Category: Class 8

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 11.3 8th Std Maths Answers Solutions Chapter 11 Statistics.

Practice Set 11.3 8th Std Maths Answers Chapter 11 Statistics

Exercise 11.3 Class 8 Question 1.
Show the following information by a percentage bar graph.

Division of standard 8	A	B	C	D
Number of students securing grade A	45	33	10	15
Total number of students	60	55	40	75

Solution:

Division of standard 8	A	B	C	D
Number of students securing grade A	45	33	10	15
Total number of students	60	55	40	75
Percentage of students securing grade A	75%	60%	25%	20%
Percentage of students not securing grade A	25%	40%	75%	80%

Statistics for Class 8 Question 2.
Observe the following graph and answer the questions.

1. State the type of the bar graph.
2. How much percent is the Tur production to total production in Ajita’s farm?
3. Compare the production of Gram in the farms of Yash and Ravi and state whose percentage of production is more and by how much?
4. Whose percentage production of Tur is the least?
5. State production percentages of Tur and Gram in Sudha’s farm.

Solution:

1. The given graph is a percentage bar graph.
2. Percent of tur production to the total production in Ajita’s farm is 60%.
3. Production of Gram in the farm of Yash = 50%
   Production of Gram in the farm of Ravi = 30%
   ∴ Difference in the production = 50% – 30% =20%
   ∴ Yash’s production of Gram is more and by 20%.
4. Sudha’s percentage production of Tur is the least.
5. Production percentages of Tur and Gram in Sudha’s farm are 40% and 60% respectively.

8th Standard Statistics Question 3.
The following data is collected in a survey of some students of 10th standard from some schools. Draw the percentage bar graph of the data.

School	1st	2nd	3rd	4th
Inclination towards science stream	90	60	25	16
Inclination towards commerce stream	60	20	25	24

Solution:

School	1st	2nd	3rd	4th
Inclination towards science stream	90	60	25	16
Inclination towards commerce stream	60	20	25	24
Total number of students	150	80	50	40
Percentage of students having inclination towards science stream	60%	75%	50%	40%
Percentage of students having inclination towards commerce stream	40%	25%	50%	60%

Maharashtra Board Class 8 Maths Chapter 11 Statistics Practice Set 11.3 Intext Questions and Activities

Statistics 8th Class Question 1.
Compare and discuss a percentage bar diagram and a subdivided bar diagram. Use it to learn the graphs in the subjects like Science, Geography. (Textbook pg, no. 74)
Solution:
[Students should attempt the above activity on their own.]

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 8.2 8th Std Maths Answers Solutions Chapter 8 Quadrilateral: Constructions and Types.

Practice Set 8.2 8th Std Maths Answers Chapter 8 Quadrilateral: Constructions and Types

Question 1.
Draw a rectangle ABCD such that l(AB) = 6.0 cm and l(BC) = 4.5 cm.
Solution:

Question 2.
Draw a square WXYZ with side 5.2 cm.
Solution:

Question 3.
Draw a rhombus KLMN such that its side is 4 cm and m∠K = 75°.
Solution:

Question 4.
If diagonal of a rectangle is 26 cm and one side is 24 cm, find the other side.
Solution:

Let ₹ABCD be the rectangle.
l(BC) = 24cm, l(AC) = 26cm
In ∆ABC,
m∠ABC = 90° …[Angle of a rectangle]
∴[l(AC)]² = [l(AB)]2 + [l(BC)]²
…[Pythagoras theorem]
∴ (26 )² = [l(AB)]² + (24)²
∴(26)² – (24)² = [l(AB)]²
∴(26 + 24) (26 – 24) = [l(AB)]²
…[∵ a² – b² = (a + b)(a – b)]
∴50 x 2 = [l(AB)]²
∴100 = [l(AB)]²
i.e. [l(AB)]² = 100
∴l(AB) = √100
…[Taking square root of both sides]
∴l(AB) =10 cm
∴The length of the other side is 10 cm.

Question 5.
Lengths of diagonals of a rhombus ABCD are 16 cm and 12 cm. Find the side and perimeter of the rhombus.
Solution:
In rhombus ABCD,

l(AC) = 16 cm and l(BD) = 12 cm.
Let the diagonals of rhombus ABCD intersect at point O.
l(AO) = \(\frac { 1 }{ 2 }\) l(AC)
…[Diagonals of a rhombus bisect each other]
∴l(AO) = \(\frac { 1 }{ 2 }\) × 16
∴l(AO) = 8 cm
Also, l(DO) = \(\frac { 1 }{ 2 }\) l(BD)
…[Diagonals of a rhombus bisect each other]
∴l(DO) = \(\frac { 1 }{ 2 }\) × 12
∴l(DO) = 6 cm
In ∆DOA,
m∠DOA = 90°
..[Diagonals of a rhombus are perpendicular to each other]
[l(AD)]² = [l(AO)]² + [l(DO)]²
…[Pythagoras theorem]

= (8)² + (6)²
= 64 + 36
∴[l(AD)]² = 100
∴l(AD) = √100
… [Taking square root of both sides]
∴l(AD) = 10 cm
∴l(AB) = l(BC) = l(CD) = l(AD) = 10 cm
…[Sides of a rhombus are congruent]
Perimeter of rhombus ABCD
= l(AB) + l(BC) + l(CD) + l(AD)
= 10+10+10+10
= 40 cm
∴The side and perimeter of the rhombus are 10 cm and 40 cm respectively.

Question 6.
Find the length of diagonal of a square with side 8 cm.
Solution:
Let ₹XYWZ be the square of side 8cm.

seg XW is a diagonal.
In ∆ XYW,
m∠XYW = 90°
… [Angle of a square]
∴ [l(XW)]² = [l(XY)]² + [l(YW)]²
…[Pythagoras theorem]
= (8)² + (8)²
= 64 + 64
∴ [l(XW)]² = 128
∴ l(XW) = √128
…[Taking square root of both sides]
= √64 × 2
= 8 √2 cm
∴ The length of the diagonal of the square is 8 √2 cm.

Question 7.
Measure of one angle of a rhombus is 50°, find the measures of remaining three angles.
Solution:
Let ₹ABCD be the rhombus.

m∠A = 50°
m∠C = m∠A
….[Opposite angles of a rhombus are congruent]
∴ m∠C = 50°
Also, m∠D = m∠B …(i)
….[Opposite angles of a rhombus are congruent]
In ₹ABCD,
m∠A + m∠B + m∠C + m∠D = 360°
….[Sum of the measures of the angles of a quadrilateral is 360°]
∴ 50° + m∠B + 50° + m∠D = 360°
∴ m∠B + m∠D + 100° = 360°
∴ m∠B + m∠D = 360° – 100°
∴ m∠B + m∠B = 260° …[From (i)]
∴ 2m∠B = 260°
∴ m∠B = \(\frac { 260 }{ 2 }\)
∴ m∠B = 130°
∴ m∠D = m∠B = 130° …[From (i)]
∴ The measures of the remaining angles of the rhombus are 130°, 50° and 130°.

Maharashtra Board Class 8 Maths Chapter 8 Quadrilateral: Constructions and Types Practice Set 8.2 Intext Questions and Activities

Question 1.
Construct a rectangle PQRS by taking two convenient adjacent sides. Name the point of intersection of diagonals as T. Using divider and ruler, measure the following lengths.
i. lengths of opposite sides, seg QR and seg PS.
ii. lengths of seg PQ and seg SR.
iii. lengths of diagonals PR and QS.
iv. lengths of seg PT and seg TR, which are parts of the diagonal PR.
v. lengths of seg QT and seg TS, which are parts of the diagonal QS.
Observe the measures. Discuss about the measures obtained by your classmates. (Textbook pg. no. 44)
Solution:
Draw a rectangle PQRS such that, l(PQ) = 3 cm and l(QR) = 4 cm.

Steps of construction:
i. As shown in the rough figure, draw seg QR of length 4 cm.
ii. Placing the centre of the protractor at point Q, draw ray QW making an angle of 90° with seg QR.
iii. By taking a distance of 3 cm on the compass and placing it at point Q, draw an arc on ray QW. Name the point as P.
iv. Draw ray PV and ray RU making an angle of 90° with seg PQ and seg QR respectively.
v. Name the point of intersection of ray PV and ray RU as S.
₹PQRS is the required rectangle.

From the figure,
i. l(QR) = l(PS) = 4 cm
ii. l(PQ) = l(SR) = 3 cm
iii. l(PR) = l(QS) = 5 cm
iv. l(PT) = l(TR) = 2.5 cm
v. l(QT) = l(TS) = 2.5 cm

From the above measures, we can say that for any rectangle,
i. Opposite sides are congruent.
ii. Diagonals are congruent.
iii. Diagonals bisect each other.

Question 2.
Draw a square by taking convenient length of side. Name the point of intersection of its diagonals as E. Using the apparatus in a compass box, measure the following lengths.
i. lengths of diagonal AC and diagonal BD.
ii. lengths of two parts of each diagonal made by point E.
iii. all the angles made at the point E.
iv. parts of each angle of the square made by each diagonal, (e.g. ∠ADB and ∠CDB).
Observe the measures. Also observe the measures obtained by your classmates and discuss about them. (Textbook pg. no. 44)
Solution:
Draw a square ABCD such that its side is 5cm

Steps of construction:
i. As shown in the rough figure, draw seg BC of length 5 cm.
ii. Placing the centre of the protractor at point B, draw ray BP making an angle of 90° with seg BC.
iii. By taking a distance of 5 cm on the compass and placing it at point B, draw an arc on ray BP. Name the point as A.
iv. Placing the centre of the protractor at point C, draw ray CQ making an angle of 90° with seg BC.
v. By taking a distance of 5 cm on the compass and placing it at point C, draw an arc on ray CQ. Name the point as D.
vi. Draw seg AD.
₹ABCD is the required square.

