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
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
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
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
(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
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
For 1 litre of petrol distance covered = 64 Km
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
∴ 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
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
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
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
Question 3.
0.05 ÷ 0.5 = ?
(i) 0.01
(ii) 0.1
(iii) 0.10
(iv) 1.0
Answer:
(ii) 0.1
Hint:
