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

Students can Download Maths Chapter 1 Number System Ex 1.4 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

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
Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System 1.4 7Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System 1.4 8

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
Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System 1.4 12Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System 1.4 13

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

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