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Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 6 Answers Solutions Chapter 1 Geometrical Constructions.

Geometrical Constructions Class 7 Maths Chapter 1 Practice Set 6 Solutions Maharashtra Board

Std 7 Maths Practice Set 6 Solutions Answers

Question 1.
Write the names of pairs of congruent line segments. (Use a divider to find them.)

i. ___
ii. ___
iii. ___
iv. ___
Solution:
i. seg BG ≅ seg CG
ii. seg NG ≅ seg MG ≅ seg EG ≅ seg RG

Question 2.
On the line below, the distance between any two adjoining points shown on it is equal. Hence, fill in the blanks.

i. seg AB ≅ seg ___
ii. seg AP ≅ seg ___
iii. seg AC ≅ seg ___
iv. seg ___ ≅ seg BY
v. seg __ ≅ seg YQ
vi. seg BW ≅ seg ___
Solution:
i. BC
ii. QW
iii. QZ
iv. AZ
v. AY
vi. AC

Note: The above problem has many solutions. Students may write solutions other than the ones given.

Maharashtra Board Class 7 Maths Chapter 1 Geometrical Constructions Practice Set 6 Intext Questions and Activities

Question 1.
Try to draw triangles with the following data. Can you draw these triangles. If not, look for the reason why you could not draw so. (Textbook pg. no. 7)
i. ∆ABC in which m∠A = 85°, m∠B = 115°, l(AB) = 5cm.
Solution:
m∠A + m∠B = 85° + 115°
= 200°>180°
But the sum of the measures of the angles of a triangle is 180°
Hence, ∆ABC cannot be drawn.

ii. ∆PQR in which l(QR) = 2cm, l(PQ) = 4cm, l(PR) = 2cm.
Solution:
l(QR) + l(PR) = 2 cm + 2cm
= 4 cm
= l(PQ)
But in a triangle, the sum of the length of any two sides of a triangle is always greater than the length of the third side.
Hence, ∆PQR cannot be drawn.

Question 2.
Draw ∆ABC such that l(BC) = 8 cm, l(CA) = 6 cm, m∠ABC = 40°.
Draw a ray to make an angle of 40° with the base BC, l(BC) = 8 cm. We have to obtain point ‘A’ on the ray. With ‘C’ as the centre, draw an arc of radius 6 cm to do so. What do we observe? The arc intersects the ray in two different points. Thus, we get two triangles of two different shapes having the given measures. (Textbook pg. no. 7)
Solution:

Here ∠B is an acute angle. ∠C can be an acute angle or an obtuse angle.
Hence we get two triangles of two different shapes.

Question 3.
Can a triangle be drawn if the three angles are given, but not any side? How many such triangles can be drawn? (Textbook pg. no. 7)
Solution:
Yes a triangle can be drawn.
Since the length of side is not given, any length of side can be selected and then triangle can be constructed. We will get different triangles for different length of sides.

Question 4.
Using the ruler, measure the lengths of seg AB and seg PQ. Are they of same length? Trace the seg AB on a sheet of transparent paper. Now place this new segment on PQ verify that if point A is placed on point P, then B falls on Q. (Textbook pg. no. 7)

l(AB) = ___
l(PQ) = ___
Solution:
l(AB) = 4 cm
l(PQ) = 4 cm
Since the length of two segments is the same, if placed on one another, they will coincide.

Question 5.
From the shape shown below, write the names of the pairs of congruent line segments. (Textbook pg. no. 8)
i. seg AB ≅ seg DC
ii. seg AE ≅ seg BH
iii. seg EF ≅ seg ___
iv. seg DF ≅ seg ___

Solution:
seg EF ≅ seg AD ≅ seg BC ≅ seg HG
seg DF ≅ seg CG ≅ seg AE ≅ seg BH

Question 6.
Take a rectangular paper. Place two opposite sides upon each Other. What do you observe? (Textbook pg. no. 7)
Solution:
Opposite sides of the rectangular paper coincide and hence are congruent.

Std 7 Maths Digest

• Practice Set 1 Class 7 Answers
• Practice Set 2 Class 7 Answers
• Practice Set 3 Class 7 Answers
• Practice Set 4 Class 7 Answers
• Practice Set 5 Class 7 Answers
• Practice Set 6 Class 7 Answers
• Practice Set 7 Class 7 Answers

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 2 Answers Solutions Chapter 1 Geometrical Constructions.

Geometrical Constructions Class 7 Maths Chapter 1 Practice Set 2 Solutions Maharashtra Board

Std 7 Maths Practice Set 2 Solutions Answers

Question 1.
Draw triangles with the measures given below:
i. In ∆ABC, l(AB) = 5.5 cm, l(BC) = 4.2 cm, l(AC) = 3.5 cm.
ii. In ∆STU, l(ST) = 7 cm, l(TU) = 4 cm, l(SU) = 5 cm.
iii. In ∆PQR, l(PQ) = 6 cm, l(QR) = 3.8 cm, l(PR) = 4.5 cm.
Solution:
i. In ∆ABC, l(AB) = 5.5 cm, l(BC) = 4.2 cm, l(AC) = 3.5 cm.

ii. In ∆STU, l(ST) = 7 cm, l(TU) = 4 cm, l(SU) = 5 cm.

iii. In ∆PQR, l(PQ) = 6 cm, l(QR) = 3.8 cm, l(PR) = 4.5 cm.

Question 2.
Draw an isosceles triangle with base 5 cm and the other sides 3.5 cm each.
Solution:

Question 3.
Draw an equilateral triangle with side 6.5 cm.
Solution:

Question 4.
Choose the lengths of the sides yourself and draw one equilateral, one isosceles and one scalene triangle.
Solution:
i. Equilateral triangle LMN, l(LM) = l(MN) = l(LN) = 4 cm.

ii. Isosceles triangle STU, l(ST) = l(TU) = 4cm, l(SU) = 6 cm

iii. Scalene triangle XYZ, l(XY) = 4.5 cm, l(XY) = 6.5 cm, l(XZ) = 5.5 cm

Maharashtra Board Class 7 Maths Chapter 1 Geometrical Constructions Practice Set 2 Intext Questions and Activities

Question 1.
Draw ∆ABC such that l(AB) = 4 cm, and l(BC) = 3 cm. (Textbook pg. no. 3)

1. Can this triangle be drawn?
2. A number of triangles can be drawn to fulfill these conditions. Try it out.
3. Which further condition must be placed if we are to draw a unique triangle using the above information?

Solution:

1. ∆ABC triangle cannot be drawn as length of third side is not given.
2. For ∆ABC to draw l(AC) > l(AB) + l(BC)
   i.e., l(AC) > 4 + 3
   i.e., l(AC) > 7 cm
   ∴ number of triangles can be drawn if l(AC) > 7 cm
3. l(AC) > l(AB) + l(BC) is the required condition to draw a unique triangle.

Std 7 Maths Digest

• Practice Set 1 Class 7 Answers
• Practice Set 2 Class 7 Answers
• Practice Set 3 Class 7 Answers
• Practice Set 4 Class 7 Answers
• Practice Set 5 Class 7 Answers
• Practice Set 6 Class 7 Answers
• Practice Set 7 Class 7 Answers

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 34 Answers Solutions Chapter 8 Algebraic Expressions and Operations on them.

Algebraic Expressions and Operations on them Class 7 Maths Chapter 8 Practice Set 34 Solutions Maharashtra Board

Std 7 Maths Practice Set 34 Solutions Answers

Question 1.
Subtract the second expression from the first.
i. (4xy – 9z); (3xy – 16z)
ii. (5x + 4y + 7z); (x + 2y + 3z)
iii. (14x² + 8xy + 3y²); (26x² – 8xy – 17y²)
iv. (6x² + 7xy + 16y²); (16x² – 17xy)
v. (4x + 16z); (19y – 14z + 16x)
Solution:
i. (4xy – 9z) – (3xy – 16z)
= 4xy – 9z – 3xy + 16z
= (4xy – 3xy) + (16z – 9z)
= xy + 7z

ii. (5x + 4y + 7z) – (x + 2y + 3z)
= 5x + 4y + 7z – x – 2y – 3z
= (5x – x) + (4y – 2y) + (7z – 3z)
= 4x + 2y + 4z

iii. (14x² + 8xy + 3y²) – (26x² – 8xy – 17y²)
= 14x² + 8xy + 3y² – 26x² + 8xy + 17y²
= (14x² – 26x²) + (8xy + 8xy) + (3y² + 17y²)
= -12x² + 16xy + 20y²

iv. (6x² + 7xy + 16y²) – (16x² – 17xy)
= 6x² + 7xy + 16y² – 16x² + 17xy
= (6x² – 16x²) + (7xy + 17xy) + 16y²
= -10x² + 24xy + 16y²

v. (4x + 16z) – (19y— 14z + 16x)
= 4x + 16z – 19y + 14z – 16x
= (4x – 16x) – 19y + (16z + 14z)
= -12x – 19y + 30z

Std 7 Maths Digest

• Practice Set 32 Class 7 Answers
• Practice Set 33 Class 7 Answers
• Practice Set 34 Class 7 Answers
• Practice Set 35 Class 7 Answers
• Practice Set 36 Class 7 Answers
• Practice Set 37 Class 7 Answers
• Practice Set 38 Class 7 Answers
• Practice Set 39 Class 7 Answers
• Practice Set 40 Class 7 Answers
• Practice Set 41 Class 7 Answers
• Practice Set 42 Class 7 Answers
• Practice Set 43 Class 7 Answers

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Miscellaneous Problems Set 1 Answers Solutions.

