Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System Ex 1.4

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System Ex 1.4

Question 1.
Simplify the following
(i) 0.6 ÷ 3
(ii) 0.90 ÷ 5
(iii) 4.08 ÷ 4
(iv) 21.56 ÷ 7
(v) 0.564 ÷ 6
(vi) 41.36 ÷ 4
(vii) 298.2 ÷ 3
Solution:
(i) 0.6 ÷ 3
= \(\frac { 6 }{ 10 } \) × \(\frac { 1 }{ 3 } \)
= \(\frac { 6 }{ 3 } \) × \(\frac { 1 }{ 10 } \)
= 2 × \(\frac { 1 }{ 10 } \)
= \(\frac { 2 }{ 10 } \)
= 0.2

(ii) 0.90 ÷ 5
= \(\frac { 90 }{ 100 } \) × \(\frac { 1 }{ 5 } \)
= \(\frac { 90 }{ 5 } \) × \(\frac { 1 }{ 100 } \)
= 18 × \(\frac { 1 }{ 100 } \) = \(\frac { 18 }{ 100 } \)
= 0.18

(iii) 4.08 ÷ 4
= \(\frac { 408 }{ 100 } \) × \(\frac { 1 }{ 4 } \)
= \(\frac { 408 }{ 4 } \) x \(\frac { 1 }{ 100 } \)
= 102 × \(\frac { 1 }{ 100 } \)
= \(\frac { 102 }{ 100 } \)
= 1.02

(iv) 21.56 ÷ 7
= \(\frac { 2156 }{ 100 } \) × \(\frac { 1 }{ 7 } \)
= \(\frac { 2156 }{ 7 } \) × \(\frac { 1 }{ 100 } \)
= 308 × \(\frac { 1 }{ 100 } \)
= \(\frac { 308 }{ 100 } \)
= 3.08

(v) 0.564 ÷ 6
= \(\frac { 564 }{ 1000 } \) × \(\frac { 1 }{ 6 } \)
= \(\frac { 564 }{ 6 } \) × \(\frac { 1 }{ 1000 } \)
= \(\frac { 94 }{ 1000 } \)
= 0.094

(vi) 41.36 ÷ 4
= \(\frac { 4136 }{ 100 } \) × \(\frac { 1 }{ 4 } \)
= \(\frac { 4136 }{ 4 } \) × \(\frac { 1 }{ 100 } \)
= \(\frac { 1034 }{ 100 } \)
= 10.34

(vii) 298.2 ÷ 3
= \(\frac { 2982 }{ 10 } \) × \(\frac { 1 }{ 3 } \)
= \(\frac { 2982 }{ 3 } \) × \(\frac { 1 }{ 10 } \)
= \(\frac { 994 }{ 10 } \)
= 99.4

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System Ex 1.4

Question 2.
Simplify the following.
(i) 5.7 ÷ 10
(ii) 93.7 ÷ 10
(iii) 0.9 ÷ 10
(iv) 301.301 ÷ 10
(v) 0.83 ÷ 10
(vi) 0.062 ÷ 10
Solution:
(i) 5.7 ÷ 10
= \(\frac { 57 }{ 10 } \) × \(\frac { 1 }{ 10 } \)
= \(\frac { 57 }{ 100 } \)
= 0.57

(ii) 93.7 ÷ 10
= \(\frac { 937 }{ 10 } \) × \(\frac { 1 }{ 10 } \)
= \(\frac { 937 }{ 100 } \)
= 9.37

(iii) 0.9 ÷ 10
= \(\frac { 9 }{ 10 } \) × \(\frac { 1 }{ 10 } \)
= \(\frac { 9 }{ 100 } \)
= 0.09

(iv) 301.301 ÷ 10
= \(\frac { 301301 }{ 1000 } \) × \(\frac { 1 }{ 10 } \)
= \(\frac { 301301 }{ 10000 } \)
= 30.1301

(v) 0.83 ÷ 10
= \(\frac { 83 }{ 100 } \) × \(\frac { 1 }{ 10 } \)
= \(\frac { 83 }{ 1000 } \)
= 0.083

(vi) 0.062 ÷ 10
= \(\frac { 62 }{ 1000 } \) × \(\frac { 1 }{ 10 } \)
= \(\frac { 62 }{ 1000 } \)
= 0.0062

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System Ex 1.4

Question 3.
Simplify the following.
(i) 0.7 ÷ 100
(ii) 3.8 ÷ 100
(iii) 49.3 ÷ 100
(iv) 463.85 ÷ 100
(v) 0.3 ÷ 100
(vi) 27.4 ÷ 100
Solution:
(i) 0.7 ÷ 100
= \(\frac { 7 }{ 10 } \) × \(\frac { 1 }{ 100 } \)
= \(\frac { 7 }{ 1000 } \)
= 0.007

(ii) 3.8 ÷ 100
= \(\frac { 38 }{ 10 } \) × \(\frac { 1 }{ 100 } \)
= \(\frac { 38 }{ 1000 } \)
= 0.038

(iii) 49.3 ÷ 100
= \(\frac { 493 }{ 10 } \) × \(\frac { 1 }{ 100 } \)
= \(\frac { 493 }{ 1000 } \)
= 0.493

(iv) 463.85 ÷ 100
= \(\frac { 46385 }{ 100 } \) × \(\frac { 1 }{ 100 } \)
= \(\frac { 46385 }{ 1000 } \)
= 4.6385