From the figure,
i. l(AC) = l(BD) ≅ 7cm
ii. l(AE) = l(EC) ≅ 3.5cm,
l(BE) = l(ED) ≅ 3.5cm
iii. m∠AED = m∠BEC = m∠CED = m∠BEA = 90°
iv. Angles made by diagonal AC:
m∠BAC = m∠DAC = 45°
m∠BCA = m∠DCA = 45°
Angles made by diagonal BD:
m∠ABD = m∠CBD = 45°
m∠ADB = m∠CDB = 45°

From the above measures, we can say that for any square,
i. Diagonals are congruent.
ii. Diagonals bisect each other.
iii. Diagonals are perpendicular to each other.
iv. Diagonals bisect the opposite angles.

Question 3.
Draw a rhombus EFGH by taking convenient length of side and convenient measure of an angle.
Draw its diagonals and name their point of Intersection as M.
i. Measure the opposite angles of the quadrilateral and angles at the point M.
ii. Measure the two parts of every angle made by the diagonal.
iii. Measure the lengths of both diagonals. Measure the two parts of diagonals made by point M.
Observe the measures. Also observe the measures obtained by your classmates and discuss about them. (Textbook pg. no. 45)
Solution:
Draw a rhombus EFGH such that its side is 5 cm and m∠F = 60°.

Steps of construction:
i. As shown in the rough figure, draw seg FG of length 5 cm.
ii. Placing the centre of the protractor at point F, draw ray FX making an angle 60° with seg FG.
iii. By taking a distance of 5 cm on the compass and placing it at point F, draw an arc on ray FX. Name the point as E.
iv. By taking a distance of 5 cm on the compass and placing it at point E and point G, draw arcs. Name the point of intersection of arcs as H. ₹EFGH is the required rhombus.

From the figure,
i. Opposite angles:
m∠EFG = m∠GHE = 60°,
m∠FEH = m∠HGF = 120°
Angles at the point M:
m∠EMF = m∠FMG = m∠GMH = m∠HME = 90°

ii. Angles made by diagonal FH:
m∠EFH = m∠GFH = 30° m∠EHF = m∠GHF = 30°
Angles made by diagonal EG:
m∠FEG = m∠HEG = 60° m∠FGE = m∠HGE = 60°

iii. l(FH) ≈ 8.6 cm
l(EG) = 5 cm
l(FM) = l(HM) ≈ 4.3 cm
l(EM) = l(GM) ≈ 2.5 cm

From the above measures, we can say that for any rhombus,
i. Opposite angles are congruent.
ii. Diagonals bisect the opposite angles.
iii. Diagonals bisect each other and they are perpendicular to each other.

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Miscellaneous Exercise 1 8th Std Maths Answers Solutions.

Miscellaneous Exercise 1 8th Std Maths Answers

Question 1.
Choose the correct alternative answer for each of the following questions.
i. In ₹PQRS, m∠P = m∠R = 108°, m∠Q = m∠S = 72°. State which pair of sides of those given below is parallel. [Chapter 8]
(A) side PQ and side QR
(B) side PQ and side SR
(C) side SR and side SP
(D) side PS and side PQ
Solution:
(B) side PQ and side SR

Hint:

In ₹PQRS,
m∠P + m∠S = 108°+ 72
= 180°
Since, interior angles are supplementary.
∴ side PQ || side SR

ii. Read the following statements and choose the correct alternative from those given below them. [Chapter 8]
a. Diagonals of a rectangle are perpendicular bisectors of each other.
b. Diagonals of a rhombus are perpendicular bisectors of each other.
c. Diagonals of a parallelogram are perpendicular bisectors of each other.
d. Diagonals of a kite bisect each other.
(A) Statements (b) and (c) are true
(B) Only statement (b) is true
(C) Statements (b) and (d) are true
(D) Statements (a), (c) and (d) are true.
Solution:
(B) Only statement (b) is true

iii. If 19³ = 6859, find \(\sqrt[3]{0.006859}\). [Chapter 3]
(A) 1.9
(B) 19
(C) 0.019
(D) 0.19
Solution:
(D) 0.19

Hint:
\(\sqrt[3]{0.006859}\)

Question 2.
Find the cube roots of the following numbers. [Chapter 3]
i. 5832
ii. 4096
Solution:
i. 5832 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
= (2 × 3 × 3) × (2 × 3 × 3) × (2 × 3 × 3)
= (2 × 3 × 3)³
= (18)³
\(\sqrt[3]{5832}=18\)

ii. 4096 = (4 × 4) × (4 × 4) × (4 × 4)
= (4 × 4)
= 16³
\( \sqrt[3]{4096}=16\)

Question 3.
m∝n,n = 15 when m = 25. Hence
i. Find m when n = 87,
ii. Find n when m = 155. [Chapter 7]
Solution:
Given that, m ∝ n
∴ m = kn …(i)
where, k is the constant of variation.
When m = 25, n = 15
∴ Substituting, m = 25 and n = 15 in (i), we get
m = kn
∴ 25 = k × 15
∴ k = \(\frac { 25 }{ 15 }\)
∴ k = \(\frac { 5 }{ 3 }\)
Substituting k = \(\frac { 5 }{ 3 }\) in (i), we get
m = kn
∴ m = \(\frac { 5 }{ 3 }n\) …(ii)

i. When n = 87, m = ?
Substituting n = 87 in (ii), we get
m = \(\frac { 5 }{ 3 }n\)
m = \(\frac { 5 }{ 3 }\) × 87
m = 5 × 29
m = 145

ii. When m = 155, n = ?
∴ Substituting m = 155 in (ii), we get
m = \(\frac { 5 }{ 3 }n\)
∴ 155 = \(\frac { 5 }{ 3 }n\)
∴ \(\frac{155 \times 3}{5}=n\)
∴ n = 31 × 3
∴ n = 93

Question 4.
y varies inversely with x. If y = 30 when x = 12, find [Chapter 7]
i. y when x = 15,
ii. x when y = 18.
Solution:
Given that,
\(y \propto \frac{1}{x}\)
∴ \(y=k \times \frac{1}{x}\)
where, k is the constant of variation.
∴ y × x = k …(i)
When x = 12, y = 30
∴ Substituting, x = 12 and y = 30 in (i), we get
y × x = k
∴ 30 × 12 = k
∴ k = 360
Substituting, k = 360 in (i), we get
y × x = k
∴ y × x = 360 ….(ii)

i. When x = 15,y = ?
∴ Substituting x = 15 in (ii), we get
y × x = 360
∴ y × 15 = 360
∴ y = \(\frac { 360 }{ 15 }\)
∴ y = 24

ii. When y = 18, x = ?
∴ Substituting y = 18 in (ii), we get
y × x = 360
∴18 × x = 360
∴ x = \(\frac { 360 }{ 18 }\)
∴ x = 20

Question 5.
Draw a line l. Draw a line parallel to line l at a distance of 3.5 cm. [Chapter 2]
Solution:
Steps of construction:

1. Draw a line l and take any two points M and N on the line.
2. Draw perpendiculars to line l at points M and N.
3. On the perpendicular lines take points S and T at a distance 3.5 cm from points M and N respectively.
4. Draw a line through points S and T. Name the line as n.

Line n is parallel to line l at a distance of 3.5 cm from it.

Question 6.
Fill in the blanks in the following statement.
The number \((256)^{\frac{5}{7}}\) is __ of __ power of __. [Chapter 3]
Solution:
The number \((256)^{\frac{5}{7}}\) is 7th root of 5th power of 256.

Question 7.
Expand.
i. (5x – 7) (5x – 9)
ii. (2x – 3y)³
iii. \(\left(a+\frac{1}{2}\right)^{3}\) [Chapter 5]
Solution:
i. (5x – 7) (5x – 9)
= (5x)² + (-7 -9) 5x + (-7) × (-9).
…[∵ (x + a) (x + b) = x² + (a + b)x + ab]
= 25x² + (-16) × 5x + 63
= 25x² – 80x + 63

ii. Here, a = 2x and b = 3y
(2x – 3y)³
= (2x)³ – 3 (2x)² (3y) + 3 (2x) (3y)² – (3y)³
…[∵ (a – b)³ = a³ – 3a²b + 3ab² – b³]
= 8x³ – 3 (4x²) (3y) + 3 (2x) (9y²) – 27y³
= 8x³ – 36x²y + 54xy² – 27p³

iii. Here, A= a and B = \(\frac { 1 }{ 2 }\)
\(\left(a+\frac{1}{2}\right)^{3}=(a)^{3}+3(a)^{2}\left(\frac{1}{2}\right)+3(a)\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{3}\)
…[(A + B)³ = A³ + 3A²B + 3AB² + B³]
\(=\mathbf{a}^{3}+\frac{3 \mathbf{a}^{2}}{2}+\frac{3 \mathbf{a}}{4}+\frac{1}{8}\)

Question 8.
Draw an obtuse angled triangle. Draw all of its medians and show their point of concurrence. [Chapter 4]
Solution:

The point of concurrence of the medians PS, RU and QV is G.

Question 9.
Draw ∆ABC such that l(BC) = 5.5 cm, m∠ABC = 90°, l(AB) = 4 cm. Show the orthocentre of the triangle. [Chapter 4]
Solution:

Here, point B is the orthocentre of ∆ABC.

