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.
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