Maharashtra Board Miscellaneous Problems Set 1 Class 7 Maths Solutions

Question 1.
Solve the following:
i. (-16) × (-5)
ii. (72) ÷ (-12)
iii. (-24) × (2)
iv. 125 ÷ 5
v. (-104) ÷ (-13)
vi. 25 × (-4)
Solution:
i. (-16) × (-5) = 80

ii. 72 ÷ (-12) = \(\frac { 72 }{ -12 }\)
= \(\frac{1}{(-1)} \times \frac{72}{12}\)
(-1) × 12
= -6

iii. (-24) × 2 = -48

iv. 125 ÷ 5 = \(\frac { 125 }{ 5 }\)
= 25

v. (-104) ÷ (-13) = \(\frac { -104 }{ -13 }\)
= \(\frac { 104 }{ 13 }\)
= 8

vi. 25 × (-4) = -100

Question 2.
Find the prime factors of the following numbers and find their LCM and HCF:
i. 75,135
ii. 114,76
iii. 153,187
iv. 32,24,48
Solution:
i. 75 = 3 × 25
= 3 × 5 × 5
135 = 3 × 45
= 3 × 3 × 15
= 3 × 3 × 3 × 5
∴ HCF of 75 and 135 = 3 × 5
= 15
LCM of 75 and 135 = 3 × 5 × 5 × 3 × 3
= 675

ii. 114 = 2 × 57
= 2 × 3 × 19
76 = 2 × 38
= 2 × 2 × 19
∴ HCF of 114 and 76 = 2 × 19
= 38
LCM of 114 and 76 = 2 × 19 × 3 × 2
= 228

iii. 153 = 3 × 51
= 3 × 3 × 17
187 = 11 × 17
∴ HCF of 153 and 187 = 17
LCM of 153 and 187 = 17 × 3 × 3 × 11
= 1683

iv. 32 = 2 × 16
= 2 × 2 × 8
= 2 × 2 × 2 × 4
= 2 × 2 × 2 × 2 × 2
24 = 2 × 12
= 2 × 2 × 6
= 2 × 2 × 2 × 3
48 = 2 × 24
= 2 × 2 × 12
= 2 × 2 × 2 × 6
= 2 × 2 × 2 × 2 × 3
∴ HCF of 32, 24 and 48 = 2 × 2 × 2
= 8
LCM of 32,24 and 48 = 2 × 2 × 2 × 2 × 2 × 3
= 96

Question 3.
Simplify:
i. \(\frac { 322 }{ 391 }\)
ii. \(\frac { 247 }{ 209 }\)
iii. \(\frac { 117 }{ 156 }\)
Solution:
i. \(\frac { 322 }{ 391 }\)

ii. \(\frac { 247 }{ 209 }\)

iii. \(\frac { 117 }{ 156 }\)

Question 4.
i. 784
ii. 225
iii. 1296
iv. 2025
v. 256
Solution:
i. 784

∴ 784 = 2 × 2 × 2 × 2 × 7 × 7
∴ √784 = 2 × 2 × 7
= 28

ii. 225

∴ 225 = 3 × 3 × 5 × 5
∴ √225 = 3 × 5
= 15

iii. 1296

∴ 1296 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3
∴ √1296 = 2 × 2 × 3 × 3
= 36

iv. 2025

∴ 2025 = 3 × 3 × 3 × 3 × 5 × 5
∴ √2025 = 3 × 3 × 5
= 45

v. 256

∴ 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
∴ √256 = 2 × 2 × 2 × 2
= 16

Question 5.
There are four polling booths for a certain election. The numbers of men and women who cast their vote at each booth is given in the table below. Draw a joint bar graph for this data.

Polling Booths	Navodaya Vidyalaya	Vidyaniketan School	City High School	Eklavya School
Women	500	520	680	800
Men	440	640	760	600

Solution:

Question 6.
Simplify the expressions:
i. 45 ÷ 5 + 120 × 4 – 12
ii. (38 – 8) × 2 ÷ 5 + 13
iii. \(\frac{5}{3}+\frac{4}{7} \div \frac{32}{21}\)
iv. 3 × {4 [85 + 5 – (15 – 3)] + 2}
Solution:
i. 45 ÷ 5 + 120 × 4 – 12
= 9 + 80 – 12
= 89 – 12
= 77

ii. (38 – 8) × 2 ÷ 5 + 13
= 30 × 2 ÷ 5 + 13
= 60 ÷ 5 + 13
= 12 + 13
= 25

iii. \(\frac{5}{3}+\frac{4}{7} \div \frac{32}{21}\)
\(\frac{5}{3}+\frac{4}{7} \times \frac{21}{32}\)
\(\frac{5}{3}+\frac{3}{8}=\frac{40}{24}+\frac{9}{24}\)
\(\frac{49}{24}\)

iv. 3 × {4 [85 + 5 – (15 – 3)] + 2}
= 3 × {4[90 – 5] + 2}
= 3 × {4 × 85 + 2}
= 3 × (340 + 2)
= 3 × 342
= 1026

Question 7.
Solve:
i. \(\frac{5}{12}+\frac{7}{16}\)
ii. \(3 \frac{2}{5}-2 \frac{1}{4}\)
iii. \(\frac{12}{5} \times \frac{(-10)}{3}\)
iv. \(4 \frac{3}{8} \div \frac{25}{18}\)
Solution:
i. \(\frac{5}{12}+\frac{7}{16}\)

ii. \(3 \frac{2}{5}-2 \frac{1}{4}\)

iii. \(\frac{12}{5} \times \frac{(-10)}{3}\)
= 4 × (-2)
= -8

iv. \(4 \frac{3}{8} \div \frac{25}{18}\)
= \(\frac{7}{4} \times \frac{9}{5}\)
= \(\frac { 63 }{ 20 }\)

Question 8.
Construct ∆ABC such that m∠A = 55°, m∠B = and l(AB) = 5.9 cm.
Solution:

Question 9.
Construct ∆XYZ such that, l(XY) = 3.7 cm, l(YZ) = 7.7 cm, l(XZ) = 6.3 cm.
Solution:

Question 10.
Construct ∆PQR such that, m∠P = 80°, m∠Q = 70°, l(QR) = 5.7 cm.
Ans:
In ∆PQR,
m∠P + m∠Q + m∠R = 180° …. (Sum of the measures of the angles of a triangle is 180°)
∴ 80 + 70 + m∠R = 180
∴ 150 + m∠R = 180
∴ m∠R = 180 – 150
∴ m∠R = 30°

Question 11.
Construct ∆EFG from the given measures. l(FG) = 5 cm, m∠EFG = 90°, l(EG) = 7 cm.
Solution:

Question 12.
In ∆LMN, l(LM) = 6.2 cm, m∠LMN = 60°, l(MN) 4 cm. Construct ∆LMN.
Solution:

Question 13.
Find the measures of the complementary angles of the following angles:
i. 35°
ii. a°
iii. 22°
iv. (40 – x)°
Solution:
i. Let the measure of the complementary
angle be x°.
35 + x = 90
∴35 + x-35 = 90 – 35
….(Subtracting 35 from both sides)
∴x = 55
∴The complementary angle of 35° is 55°.

ii. Let the measure of the complementary angle be x°.
a + x = 90
∴a + x – a = 90 – a
….(Subtracting a from both sides)
∴x = (90 – a)
∴The complementary angle of a° is (90 – a)°.

iii. Let the measure of the complementary angle be x°.
22 + x = 90
∴22 + x – 22 = 90 – 22
….(Subtracting 22 from both sides)
∴x = 68
∴The complementary angle of 22° is 68°.

iv. Let the measure of the complementary angle be a°.
40 – x + a = 90
∴40 – x + a – 40 + x = 90 – 40 + x
….(Subtracting 40 and adding x on both sides)
∴a = (50 + x)
∴The complementary angle of (40 – x)° is (50 + x)°.

Question 14.
Find the measures of the supplements of the following angles:
i. 111°
ii. 47°
iii. 180°
iv. (90 – x)°
Solution:
i. Let the measure of the supplementary
angle be x°.
111 + x = 180
∴ 111 + x – 111 = 180 – 111
…..(Subtracting 111 from both sides)
∴ x = 69
∴ The supplementary angle of 111° is 69°.

ii. Let the measure of the supplementary angle be x°.
47 + x = 180
∴47 + x – 47 = 180 – 47
….(Subtracting 47 from both sides)
∴x = 133
∴The supplementary angle of 47° is 133°.

iii. Let the measure of the supplementary angle be x°.
180 + x = 180
∴180 + x – 180 = 180 – 180
….(Subtracting 180 from both sides)
∴x = 0
∴The supplementary angle of 180° is 0°.

iv. Let the measure of the supplementary angle be a°.
90 – x + a = 180
∴90 – x + a – 90 + x = 180 – 90+ x
….(Subtracting 90 and adding x on both sides)
∴a = 180 – 90 + x
∴a = (90 + x)
∴The supplementary angle of (90 – x)° is (90 + x)°.

Question 15.
Construct the following figures:
i. A pair of adjacent angles
ii. Two supplementary angles which are not adjacent angles.
iii. A pair of adjacent complementary angles.
Solution:
i.

ii.

iii.

Question 16.
In ∆PQR the measures of ∠P and ∠Q are equal and m∠PRQ = 70°, Find the measures of the following angles.

1. m∠PRT
2. m∠P
3. m∠Q

Solution:
Here, ∠PRQ and ∠PRT are angles in a linear pair.
m∠PRQ + m∠PRT = 180°
∴70 + m∠PRT = 180
∴m∠PRT = 180 – 70
∴m∠PRT = 110°
Now, ∠PRT is the exterior angle of ∆PQR.
∴m∠P + m∠Q = m∠PRT
∴m∠P + m∠P = m∠PRT ….(The measures of ∠P and ∠Q is same)
∴2m∠P = 110
∴m∠P = \(\frac { 110 }{ 2 }\)
∴m∠P = 55°
∴m∠Q =

Question 17.
Simplify
i. 5⁴ × 5³
ii. \(\left(\frac{2}{3}\right)^{6} \div\left(\frac{2}{3}\right)^{9}\)
iii. \(\left(\frac{7}{2}\right)^{8} \times\left(\frac{7}{2}\right)^{-6}\)
iv. \(\left(\frac{4}{5}\right)^{2} \div\left(\frac{5}{4}\right)\)
Solution:
Simplify
i. 5⁴ × 5³
= 5⁴⁺³
= 5⁷

ii. \(\left(\frac{2}{3}\right)^{6} \div\left(\frac{2}{3}\right)^{9}\)

iii. \(\left(\frac{7}{2}\right)^{8} \times\left(\frac{7}{2}\right)^{-6}\)

iv. \(\left(\frac{4}{5}\right)^{2} \div\left(\frac{5}{4}\right)\)

Question 18.
Find the value:
i. 17¹⁶  ÷ 17¹⁶
ii. 10^-3
iii. (2³)²
iv. 4⁶ × 4^-4
Solution:
i. 17¹⁶  ÷ 17¹⁶
= 17⁰
= 1

ii. 10^-3
= \(\frac{1}{10^{3}}\)
= \(\frac{1}{1000}\)

iii. (2³)²
= 2^3×2
= 2⁶
= 2 × 2 × 2 × 2 × 2 × 2
= 64

iv. 4⁶ × 4^-4
= 4^6+(-4)
= 4²
= 4 × 4
= 16

Question 19.
Solve:
i. (6a – 5b – 8c) + (15b + 2a – 5c)
ii. (3x + 2y) (7x – 8y)
iii. (7m – 5n) – (-4n – 11m)
iv. (11m – 12n + 3p) – (9m + 7n – 8p)
Solution:
i. (6a – 5b – 8c) + (15b + 2a – 5c)
= (6a + 2a) + (-5b + 15b) + (-8c – 5c)
= 8a + 10b – 13c

ii. (3x + 2y) (7x – 8y)
= 3x × (7x – 8y) + 2yx (7x – 8y)
= 21x² – 24xy + 14xy – 16y²
= 21x² – 10xy – 16y²

iii. (7m – 5n) – (-4n – 11m)
= 7m – 5n + 4n + 11m
= (7m + 11m) + (-5n + 4n)
= 18m – n

iv. (11m – 12n + 3p) – (9m + 7n – 8p)
= 11m – 12n + 3p – 9m – 7n + 8p
= (11m – 9m) + (-12n – 7n) + (3p + 8p)
= 2m – 19n + 11p

Question 20.
Solve the following equations:
i 4(x + 12) = 8
ii. 3y + 4 = 5y – 6
Solution:
i. 4(x + 12) = 8
∴4x + 48 = 8
∴4x + 48 – 48 = 8 – 48
….(Subtracting 48 from both sides)
∴ 4x = -40
∴ x = \(\frac { -40 }{ 4 }\)
∴ x = -10

ii. 3y + 4 = 5y – 6
∴ 3y + 4 + 6 = 5y – 6 + 6
….(Adding 6 on both sides)
∴ 3y + 10 = 5y
∴ 3y + 10 – 3y = 5y – 3y
….(Subtracting 3y from both sides)
∴ 10 = 2y
∴ 2y = 10
∴ y = \(\frac { 10 }{ 2 }\)
∴ y = 5

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Miscellaneous Problems Set 2 Answers Solutions.