Question 10.
Identify the variation and solve.
It takes 5 hours to travel from one town to the other if speed of the bus is 48 km/hr. If the speed of the bus is reduced by 8 km/hr, how much time will it take for the same travel? [Chapter 7]
Solution:
Let, v represent the speed of the bus and t represent the time required to travel from one town to the other.
The speed of the bus varies inversely with the time required to travel from one town to the other.
∴ \(\mathbf{v} \propto \frac{1}{\mathbf{t}}\)
∴ \(\mathbf{v}=\mathbf{k} \times \frac{1}{\mathbf{t}}\)
where, k is the constant of variation.
∴ v × t = k …(i)
It takes 5 hours to travel from one town to the other if speed of the bus is 48 km/hr.
i.e., when v = 48, t = 5
∴ Substituting v = 48 and t = 5 in (i), we get
v × t = k
∴ 48 × 5 = k
∴ k = 240
Substituting k = 240 in (i), we get
v × t = k
∴ v × t = 240 …(ii)
Since, the speed of the bus is reduced by 8 km/hr,
∴ Speed of the bus in second case (v)
= 48 – 8 = 40 km/hr
∴ When v = 40, t = ?
∴ Substituting v = 40 in (ii), we get
v × t = 240
∴ 40 × t = 240
∴ \(t=\frac { 240 }{ 40 }\)
∴ t = 6
∴ The problem is of inverse variation and the bus would take 6 hours to travel the distance if its speed is reduced by 8 km/hr.

Question 11.
Seg AD and seg BE are medians of ∆ABC and point G is the centroid. If l(AG) = 5 cm, find l(GD). If l(GE) = 2 cm, find l(BE). [Chapter 4]
Solution:
The centroid of a triangle divides each median in the ratio 2:1.
i. Point G is the centroid and seg AD is the median.

ii. Point G is the centroid and seg BE is the median.

∴ l(BG) × 1 = 2 × 2
∴ l(BG) = 4 cm
Now, l(BE) = l(BG) + l(GE)
∴ l(BE) = 4 + 2
∴ l(BE) = 6 cm

Question 12.
Convert the following rational numbers into decimal form. [Chapter 1]
i. \(\frac { 8 }{ 13 }\)
ii. \(\frac { 11 }{ 7 }\)
iii. \(\frac { 5 }{ 16 }\)
iv. \(\frac { 7 }{ 9 }\)
Solution:
i. \(\frac { 8 }{ 13 }\)

ii. \(\frac { 11 }{ 7 }\)

iii. \(\frac { 5 }{ 16 }\)

iv. \(\frac { 7 }{ 9 }\)

Question 13.
Factorize.
i. 2y² – 11y + 5
ii. x² – 2x – 80
iii. 3x² – 4x + 1
Solution:
i. 2y² – 11y + 5
= 2y² – 10y – y + 5
= 2y(y – 5) – 1(y – 5)
= (y – 5)(2y – 1)

ii. x² – 2x – 80
= x² – 10x + 8x – 80
= x (x – 10) + 8 (x – 10)
= (x – 10)(x + 8)

iii. 3x² – 4x + 1
= 3x² – 3x – x + 1
= 3x(x – 1) – 1(x – 1)
= (x – 1) (3x – 1)

Question 14.
The marked price of a T.V. set is Rs 50,000. The shopkeeper sold it at 15% discount. Find the price of it for the customer. [Chapter 9]
Solution:
Here, marked price = Rs 50,000,
discount = 15%
Let the discount percent be x
∴x = 15%
i. Discount

= 500 × 15
= Rs 7,500

ii. Selling price = Marked price – Discount
= 50,000 – 7,500
= Rs 42,500
∴The price of the T.V. set for the customer is Rs 42,500.

Question 15.
Rajabhau sold his flat to Vasantrao for Rs 88,00,000 through an agent. The agent charged 2 % commission for both of them. Find how much commission the agent got. [Chapter 9]
Solution:
Here, selling price of the flat = Rs 88,00,000
Rate of commission = 2%
Commission = 2% of selling price
= \(\frac { 2 }{ 100 }\) × 88,00,000
= 2 × 88,000
= Rs 1,76,000
∴ Total commission = Commission from Rajabhau + Commission from Vasantrao
= Rs 1,76,000 + Rs 1,76,000
= Rs 3,52,000
∴ The agent got a commission of Rs 3,52,000.

Question 16.
Draw a parallelogram ABCD such that l(DC) = 5.5 cm, m∠D = 45°, l(AD) = 4 cm. [Chapter 8]
Solution:
Opposite sides of a parallelogram are congruent.
∴ l(AD) = l(BC) = 4 cm and
l(DC) = l(AB) = 5.5 cm

Question 17.
In the figure, line l || line m and line p || line q. Find the measures of ∠a, ∠b, ∠c and ∠d. [Chapter 2]

Solution:
i. line l|| line m and line p is a transversal.
∴m∠a = 78° …(i) [Corresponding angles]

ii. line p || line q and line m is a transversal.
∴m∠d = m∠a …[Corresponding angles]
∴m∠d = 78° …(ii)[From (i)]

iii. m∠b = m∠d …[Vertically opposite angles]
∴m∠b = 78° …[From (ii)]

iv. line l|| line m and line q is a transversal.
∴m∠c + m∠d = 180° …[Interior angles]
∴m∠c + 78° = 180° … [From (ii)]
∴m∠c =180° – 78°
∴m∠c = 102°
∴m∠a = 78°, m∠b = 78°, m∠c = 102°, m∠d = 78°

Maharashtra Board Class 8 Maths Solutions

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 17.1 8th Std Maths Answers Solutions Chapter 17 Circle: Chord and Arc.

Practice Set 17.1 8th Std Maths Answers Chapter 17 Circle: Chord and Arc

Question 1.
In a circle with centre P, chord AB is drawn of length 13 cm, seg PQ ⊥ chord AB, then find l(QB)

Solution:
seg PQ ⊥ chord AB … [Given]
∴l(QB) = \(\frac { 1 }{ 2 }\) l(AB)… [Perpendicular drawn from the centre of a circle to its chord bisects the chord]
∴l(QB) = \(\frac { 1 }{ 2 }\) x 13 …[∵ l(AB) = 13 cm]
∴l(QB) = 6.5 cm

Question 2.
Radius of a circle with centre O is 25 cm. Find the distance of a chord from the centre if length of the chord is 48 cm.

Solution:
seg OP ⊥ chord CD … [Given]
∴l(PD) = \(\frac { 1 }{ 2 }\) l(CD) … [Perpendicular drawn from the centre of a circle to its chord bisects the chord]
∴l(PD) = \(\frac { 1 }{ 2 }\) x 48 …[∵ l(CD) = 48 cm]
∴l(PD) = 24 cm …(i)
In ∆OPD, m∠OPD = 90°
∴[l(OD)]² = [l(OP)]² + [l(PD)]² … [Pythagoras theorem]
∴(25)² = [l(OP)]² + (24)² … [From (i) and l(OD) = 25 cm]
∴(25)² – (24)² = [l(OP)]²
∴(25 + 24) (25 – 24) = [l(OP)]² …[∵ a² – b² = (a + b) (a – b)]
∴49 x 1 = [l(OP)]²
∴[l(OP)]² = 49
∴l(OP) = √49 …[Taking square root of both sides]
∴l(OP) = 7 cm
∴The distance of the chord from the centre of the circle is 7 cm.

Question 3.
O is centre of the circle. Find the length of radius, if the chord of length 24 cm is at a distance of 9 cm from the centre of the

Solution:
Let seg OP ⊥ chord AB
∴ l(AP) = \(\frac { 1 }{ 2 }\) l(AB) … [Perpendicular drawn from the centre of a circle to its chord bisects the chord]

∴l(AP) = \(\frac { 1 }{ 2 }\) x 24 …[∵ l(AB) = 24 cm]
∴l(AP) = 12 cm …(i)
In ∆OPA, m∠OPA = 90°
∴[l(AO)]² = [l(OP)]² + [l(AP)]² … [Pythagoras theorem]
∴[l(AO)]² = (9)² + (12)² … [From (i) and l(OP) = 9 cm]
= 81 + 144
∴[l(AO)]² = 225
∴l(AO) = √225 …[Taking square root of both sides]
∴l(AO) = 15 cm
∴The length of radius of the circle is 15 cm.

Question 4.
C is the centre of the circle whose radius is 10 cm. Find the distance of the chord from the centre if the length of the chord is 12 cm.
Solution:
Let seg AB be the chord of the circle with centre C.
Draw seg CD ⊥ chord AB.

∴l(AD) = \(\frac { 1 }{ 2 }\) l(AB) …[Perpendicular drawn from the centre of a circle to its chord bisects the chord]
= \(\frac { 1 }{ 2 }\) x 12 …[∵ l(AB) = 12 cm]
∴l(AD) = 6 cm …(i)
∴In ∆ACD, m∠ADC = 90°
∴[l(AC)]² = [l(AD)]² + [l(CD)]² … [Pythagoras theorem]
∴(10)² = (6)² + [l(CD)]² … [From (i) and l(AC) = 10 cm]
∴(10)² – (6)² = [l(CD)]²
∴100 – 36 = [l(CD)]²
∴64 = [l(CD)]²
i. e. [l(CD)]² = 64
∴l(CD) = √64 …[Taking square root of both sides]
∴l(CD) = 8 cm
∴The distance of the chord from the centre of the circle is 8 cm.

Maharashtra Board Class 8 Maths Chapter 17 Circle: Chord and Arc Practice Set 17.1 Intext Questions and Activities

Question 1.
In the given figure, O is the centre of the circle. With reference to the figure fill in the blanks. (Textbook pg. No. 114)

Solution:

1. Seg OD is radius of the circle.
2. Seg AB is diameter of the circle.
3. Seg PQ is chord of the circle.
4. ∠DOB is the central angle.
5. Minor arc : arc AXD, arc BD, arc AP, arc PQ, arc BQ, etc.
6. Major arc : arc PAB, arc PDQ, arc PDB, arc ADQ, etc.
7. Semicircular arc : arc ADB, arc AQB.
8. m (arc DB) = m∠DOB
9. m (arc DAB) = 360° – m∠DOB

Question 2.
Draw chord AB of a circle with centre O. Draw perpendicular OP to chord AB. Measure seg AP and seg PB. What do you observe. (Textbook pg. no. 114)
Solution:
l(AP) = l(PB) = 0.9 cm
∴the perpendicular drawn from the centre of the circle to its chord bisects the chord.

