Students can Download Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

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

Learn how to calculate if 596/652 and 365/838 are proportional ratios.

Question 1.

Solve:

I) 2 – \(\frac{3}{5}\)

Solution:

ii) 4 + \(\frac{7}{8}\)

Solution:

iii) \(\frac{3}{5}+\frac{2}{7}\)

Solution:

iv) \(\frac{9}{11}-\frac{4}{15}\)

Solution:

v)

Solution:

vi)

Solution:

⇒ convert mixed fractions to improper fraction

Multiplying fractions with whole numbers calculator is a free online tool that gives the product of fractional and whole numbers for the given input.

vii)

Solution:

convert mixed fractions to improper fraction

Question 2.

Arrange the following in descending order:

I)

Solution:

= We need to arrange these in descending order, To find which number is greater or smaller, we make their denominators equal.

ii)

Solution:

⇒ 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,

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)

∴ Perimeter of the rectangular sheet of paper is \(46 \frac{1}{3}\) cm.

Question 5.

Find the perimeters of

(a) ∆ ABE

(b) the rectangle BCDE in this figure. Whose perimeter is greater?

Solution:

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:

We have,

Hence, \(\frac{3}{10}\) cm width of the picture should be trimmed.

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 of the apple eaten by Somu = \(1-\frac{3}{5}\)

= \(\frac{5}{5}-\frac{3}{5}=\frac{5-3}{5}=\frac{2}{5}\)

Since \(\frac{3}{5}>\frac{2}{5}\) , Ritu had the larger share.

Now, \(\frac{3}{5}-\frac{2}{5}=\frac{3-2}{5}=\frac{1}{5}\)

∴ Ritu had( \(\frac{1}{5}\) ) part more then Somu.

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

Hence, Vaibhav took \(\frac{1}{6}\) hour more then minchel.