KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1
Karnataka State Syllabus Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1
Question 1.
Solve:
i)
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Solution:
![KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 2](https://i2.wp.com/www.kseebsolutions.com/wp-content/uploads/2020/10/KSEEB-Solutions-for-Class-7-Maths-Chapter-2-Fractions-and-Decimals-Ex-2.1-2.png?zoom=1.5&resize=137%2C142&ssl=1)
ii)
Solution:
iii)
Solution:
L.C.M = 5 × 7 × 1 × 1 = 35
iv)
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
9/11 – 4/5
The LCM of 11 and 5 (denominators of the fractions) is 55
= 9 x 5 – 4 x 11/55
= 45 – 44/55
= 1/55
Result: 1/55
v)
Solution:
vi)
⇒ convert mixed fractions to improper fraction
vii)
convert mixed fractions to improper fraction
Question 2.
Arrange the following in descending order:
i)
= We need to arrange these in descending order,
To find which number is greater or smaller, we make their denoinators equal.
ii)
⇒ We make their denominators equal, to find the descending order.
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 ?
Solution:
For Row,
For diagonals,
Since Sum of air rows, columns and diagonals are equal.
Question 4.
A rectangular sheet of paper is 121/2 cm long and 102/3 cm wide. Find its perimeter.
Solution:
Length of rectangular sheet of paper = 121/2 cm
(breadth) width of rectangular sheet paper = 102/3 cm
Perimeter of rectangle = 2 (length + breadth)
converting the above fraction to mixed fraction,
Question 5.
Find the perimeters of
(i) ∆ ABE
(ii) the rectangle BCDE in this figure. Whose perimeter is greater ?
Solution:
(i) ∆ ABE
perimeter of ∆ ABE
perimeter = AB + AE + BE
ii) The rectangle BCDE in this figure
Perimeter of rectangle BCDE,
As it is a rectangle, opposite sides are equal
BC = DE CD = BE
perimeter of rectangle = 2(l + b)
Also,
We have to find which perimeter is greater
To find which fraction is greater, we make its denominator equal.
Perimeter of ∆ ABE > Perimeter of BCDE
(Thus, Perimeter of ∆ ABE is greater)
Question 6.
Salil wants to put a picture in a frame. The picture is 73/10 cm wide. How much should the picture be trimmed ?
Solution:
There are two things here – picture, and frame
so, picture has to be trimmed
=3/10
∴ Picture has to trimmed by = =3/10 cm
Question 7.
Ritu ate 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 = 3/5
Part eaten by Somu = Total part – part eaten by Ritu
Now,
We have to tell who ate the larger share so, we have to compare,
∴ Ritu ate the larger share.
Ritu ate large share by 1/5
Question 8.
Michael finished colouring in 7/12 hour. Vaibhav finished colouring the same picture in 3/4 hour. Who worked longer ? By what fraction was it longer ?
Solution:
Michael finished work in = 7/12 hour
Vaibhav finished work in = 3/4
We need to find who worked longer.
i.e., we need to find greater of = 7/12 & 3/4
We make their denominators equal
∴ Vaibhav worked longer.
We also need to find by how much
Vaibhav worked longer by 1/6 hours.