Question 3.
Draw five circles with different radii. Draw a chord and perpendicular from the centre to each chord in each circle. Verify with a divider that the two parts of the chords are equal. (Textbook pg. no. 114)
Solution:
[Students should attempt the above activities on their own.]

Question 4.
Draw five circles of different radii on a paper. Draw a chord in each circle. Find the midpoint of each chord. Join the centre of the circle and midpoint of the chord as shown in the figure. Name the chord as AB and midpoint of the chord as P. Check with set-square or protractor that ∠APO or ∠BPO are right angles.
Check whether the same result is observed for the chord of each circle. (Textbook pg, no. 115)
Solution:
[Students should attempt the above activities on their own.]

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 14.2 8th Std Maths Answers Solutions Chapter 14 Compound Interest.

Practice Set 14.2 8th Std Maths Answers Chapter 14 Compound Interest

Compound Interest class 8 practice set 14.2 Question 1. On the construction work of a flyover bridge there were 320 workers initially. The number of workers were increased by 25% every year. Find the number of workers after 2 years.
Solution:
Here, P = Initial number of workers = 320
R = Increase in the number of workers per year = 25%
N = 2 years
A = Number of workers after 2 years

∴ The number of workers after 2 years would be 500.

Question 2.
A shepherd has 200 sheep with him. Find the number of sheeps with him after 3 years if the increase in number of sheeps is 8% every year.
Solution:
Here, P = Present number of sheeps = 200
R = Increase in number of sheeps per year = 8%
N = 3 years
A = Number of sheeps after 3 years

= \(\frac{0.32}{25} \times 27 \times 27 \times 27\)
= 0.0128 × 27 × 27 × 27
= 251.9424
= 252
∴ The number of sheeps with the shepherd after 2 years would be 252 (approx).

8th Class Math Practice Set 14.2 Question 3.
In a forest there are 40,000 trees. Find the expected number of trees after 3 years if the objective is to increase the number at the rate 5% per year.
Solution:
Here, P = Present number of trees in the forest = 40,000
R = Increase in the number of trees per year = 5%
N = 3 years
A = Number of trees after 3 years

= 5 × 21 × 21 × 21
= 5 × 9261
= 46,305
∴ The expected number of trees in the forest after 3 years is 46,305.

Std 8 Maths Practice Set 14.2 Question 4.
The cost price of a machine is Rs 2,50,000. If the rate of depreciation is 10% per year, find the depreciation in price of the machine after two years.
Solution:
Here, P = Cost price of machine = Rs 2,50,000
R = Rate of depreciation per year = 10%
N = 2 years
A = Depreciated price of the machine after 2 years

= 2,500 × 81
= Rs 2,02,500
Depreciation in price = Cost price (P) – Depreciated price (A)
= 2,50,000 – 2,02,500
= Rs 47,500
∴ The depreciation in price of the machine after 2 years would be Rs 47,500.

Question 5.
Find the compound interest if the amount of a certain principal after two years is Rs 4036.80 at the rate of 16 p.c.p.a.
Solution:
Here, A = Rs 4036.80, R = 16 p.c.p.a. and N = 2 years
i. \(\mathbf{A}=\mathbf{P}\left[1+\frac{\mathbf{R}}{100}\right]^{N}\)

ii. Interest = Amount (A) – Principal (P)
= 4036.80 – 3000
= Rs 1036.80
∴ The compound interest after 2 years would be Rs 1036.80.

Question 6.
A loan of Rs 15,000 was taken on compound interest. If the rate of compound interest is 12 p.c.p.a. find the amount to settle the loan after 3 years.
Solution:
Here, P = Rs 15,000, R = 12 p.c.p.a, and
N = 3 years

∴ The amount required to settle the loan after 3 years is Rs 21,073.92.

Practice Set 14.2 Class 8 Question 7.
A principal amounts to Rs 13,924 in 2 years by compound interest at 18 p.c.p.a. Find the principal.
Solution:
Here, A = Rs 13,924, R = 18 p.c.p.a., and N = 2 years

∴ P = 4 x 50 x 50
∴ P = Rs 10,000
∴ The principal is Rs 10,000.

Question 8.
The population of a suburb is 16,000. Find the rate of increase in the population if the population after two years is 17,640.
Solution:
Here, P = Population of a suburb = 16,000
N = 2 years
A = Increase in the population after 2 years = 17,640
R = Rate of increase in population


∴5 = R
i.e., R = 5%
∴The rate of increase in the population is 5 p.c.p.a.

Compound Interest Practice Set 14.2 Question 9.
In how many years Rs 700 will amount to Rs 847 at a compound interest rate of 10 p.c.p.a.
Solution:
Here, P = Rs 700, R = 10 p.c.p.a., A = Rs 847

∴Rs 700 will amount to Rs 847 in 2 years.

Practice Set 14.2 Question 10.
Find the difference between simple interest and compound interest on Rs 20,000 in 2 years at 8 p.c.p.a.
Solution:
Here, P = Rs 20,000, R = 8 p.c.p.a.,
N = 2 years
i. Simple interest (I)

Simple interest (I) = Rs 3200

ii. Compound Interest (I):


= 32 × 27 × 27
= Rs 23,328
Compound interest (I)
= Amount (A) – Principal (P)
= 23,328 – 20,000
= Rs 3328 ,..(ii)

iii. Difference
= Compound interest – Simple interest
= 3328 – 3200 … [Form (i) and (ii)]
= Rs 128
∴ The difference between compound interest and simple interest is Rs 128.
[Note: The question is modified as per the answer given in the textbook.]

Maharashtra Board Class 8 Maths Chapter 14 Compound Interest Practice Set 14.2 Intext Questions and Activities

8th Standard Maths Practice Set 14.2 Question 1.
Visit the bank nearer to your house and get the information regarding the different schemes and rates of interests. Make a chart and display in your class. (Textbook pg. no. 90)
Solution:
(Students should attempt this activity at their own.)

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 13.1 8th Std Maths Answers Solutions Chapter 13 Congruence of Triangles.

Practice Set 13.1 8th Std Maths Answers Chapter 13 Congruence of Triangles

Congruence of Triangles Practice Set 13.1 Question 1.
In each pair of triangles in the following figures, parts bearing identical marks are congruent. State the test and correspondence of vertices by which triangles in each pair are congruent.
i.

ii.

iii.

iv.

v.

Solution:
i.

The two triangles are congruent by SAS test in the correspondence XWZ ↔ YWZ.

ii.

The two triangles are congruent by hypotenuse-side test in the correspondence KJI ↔ LJI.

iii.

The two triangles are congruent by SSS test in the correspondence HEG ↔ FGE.

iv.
The two triangles are congruent by ASA test is the correspondence SMA ↔ OPT.

v.

The two triangles are congruent by ASA test or SAS test or SAA test in the correspondence MTN ↔ STN.

Maharashtra Board Class 8 Maths Chapter 13 Congruence of Triangles Practice Set 13.1 Intext Questions and Activities

Practice Set 13.1 Question 1.
Write answers to the following questions referring to the given figure.

1. Which is the angle opposite to the side DE?
2. Which is the side opposite to ∠E?
3. Which angle is included by side DE and side DF?
4. Which side is included by ∠E and ∠F?
5. State the angles adjacent to side DE. (Textbook pg, no. 81)

Solution:

1. ∠DFE i.e. ∠F is the angle opposite to side DE.
2. Side DF is the side opposite to ∠E.
3. ∠EDF i.e. ∠D is included by side DE and side DF.
4. Side EF is included by ∠E and ∠F.
5. ∠DEF and ∠EDF i.e. ∠E and ∠D are adjacent to side DE.

Congruence of Triangles Class 8th Practice Set 13.1 Question 2.
In the given figure, parts of triangles indicated by identical marks are congruent.
a. Identify the one-to-one correspondence of vertices in which the two triangles are congruent and write the congruence.
b. State with reason, whether the statement, ∆XYZ ≅ ∆STU is right or wrong. (Textbook pg. no. 82)

Solution:
a. From the figure,
S ↔ X, T ↔ Z, U ↔ Y i.e.,
STU ↔ XZY, or SUT ↔ XYZ, or
TUS ↔ ZYX, or TSU ↔ ZXY, or
UTS ↔ YZX, or UST ↔ YXZ

∴ ∆STU ≅ ∆XZY, or ∆SUT ≅ ∆XYZ, or
∆TUS ≅ ∆ZYX, or ∆TSU ≅ ∆ZXY, or
∆UTS ≅ ∆YZX, or ∆UST ≅ ∆YXZ

b. If ∆XYZ ≅ ∆STU, then
∠Y ≅ ∠T, ∠Z ≅ ∠U,
seg XY ≅ seg ST, seg XZ ≅ seg SU
∴ But, all the above statements are wrong. The statement AXYZ ≅ ASTU is wrong.

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 1.4 8th Std Maths Answers Solutions Chapter 1 Rational and Irrational Numbers.

Practice Set 1.4 8th Std Maths Answers Chapter 1 Rational and Irrational Numbers

Question 1.
The number √2 is shown on a number line. Steps are given to show √3 on the number line using √2. Fill in the boxes properly and complete the activity.