Maharashtra Board Miscellaneous Problems Set 2 Class 7 Maths Solutions

Question 1.
Angela deposited Rs 15000 in a bank at a rate of 9 p.c.p.a. She got simple interest amounting to Rs 5400. For how many years had she deposited the amount?
Solution:
Here, P = Rs 15000, R = 9 p.c.p.a., I = Rs 5400

∴ T = 4
∴ Angela had deposited the amount for 4 years.

Question 2.
Ten men take 4 days to complete the task of tarring a road. How many days would 8 men take?
Solution:
Let us suppose that 8 men require x days to tar the road.
Number of days required by 10 men to tar the road = 4
The number of men and the number of days required to tar the road are in inverse proportion.
∴ 8 × x = 10 x 4
∴ \(x=\frac{10 \times 4}{8}\)
∴ x = 5
∴ 8 men will require 5 days to tar the road.

Question 3.
Nasruddin and Mahesh invested Rs 40,000 and Rs 60,000 respectively to start a business. They made a profit of 30%. How much profit did each of them make?
Solution:
Total amount invested = Rs 40,000 + Rs 60,000
= Rs 1,00,000
Profit earned = 30%
∴ Total profit = 30% of 1,00,000
= \(\frac { 30 }{ 100 }\) × 100000
= Rs 30000
Proportion of investment = 40000 : 60000
= 2:3 …. (Dividing by 20000)
Let Nasruddin’s profit be Rs 2x and Mahesh’s profit be Rs 3x.
∴ 2x + 3x = 30000
∴ 5x = 30000
∴ x = \(\frac { 30000 }{ 5 }\).
∴ x = 6000
∴ Nasruddin’s profit = 2x = 2 × 6000 = Rs 12000
Mahesh’s profit = 3x = 3 × 6000 = Rs 18000
∴ The profits of Nasruddin and Mahesh are Rs 12000 and Rs 18000 respectively.

Question 4.
The diameter of a circle is 5.6 cm. Find its circumference.
Solution:
Diameter of the circle (d) = 5.6 cm
Circumference = πd
= \(\frac{22}{7} \times 5.6\)
= \(\frac{22}{7} \times \frac{56}{10}\)
= 17.6 cm
∴ The circumference of the circle is 17.6 cm.

Question 5.
Expand:
i. (2a – 3b)²
ii. (10 + y)²
iii. \(\left(\frac{p}{3}+\frac{q}{4}\right)^{2}\)
iv. \(\left(y-\frac{3}{y}\right)^{2}\)
Solution:
i. Here, A = 2a and B = 3b
∴ (2a – 3b)² = (2a)² – 2 × 2a × 3b + (3b)²
…. [(A – B)² = A² – 2AB + B²]
= 4a² – 12ab + 9b²

ii. Here, a = 10 and b = y
(10 + y)² = 102 + 2 × 10xy + y²
…. [(a + b)² = a² + 2ab + b²]
= 100 + 20y + y²

iii. Here, a = \(\frac { p }{ 3 }\) and b = \(\frac { q }{ 4 }\)
\(\left(\frac{p}{3}+\frac{q}{4}\right)^{2}=\left(\frac{p}{3}\right)^{2}+2 \times \frac{p}{3} \times \frac{q}{4}+\left(\frac{q}{4}\right)^{2}\)
…. [(a + b)² = a² + 2ab + b²]
\(\frac{p^{2}}{9}+\frac{p q}{6}+\frac{q^{2}}{16}\)

iv. Here, a = y and b = \(\frac { 3 }{ y }\)
\(\left(y-\frac{3}{y}\right)^{2}=y^{2}-2 \times y \times \frac{3}{y}+\left(\frac{3}{y}\right)^{2}\)
…. [(a – b)² = a² – 2ab + b²
= \(y^{2}-6+\frac{9}{y^{2}}\)

Question 6.
Use a formula to multiply:
i. (x – 5)(x + 5)
ii. (2a – 13)(2a + 13)
iii. (4z – 5y)(4z + 5y)
iv. (2t – 5)(2t + 5)
Solution:
i. Here, a = x and b = 5
(x – 5)(x + 5) = (x)² – (5)²
…. [(a + b)(a – b) = a² – b²]
= x² – 25

ii. Here, A = 2a and B = 13
(2a – 13)(2a + 13) = (2a)² – (13)²
…. [(A + B)(A – B) = A² – B²]
= 4a² – 169

iii. Here, a = 4z and b = 5y
(4z – 5y)(4z + 5y) = (4z)² – (5y)²
…. [(a + b)(a – b) = a² – b²]
= 16z² – 25y²

iv. Here, a = 2t and b = 5
(2t – 5)(2t + 5) = (2t)² – (5)²
…. [(a + b)(a – b) = a² – b²]
= 4t² – 25

Question 7.
The diameter of the wheel of a cart is 1.05 m. How much distance will the cart cover in 1000 rotations of the wheel?
Solution:
Diameter of the wheel (d) = 1.05 m
∴ Distance covered in 1 rotation of wheel = Circumference of the wheel
= πd
= \(\frac{22}{7} \times 1.05\)
= 3.3 m
∴ Distance covered in 1000 rotations = 1000 x 3.3 m
= 3300 m
= \(\frac { 3300 }{ 1000 }\) km …[1m = \(\frac { 1 }{ 1000 }\)km]
= 3.3 km
∴ The distance covered by the cart in 1000 rotations of the wheel is 3.3 km.

Question 8.
The area of a rectangular garden of length 40 m, is 1000 sq m. Find the breadth of the garden and its perimeter. The garden is to be enclosed by 3 rounds of fencing, leaving an entrance of 4 m. Find the cost of fencing the garden at a rate of Rs 250 per metre.
Solution:
Length of the rectangular garden = 40 m
Area of the rectangular garden = 1000 sq. m.
∴ length × breadth = 1000
∴ 40 × breadth = 1000
∴ breadth = \(\frac { 1000 }{ 40 }\)
= 25 m
Now, perimeter of the rectangular garden = 2 × (length + breadth)
= 2 (40 + 25)
= 2 × 65
= 130 m
Length of one round of fence = circumference of garden – width of the entrance
= 130 – 4
= 126 m
∴ Total length of fencing = length of one round of wire × number of rounds = 126 × 3
= 378 m
∴ Total cost of fencing = Total length of fencing × cost per metre of fencing
= 378 × 250
= 94500
∴ The cost of fencing the garden is Rs 94500.

Question 9.
From the given figure, find the length of hypotenuse AC and the perimeter of ∆ABC.
Solution:

In ∆ABC, ∠B = 90°, and l(BC) = 21, and l(AB) = 20
∴ According to Pythagoras’ theorem,
∴ l(AC)² = l(BC)² + l(AB)²
∴ l(AC)² = 21² + 20²
∴ l(AC)² = 441 + 400
∴ l(AC)² = 841
∴ l(AC)² = 29²
∴ l(AC) = 29
Perimeter of ∆ABC = l(AB) + l(BC) + l(AC)
= 20 + 21 + 29
= 70
∴ The length of hypotenuse AC is 29 units, and the perimeter of ∆ABC is 70 units.

Question 10.
If the edge of a cube is 8 cm long, find its total surface area.
Solution: ,
Total surface area of the cube = 6 × (side)²
= 6 × (8)²
= 6 × 64
= 384 sq. cm
The total surface area of the cube is 384 sq.cm.

Question 11.
Factorize: 365y⁴z³ – 146y²z⁴
Solution:
= 365y⁴z³ – 146y²z⁴
= 73 (5y⁴z³ – 2y²z⁴)
= 73y² (5y²z³ – 2z⁴)
= 73y²z³(5y² – 2z)

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Miscellaneous Exercise 2 8th Std Maths Answers Solutions.

Miscellaneous Exercise 2 8th Std Maths Answers

Question 1.
Questions and their alternative answers are given. Choose the correct alternative answer.
i. Find the circumference of a circle whose area is 1386 cm²? [Chapter 15]
(A) 132 cm²
(B) 132 cm
(C) 42 cm
(D) 21 cm²
Solution:
(B) 132 cm

Hint:
i. Area of the circle = πr²
1386 = \(\frac { 22 }{ 7 }\) x r²
r² = 1386 x \(\frac { 7 }{ 22 }\)
= 63 x 7
= 441
r = √441 … [Taking square root of both sides]
= 21 cm
Circumference of the circle = 2πr
= 2 x \(\frac { 22 }{ 7 }\) x 21
= 132 cm

ii. The side of a cube is 4 m. If it is doubled, how many times will be the volume of the new cube, as compared with the original cube? [Chapter 16]
(A) Two times
(B) Three times
(C) Four times
(D) Eight times
Solution:
(D) Eight times

Hint:
ii. Original volume = (4)³ = 64 cu.m
New side = 8 m
∴ New volume = (8)² = 512 cu.m
Now, \(\frac{\text { new volume }}{\text { original volume }}=\frac{512}{64}\) = 8
original volume 64
∴ volume of new cube will increase 8 times as compared to the volume of original cube.

Question 2.
Pranalee was practicing for a 100 m running race. She ran 100 m distance 20 times. The time required, in seconds, for each attempt was as follows. [Chapter 11]
18, 17, 17, 16,15, 16, 15, 14,16, 15, 15, 17, 15, 16,15, 17, 16, 15, 14,15
Find the mean of the time taken for running.
Solution:

∴ The mean of the time taken for running 100 m race is 15.7 seconds.

Question 3.
∆DEF and ∆LMN are congruent in the correspondence EDF ↔ LMN. Write the pairs of congruent sides and congruent angles in the correspondence. [Chapter 13]
Solution:

∆EDF ≅ ∆LMN
∴side ED ≅ side LM
side DF ≅ side MN
side EF ≅ side LN
∠E ≅∠L
∠D ≅∠M
∠F ≅∠N

Question 4.
The cost of a machine is Rs 2,50,000. It depreciates at the rate of 4% per annum. Find the cost of the machine after three years. [Chapter 14]
Solution:
Here, P = Cost of the machine = Rs 2,50,000
R = Rate of depreciation = 4%
N = 3 Years
A = Depreciated price of the machine

∴The cost of the machine after three years will be Rs 2,21,184.

