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

9/11 – 4/5

## 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.