The point Q on the number line shows the number ……….
A line perpendicular to the number line is drawn through the point Q. Point R is at unit distance from Q on the line.
Right angled ∆OQR is obtained by drawing seg OR.
l(OQ) = √2, l(QR) = 1
∴By Pythagoras theorem,
[l(OR)]² = [l(OQ)]² + [l(QR)]²

Draw an arc with centre O and radius OR. Mark the point of intersection of the line and the arc as C. The point C shows the number √3
Solution:
The point Q on the number line shows the number √2
A line perpendicular to the number line is drawn through the point Q. Point R is at unit distance from Q on the line.
Right angled ∆OQR is obtained by drawing seg OR.
l(OQ) = √2, l(QR) = 1
∴By Pythagoras theorem,
[l(OR)]² = [l(OQ)]² + [l(QR)]²

.. .[Taking square root of both sides]
Draw an arc with centre O and radius OR. Mark the point of intersection of the line and the arc as C. The point C shows the number √3.

Question 2.
Show the number √5 on the number line.
Solution:
Draw a number line and take a point Q at 2
such that l(OQ) = 2 units.
Draw a line QR perpendicular to the number line through the point Q such that l(QR) = 1 unit.
Draw seg OR.
∆OQR formed is a right angled triangle.
By Pythagoras theorem,
[l(OR)]² = [l(OQ)]² + [l(QR)]²
= 2² + 1²
= 4 + 1
= 5
∴l(OR) = √5 units
…[Taking square root of both sides]
Draw an arc with centre O and radius OR. Mark the point of intersection of the number line and arc as C. The point C shows the number √5.

Question 3.
Show the number √7 on the number line.
Solution:
Draw a number line and take a point Q at 2 such that l(OQ) = 2 units.
Draw a line QR perpendicular to the number line through the point Q such that l(QR) = 1 unit.
Draw seg OR.
∆OQR formed is a right angled triangle.
By Pythagoras theorem,
[l(OR)]² = [l(OQ)]² + [l(QR)]²
= 2² + 1²
= 4 + 1
= 5
∴ l(OR) = √5 units
… [Taking square root of both sides]
Draw an arc with centre O and radius OR.
Mark the point of intersection of the number line and arc as C. The point C shows the number √5.
Similarly, draw a line CD perpendicular to the number line through the point C such that l(CD) = 1 unit.
By Pythagoras theorem,
l(OD) = √6 units
The point E shows the number √6 .
Similarly, draw a line EP perpendicular to the number line through the point E such that l(EP) = 1 unit.
By Pythagoras theorem,
l(OP) = √7 units
The point F shows the number √7.

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 13.2 8th Std Maths Answers Solutions Chapter 13 Congruence of Triangles.

Practice Set 13.2 8th Std Maths Answers Chapter 13 Congruence of Triangles

Congruence of Triangles Class 8th Practice Set 13.2 Question 1.
In each pair of triangles given below, parts shown by identical marks are congruent. State the test and the one-to-one correspondence of vertices by which triangles in each pair are congruent. Also state the remaining congruent parts.


Solution:
i. In ∆MST and ∆TBM,
∴ side MS ≅ side TB … [Given]
m∠MST = m∠TBM = 90° … [Given]
hypotenuse MT ≅ hypotenuse MT
…[Common side]
∴ ∆MST ≅ ∆TBM …[by hypotenuse-side test]
∴ side ST ≅ side BM …[Corresponding sides of congruent triangles]
∠SMT ≅ ∠BTM …[Corresponding sides of congruent triangles]
∠STM ≅ ∠BMT …[Corresponding sides of congruent triangles]

ii. In ∆PRQ and ∆TRS,
side PR ≅ side TR … [Given]
∠PRQ ≅ ∠TRS …[Vertically opposite angles]
side RQ ≅ side RS … [Given]
∴ ∆PRQ ≅ ∆TRS …[by SAS test]
∴ side PQ ≅ side TS …[Corresponding sides of congruent triangles]
∠RPQ ≅ ∠RTS …[Corresponding sides of congruent triangles]
∠PQR ≅ ∠TSR …[Corresponding sides of congruent triangles]

iii. In ∆DCH and ∆DCF,
∠DCH ≅ ∠DCF …[Given]
∠DHC ≅ ∠DFC …[Given]
side DC ≅ side DC …[Common side]
∴ ∆DCH ≅ ∆DCF …[by AAS test]
∴ side HC ≅ side FC …[Corresponding sides of congruent triangles]
side DH ≅ side DF…[Corresponding sides of congruent triangles]
∠HDC ≅ ∠FDC ….[Corresponding sides of congruent triangles]

Congruence of Triangles Practice Set 13.2 Question 2.
In the given figure, seg AD ≅ seg EC. Which additional information is needed to show that ∆ABD and ∆EBC will be congruent by AAS test?

Solution:
In ∆ABD and ∆CBE,
∴ seg AD ≅ seg CE …[Given]
∠ABD ≅ ∠CBE …[Vertically opposite angles]
∴ The necessary condition for the two triangles to be congruent by AAS test is
∠ADB ≅ ∠CEB, or
∠DAB ≅ ∠ECB

Maharashtra Board Class 8 Maths Chapter 13 Congruence of Triangles Practice Set 13.2 Intext Questions and Activities

Practice Set 13.2 Class 8 Question 1.
Draw ∆ABC and ∆LMN such that two pairs of their sides and the angles included by them are congruent.
Draw ∆ABC and ∆LMN, l(AB) = l(LM), l(BC) = l(MN), m∠ABC = m∠LMN.
Copy ∆ABC on a tracing paper. Place the paper on ∆LMN in such a way that point A coincides with point L, side AB overlaps side LM. What do you notice?(Textbook pg. no. 83)

Solution:
We notice that ∆ABC ≅ ∆LMN.

Congruence of Triangles Class 8 Solutions Question 2.
Draw ∆PQR and ∆XYZ such that l(PQ) = l(X Y), l(Q R) = l(YZ), l(RP) = l(ZX). Copy ∆PQR on a tracing paper. Place it on ∆XYZ observing the correspondence P ↔ X, Q ↔ Y, R ↔ Z. What do you notice? (Textbook pg. no. 84)

Solution:
We notice that ∆PQR ≅ ∆XYZ.

Congruence of Triangles Class 8 Question 3.
Draw ∆XYZ and ∆DEF such that, l(XZ) = l(DF), ∠X ≅ ∠D and ∠Z ≅ ∠F.
Copy ∆XYZ on a tracing paper and place it over ∆DEF. What do you notice?(Textbook pg. no. 84)

Solution:
We notice that ∆XYZ ≅ ∆DEF in the correspondence X ↔ D, Y ↔ E, Z ↔ F.

Question 4.
Draw two right angled triangles such that a side and the hypotenuse of one is congruent with the corresponding parts of the other. Copy one triangle on tracing paper and place it over the other. What do you notice? (Textbook pg. no. 84)

Solution:
We notice that the two triangles are congruent.
(Students should draw figures and verify the answers.)

Balbharti Maharashtra State Board Class 8 Hindi Solutions Sulabhbharati Chapter 6 अंधायुग Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 8 Hindi Solutions Chapter 6 अंधायुग

Hindi Sulabhbharti Class 8 Solutions Chapter 6 अंधायुग Textbook Questions and Answers

सूचना के अनुसार कृतियाँ करो:

कृति करो:

Question 1.

Answer:

संजाल पूर्ण करो:

Question 1.

Answer:

उत्तर लिखो:

Question 1.

Answer:

कविता में प्रयुक्त पात्र:

1. युयुत्सु
2. अश्वत्थामा
3. वृद्ध

भाषा बिंद

पाठों में आए मुहावरों का अर्थ लिखकर उनका अपने स्वतंत्र वाक्यों में प्रयोग करो:
Answer:

पढ़ो और समझो:

स्वयं अध्ययन

‘कर्म ही पूजा है’, विषय पर अपने विचार सौ शब्दों में लिखो।

उपयोजित लेखन:

निम्नलिखित मुद्दों के आधार पर कहानी लिखकर उसे उचित शीर्षक दो।

Answer:
एक गाँव – दर्जी की दुकान – प्रतिदिन हाथी का दुकान से ,होकर नदी पर नहाने जाना – दर्जी का हाथी – को केला, खिलाना – एक दिन दर्जी को मजाक सूझना – दर्जी दवारा हाथी को सुई चुभाना- परिणाम – शीर्षक

जैसी करनी, वैसी भरनी

गणेशपुर नामक गाँव में एक किसान के घर एक पालत हाथी रहता था। वह बड़ा चालाक था। उसका नियम था कि प्रतिदिन सबेरे के समय तालाब पर पीने जाता था। उस रास्ते में एक दर्जी की दुकान पड़ती थी। वह रोज उसे खाने के लिए केले देता था। हाथी इस उपकार का बदला चुकाने के लिए तालाब से लौटते समय दर्जी को एक कमल का फूल देता था। इस प्रकार दोनों में गहरी दोस्ती हो गई। एक दिन दर्जी ने मन में सोचा क्यों न आज हाथी के साथ मजाक करूँ।

जब हाथी प्रतिदिन के नियमानुसार केले लेने आया, तो दर्जी ने केला के बदले हाथी की सूंड में सुई चुभा दी। हाथी खून का चूंट पीकर तालाब पर चला गया। रास्ते में मन-ही-मन बदला लेने की युक्ति सोचने लगा। तालाब से लौटते समय वह अपनी सूंड में कीचड़ भर लाया और दर्जी की दुकान में डाल दिया। दुकान में रखे सारे नए कपड़े खराब हो गए। दर्जी अपनी हानि देखकर पछताने लगा। सीख : इस कहानी से हमें यह सीख मिलती है कि जैसा बीज बोओगे, वैसा फल पाओगे।