Question 5.
In ☐ABCD, side AB || side DC, seg AE ⊥ seg DC. If l(AB) = 9 cm, l(AE) = 10 cm, A(☐ABCD) = 115 cm² , find l(DC). [Chapter 15]
Solution:
Given, side AB || side DC.
∴ ☐ABCD is a trapezium.

Given, l(AB) = 9 cm, l(AE) = 10 cm,
A(☐ABCD) = 115 cm²
Area of a trapezium
= \(\frac { 1 }{ 2 }\) x sum of lengths of parallel sides x height
∴ A(☐ABCD) = \(\frac { 1 }{ 2 }\) x [l(AB) + l(DC) x l(AE)]
∴ 115 = \(\frac { 1 }{ 2 }\) x [9 + l(DC)] x 10
∴ \(\frac { 115 \times 2 }{ 10 }\) = 9 + l(DC)
∴ 23 = 9 + l(DC)
∴ l(DC) = 23 – 9
∴ l(DC) = 14cm

Question 6.
The diameter and height of a cylindrical tank is 1.75 m and 3.2 m respectively. How much is the capacity of tank in litre?
[π = \(\frac { 22 }{ 7 }\)] [Chapter 16]
Solution:
Given: For cylindrical tank:
diameter (d) = 1.75 m, height (h) = 3.2 m
To Find: Capacity of tank in litre
diameter (d) = 1.75 m
= 1.75 x 100
….[∵ 1 m = 100cm]
= 175 cm
∴ radius (r) = \(=\frac{\mathrm{d}}{2}=\frac{175}{2}\) cm
h = 3.2 cm
= 3.2 x 100
= 320 cm
Capacity of tank = Volume of the cylindrical tank
= πr²h

∴ The capacity of the tank is 7700 litre.

Question 7.
The length of a chord of a circle is 16.8 cm, radius is 9.1 cm. Find its distance from the centre. [Chapter 17]
Solution:

Let CD be the chord of the Circle with centre O.
Draw seg OP ⊥ chord CD
∴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 16.8 …[l(CD) = 16.8cm]
∴l(PD) = 8.4 cm …(i)
∴In ∆OPD, m∠OPD = 90°
∴[l(OD)]² = [l(OP)]² + [l(PD)]² …..[Pythagoras theorem]
∴(9.1)² = [l(OP)]² + (8.4)² … [From (i) and l(OD) = 9.1 cm]
∴(9.1)² – (8.4)² = [l(OP)]²
∴(9.1 + 8.4) (9.1 – 8.4) = [l(OP)]²
…[∵ a² – b² = (a + b) (a – b)]
∴17.5 x (0.7) = [l(OP)]²
∴12.25 = [l(OP)]²
i.e., [l(OP)]² = 12.25
∴l(OP) = √12.25
…[Taking square root of both sides]
∴l(OP) = 3.5 cm
∴The distance of the chord from the centre is 3.5 cm.

Question 8.
The following tables shows the number of male and female workers, under employment guarantee scheme, in villages A, B, C and D.

Villages	A	B	C	D
No. of females	150	240	90	140
No. of males	225	160	210	110

i. Show the information by a sub-divided bar-diagram.
ii. Show the information by a percentage bar diagram. [Chapter 11]
Solution:
i.

Villages	A	B	C	D
No. of females	150	240	90	140
No. of males	225	160	210	110
Total	375	400	300	250

ii.

Villages	A	B	C	D
No. of females	150	240	90	140
No. of males	225	160	210	110
Total	375	400	300	250
Percentage of females	40%	60%	30%	56%
Percentage of males	60%	40%	70%	44%

Question 9.
Solve the following equations.
i. 17 (x + 4) + 8 (x + 6) = 11 (x + 5) + 15 (x + 3)
ii. \(\frac{3 y}{2}+\frac{y+4}{4}=5-\frac{y-2}{4}\)
iii. 5(1 – 2x) = 9(1 -x)
[Chapter 12]
Solution:
i. 17 (x + 4) + 8 (x + 6) = 11 (x + 5) + 15 (x + 3)
∴ 17x + 68 + 8x + 48 = 11x + 55 + 15x + 45
∴ 17x + 8x + 68 + 48 = 11x + 15x + 55 + 45
∴ 25x + 116 = 26x + 100
∴ 25x + 116 – 116 = 26x + 100 – 116
… [Subtracting 116 from both the sides]
∴ 25x = 26x – 16
∴ 25x – 26x = 26x – 16 – 26x
… [Subtracting 26x from both the sides]
∴ -x = -16
∴ \(\frac{-x}{-1}=\frac{-16}{-1}\)
∴ x = 16

ii. \(\frac{3 y}{2}+\frac{y+4}{4}=5-\frac{y-2}{4}\)
∴ \(\frac{3 y \times 2}{2 \times 2}+\frac{y+4}{4}=5-\frac{y-2}{4}\)
∴ \(\frac{6 y}{4}+\frac{y+4}{4}=5-\frac{y-2}{4}\)
∴ \(\frac{6 y}{4} \times 4+\frac{y+4}{4} \times 4=5 \times 4-\frac{y-2}{4} \times 4\)
……[Multiplying both the sides by 4]
∴ 6y + y + 4 = 20 – (y – 2)
∴ 7y + 4 = 20 – y + 2
∴ 7y + 4 = 22 – y
∴ 7y + 4 – 4 = 22 – y – 4
…..[Subtracting 4 from both the sides]
∴ 7y = 18 – y
∴ 7y + y = 18 – y + y
…[Adding y on both the sides]
∴ 8y = 18
∴ \(\frac{8 y}{8}=\frac{18}{8}\) … [Dividing both the sides by 8]
∴ \(y=\frac { 9 }{ 4 }\)

iii. 5(1 – 2x) = 9(1 – x)
∴ 5 – 10x = 9 – 9x
∴ 5 – 10x – 5 = 9 – 9x – 5
….[Subtracting 5 from both the sides]
∴ -10x = 4 – 9x
∴ -10x + 9x = 4 – 9x + 9x
… [Adding 9x on both the sides]
∴ -x = 4
∴ -x x (- 1) = 4 x (- 1)
… [Multiplying both the sides by – 1]
∴ x = – 4

Question 10.
Complete the activity according to the given steps.
i. Draw rhombus ABCD. Draw diagonal AC.
ii. Show the congruent parts in the figure by identical marks.
iii. State by which, test and in which correspondence ∆ADC and ∆ABC are congruent.
iv. Give reason to show ∠DCA ≅ ∠BCA, and ∠DAC ≅ ∠BAC
v. State which property of a rhombus is revealed from the above steps. [Chapter 13]
Solution:
a.

b. In ∆ADC and ∆ABC,
side AD ≅ side AB …..[Sides of a rhombus]
side DC ≅ side BC …..[Sides of a rhombus]
side AC ≅ side AC … [Common side]
∆ADC ≅ ∆ABC … [By SSS test]
∠DCA ≅ ∠BCA …[Corresponding angles of congruent triangles]
∠DAC ≅ ∠BAC …[Corresponding angles of congruent triangles]
From the above steps, property of rhombus revealed is ‘diagonal of a rhombus bisect the opposite angles’.

Question 11.
The shape of a farm is a quadrilateral. Measurements taken of the farm, by naming its corners as P, Q, R, S in order are as follows. l(PQ) = 170 m,
l(QR) = 250 m, l(RS) = 100 m, l(PS) = 240 m, l(PR) = 260 m.
Find the area of the field in hectare (1 hectare = 10,000 sq.m). [Chapter 15]
Solution:

Area of the field = A(∆PQR) + A(∆PSR)
In ∆PQR, a = 170 m, b = 250 m, c = 260 m
Semiperimeter of ∆PQR = s

Area of the field = A(∆PQR) + A(∆PSR)
= 20400 + 12000
= 32400 sq.m
= \(\frac { 32400 }{ 10000 }\)
…[1 hectare = 10,000 sq.m]
= 3.24 hectares
∴ The area of the field is 3.24 hectares.

Question 12.
In a library, 50% of total number of books is of Marathi. The books of English are \(\frac { 1 }{ 3 }\) of Marathi books. The books on Mathematics are 25% of the English books. The remaining 560 books are of other subjects. What is the total number of books in the library? [Chapter 12]
Solution:
Let the total number of books in the library be x
50% of total number of books is of Marathi.
Number of Marathi books = 50% of x
= \(\frac { 50 }{ 100 }x\)
= \(\frac { x }{ 2 }\)
The books of English are \(\frac { 1 }{ 3 }\) of Marathi books.
Number of books of English = \(\frac{1}{3} \times \frac{x}{2}\)
= \(\frac { x }{ 6 }\)
The books on Mathematics are 25% of the English books.
Number of books of Mathematics
= 25% of \(\frac { x }{ 6 }\)
= \(\frac{25}{100} \times \frac{x}{6}\)
= \(\frac { x }{ 24 }\)
Since, there are 560 books of other subjects, the total number of books in the library are


∴ 24x – 17x = 17x + 13440 – 17x
∴ 7x = 13440
∴ \(\frac{7 x}{7}=\frac{13440}{7}\)
∴ x = 1920
∴ The total number of books in the library are 1920.

Question 13.
Divide the polynomial (6x³ + 11x² – 10x – 7) by the binomial (2x + 1). Write the quotient and the remainder. [Chapter 10]
Solution:

∴ Quotient = 3x² + 4x – 7,
remainder = 0
Explanation:

Maharashtra Board Class 8 Maths Solutions

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 9 Sanskrit Solutions Aamod Chapter 8 पिनकोड्प्रवर्तकः महान् संस्कृतज्ञः Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 9 Sanskrit Aamod Solutions Chapter 8 पिनकोड्प्रवर्तकः महान् संस्कृतज्ञः

Sanskrit Aamod Std 9 Digest Chapter 8 पिनकोड्प्रवर्तकः महान् संस्कृतज्ञः Textbook Questions and Answers

भाषाभ्यास:

1. उचितं पर्यायं चिनुत।

प्रश्न अ.
पत्रवाहक: किं पठित्वा पत्रं वितरति ?
1. देवनागरीलिपीम्
2. कन्नडलिपीम्
3. आङ्ग्लभाषाम्
4. सङ्केतक्रमाङ्कम्
उत्तरम् :
4. सङ्केतक्रमाङ्कम्

प्रश्न आ.
पिन्कोक्रमाङ्कः अत्र न आवश्यकः।
1. रुग्णालये
2. सद्यस्क-क्रयणे
3. ATM शाखा-अन्वेषणे
4. पुस्तकपठने
उत्तरम् :
4. पुस्तकपठने

प्रश्न इ.
पिनकोड्-प्रणालिः केन समारब्धा ?
1. इन्दिरामहोदयया
2. गान्धीमहोदयेन
3. प्रधानमन्त्रिणा
4. वेलणकरमहोदयेन
उत्तरम् :
4. वेलणकरमहोदयेन

प्रश्न ई.
पिनकोड्क्रमाङ्के कति सङ्ख्या:?
1. 6
2. 4
3. 3
4. 2
उत्तरम् :
1. 6

2. वेलणकरमहोदयस्य कार्यवैशिष्ट्यानि माध्यमभाषया लिखत।

प्रश्न 1.
वेलणकरमहोदयस्य कार्यवैशिष्ट्यानि माध्यमभाषया लिखत।
उत्तरम् :
‘पिनकोड् प्रवर्तकः, महान् संस्कृतज्ञः’ या पाठात पत्रव्यवहार बिनचूक होण्यासाठी आवश्यक अशा पिन-कोड् प्रणालीविषयी माहिती मिळते. श्रीराम भिकाजी वेलणकर हे कलकत्ता पत्रविभागाचे संचालक म्हणून कार्यरत होते. त्यांच्याकडे एक सैनिक एक समस्या घेऊन आला.

त्याला आपल्या आई-वडिलांच्या तब्येतीविषयी माहिती कळत नव्हती कारण तो अरुणाचल प्रदेशात राहत होता व त्याचे आई – वडील केरळमध्ये, केरळहून अरुणाचलला पत्र पोहोचण्यास एक महिन्याहून अधिक काळ लागत असे. त्याची समस्या ऐकून वेलणकर महोदयांनी त्याची दखल घेतली व विचार सुरु केला.

त्यांनी सर्वप्रथम पत्रवाटप कार्यातील नेमकी समस्या जाणून घेतली. लिखाणातील संदिग्धतेमुळे समस्या निर्माण होते हे लक्षात घेऊन त्यांनी अंकाधारित पिनकोड पद्धत निर्माण केली. त्यासाठी त्यांनी देशाचे आठ भाग करून त्यांना राज्यनिहाय क्रमांक दिले.

अशाप्रकारे सहा अंकांचा स्थानसंकेतांक असणारी पिनकोड पद्धत त्यांनी भारतात सुरु केली. वेलणकर महोदय गणितज्ञ असल्यामुळे पत्रविभागातील कार्यातसुद्धा त्यांनी त्यांच्या गणित ज्ञानाचा कौशल्याने वापर केला. ते आपल्या कार्याशी प्रामाणिक व कर्तव्यदक्ष होते.

In the lesson ‘पिनकोड् प्रवर्तकः, महान संस्कृतज्ञ’ we get to know about a great mathematician and a dedicated sanskrit scholar Mr. Velankar who invented PIN Code System. Mr. Velankar was director of Kolkata post office.

Once a soldier came to him and told that he was facing a problem in getting information regarding his parents’ health as it used to take more than a month for a letter to reach Kerala from Arunachal Pradesh.

Hearing his problem Mr. Velankar immediately started to think about the solution. He first tried to analyse the reasons behind the problem. He realised that due to bad handwriting, incomplete address etc. such problems arise.

He then divided the country in eight and numbered these divisions based on the states. This number has six-digits. This sixdigit number is PIN code. He was a very efficient post office director and was very devoted to his work.

3. एकवाक्येन उत्तरत।

प्रश्न अ.
वेलणकरमहोदयस्य कः प्रियः विषयः ?
उत्तरम् :
वेलणकरमहोदयस्य सङ्गीतं प्रिय: विषयः।

प्रश्न आ.
वेलणकरमहोदयः के वाद्ये वादयति स्म ?
उत्तरम् :
वेलणकरमहोदयः व्हायोलिन तबला च इति वाद्ये वादयति स्म।

प्रश्न इ.
वेलणकरमहोदयेन रचितः सङ्गीतविषयकः ग्रन्थः कः ?
उत्तरम् :
वेलणकरमहोदयः रचितः सङ्गीतविषयक: ग्रन्च: गीतगीर्वाणम्।

प्रश्न ई.
पिनकोड्प्रणालिनिमित्तं वेलणकरमहोदयेन देशस्य कति विभागाः कृताः ?
उत्तरम् :
पिनकोप्रणालिनिमित्तं वेलणकरमहोदयेन देशस्य अष्ट विभागा: कृताः ।

प्रश्न उ.
वेलणकरमहोदयः कस्य विभागस्य निर्देशकः आसीत् ?
उत्तरम् :
वेलणकरमहोदयः डाकतारविभागस्य निर्देशकः आसीत्।

प्रश्न ऊ.
वेलणकरमहोदयस्य अभिमतः साहित्यप्रकार: कः?
उत्तरम् :
वेलणकरमहोदयस्य अभिमतः साहित्यप्रकार: नाट्यलेखनम्।

5. विशेषणानि अन्विष्य लिखत।

प्रश्न 1.
अ. ………………. प्रणालिः।
आ. …………… सैनिकः ।
इ. ………………. भाषाः।
ई. ………………. हस्ताक्षरम् ।
ए. ………………. विषयः।
ऐ. ………………. रचनाः।

उत्तरम् :
आ. दुःखितः, केरलं प्रदेशीयः
इ. भिन्ना:
ई. दुबोधम्
ए. प्रियः
ऐ. विविधाः

अन्विष्यत लिखत च।

1. एताः सङ्ख्याः किं निर्दिशन्ति।

2. विविधदेशेषु अपि पत्रसङ्केताङ्कप्रणालिः वर्तते वा?

Sanskrit Aamod Class 9 Textbook Solutions Chapter 8 पिनकोड्प्रवर्तकः महान् संस्कृतज्ञः Additional Important Questions and Answers

उचितं पर्यायं चिनुत ।

प्रश्न 1.
…………. प्रधानमन्त्रिमहोदयायै कदा उपहारीकृता एषा पिन्कोड् महत्त्वपूर्णा प्रणालि:?
(अ) ऑगस्ट 1972
(आ) ऑगस्ट 1929
(इ) सप्टेंबर 1972
(ई) ऑगस्ट 1927
उत्तरम्
(अ) ऑगस्ट 1972

प्रश्ननिर्माणं कुरुत।

प्रश्न 1.
1. पत्रालय-अधिकारी संस्कृतपण्डितः गणितज्ञः श्रीराम भिकाजी वेलणकरमहोदयः।
2. वयं पत्रपेटिकायां पत्रं क्षिपामः।
उत्तरम्
1. पत्रालय-अधिकारी क: अस्ति?
2. वयं कुत्र पत्रं क्षिपाम:?

उचितं पर्यायं चित्वा वाक्यं पुनर्लिखत ।

प्रश्न 1.

1. वेलणकरमहोदय: डाकतार-विभागस्य निर्देशकरूपेण ………. आसीत्। (कार्यविभागे / कार्यमग्नः)
2. पत्रस्य प्रवासकाल: ……… आसीत्। (मासः / मासाधिक:)
3. वेलणकरमहोदयेन ……… अङ्कानाम् उपयोगः कृतः? (शास्त्रज्ञेन / गणितज्ञेन)
4. वेलणकरमहोदयस्य प्रियविषयः ……… आसीत्। (सङ्गीतम् / इतिहास:)
5. ……….. ख्याति: नाटकानाम् आसीत्। (सबकुछ श्रीः / सबकुछ श्रीभिः)

उत्तरम् :

1. कार्यमग्नः
2. मासाधिकः
3. गणितज्ञेन
4. सङ्गीतम्
5. सबकुछ श्रीभिः

एकवाक्येन उत्तरत।

प्रश्न 1.
कः कोलकतानगरे कार्यरतः आसीत्?
उत्तरम् :
श्रीराम-भिकाजी-वेलणकरमहोदय : कोलकतानगरे कार्यरतः आसीत्।

प्रश्न 2.
वेलणकरमहोदयस्य कार्यकाले क: आगतः?
उत्तरम् :
वेलणकरमहोदयस्य कार्यकाले केरल-प्रदेशीयः सैनिक: मेलितुम् आगतः।

प्रश्न 3.
वेलणकरमहोदयस्य चिन्तनं किम् आसीत्?
उत्तरम् :
“कश्यं पत्रस्य प्राप्तिकालं न्यूनं कर्तुं शक्नोमि?” इति वेलणकरमहोदयस्य चिन्तनम् आसीत्।

प्रश्न 4.
वेलणकरमहोदयेन केषाम् उपयोगः कृतः?
उत्तरम् :
वेलणकरमहोदयेन अङ्कानाम् उपयोगः कृतः।

प्रश्न 5.
वेलणकरमहोदयः कस्मिन् वादने निपुण: आसीत्?
उत्तरम् :
वेलणकरमहोदयः व्हायोलिनवादने तथा तबलावादनेऽपि निपुणः आसीत्।

प्रश्न 6.
वेलणकरमहोदयः किं रचितवान् ?
उत्तरम् :
वेलणकरमहोदयः गीतगीर्वाणम् इति सङ्गीतविषयकं शास्त्रीयग्रन्थं रचितवान्।

प्रश्न 7.
पत्राणाम् वितरणव्यवस्था कथं शक्या भवति?
उत्तरम् :
पत्राणां वितरणव्यवस्था पिनकोड्प्रणाल्या शक्या भवति।

प्रश्न 8.
एषा महत्त्वपूर्णा प्रणालिः कस्यै उपहारीकृता?
उत्तरम् :
एषा महत्त्वपूर्णा प्रणालि: प्रधानमन्त्रि-इन्दिरा-गान्धी महोदयायै उपहारीकृता।

प्रश्न 9.
पिन्कोक्रमाङ्के कति सङ्ख्या:?
उत्तरम् :
पिन्कोक्रमाङ्के षट् सङ्ख्याः ।

सत्यं वा असत्यं लिखत।

प्रश्न 1.

1. पत्रस्य प्रवासकाल: मासाधिको नासीत्।
2. अधुनापि वयं विनायासं शीघ्र पत्र लभामहे।
3. वेलणकर: महोदयः न अनुशासनप्रियः ।
4. वेलणकरमहोदयः सर्वमेव कर्तुं न समर्थः ।
5. पत्रालयात् निर्दिष्टं स्थानं प्रति पत्राणि गच्छन्ति।
6. पत्रलस्य अधिकारी संस्कृतपण्डितः शास्त्रज्ञः च।

उत्तरम् :

1. असत्यम्
2. सत्यम्
3. असत्यम्
4. असत्यम्
5. सत्यम्
6. असत्यम्

शब्दस्य वर्णविग्रहं कुरुत।

• दुर्बोधम् – द् + उ + र् + ब् + ओ + ध् + अ + म् ।
• विनायासम् – व् + इ + न् + आ + य् + आ + स् + अ + म्।
• क्रमाङ्कनम् – क् + र् + अ + म् + आ + ङ् + क् + अ + न् + अ + म्।
• कर्तुम् – क् + अ + र + त् + उ + म्।
• विद्वद्वरेण्यः – व् + इ + द् + व् + अ + द्  + व् + अ + र + ए + ण् + य् + अः।
• प्राप्नोति – प् + र + आ + प् + न् + ओ + त् + इ।
• महोदयायै – म् + अ + ह् + ओ + द् + अ + य् + आ + य् + ऐ।
• प्रणालिः – प् + र् + अ + ण् + आ + ल् + इ:।

विभक्त्यन्तरूपाणि।

• प्रथमा – मित्राणि, वयम्, एषा, सः, विभागाः, सैनिकः, वयम्, प्रत्यूहाः, नैके, वयम्, रचनाः, गीतानि, कथाः, साहित्यप्रकारः।
• द्वितीया – पत्रालयः, स्थानम्, पत्रम्, नाट्यलेखनम्, सर्वम्, दिग्दर्शनम्।
• तृतीया – केन, निर्देशकरूपेण, तेन, संस्कृतेन, तेन, येन।
• षष्ठी – अस्माकम्, एतस्य, प्रश्नस्य, डाकतारविभागस्य, पत्रस्य, तेषाम् पित्रोः, तस्य, नाटकानाम्।
• सप्तमी – देशे, पत्रपेटिकायाम्, कोलकतानगरे, एकस्मिन, दिने।

त्वान्त/ल्यबन्त/तुमन्त अव्ययानि।

त्वान्त अव्यय धातु + त्वा / ध्वा / ट्वा / ढ्वा / इत्वा अयित्वा	ल्यबन्त अव्यय उपसर्ग + धातु + य / त्य	तुमन्त अव्यय   थातु + तुम् / धुम् / टुम् / ढुम् / इतुम् / अयितुम्
–	–	मेलितुम्
–	–	ज्ञातुम्
–	–	कर्तुम्

विशेषण – विशेष्य – सम्बन्धः

प्रश्न 1.
……… साहित्यप्रकारः।
उत्तरम् :
अभिमतः

धातुसाधितविशेषणानि।

धातुसाधित – विशेषणम्	विशेष्यम्
प्रेषितानि	पत्राणि
निर्दिष्टम्	स्थानम्
शक्या	वितरणव्यवस्था
लिखिताः	सङख्या:
उपहारीकृता	प्रणालि:
प्रवर्तिता	सङ्केतप्रणालिः
आगतः	सैनिक:
आरब्धम्	चिन्तनम्
कृतः	उपयोग:
कृताः	विभागाः, रचनाः
कृतम्	क्रमाङ्कनम्
उपकृताः	वयम्

पिनकोड्प्रवर्तकः महान् संस्कृतज्ञः Summary in Marathi and English

प्रस्तावना :

आजच्या इंटरनेट युगात आपण पत्रव्यवहार कमी करत असलो तरीसुद्धा आजही भारतामध्ये पत्रविभाग हा महत्वाचा प्रशासकीय विभाग आहे. ह्य विभाग ज्या प्रणालीमुळे गेली अनेक वर्षे अव्याहतपणे चालू आहे ती पिनकोड प्रणाली पत्र-व्यवहाराचा आधार आहे. प्रस्तुत गद्यांश वेलणकर महोदयांनी प्रचलित केलेल्या पिनकोड प्रणाली विषयी माहिती देतो.

वेलणकर महोदय हे श्रेष्ठ गणितज्ञ तसेच संस्कृत विषयाचे गाढे अभ्यासक होते. त्यांनी अनेक संस्कृत नाटके, काव्ये रचली. सङ्गीतसौभद्रम् कालिदासचरितम्, जन्म रामायणस्य, तनयो राजा भवति कथं मे? राज्ञी दुर्गावती, रणश्रीरङ्गम्, कालिन्दी, स्वातन्त्र्यलक्ष्मीः, कौटिलीयार्थनाट्यम् या त्यांच्या काही सुप्रसिद्ध रचना आहेत.

Even though in the era of ‘e-mail’, we are not much acquainted with writing the traditional letter, in India postal service is still one of the most important pillars of the administrative system. PIN-code system is unique feature of this service.

This system was invented by Mr. Velankar, He was a great mathematician and Sanskrit scholar. He has composed several Sanskrit poems, plays etc. Torname a fear सङ्गीतसौभद्रम्,कालिदासचरितम्, जन्म रामायणस्य,तनयो राजा भवति कथं मे? राज्ञी दुर्गावती, रणबीरङ्गम् कालिन्दी, स्वातन्त्र्यलक्ष्मी:, कौटिलीयार्थनाट्यम् are his popular compositions.

उपोद्घात:

• गौरवः – अरे अमेय, कुछ गच्छसि?
• अमेयः – समीपे एव पत्रपेटिका वर्तते। मातुलं प्रति पत्रं प्रेषयितुं तत्र गच्छामि।
• गौरवः – दर्शय पत्रम्। अपि बेङ्गलुरुनगरे निवसति तव मातुल:? तत् तु अतीव दूरे किल? कदा पत्रं विन्देत् स:?
• अमेयः – सामान्यत: त्रीणि दिनानि अपेक्षितानि तदर्थम्।
• गौरवः – जीणि दिनानि एव? एतावत् शीघ्रम्?
• अमेयः – सत्यम्।
• गौरवः – किन्तु सङ्केत: तु देवनागरीलिप्यां लिखितः। कर्णाटके तु कन्नडलिपिः प्रयुज्यते नु? तहि कथं पत्रवाहकः तत् पठितुं शक्नुयात्?
• अमेयः – पश्य एतत्। सङ्केतस्य अन्ते आङ्ग्लभाषया कश्चन क्रमाङ्क: लिखितः अस्ति। जानासि खलु तद्विषये?
• गौरवः – आम, सः तु पिन्कोक्रमाङ्कः । अहं पत्रं न लिखामि किन्तु सद्यस्क – क्रयणसमये (Online Shopping) पिन्कोक्रमाङ्क: अनिवार्यः एव। इतोऽपि च अस्माकं विभागे ATM शाखाः, रुग्णालयाः, विशिष्टसेवा: वा कुत्र सन्ति इति ज्ञातुं पिन्कोक्रमाङ्क: आवश्यक: एव।
• गौरव – अरे अमये, कुठे जात आहेस?
• अमेय – जवळच टपालपेटी आहे. मामाला पत्र पाठविण्यासाठी तिथे जात आहे.
• गौरव  – पत्र दाखव, तुझा मामा बंगळुरूला राहतो का? ते तर खूप दूर आहे ना? त्याला पत्र केव्हा मिळेल?
• अमेय – साधारणपणे त्यासाठी तीन दिवस लागतात.
• गौरव – केवळ तीनच दिवस? एवढ्या लवकर?
• अमेय – खरेच.
• गौरव – परंतु पत्ता तर देवनागरी लिपीमध्ये लिहिलेला आहे. कर्नाटकात तर कन्नड लिपी वापरली जाते ना? तर मग टपालवाहकाला ते वाचणे कसे शक्य होईल?
• अमेय – हे बघ. पत्त्याच्या शेवटी इंग्रजी भाषेत काही क्रमांक लिहिलेले आहेत. त्याविषयी काही माहिती आहे का?
• गौरव – हो, तो तर पिनकोड क्रमांक आहे. मी पत्र लिहीत नाही परंतु ऑनलाईन शॉपिंग करतेवेळी पिनकोड क्रमांक अनिवार्यच आहे. आपल्या विभागातील ATM शाखा, रुग्णालये अथवा विशिष्ट सेवा कुठे आहेत हे जाणून घेण्यासाठी पिनकोड क्रमांक आवश्यकच आहे.
• Gaurav – Ameya, where are you going?
• Ameya – I have a post box near by. I am going there to send the letter to my uncle.
• Gaurav – Show me the letter! Does your uncle live in Bengaluru city? That is really far, isn’t it? When will he get the letter?
• Ameya – Generally two three days are expected for that.
• Gaurav – Just three days? So fast?
• Ameya – Yes, true!
• Gaurav – But the address is written in Devnagari. But in Karnataka, Kannada script is used, right? Then how would the postman be able to read it?
• Ameya – See this. At the end of the address, a certain number is written. Do you know about that?
• Gaurav – Yes, that is the PIN code number. I do not write a letter, but while shopping online, PIN-code number is necessary. Apart from this, to know where ATM branches, hospitals or any special services are, it is essential to know the PIN code number.

परिच्छेद : 1

अयि मित्राणि, ………… वेलणकरमहोदयः।

अयि मित्राणि, भारतसदृशे विशाले देशे सुदूरपान्तेभ्यः अन्यप्रान्तं प्रति प्रेषितानि पत्राणि बिना विलम्बं विना प्रमादं कथं लभ्यन्ते ? वयम् अस्माकं निवासविभागे पत्रपेटिकायां पत्रं क्षिपामः । तत: तत् पत्रालयं प्रति गच्छति। तत: च निर्दिष्टं स्थानं प्रति प्राप्नोति। कथं शक्या भवति एषा वितरणव्यवस्था? “पिन्कोमणालिः’ इत्येव एतस्य प्रश्नस्य उत्तरम्। पिन्कोक्रमाङ्कः नाम पत्रसङ्केतस्य अन्ते लिखिताः षट् सङ्ख्याः । पिनकोड् नाम स्थानसङ्केताङ्कः ।

(PIN – Postal Index Number), ऑगस्ट 1972 – तमे खिस्ताब्दे स्वातन्त्र्यदिनसमारोहे प्रधानमन्त्रि-इन्दिरा-गान्धी-महोदयायै उपहारीकृता एषा महत्वपूर्णा प्रणालिः । केन खलु प्रवर्तिता इयं सङ्केताङ्कप्रणालिः? स: पत्रालय-अधिकारी आसीत् संस्कृतपण्डित: गणितज्ञः श्रीराम-भिकाजी-वेलणकरमहोदयः।

अनुवादः

अहो मित्रहो, भारतासारख्या विशालदेशात दूरवरच्या प्रांतातून इतर प्रांतात पाठवलेली पत्रे उशीर न होता व कोणत्याही चुकीशिवाय कशी बरे मिळतात? निवास विभागातील टपालपेटीत पत्र टाकतो. त्यानंतर ते पत्रालयात जाते. आणि तिथून निर्दिष्ट केलेल्या स्थळी पोहोचते. ही वितरणव्यवस्था कशी शक्य होते?

“पिनकोड्पद्धत’ हेच या प्रश्नाचे उत्तर, पिनकोड क्रमांक म्हनजे पत्रावरील पत्त्याच्या शेवटी लिहिलेल्या सहा संख्या होय, पिनकोड म्हणजे स्थानसंकेतांक (स्थानाचा निर्देश करणारे विशिष्ट अंक) ऑगस्ट 1972 रोजी स्वातंत्र्यदिनाच्या समारंभाला भेट केलेली ही महत्त्वपूर्ण प्रणाली आहे. ही संकेतांक प्रणाली कोणी बरे निर्माण केली? ते पत्रालय-अधिकारी होते. संस्कृतपंडित, गणितज्ञ श्रीराम भिकाजी वेलणकरमहोदय.

O friends, in a huge country like India, how do letters sent from far-flunged region to other places reach without delay and without errors? We drop the letter in a post box, which is there in our residential area. Then it goes to the post office.

And then reaches the place indicated. How does this distribution system become possible? ‘PIN-code system’ is the answer to this question. PIN-code number means the six digits written at the end of the letter’s address. PIN-code means Postal Index Number.

This system handed over to the Prime Minister Indira Gandhi in August 1972 on the occasion of the Independence Day, However, who invented this PIN-code system? That post-office officer was a Sanskrit scholar and mathematician Mr. Shriram Bhikaji Velankar.

परिच्छेद : 2

स: कोलकतानगरे ……………… लभामहे ।

स: कोलकतानगरे डाकतारविभागस्य निर्देशकरूपेण कार्यरतः आसीत् । तस्य कार्यकाले एकस्मिन् दिने कोऽपि दुःखितः केरल-प्रदेशीयः सैनिक: महोदयं मेलितुम् आगतः । सः तस्य पित्रो: स्वास्थ्यविषये किञ्चिदपि ज्ञातुम् असमर्थः । यतः तदा पत्रस्य प्रवासकालः केरलत:अरुणाचलप्रदेशपर्यन्त क्वचित् मासाधिकोऽपि आसीत्।

तेन पं. वेलणकरमहोदयस्य चिन्तनम् आरब्धम्, ‘कथं पत्रस्य प्राप्तिकालं न्यून कर्तुं शक्नोमि ? दुर्बोध हस्ताक्षरं, भिन्नाः प्रान्तीया: भाषा:, अपूर्णः पत्रसङ्केत: इति नैके प्रत्यूह्यः । अत: गणितज्ञेन तेन अङ्कानाम् उपयोगः कृतः । तेन देशस्य अष्ट विभागाः कृताः । तेषां च उपविभाग-जनपदाधारण क्रमाङ्कनं कृतम्। तेन अधुनापि वयं विनायासं शीघ्र पत्रं लभामहे।

अनुवादः

ते कोलकतानगरात डाकतार विभागाचे निर्देशक म्हणून कार्यरत होते. त्यांच्या कार्यकाळात एके दिवशी कोणी एक दुःखी केरळ प्रदेशातील सैनिक (त्या) महोदयांना भेटण्यासाठी आला. तो त्याच्या आई-वडिलांच्या स्वास्थाविषयी अजिबात जाणून घेऊ शकत नव्हता. कारण तेव्हा पत्राचा प्रवासकाळ केरळपासून अरुणाचलप्रदेशापर्यंत एक महिन्याहूनही अधिक होता.

त्यामुळे पं, वेलणकर महोदयांनी विचार करण्यास सुरुवात केली, पत्र पोहचण्याचा काळ कसा बरे कमी करता येईल?’ दुर्बोध (समजण्यासाठी कठीण) हस्ताक्षर, वेगवेगळ्या प्रांतातील वेगवेगळ्या भाषा, अपूर्ण पत्ते अशा अनेक समस्या होत्या.

म्हणून गणितज्ञ असलेल्या त्यांनी अंकांचा उपयोग केला. त्यांनी देशाचे आठ विभाग पाडले. आणि त्यांचे उपविभाग, राज्यांच्या आधाराने क्रमांकन केले. त्यामुळे आत्तासुद्धा आपल्याला विनासायास (सहजपणे) त्वरित पत्र मिळते.

He worked as the director of postal department at Kolkata. During his tenure, a certain grieved soldier who was native of Kerala came to meet Mr. Velankar. He was totally unable to know about his parent’s health Because at that time, travelling time of a letter from Kerala to Arunachal Pradesh was sometimes even more than a month.

Hence, Mr. Velankar started thinking ‘how can I reduce the time for the letter to reach? There are many obstacles like bad handwriting, different regional languages, incomplete address etc. Hence, the digits were used by that mathematician.

He made eight divisions of the country, They were numbered on the basis of subdivision states. Because of that, we still get the letter fast without much effort.

परिच्छेद : 3

अनुशासनप्रियस्य …………… उपकृताः ।

अनुशासनप्रियस्य वेलणकरमहोदयस्य सङ्गीतमपि प्रियः विषयः । स: व्हायोलिनवादने तथा तबलावादनेऽपि अतीव निपुणः। अतः एव सः ‘गीतगीर्वाणम् इति सङ्गीतविषयक शास्त्रीयग्रन्थं संस्कृतेन रचितवान्। कथाः, गीतानि, बालगीतानि, नृत्यनाट्यम् इति बहुविधाः रचना: तेन कृताः तथापि नाट्यलेखन तस्य अभिमत: साहित्यप्रकार: ।

लेखन, दिग्दर्शनं, नाट्यनिर्मितिः, व्यवस्थापन, सङ्गीतसंयोजनम् इति सर्वमेव कर्तुं सः समर्थः अत: “सबकुछ श्रीभिः” इति तस्य नाटकानां ख्यातिः आसीत्। धन्यः सः विद्रद्वरेण्यः येन वयं पिनकोड्प्रणाल्या उपकृताः।

अनुवादः

शिस्तप्रिय वेलणकर महोदयांचा संगीत सुद्धा आवडता विषयोता. ते व्हायोलिनवादनात तसेच तबला वादनातही अतिशय पारंगत होते. म्हणूनच त्यांनी ‘गीतगीर्वाणम्’ हा संगीत विषयक शास्त्रीय ग्रंथ संस्कृतात रचला. कथा, गाणी, बालगीते, नृत्यनाट्य अशा पुष्कळ रचना त्यांनी केल्या. तरी नाट्यलेखन त्यांचा आवडता साहित्य प्रकार होता.

ते लेखन, दिग्दर्शन, नाट्यनिर्मिती, व्यवस्थापन, संगीतसंयोजन हे सर्व काही करु शकत म्हणून “सबकुछ श्रीभिः” अशी त्यांच्या नाटकांची ख्याती होती. धन्य ते श्रेष्ठ विद्वान (वेलणकर महोदय) ज्यांनी आपल्याला पिनकोडप्रणालीने उपकृत केले.

Mr. Velankar who liked discipline was also fond of music He was expert in playing the violin and tabla. Hence, he even composed a book, in Sanskrit “गीतगीर्वाणम्” which is related to music. He has composed many songs, songs for kids, danceplays and many compositions, yet play-writing was his favorite literature genre.

He was able to do writing, direction, production of dramas, management and music composition and hence his plays were famous as ‘सबकुछ श्रीभि:’. He was a great scholar who has indebted us with PINcode system.

सन्धिविग्रहः

• इत्येव – इति + एव ।
• कोऽपि – कः + अपि ।
• किञ्चिदपि – किचित् + अपि ।
• मासाधिकोऽपि – मासाधिक: + अपि ।
• अधुनापि – अधुना + अपि ।
• तबलावादनेऽपि – तबलावादने + अपि ।
• तथापि – तथा + अपि ।
• सर्वमेव – सर्वम् + एव ।

समानार्थकशब्दाः

1. मित्राणि – वयस्याः।
2. नाम – अभिधानम्।
3. देशे – राष्ट्रे।
4. ख्रिस्ताब्दे – संवत्सरे।
5. रतः – मग्नः ।
6. दिने – दिवसे।
7. जनकः – पिता।
8. समयम् – कालः।
9. अधुना – इदानीम्।
10. नुिपणः – पारङ्गतः।
11. ग्रन्थः – पुस्तकम्।
12. ख्यातिः – प्रसिद्धिः।

विरुद्धार्थकशब्दाः

1. मित्राणि × रिपवः।
2. पण्डितः × मूढः।
3. अन्ते × आरम्भे।
4. विशाल: × लघुः।
5. शक्या × अशक्या।
6. दुःखितः × आनन्दितः।
7. असमर्थः × समर्थः।
8. न्यूनम् × बहु, दीर्घम्।
9. उपयोग: × निरुपयोगः।
10. शीघ्रम् × शनैः शनैः।
11. प्रियः × अप्रियः।
12. समर्थम् × असमर्थम्।

शब्दार्थाः

1. विलम्ब: – delay – उशीर
2. प्रमादः – mistake – चूक
3. पत्रपेटिकायाम् – in letter box – पत्रपेटीत
4. क्षिपाम: – we drop – टाकतो
5. उपहारीकृता – handed over to – सुपूर्द करणे
6. प्रणालिः – system – प्रणाली, पद्धती
7. गणितज्ञः – mathematician – गणितज्ञ
8. कार्यरतः – engrossed in work – कार्यरत
9. पित्रोः – parents – आई-वडिलांचे
10. असमर्थः – unable – असमर्थ
11. दुर्बोधम् – difficult to understand – अवघड
12. विनायास – easily, without effort – सहजपणे
13. निपुणः – expert – निपुण
14. अभिमतः – favourite – आवडता
15. ख्यातिः – fame, popularity – प्रसिद्धी
16. विद्वद्वरेण्यः – great scholar – श्रेष्ठ विद्वान
17. उपकृताः – blessed, indebted – उपकृत
18. रचितवान् – composed – रचले

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 45 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 45

Question 1.
Add the following :
(1) 2 hours 30 minutes + 4 hours 55 minutes
Solution:

Hrs.	Min.
1
2 + 4	3 0 5 5
7	2 5

85 minutes = 1 hr 25 min

(2) 3 hours 50 minutes + 4 hours 20 minutes
Solution:

Hrs.	Min.
3 + 4	5 0 2 0
7	7 0
8	1 0

70 minutes = 1 hr 10 min.

(3) 3 hours 45 minutes + 1 hour 35 minutes
Solution:

Hrs.	Min.
3 + 1	4 5 3 5
4	8 0
5	2 0

80 minutes = 1 hr 20 min

(4) 4 hours 15 minutes + 2 hours 50 minutes
Solution:

	Min.
4 + 2	1 5 5 0
6	6 5
7	0 5

65 minutes = 1 hr 05 min

Question 2.
Subtract the following :
(1) 3 hours 10 minutes – 2 hours 40 minutes

Hrs.	Min.
2	60 + 10
3 – 2	1 0 4 0
0	3 0

(2) 5 hours 20 minutes – 2 hours 35 minutes
Solution:

Hrs.	Min.
4	60 + 20
5 – 2	20 3 5
2	4 5

(3) 4 hours 25 minutes – 1 hour 55 minutes

Hrs.	Min.
3	60 + 25
4 – 1	25 5 5
2	3 0

(4) 6 hours 15 minutes – 2 hours 45 minutes

Hrs.	Min.
5	60 + 15
6 – 2	15 4 5
3	3 0

Question 3.
A government office opens at 7 in the morning and closes at 3 in the afternoon. How long is this office open?
Solution:

Hrs.	Min.
1 5 – 7	0 0 Closing time 0 0 Opening time
8	0 0

∴ Office remain open for 8 hours

Question 4.
A movie starts at 45 minutes past 3 in the afternoon and finishes two and a half hours later. At what time does the movie end?
Solution:

∴ Movie ends at 6:15 in the evening

Question 5.
Sakharam was ploughing the field from 8 in the morning. At 12:30 in the afternoon, he stopped and started for home. He reached home at 1:30. How long was he ploughing the field? How long did it take him to reach home from the field?
Solution:

∴ He ploughed for 4:30 hrs. He took 1 hour to reach home.

