KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1

Students can Download Chapter 2 Fractions and Decimals Ex 2.1, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1

Question 1.
Solve:
i)
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 1
Solution:
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 2

ii)
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 3
Solution:
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 4

iii)
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 5
Solution:
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 6
L.C.M = 5 × 7 × 1 × 1 = 35
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 7
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 8

iv)
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 9
Solution:

Online Subtracting Fractions Calculator subtracts the fractions 9/11 and 4/5 i.e. 1/55

The given fractions are 9/11 and 4/5

Firstly the L.C.M should be done for the denominators of the two fractions 9/11 and 4/5

9/114/5

The LCM of 11 and 5 (denominators of the fractions) is 55

Given numbers has no common factors except 1. So, there LCM is their product i.e 55

 

= 9 x 5 – 4 x 11/55

= 45 – 44/55

= 1/55

Result: 1/55

v)
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 12
Solution:
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 13
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 14

vi)
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 15
⇒ convert mixed fractions to improper fraction
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vii)
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 17
convert mixed fractions to improper fraction
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 18

Question 2.
Arrange the following in descending order:
i)
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 19
= We need to arrange these in descending order,
To find which number is greater or smaller, we make their denominators equal.
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 20

ii)
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 21
⇒ We make their denominators equal, to find the descending order.
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 22

Question 3.
In a “magic square”, the sum of the numbers in each row, in each column and along the diagonals is the same. Is this a magic square ?
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 23
Solution:
For Row,
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 24
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 25
For diagonals,
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 26
Since Sum of air rows, columns and diagonals are equal.

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Question 4.
A rectangular sheet of paper is 12\(\frac{1}{2}\) cm long and 10\(\frac{2}{3}\) cm wide. Find its perimeter.
Solution:
Length of rectangular sheet of paper = 12\(\frac{1}{2}\) cm
(breadth) width of rectangular sheet paper = 10\(\frac{2}{3}\) cm
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Perimeter of rectangle = 2 (length + breadth)
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converting the above fraction to mixed fraction,
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 29

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Question 5.
Find the perimeters of
(i) ∆ ABE
(ii) the rectangle BCDE in this figure. Whose perimeter is greater ?
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 30
Solution:
(i) ∆ ABE
perimeter of ∆ ABE
perimeter = AB + AE + BE
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 31
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 311

ii) The rectangle BCDE in this figure
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 32
Perimeter of rectangle BCDE,
As it is a rectangle, opposite sides are equal
BC = DE CD = BE
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perimeter of rectangle = 2(l + b)
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 34
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 35
Also,
We have to find which perimeter is greater
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To find which fraction is greater, we make its denominator equal.
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 37
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 38
Perimeter of ∆ ABE > Perimeter of BCDE
(Thus, Perimeter of ∆ ABE is greater)

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Question 6.
Salil wants to put a picture in a frame. The picture is 7\(\frac{3}{10}\) cm wide. How much should the picture be trimmed ?
Solution:
There are two things here – picture, and frame
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 39
so, picture has to be trimmed
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 40
\(=\frac{3}{10}\)
∴ Picture has to trimmed by = \(=\frac{3}{10}\) cm

Question 7.
Ritu ate \(\frac{3}{5}\) part of an apple and the remaining apple was eaten by her brother Somu. How much part of the apple did Somu eat ? Who had the larger share ? By how much ?
Solution:
Total part = 1
Part eaten by Ritu = \(\frac{3}{5}\)
Part eaten by Somu = Total part – part eaten by Ritu
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Now,
We have to tell who ate the larger share so, we have to compare,
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 42
∴ Ritu ate the larger share.
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 43
Ritu ate large share by \(\frac{1}{5}\)

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Question 8.
Michael finished colouring in \(\frac{7}{12}\) hour. Vaibhav finished colouring the same picture in \(\frac{3}{4}\) hour. Who worked longer ? By what fraction was it longer ?
Solution:
Michael finished work in = \(\frac{7}{12}\) hour
Vaibhav finished work in = \(\frac{3}{4}\)
We need to find who worked longer.
i.e., we need to find greater of = \(\frac{7}{12}\) & \(\frac{3}{4}\)
We make their denominators equal
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 44
∴ Vaibhav worked longer.
We also need to find by how much
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 45
Vaibhav worked longer by \(\frac{1}{6}\) hours.

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