कल्पना पल्लवन

Question 1.
‘मनुष्य का भविष्य उसके हाथों में है।’ अपने विचार लिखिए।
Answer:
मनुष्य स्वयं अपने भाग्य का निर्माता होता है। वह चाहे तो अच्छे कार्यों द्वारा अपने जीवन को स्वर्ग के जैसा सुंदर बना सकता है और यदि चाहे तो अपने जीवन को नर्क बना सकता है। मनुष्य को जो मानव जन्म मिला है वह बहुत ही दुर्लभ जन्म है। अत: मनुष्य को अपने जीवन में मनुष्यता का पालन करते हुए स्वयं के जीवन को खुशहाल व सफल बनाना चाहिए। मनुष्य अपने जीवन में सद्गुण एवं जीवनमूल्यों को अपनाकर स्वयं का चरित्र उज्ज्व ल बना सकता है। संसार में ऐसे कई महापुरुष हुए हैं जिन्होंने स्वयं के जीवन को अपने कार्य द्वारा महान बनाया है।

विद्यार्थी जीवन में प्रत्येक विद्यार्थी को पढ़ाई में ध्यान लगाना चाहिए और अच्छे संस्कारों को अपनाना और उन पर चलना चाहिए। इससे उनका भविष्य उज्ज्वल हो सकता है। यदि वे बचपन से कुसंगति में फंस गए तो स्वयं के जीवन को नर्क के समान यातना व पीड़ादायी बनाएंगे। यह मनुष्य के ऊपर निर्भर करता है कि वह फूल को चुने या काँटी को।

Question 2.
कर्म ही पूजा है। इस विषय पर अपने विचार लिखिए।
Answer:
गीता में लिखा है कि कर्म ही पूजा है। कर्म से बढ़कर व्यक्ति का अन्य कोई धर्म नहीं है। इसलिए कर्म करना मनुष्य का पहला लक्ष्य होना चाहिए। कर्म करने से व्यक्ति को बड़ा आनंद मिलता है। पक्के इरादे से किया गया कर्म ज्यादा सफल होता है। मनुष्य के कर्म को ही संसार में याद किया जाता है। उसकी मृत्यु के उपरांत वह सिर्फ अपने अच्छे कर्मों के कारण याद किया जाता है। इसलिए छात्रों को भी आलस्य त्यागकर अध्ययन का कर्म करते रहना चाहिए। हमें स्वामी विवेकानंद, महात्मा गांधी आदि की तरह जीवन में कर्म करते रहना चाहिए। कर्म सफलता का आधार है। कर्म ही श्रेष्ठ है।

आज संसार में कई ऐसे देश हैं, जो समृद्ध व संपन्न हैं; प्रगत एवं विकसित हैं। वहाँ पर विद्या, धन, बुद्धि व ऐश्वर्य हैं। इसका एक ही कारण है, वह यह है कि वहाँ पर रहने वाले लोगों ने कर्म को ही अपना लक्ष्य मान लिया है। कर्मवीरों के लिए समय बहुत ही महत्त्वपुर्ण होता है। इसलिए वे समय को कभी भी व्यर्थ नहीं गंवाते। जिस काम को जिस समय करना हैं, उसे उसी समय कर देते हैं। उसे कल पर नहीं छोड़ते हैं। जहाँ काम करना हो, वहाँ काम करते हैं, बातों में समय व्यर्थ नहीं करते। आज का काम कल पर नहीं छोड़कर वे अपने दिनों को व्यर्थ नहीं करते। समय का सदुपयोग करना यही उनका कर्तव्य होता है। कोशिश या मेहनत करने से वे कभी-भी जी नहीं चुराते हैं। कर्म करने से ही व्यक्ति के जीवन को सौंदर्य प्राप्त हो जाता है। सबकी भलाई के लिए कर्म करते जीना ही जीवन का सच्चा मूलमंत्र है।

Hindi Sulabhbharti Class 8 Solutions Chapter 6 अंधायुग Additional Important Questions and Answers

समझकर लिखिए

Question 1.

Answer:

संजाल पूर्ण कीजिए।

Question 1.

Answer:

निम्नलिखित पद्यांशों का भावार्थ लिखिए।

Question 1.
युयुत्सु ……………………. मरण के बाद।
Answer:
प्रस्तुत पंक्तियाँ कवि डॉ. धर्मवीर भारती लिखित ‘अंधायुग’ गीतिनाट्य से ली गई हैं। यह प्रसंग उस समय का है जब श्रीकृष्ण की जीवन-यात्रा समाप्त हो गई थी। युयुत्सु अश्वत्थामा से कहता है कि प्रभु की मृत्यु के बाद दूसरों का वध करने वालों को मुक्ति जरूर मिली होगी। उनका उद्धार हो जाएगा, लेकिन अब जो यह अंधा युग आने वाला हैं. इसमें मानव की रक्षा कौन करेगा? आखिर भगवान
कृष्ण ने एक कायर की भाँति मृत्यु को स्वीकार किया है।

Question 2.
अश्वत्थामा मस्तक पर।
Answer:
प्रस्तुत पंक्तियाँ कवि डॉ. धर्मवीर भारती लिखित ‘अंधायुग’ गीतिनाट्य से ली है। अश्वत्थामा युयुत्सु से कहता है, “भगवान श्रीकृष्ण मेरे शत्रु थे। मुझे इस बात का पता नहीं कि उन्होंने कायर की भाँति मृत्यु को स्वीकार किया था या नहीं? लेकिन मृत्यु के समय उनके स्वर्ण मस्तक पर शांति छाई हुई थी।

निम्नलिखित वाक्य सत्य है या असत्य लिखिए।

Question 1.
श्रीकृष्ण ने सभी का दायित्व अपने सिर पर लिया था।
Answer:
सत्य

Question 2.
श्रीकृष्ण ने सभी पर उत्तरदायित्व नहीं सौंपा।
Answer:
असत्य

कविता की पंक्तियाँ पूर्ण कीजिए।

Question 1.
अब तक मानव भविष्य को मैं जिलाता था …..
Answer:
लेकिन इस अंधे युग में मेरा एक अंश निष्क्रिय रहेगा, आत्मघाती रहेगा और विगलित रहेगा।

कृति पूर्ण कीजिए।

Question 1.

Answer:

समझकर लिखिए।

Question 1.
प्रभु अंत समय इससे बात करे रहे थे।
Answer:
शिकारी से

Question 2.
प्रभु ने अंत समय जो बाते की वह बातें इसने सुनी थी।
Answer:
वृद्ध ने

निम्नलिखित पद्यांशों का भावार्थ लिखिए।

Question 1.
वृद्ध ……………………. उठूगाम बारम्बार……. उलूंगा मैं बार-बार।
Answer:
‘प्रस्तुत पंक्तियाँ कवि डॉ. धर्मवीर भारती के ‘अंधायुग’ गीतिनाट्य से ली गई हैं। वृद्ध ने कहा, “ अंत समय में श्रीकृष्ण ने सभी से कहा कि यह उनका मरण नहीं है बल्कि यह तो शरीर का रूपांतरण है। अब तक उन्होंने सभी की जिम्मेदारी अपने सिर पर ली थी और अब वे सभी को अपनी-अपनी जिम्मेदारी सौंपकर इस संसार से अपने धाम चले जाएंगे। अब तक वे मानव भविष्य को जिलाते रहे, लेकिन आने वाले इस अंधे युग में उनका एक अंश संजय, युयुत्सु व अश्वत्थामा की भाँति निष्क्रिय रहेगा; आत्मघाती रहेगा; पिघला हुआ रहेगा क्योंकि इन सबकी उन्होंने अपने ऊपर जिम्मेदारी ली हैं।

समझकर लिखिए।

Question 1.
प्रभु का दायित्व इसमें स्थित रहेगा।
Answer:
मानव मन के वृत्त में।

Question 2.
प्रभु के दायित्व को आधार बनाकर मानव यह करेगा।
Answer:
नूतन निर्माण करेगा।

संजाल पूर्ण कीजिए।

Question 1.

Answer:

पद्यांश के आधार पर वाक्य पूर्ण कीजिए।

Question 1.
मानव पिछले ध्वसों पर………..
Answer:
नूतन निर्माण करेगा।

Question 2.
निर्भयता साहस ममता व रस के क्षण में……..
Answer:
प्रभु बार बार सक्रिय व जीवित हो उठेंगे।

समझकर लिखिए।

Question 1.
श्रीकृष्ण ने अपना दायित्व इन्हें सौंप दिया है
Answer:
पृथ्वी के हर प्राणी को ।

कृति ग (३) निम्नलिखित पदयांशों का भावार्थ लिखिए।

Question 1.
बोले ……………………. उलूंगा मैं बार बार।
Answer:
प्रस्तुत पंक्तियाँ कवि डॉ. धर्मवीर भारती के ‘अंधायुग’ गीतिनाट्य से ली गई हैं। प्रभू ने कहा कि अब उनका दायित्व सभी मानव लेंगे। उनका दायित्व मानव मन में स्थित रहेगा। इसके सहारे मानव सभी परिस्थितियों की मर्यादा को तोड़ते हुए अपने पिछले विनाश के स्थान पर नवनिर्माण करेगा। मानव मर्यादायुक्त आचरण का पालन करते हुए हमेशा नया निर्माण करना होगा। सभी को सृजन, साहस, निर्भयता एवं ममत्वपूर्ण व्यवहार करने होंगे। जब ऐसा होगा तब श्रीकृष्ण बार-बार सक्रिय व जीवित हो उठेंगे।

संजाल पूर्ण कीजिए।

Question 1.

Answer:

कविता की पंक्तियाँ पूर्ण कीजिए।

Question 1.
उसके इस ………….. सार्थकता पा जाएगा?
Answer:
उसके इस नये अर्थ में क्या हर छोटे-से-छोटा व्यक्ति विकृत, अर्धबर्बर, आत्मघाती, अनास्थामय अपने जीवन की सार्थकता पा जाएगा?

समझकर लिखिए।

Question 1.
इसके हाथ में हैं मानव जीवन
Answer:
मनुष्य के

Question 2.
मनुष्य यदि चाहे तो
Answer:
जीवन को नष्ट करे अथवा जीवन प्रदान करें।

कृति पूर्ण कीजिए।

Question 1.

