Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System Ex 1.2

Students can Download Maths Chapter 1 Number System Ex 1.2 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 1.2

Question 1.
Add by using grid 0.51 + 0.25.
Solution:
Here 0.51 = \(\frac { 51 }{ 100 } \) and 0.25 = \(\frac { 25 }{ 100 } \)
Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System 1.1 1
First we shade the region 0.51 and then 0.25.
The sum is the total shaded area. 0.51 + 0.25 = 0.76

Question 2.
Add the following by using place value grid.
(i) 25.8 + 18.53
(ii) 17.4 + 23.435
Solution:
(i) 25.8 + 18.53.
Using place value grid.
Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System 1.1 2
Therefore 25.8 + 18.53 = 44.33

(ii) 17.4 + 23.435
Lets use the place value grid.
Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System 1.1 3
Therefore 17.4 + 23.435 = 40.835

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System Ex 1.2

Question 3.
Find the value of 0.46 – 0.13 by grid model.
Solution:
Here 0.46 = \(\frac { 46 }{ 100 } \) and 0.13 = \(\frac { 13 }{ 100 } \)
Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System 1.1 4
Shading the region 0.46 and then crossing out 0.13 from the shaded area. The left out shaded region without cross marks is the difference. So 0.46 – 0.13 = 0.33

Question 4.
Subtract the following by using place value grid, (i) 6.567 from 9.231 (ii) 3.235 from 7
Solution:
(i) Let as use place value grid
Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System 1.1 5
Therefore 9.231 – 6.567 = 2.664

(ii) Let as use place value grid.
Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System 1.1 6
Therefore 7 – 3.235 = 3.765

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System Ex 1.2

Question 5.
Simplify: 23.5 – 27.89 + 35.4 – 17.
Solution:
23.5 – 27.89 + 35.4 – 17 = 14.01
Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System 1.1 7

Question 6.
Sulaiman bought 3.350 kg of Potato, 2.250 kg of Tomato and some onions. If the weight of the total items are 10.250 kg, then find the weight of onions?
Solution:
Weight of Potato = 3.350 kg
Weight of Tomato = 2.250 kg
Total weight of Potato and Tomato = (3.350 + 2.250 kg)
= 5.600 kg
Weight of potato, tomato and onions = 10.250
Weight of potato and tomato = 5.600
∴ Weight of onions = (10.250 – 5.600) Kg = 4.650 Kg
Weight of onions = 4.650 Kg

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System Ex 1.2

Question 7.
What should be subtracted from 7.1 to get 0.713?
Solution:
To get the number to be subtracted
Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System 1.1 8
We have 7.1 – 0.713 = 6.387
∴ The number to be subtracted = 6.387

Question 8.
How much is 35.6 km less than 53.7 km?
Solution:
To get the answer we must subtract 53.7km – 35.6 km = 18.1 km
Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System 1.1 9
So 35.6 km is 18.1 km less than 53.7 km.

Question 9.
Akilan purchased a geometry box for ₹ 25.75, a pencil for ₹ 3.75 and a pen for ₹ 17.90. He gave ₹ 50 to the shopkeeper. What amount did he get back?
Solution:
Cost of geometry box = ₹ 25.75 (+)
Cost of Pencil box = ₹ 3.75
Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System 1.1 10

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System Ex 1.2

Question 10.
Find the perimeter of an equilateral triangle with a side measuring 3.8 cm.
Solution:
Perimeter of an equilateral triangle = (Side + Side + Side) Sq. units.
Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System 1.1 11
Given side = 3.8
∴ Perimeter = 3.8 + 3.8 + 3.8
Perimeter of the triangle = 11.4 cm

Objective Type Questions

Question 1.
1.0 + 0.83 = ?
(i) 0.17
(ii) 0.71
(iii) 1.83
(iv) 1.38
Answer:

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System 1.1 12
(iii) 1.83

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System Ex 1.2

Question 2.
7.0 – 2.83 = ?
(i) 3.47
(ii) 4.17
(iii) 7.34
(iv) 4.73
Answer:
Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System 1.1 13
(ii) 4.17

Question 3.
Subtract 1.35 from 3.51
(i) 6.21
(ii) 4.86
(iii) 8.64
(iv) 2.16
Answer:
Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System 1.1 14
(iv) 2.16

Question 4.
Sum of two decimals is 4.78 and one decimal is 3.21 then the other one is
(i) 1.57
(ii) 1.75
(iii) 1.59
(iv) 1.58
Answer:
Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System 1.1 15
(i) 1.57

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System Ex 1.2

Question 15.
The difference of two decimals is 86.58 and one of the decimal is 42.31 Find the other one
(i) 128.89
(ii) 128.69
(iii) 128.36
(iv) 128.39
Answer:
Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System 1.1 16
(i) 128.89

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