(v) 0.3 ÷ 100
= \(\frac { 3 }{ 10 } \) × \(\frac { 1 }{ 100 } \)
= \(\frac { 3 }{ 1000 } \)
= 4.6385

(vi) 27.4 ÷ 100
= \(\frac { 274 }{ 10 } \) × \(\frac { 1 }{ 100 } \)
= \(\frac { 274 }{ 1000 } \)
= 0.274

Question 4.
Simplify the following.
(i) 18.9 ÷ 1000
(ii) 0.87 ÷ 1000
(iii) 49.3 ÷ 1000
(iv) 0.3 ÷ 1000
(v) 382.4 ÷ 1000
(vi) 93.8 ÷ 1000
Solution:
(i) 18.9 ÷ 1000
= \(\frac { 189 }{ 10 } \) × \(\frac { 1 }{ 1000 } \)
= \(\frac { 189 }{ 10000 } \)
= 0.0189

(ii) 0.87 ÷ 1000
= \(\frac { 87 }{ 100 } \) × \(\frac { 1 }{ 1000 } \)
= \(\frac { 87 }{ 100000 } \)
= 0.00087

(iii) 49.3 ÷ 1000
= \(\frac { 493 }{ 10 } \) × \(\frac { 1 }{ 100 } \)
= \(\frac { 493 }{ 1000 } \)
= 0.493

(iv) 0.3 ÷ 1000
= \(\frac { 3 }{ 10 } \) × \(\frac { 1 }{ 1000 } \)
= \(\frac { 3 }{ 10000 } \)
= 0.0003

(v) 382.4 ÷ 1000
= \(\frac { 3824 }{ 10 } \) × \(\frac { 1 }{ 1000 } \)
= \(\frac { 3824 }{ 10000 } \)
= 0.3824

(vi) 93.8 ÷ 1000
= \(\frac { 938 }{ 10 } \) × \(\frac { 1 }{ 1000 } \)
= \(\frac { 938 }{ 10000 } \)
= 0.0938

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System Ex 1.4

Question 5.
Simplify the following.
(i) 19.2 ÷ 2.4
(ii) 4.95 ÷ 0.5
(iii) 19.11 ÷ 1.3
(iv) 0.399 ÷ 2.1
(v) 5.4 ÷ 0.6
(vi) 2.197 ÷ 1.3
Solution:
(i) 19.2 ÷ 2.4
Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System 1.4 1

(ii) 4.95 ÷ 0.5
Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System 1.4 2

(iii) 19.11 ÷ 1.3
Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System 1.4 3

(iv) 0.399 ÷ 2.1
Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System 1.4 4Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System 1.4 5

(v) 5.4 ÷ 0.6
Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System 1.4 6

(vi) 2.197 ÷ 1.3
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Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System Ex 1.4

Question 6.
Divide 9.55 kg of sweet among 5 children. How much will each child get?
Solution:
Weight of the sweet = 9.55 Kg
Weight of sweet for 5 children = \(\frac { 955 }{ 100 } \) Kg
Weight of sweet for 1 child = \(\frac{\left(\frac{955}{100}\right)}{5}\) = \(\frac { 955 }{ 100 } \) × \(\frac { 1 }{ 5 } \) = \(\frac { 955 }{ 5 } \) × \(\frac { 1 }{ 100 } \)
= \(\frac { 191 }{ 100 } \) = 1.91
Each child will get 1.91 kg sweet.

Question 7.
A vehicle covers a distance of 76.8 km for 1.2 litre of petrol. How much distance will it cover for one litre of petrol?
Solution:
For 1.2 litre of petrol the distance covered = 76.8 Km = \(\frac { 768 }{ 10 } \) Km
Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System 1.4 9
For 1 litre of petrol distance covered = 64 Km

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System Ex 1.4

Question 8.
Cost of levelling a land at the rate of ₹ 15.50 sq. ft is ₹ 10,075. Find the area of the land.
Solution:
Cost of levelling the entire land = ₹ 10,075
Cost of levelling 1 sq. ft = ₹ 15.50
Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System 1.4 10
∴ Area of the land = 650 sq.ft.

Question 9.
The cost of 28 books are ₹ 1506.4. Find the cost of one book.
Solution:
Cost of 28 books = ₹ 1506.4
Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System 1.4 11
Cost of 1 book = ₹ 53.80

Question 10.
The product of two numbers is 40.376. One number is 14.42. Find the other number.
Product of two numbers = 40.376
One number = 14.42
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Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System Ex 1.4

Objective Type Questions

Question 1.
5.6 ÷ 0.5 = ?
(i) 11.4
(ii) 10.4
(iii) 0.14
(iv) 11.2
Answer:
(iv) 11.2
Hint:
\(\frac { 5.6 }{ 0.5 } \) = \(\frac { 56 }{ 5 } \)
= 11.2
Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System 1.4 14

Question 2.
2.01 ÷ 0.03 = ?
(i) 6.7
(ii) 67.0
(iii) 0.67
(iv) 0.067
Answer:
(ii) 67.0
Hint: \(\frac { 2.01 }{ 0.03 } \) = \(\frac { 201 }{ 3 } \)
= 67
Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System 1.4 15

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System Ex 1.4

Question 3.
0.05 ÷ 0.5 = ?
(i) 0.01
(ii) 0.1
(iii) 0.10
(iv) 1.0
Answer:
(ii) 0.1
Hint:
Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System 1.4 16