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
Multiplying and Dividing Rational Expressions Calculator.
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
Using a fraction and whole number calculator in situations where you struggle to solve your arithmetic assignments can be useful.
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.
Free quadratic equation completing the square calculator – Solve quadratic equations using completing the square step-by-step.
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,
2 2/3 as an improper fraction in its simplest form.
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.