Question 6.
Rambhau started the water pump at ten-thirty at night and put it off the same night at a quarter to twelve. How long was the water pump on?
Solution:
Quater to 12 is 11:45 pm

∴ pump was on for 1 hour 15 minutes

Question 7.
Geeta taught in the classroom for 2 hours and 25 minutes in the morning and 1 hour and 45 minutes in the afternoon. How long was she teaching in all?
Solution:

∴ Total teaching of Geeta was for 4 hrs 10 min.

Question 8.
If a bank is open for business from 10 in the morning to 4:30 in the evening, how long is it open?
Solution:
Here, in 24 hours clock, 4:30 in the evening = 16:30

∴ Bank opens for 6 hrs 30 min.

Question 9.
If a shop is open from 9:30 am to 10 pm, how long is it open?
Solution:
Here, 10 pm in 24 hours clock is 22:00

∴ Shop opens for 12 hours 30 minutes

Question 10.
If the Maharashtra Express leaving from Kolhapur at 15:30 arrives at Gondia the next day at 20:15, how long is the journey from Kolhapur to Gondia?
Solution:
15:30 to next 15:30 is 24 hours

24 hours + 4 hr. 45 min. = 28 hr. 45 min.
∴ jurney from Koihapur to Gondiya is 28 hours and 45 minutes.

Measuring Time Problem Set 45 Additional Important Questions and Answers

Question 1.
Add the following.

(1) 5 hours 25 minutes + 2 hours 35 minutes
Solution:

Hrs.	Min.
5 + 2	2 5 3 5
7	6 0
8	0 0

60 minutes = 1 hr

(2) 6 hours 55 minutes + 2 hours 15 minutes
Solution:

Hrs.	Min.
6 + 2	5 5 1 5
8	7 0
9	1 0

70 minutes = 1 hr. 10 min

Question 2.
Subtract the following.

(1) 7 hours 30 minutes – 4 hours 50 minutes
Solution:

Hrs.	Min.
6	60 + 30
7 – 4	3 0 5 0
2	4 0

(2) 2 hours 35 minutes – 1 hour 40 minutes
Solution:

Hrs.	Min.
1	60 + 35
2 – 1	3 5 4 0
0	5 5

Question 3.
Solve the following:

(1) Supriya left for a picnic at 7:15 am. She came back at 6:45 pm. How long was she out for the picnic?
Here 6:45 pm = 18:45 (In 24 hours clock)
Solution:

Hrs.	Min.
1 8 – 7	4 5 1 5
1 1	3 0

∴ Total time of picnic is 11:30 hrs.

(2) In Dave’s school, the tree planting ceremony started at 10:00 in the morning and got over at 13:45. How long did the ceremony go on?
Solution:

Hrs.	Min.
1 3 – 1 0	4 5 1 5
3	4 5

∴ Ceremony of planting tree go on for 3 hrs 45 min

Question 4.
Write the time shown in each clock in the box given below it.

Answer:
35 minutes past 3

Answer:
Five minutes to 5

Answer:
Quarter to 2

Answer:
Half past eight

Question 5.
Draw the hands of the clock to show the time given in the box.

Answer:
Quater past 6

Answer:
50 minutes past 1

Answer:
Half past 3

Answer:
5 minutes to 5

Question 6.
The time below is given by the 12 hour clock. Write the same by the 24 hour clock.
(1) 45 minutes past 8 in the morning.
(2) 30 minutes past 2 in the evening.
(3) 50 minutes past 7 in the evening.
(4) 15 minutes past 11 in the evening.
(5) 25 minutes past after midnight.
(6) 25 minutes past 12 in afternoon.
Answer:
(1) 8:45
(2) 114:30
(3) 19:50
(4) 23:15
(5) 00:25
(6) 12:25

Question 7.
Match the following.

‘A’	‘B’
(1) 7:20 am	(a) 13:20
(2) 1:20 pm	(b) 22:10
(3) 6:10 pm	(c) 7:20
(4) 10:10 pm	(d) 6:10
(5) 6:10 am	(e) 18:10

Answer:
(1 – c),
(2 – a),
(3 – e),
(4 – b),
(5 – d)

Question 8.
Add the following.
(1) 3 hours 40 minutes + 2 hours 55 minutes
(2) 5 hours 25 minutes + 4 hours 35 minutes
(3) 6 hours 45 minutes + 1 hour 30 minutes
(4) 7 hours 50 minutes + 2 hours 30 minutes
(5) 9 hours 10 minutes + 3 hours 20 minutes
(6) 15 hours 45 minutes + 20 hours 15 minutes
Answer:
(1) 6 hrs 35 min
(2) 10 hrs
(3) 8 hrs 15 min
(4) 10 hrs 20 min
(5) 12 hrs 30 min
(6) 36 hours

Question 9.
Subtract the following.
(1) 4 hours 20 minutes – 1 hour 30 minutes
(2) 3 hours 25 minutes – 1 hour 45 minutes
(3) 5 hours 10 minutes – 2 hours 40 minutes
(4) 2 hours 15 minutes – 50 minutes
(5) 9 hours 10 minutes – 6 hours 10 minutes
(6) 17 hours 30 minutes – 5 hours 25 minutes
Answer:
(1) 2 hours 50 minutes
(2) 1 hour 40 minutes
(3) 2 hours 30 minutes
(4) 1 hour 25 minutes
(5) 3 hours
(6) 12 hours 05 minutes

Question 10.
Solve the following word problems.
(1) A play started at 9:50 at night and finished at 11:45 the same night. What was the duration of the play?
(2) Ramu went out at 10:45 in the morning and came back home at 7 pm. How long he was out of the home?
(3) Local train started from the Virar station at 8:35 am and reached at Churchgate at 10:30 am. Find the journey time taken by the train.
(4) Anita started her homework at 5:45 pm and completed the work at 7:30 pm. How much time is taken by Anita for the homework?
(5) One day test started at 9:15 am and the test ends at 4-.10 in the evening. How much time was taken for this test?
(6) Seema travelled for 2 hours and 20 minutes by train and 1 hour 30 niinutes by bus. What the total time of her journey?
(7) A train that starts from Mumbai at 17:50 reaches Nira at 2:10. How long does the Mumbai – Nira journey take?
Answer:
(1) 1 hour 55 minutes
(2) 8 hrs 15 min
(3) 1 hr 55 min
(4) 1 hr 45 min
(5) 6 hrs 55 min
(6) 3 hrs 50 min
(7) 8 hrs 20 min

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 2 Number Work Problem Set 4 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 2 Number Work Problem Set 4

Question 1.
Read the numbers and write them in words.
(1) 25,79,899
(2) 30,70,506
(3) 45,71,504
(4) 21,09,900
(5) 43,07,854
(6) 50,00,000
(7) 60,00,010
(8) 70,00,100
(9) 80,01,000
(10) 90,10,000
(11) 91,00,000
(12) 99,99,999
Answer:
(1) Twenty-five lakh, seventy-nine thousand, eight hundred and ninety-nine.
(2) Thirty lakh, seventy thousand, five hundred and six.
(3) Forty-five lakh, seventy-one thousand, five hundred and four.
(4) Twenty-one lakh, nine thousand, nine hundred.
(5) Forty-three lakh, seven thousand, eight hundred and fifty-four.
(6) Fifty lakh.
(7) Sixty lakh and ten.
(8) Seventy lakh and one hundred.
(9) Eighty lakh and one thousand
(10) Ninety lakh and ten thousand
(11) Ninety-one lakh
(12) Ninety-nine lakh, ninety-nine thousand, nine hundred and ninety-nine.

Question 2.
Given below are the deposits made in the Women’s Co-operative Credit Societies of some districts. Read those figures.
Pune : ₹ 94,29,408
Nashik : ₹ 61,07,187
Nagpur : ₹ 46,53,570
Ahmadnagar : ₹ 45,43,159
Aurangabad : ₹ 37,01,282
Yavatmal : ₹ 27,72,348
Sindhudurg : ₹ 58,49,651
Answer:
Rupees ninety-four lakh, twenty-nine thousand, four hundred and eight.
Rupees sixty-one lakh, seven thousand, one hundred and eighty-seven
Rupees forty-six lakh, fifty-three thousand, five hundred and seventy.
Rupees forty-five lakh, forty-three thousand one hundred and fifty-nine.
Rupees thirty-seven lakh, one thousand, two hundred and eighty-two.
Rupees twenty-seven lakh, seventy two thousand, three hundred and forty-eight.
Rupees fifty-eight lakh, forty-nine thousand, six hundred and fifty-one.

The expanded form of a number and the place value of digits

Teacher : Look at the place value of each of the digits in the number 27,65, 043.

Hamid : When we write the place values of the digits as an addition, we get the expanded form of the number. So, the expanded form of the number 27,65,043 is 20,00,000 + 7,00,000 + 60,000 + 5,000 + 0 + 40 + 3.

Teacher : Now tell me the expanded form of 95,04,506.

Soni : 90,00,000 + 5,00,000 + 0 + 4,000 + 500 + 0 + 6.

Teacher : Good! It can also be written as 90,00,000 + 5,00,000 + 4,000 + 500 + 6. Now write the number from the expanded form that I give you. 4,00,000 + 90,000 + 200

Asha : Here, we have 4 in the lakhs place, 9 in the ten thousands place and 2 in the hundreds place. There are no digits in the ten thousands place and in the tens and units places. Hence, we write 0 in those places. Therefore, the number is 4,90,200.

Teacher : Tell me the place value of the underlined digit in the number 59,30,478.
Soni : The underlined digit is 5. The digit is in the ten lakhs place. Hence, its place value is 50,00,000 or fifty lakhs.

Roman Numerals Problem Set 4 Additional Important Questions and Answers

Question 1.
Read the numbers and write them in words:

(1) 80,91,001
Answer:
Eighty lakh, ninety-one thousand and one.

(2) 50,50,505
Answer:
Fifty lakh, fifty thousand, five hundred and five.

(3) 68,06,086
Answer:
Sixty-eight lakh, six thousand and eighty- six.

Question 2.
Given below are the deposits made in the Women’s Co-operative Credit Societies of some districts. Read those figures.

(1) Thane : 75,14,365
Answer:
Rupees seventy-five lakh, fourteen thousand, three hundred and sixty-five.

(2) Jalgaon : 39,42,180
Answer:
Rupees thirty-nine lakh, forty-two thousand, one hundred and eighty.

(3) Kalyan : 37,40,509
Answer:
Rupees thirty-seven lakh, forty thousand, five hundred and nine.

(4) Kolhapur: 16,05,430
Answer:
Rupees sixteen lakh, five thousand, four hundred and thirty.