Answer:

निम्नलिखित पद्यांशों का भावार्थ लिखिए।

Question 1.
अश्वत्थामा …………. पा जाएग।
Answer:
प्रस्तुत पंक्तियाँ कवि डॉ.धर्मवीर भारती के ‘अंधायुग’ गीतिकाव्य से ली गई हैं। अश्वत्थामा वृद्ध से पूछता है कि जैसे श्रीकृष्ण ने बताया वैसे उनके इस नए अर्थ के अनुसार छोटे से छोटा व्यक्ति क्या अपने जीवन की सार्थकता पाने में सफल सिद्ध हो सकता है? आखिर, भविष्य का मानव विकृत, हिंसक, आत्मघाती व ईश्वर के प्रति अनास्था रखने वाला होगा।

Question 2.
वृद्ध ………जीवन लो।
Answer:
प्रस्तुत पंक्तियाँ कवि डॉ. धर्मवीर भारती के ‘अंधायुग’ गीतिकाव्य से ली गई हैं। वृद्ध अश्वत्थामा को इस प्रश्न का उत्तर देते हुए कहता है कि निश्चय ही, मनुष्य भले ही अच्छा रहे या बुरा आखिर मानव जीवन उसी के ही हाथ में है। वह चाहे तो उसे नष्ट कर दे अथवा जीवन प्रदान करें।

Balbharti Maharashtra State Board Class 8 Geography Solutions Chapter 1 Local Time and Standard Time Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 8 Geography Solutions Chapter 1 Local Time and Standard Time

Class 8 Geography Chapter 1 Local Time and Standard Time Textbook Questions and Answers

1. Complete the sentence by selecting the correct option:

Question a.
The earth requires 24 hours for one rotation. In one hour, ……………. .
(a) 5 longitudes will face the sun
(b) 10 longitudes will face the sun
(c) 15 longitudes will face the sun
(d) 20 longitudes will face the sun
Answer:
(c) 15 longitudes will face the sun

Question b.
To calculate the difference between the local times of any two places on the earth, …………. .
(a) the noon time at both the places should be known
(b) the difference in degrees of their longitudes should be known
(c) the difference in standard times of both the places should be known
(d) changes need to be made according to International Date Line.
Answer:
(b) the difference in degrees of their longitudes should be known

Question c.
The difference between the local time of any two consecutive longitudes is …………… .
(a) 15 minutes
(b) 04 minutes
(c) 30 minutes
(d) 60 minutes
Answer:
(b) 04 minutes

2. Give geographical reasons for the following:

Question a.
The local time is decided by the noontime.
Answer:
1. When at a particular place, the sun reaches the maximum height in the sky it is assumed that almost half of the daytime is over and this time is considered 12 noon.

2. The time of a particular place as decided by the overhead position of the sun in the sky is considered as the local time of that place.

3. During the rotation of the earth, when a particular longitude comes exactly in front of the sun, it is considered as noontime (12 noon) on that longitude. This noontime is considered as local time of that longitude.

In this way, the local time is decided by the noontime.

Question b.
The local time at Greenwich is considered to be the International Standard Time.
Answer:

1. The International Standard Time has been decided according to 0° longitude.
2. 0° longitude passes through Greenwich, England.
3. For an international coordination, it is essential to bring compatibility between the standard times of various countries in the world.

For this purpose, the local time at Greenwich is considered to be the International Standard Time.

Question c.
The standard time of India has been decided by the local time at 82°30′ E longitude.
Answer:

1. With respect to the longitudinal extent, 82°30’E longitude passes through the middle of India.
2. The difference between the local time at 82°30’E longitude and the local time at the extreme east and west longitude passing through India is not more than one hour.

Therefore, the standard time of India has been decided by the local time at 82°30’E longitude.

Question d.
Canada has 6 different standard times.
Answer:

1. The longitudinal extent of Canada is between 52°37′ W and 141° W.
2. Thus, the difference between the extreme east and west longitude passing through Canada is of 88 degrees.
3. The difference between the local time at extreme east and west longitude passing through Canada is of 352 minutes i.e. 5 hours and 52 minutes.

Therefore, it is not practically helpful to consider single standard time in Canada. Therefore, for synchronizing the routine activities in the country, Canada has 6 different standard times.

3. Answer in brief:

Question a.
If it is 12 noon at 60° E longitude, then explain what would be the time at 30° W longitude.
Answer:
1. The difference between 60° E longitude and 30° W longitude will be of 90 degrees. (The difference of 60 degrees between 0° and 60° E + the difference of 30 degrees between 0° and 30° W = 90 degrees.)

2. Difference in local time = 90 × 4
= 360 minutes.
= 360 minutes ÷ 60 minutes = 6 hours.
3. Longitudes lying to the east of any longitude are ahead of the time of that longitude while those lying to the west are behind.
4. Therefore, if it is 12 noon at 60° E longitude, then it would be 6 a.m. at 30° W longitude, (behind by 6 hours)

Question b.
How is the standard time of a place determined?
Answer:
1. The local time at the longitude passing through the middle of a country/ place is generally considered as a standard time of that country/place.

2. If the difference between the local time at the extreme east and west longitude passing through a country is less than one or two hours, one standard time is considered for a country. Thus, there exists only one standard time in a country having comparatively less longitudinal (east-west) extent.

3. If the difference between the local time at the extreme east and west longitude passing through a country is more than one or two hours, more than one standard time zones are considered for a country. Thus, there exists more than one standard time zones in a country having comparatively more longitudinal (east-west) extent.

Question c.
A football match being played at Sao Paulo, Brazil started in India at 6 a.m. IST. Explain what would be the local time at Sao Paulo?
Answer:
1. Statement: For any longitude lying to the west of particular longitude, the local time decreases by 4 minutes for every longitude. (Sao Paulo is located to the west of India)

2. The difference between the longitudes of Sao Paulo and India = 127°30’.

3. Difference in local time = 127.5 × 4
= 510 minutes.
= 510 minutes ÷ 60 minutes
= 8 hours 30 minutes.

4. Thus, if it is 6 a.m. at India, it would be 9.30 p.m. of the previous day at Sao Paulo. Therefore, if a football match being played at Sao Paulo, Brazil started in India at 6 a.m. 1ST, the local time at Sao Paulo would be 9.30 p.m. of previous day.

4. Complete the following table:

If it is 10 pm on 21st June at Prime Meridian, write the dates and time at A, B and C in the following table:
(Note: The answer is given directly.)

Question 1.

Answer:

5. Write the situation of place A shown in these diagrams in the boxes below them:

Question 1.
Write the situation of place A shown in these diagrams in the boxes below them
(i) Sunrise
(ii) Midnight
(iii) Noon
(iv) Sunset

Answer:

Let’s recall: 

Question 1.
Why does the duration of day and night keep changing?
Answer:
The duration of day and night keeps changing as the earth is tilted at an angle of 23.5 degrees.

Question 2.
How many longitudes can be drawn on a world map keeping an interval of 1° each?
Answer:
360 longitudes can be drawn on a world map keeping an interval of 10 each.

Question 3.
The apparent movement of the sun from east to west is a result of what?
Answer:
The apparent movement of the sun from east to west is a result of the rotation of the earth from west to east.

Question 4.
What is the direction of the rotation of the earth?
Answer:
The direction of the rotation of the earth is from west to east.

Question 5.
While the earth rotates, how many longitudes face the sun daily?
Answer:
While the earth rotates, 360 longitudes face the sun dally.

Question 6.
At which longitude does the date change?
Answer:
The date changes at 1800 longitude.

Question 7.
How was the time measured in olden days?
Answer:
In olden days, the time was measured with the help of the natural events of sunrise and sunset and the instruments like Ghatikapaatra, sand timer, etc.

Question 8.
In present tunes, what are the instruments used for time measurement?
Answer:
In present times, the instruments like watches, calendars, etc. are used for time measurement.

Activity:

Question 1.
Look for the actual granny’s clock in Shri Acharya Atre’s poem: “Aajiche Ghadyal” (Granny’s clock). Look for this poem on the internet or in reference book.

Question 2.
Find out the velocity of the earth’s rotation in km/hour.

Class 8 Geography Chapter 1 Local Time and Standard Time Additional Important Questions and Answers

Complete the sentence by selecting the correct option:

Question a.
There are …………. time zones in the world.
(a) 360
(b) 24
(c) 100
(d) 365
Answer:
(b) 24

Question b.
Indian Standard Time is …………..
(a) ahead of Greenwich Mean Time by 5 hours 30 minutes
(b) ahead of Greenwich Mean Time by 3 hours 50 minutes
(c) behind Greenwich Mean Time by 5 hours 30 minutes
(d) behind Greenwich Mean Time by 3 hours 50 minutes
Answer:
(a) ahead of Greenwich Mean Time by 5 hours 30 minutes

Examine the following statements and correct the incorrect ones:

Question 1.
Mumbai is located at 73° E longitude.
Answer:
Correct.

Question 2.
In India, three standard times exist.
Answer:
Incorrect.
Correct statement: In India, one standard time exists.

Question 3.
If the difference between two longitudes is of 20 degrees, the difference in their local times will be of four hours.
Answer:
Incorrect.
Correct statement: If the difference between two longitudes is of 20 degrees, the difference in their local times will be of 1 hour and 20 minutes.

Answer the following questions in one sentence each:

Question 1.
Think about it.
Question a.
What is the maximum number of local times that can there be in the world?
Answer:
The maximum number of local times that can be there in the world is 360.

Question b.
How many longitudes pass the sun in one hour?
Answer:
15 longitudes pass the sun in one hour.

Question 2.
How much time does the earth take to complete one rotation?
Answer:
The earth takes nearly 24 hours to complete one rotation.

Question 3.
What is meant by the local time of a place?
Answer:
When the sun is directly overhead at a particular place, it is considered as noon at that place and that time is considered as a local time of that place.

