as popular and scholarly.

• If a periodical aims at specialists and researchers, it is a ‘journal1. Articles are generally written by experts in the subject.
• Popular periodicals are magazines published with variety of content. They can be on fashion, sports, entertainment and films.
• Bharatiya Itihas ani Sanskruti and Marathwada Itihas Parishad Patrika are periodicals of present times. Periodicals are an important source to study history.

(c) Write about the important role of newspaper in the freedom struggle.
Answer:

1. The press was the chief instrument for carrying out the political tasks and propagation of nationalist ideology.
2. Both English and Vernacular press started by prominent “leaders like Gopal Ganesh Agarkar and Lokmanya Tilak acted as catalyst to the freedom struggle. They started ‘Kesari’ and ‘Maratha’ in 1881.
3. Newspapers played a great role in building up an increasingly strong national sentiment and consciousness among people. It was an instrument to arouse, train, mobilise and consolidate nationalist public opinion.
4. The newspapers were an effective tool in the hands of social reformers. They exposed social evils such as child marriage, ban on remarriage of the widows, inhuman institution as untouchability, caste fetters, etc. It became a weapon in their hands to educate masses.
5. A comparative study was presented in newspaper on western education, knowledge and national education.
6. Newspapers also discussed political institutions in India and the west. The main aim of these newspapers was not to gain profit but to serve the people.

(d) Give a short account of the development of Indian television.
Answer:

1. Television service started in India in 1959. Dr. Rajendra Prasad, the first Indian President, inaugurated the Delhi Doordarshan centre.
2. Mumbai centre began to telecast its programmes on 2nd October 1972. Colour television was introduced in India on 15th August 1982.
3. The national telecast began in 1983. Doordarshan started Channels like DD Sports, DD Metro, news, etc. along with 10 regional channels.
4. In the year 1991, the Indian government gave permission to private, national and international channels to telecast in India.
5. Presently there are more than 800 national and regional channels. Some of them are exclusive news, sports, music, movies and religious channels which telecast programmes 24 hours a day.

(e) Distinguish between Newspapers and Magazines.
Answer:
Newspapers and magazines differ in their format, objectives and duration of getting published. The differences are noted below:

Newspapers	Magazines
1. Newspapers document the current events.	1. Magazines give no importance to current news.
2. News, articles, columns, editorials have importance in a newspaper.	2. Magazines give preferences to particular subject and publish articles on it.
3. Newspapers are also called ‘Dailies’ as they are published every day.	3. Duration of publication of magazine varies. Some are published weekly, fortnightly, monthly and annually.
4.. The main purpose of newspaper is to report local, national and international news.	4. Instead of providing news, their content is entertaining and knowledge-based.
5. Newspapers make the people aware of the events happening in the society. They do not stick to any one subject.	5. Magazines are about a specific topic. On the basis of their appearance, size, readability, content and audience, magazines differ from newspapers.
6. Newspapers mostly write about current news. Whatever happens in the world appears in the newspaper within a span of 24 hours. It shapes public opinion and keep people updated about the activities of the government.	6, As magazines have lot of detailed information on specific topics they are considered as historical source.

Brain Teaser

Across:

• Referred to as the ‘First Editor’.
• A newspaper representing masses of the Indian society (Bahujan Samaj).
• Tryambak Shankar Shejwalkar edited this journal.
• Letters by Lokhitvadi.

Down:

• The history of French Revolution was published in this newspaper.
• Newspaper started by James Augustus Hickey.
• First monthly magzine in Marathi.
• Pandit Narendra Sharma suggested this name for AIR.

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