# 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)

Solution:

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

9/114/5

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)

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 denominators 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 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

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 7$$\frac{3}{10}$$ cm wide. How much should the picture be trimmed ?
Solution:
There are two things here – picture, and frame

so, picture has to be trimmed

$$=\frac{3}{10}$$
∴ 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

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 $$\frac{1}{5}$$

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

∴ Vaibhav worked longer.
We also need to find by how much

Vaibhav worked longer by $$\frac{1}{6}$$ hours.