Solve the following:

1. Try this:

Question A.
Mashad, a town in Iran, is located on the 60° E longitude. When it is 12 noon at Greenwich, calculate the local time of Mashad town.
Answer:
1. Statement: As we move towards the east of the Prime Meridian, the local time increases by four minutes for every longitude. (Mashad is located to the east of Greenwich.)

2. The difference between the longitudes of Greenwich and Mashad = 60°.

3. Difference in local time
= 60 × 4
= 240 minutes.
= 240 minutes ÷ 60 minutes
= 4 hours. ‘

4. Therefore, when it is noon at Greenwich, it would be 4 p.m. at Mashad. (Ahead of Greenwich Mean Time by 4 hours.)

Question B.
Manaus city in Brazil is located on 60° W longitude. What would be the local time at Manaus when it is 12 noon at Greenwich?
Answer:
1. Statement: As we move towards the west of the Prime Meridian, the local time decrease by four minutes for every longitude. (Manaus is located to the west of Greenwich.)

2. The difference between the longitudes of Greenwich and Manaus = 60°.

3. Difference in local time = 60 × 4
= 240 minutes.
= 240 minutes ÷ 60 minutes
= 4 hours.

4. Therefore, when it is noon at Greenwich, it would be 8 a.m. at Manaus. (Behind Greenwich Mean Time by 4 hours.)

2. Can you tell?

Question A.
Mumbai is located at 73° E longitude. Kolkata is located at 88° E longitude. Find the difference between the longitudes of these two cities.
Answer:
Mumbai is located at 73° E longitude. Kolkata is located at 88° E longitude. Therefore, the difference between the longitudes of these two cities would be of (88 – 73 = 15) 15°.

Question B.
If the local time at Mumbai is 3 p.m. then what would be the local time at Kolkata?
Answer:
1. Statement: For any longitude lying to the east of particular longitude, the local time increases by 4 minutes for every longitude.

2. The difference between the longitudes of Mumbai and Kolkata = 15°.

3. Difference in local time = 15 × 4
= 60 minutes.
= 60 minutes -r- 60 minutes = 1 hour.

4. Kolkata is located to the east of Mumbai. Therefore, if it is 3 p.m. at Mumbai, then it would be 4 p.m. at Kolkata. (ahead by 1 hour)

Answer the following questions in brief:

Question a.
Give brief information about the Indian Standard Time.
Answer:
1. The Indian Standard Time (IST) is calculated on the basis of 82°30’E longitude, passing through Mirzapur, near Allahabad in Uttar Pradesh.

2. With respect to longitudinal extent, 82°30′ E longitude passes through the middle of India.

3. When the sun is directly overhead on this longitude, it is considered that it is 12 noon at every place in India. Thus, the local time at 82°30’E longitude is considered as the Standard time of India.

4. The difference between the local time at 82°30′ E longitude and the local time at the extreme east and west longitude passing through India is not more than one hour.

Question b.
Give brief information about Universal Standard Time.
Answer:
1. As the standard time is essential for synchronizing routine activities in a country with comparatively more j longitudinal extent, the universal standard i time is essential for an international coordinating between the countries in the world.

2. For this purpose, the local time at Greenwich (Greenwich Mean Time) in England is considered to be the International/Universal Standard Time (UST).

3. With reference to GMT, the differences in standard times of various countries in the world are calculated.

4. The Indian Standard Time is ahead of GMT by 5 hours and 30 minutes. For example, if it is 12 noon at Greenwich, then it would be 5.30 p.m. in India.

Give geographical reasons for the following:

Question a.
In the countries with larger area, it is convenient to use one standard time instead of more than one local times.
Answer:
1. In the countries with larger area, much difference between the local times at extreme east and west longitude passing through the countries is found.

2. In such countries, if more than one local times are followed, it could lead to lots of confusion and chaos.

3. For example, if more than one local times are followed, it will become difficult to synchronize the timetable of railways, airways, etc. Therefore, in countries with comparatively larger area, it is convenient to use one standard time instead of more than one local times.

Study the following map/ figure/graph and answer the following questions:

1. Can you tell?
Study the figure 1.3 given on page 4 of the textbook and answer the following questions:

Question 1.
Between which longitudes does the region experience daytime?
Answer:
The region shown between 90° W and 90° E in the figure experiences daytime.

Question 2.
Which longitudes experience noon and the midnight respectively?
Answer:
0° longitude and 180° longitude experience noon and the midnight respectively.

Question 3.
Edward from New Orleans is on which longitude?
Answer:
Edward from New Orleans is on 90° W longitude.

Question 4.
What is the time at Accra city?
Answer:
It is 12 noon at Accra city.

Question 5.
At the same time, what is Sharad from Patna and Yakaito from Japan doing? What time is it in these cities?
Answer:
At the same time, Sharad is returning home from school and Yakaito from Japan is preparing for the night’s sleep. At the same time, it is 5.30 p.m. at Patna and 9.20 p.m. at Japan.

Question 6.
Select any one longitude. Calculate the local time of the longitudes lying 1° to the west and east of this longitude.
Answer:

1. The selected longitude: 60° E. The local time of the selected longitude is 4 p.m.
2. The local time of the longitude lying 1° to the west of the selected longitude (59° E) will be 3.56 p.m.
3. The local time of the longitude lying 1° to the east of the selected longitude (61° E) will be 4.04 p.m.

2. Can you tell?

Study the figure 1.4 given on page 6 of the textbook and answer the following questions:

Question 1.
Considering the longitudinal extent of India, how many longitudes with a difference of 1° can be drawn on a map?
Answer:
Considering the longitudinal extent of India, 29 longitudes with a difference of 1° can be drawn on a map.

Question 2.
By how many minutes do two consecutive longitudes differ?
Answer:
Two consecutive longitudes differ by 4 minutes.

Question 3.
What is the value of degrees of longitude at Mirzapur?
Answer:
The value of longitude at Mirzapur is 82°30′.

Question 4.
If it is 8 a.m. at 82°30′ E, what would be the time in their clocks at the following places: Jammu, Madurai, Jaisalmer, Guwahati.
Answer:
If it is 8 a.m. at 82°30′ E, it would be 8 a.m. in their clocks at Jammu, Madurai, Jaisalmer, Guwahati.
(Note: In India, the time at 82°30′ E is considered as standard time of India and therefore the same time will be considered everywhere in India.)

Question 5.
Though the distance between them is more why doesn’t the standard time differ in these places?
Answer:
In India, the time at 82°30′ E is considered as standard time of India and therefore the same time will be considered everywhere in India. Therefore, though the distance between them is more, the standard time doesn’t differ in these places.

Thought-Provoking Questions:

1. Think about it.

Question 1.
At the poles, sunrise occurs on one equinox and sunsets on the next equinox. If you happen to be at any of the poles during this time, then what would be the route of the sun in the daytime?
Answer:
At the poles, sun rises on one equinox and sets on the next equinox. If we happen to be at any of the poles during this time, then the following would be the route of the sun in the daytime:
(A) North Pole:
1. The sun will rise at the North Pole approximately on March 21.

2. Approximately, from March 21 to June 21, the sun will rise higher in the sky with each advancing day. It will reach the maximum height approximately on June 21. Approximately, from June 21 to September 21, the sun will start sinking towards the horizon in the sky with each advancing day.

3. The sun will set at the North Pole approximately on September 21.

(B) South Pole:
1. The sun will rise at the South Pole approximately on September 21.

2. Approximately, from September 21 to December 21, the sun will rise higher in the sky with each advancing day. It will reach the maximum height approximately on December 21.

3. Approximately, from December 21 to March 21, the sun will start sinking towards the horizon in the sky with each advancing day.

4. The sun will set at the South Pole approximately on March 21.

Question 2.
On which day, would the sun appear at the highest point in the sky?
Answer:

1. On the North Pole, approximately on June 21, the sun would appear at the highest point in the sky.
2. On the South Pole, approximately on December 21, the sun would appear at the highest point in the sky.

2. Use your brainpower:

Question 1.
Tick ✓ the time in the boxes which you can tell without using clock.
(Note: The answer is given directly.)

Answer:

3. Give it a try.

Question 1.
If it is 8 a.m. in India, what is the time in Greenwich?
Answer:
If it is 8 a.m. in India, it would be 2.30 a.m. in Greenwich.

Question 2.
When it is 2 p.m. in India, in which countries would it be 2 p.m. exactly?
Answer:
When it is 2 p.m. in India, it would be 2 p.m. exactly in Sri Lanka.

Question 3.
When it is 9 a.m. in India, what would be the time at 82°30′ W longitude?
Answer:
When it is 9 a.m. in India, it would be 10 p.m. of the previous day at 82°30′ W longitude.

Question 4.
What would be the time at Prime Meridian when a new day starts at 180° longitude?
Answer:
It would be 12 noon at Prime Meridian when a new day starts at 180° longitude.

4. Think about it.

Question 1.
In which of the following countries, does only one standard time exist? (Mexico, Sri Lanka, New Zealand, China)
Answer:
From the given countries, only one time exists in Sri Lanka, New Zealand and China.

Question 2.
Why does a country having a large latitudinal extent have only one standard time?
Answer:
The places lying on the same longitude have the same local time. Therefore, a country having a large latitudinal extent have only one standard time.

Question 3.
Which discrepancies will emerge if more than one local times, instead of one standard time are followed in a country with huge longitudinal extent?
Answer:
The following discrepancies will emerge if more than one local times, instead of one standard time are followed in a country with huge longitudinal extent:

1. It will become difficult to synchronize the timetable of rail transportation, air transportation, etc. in a country.
2. It will become difficult to synchronize the timings of schools, colleges, banks, libraries, etc. in a country.
3. We will always have to adjust the time in watch as we move from one place to another in a country.

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