SCERT Solutions for Class 7 Maths Chapter 2 Fraction and Decimal Exercise 2.1
SCERT Assam Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Exercise 2.1 Fractions and Decimals in English and Assamese Medium modified and updated for academic year 2024-25. All the question-answers and solutions are revised on the basis of new syllabus and latest SEBA SCERT textbooks issued for curriculum 2024-25.
All the question answers with solutions are done according to latest SCERT Assam Books for 2024-25.
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SCERT Assam Solutions for Class 7 Maths Chapter 2 Fraction and Decimal Exercise 2.1
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Class: 7 |
Mathematics |
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Chapter: 2 |
Exercise2.1 |
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Session: |
SEBA 2024 - 25 |
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Medium: |
English, Assamese & Hindi |
Class 7 Maths Solutions: Chapter 2 Fraction and Decimal solutions can be viewed by clicking on the Exercise Solutions Link below.
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SCERT Assam Solutions for Class 7 Maths Chapter 2 Fraction and Decimal Exercise 2.1
Exercise – 2.1
1. Find the product –
A. (i) `6\times\frac{2}{3}` (ii) `7\times\frac{1}{5}` (iii) `5\times2\frac{3}{4}` (iv) `3\frac{5}{7}\times28` (v) `2\frac{3}{4}\times5`
B. (i) `\frac{1}{7}\times\frac{1}{9}` (ii) `\frac{1}{45}\times\frac{9}{39}` (iii) `\frac{4}{15}\times\frac{9}{10}` (iv) `\frac{51}{40}\times\frac{64}{34}` (v) `\frac{4}{5}\times\frac{12}{7}`
C. (i) `4\frac{2}{7}\times11\frac{2}{3}` (ii) `9\frac{2}{3}\times4\frac{4}{5}` (iii) `5\frac{5}{6}\times6\frac{3}{7}` (iv) `4\frac{1}{8}\times2\frac{10}{11}` (v) `2\frac{2}{17}\times7\frac{2}{9}\times1\frac{33}{52}`
Solution:
A. (i) `6\times\frac{2}{3}`
`=\frac{12}{3}`
`=4`
(ii) `7\times\frac{1}{5}`
`=\frac{7}{5}`
`=1\frac{2}{5}`
(iii)`5\times2\frac{3}{4}`
`=5\times\frac{11}{4}`
`=\frac{55}{4}`
`=13frac{3}{4}`
(iv) `3\frac{5}{7}\times28`
`=\frac{26}{7}\times28`
`\color{fuchsia} {=26\times4}`
`\color{fuchsia} {=104}`
(v)`2\frac{3}{4}\times5`
`=\frac{11}{4}\times5`
`=\frac{55}{4}`
`=13\frac{3}{4}`
B. (i) `\frac{1}{7}\times\frac{1}{9}`
`=\frac{1}{63}`
(ii) `\frac{1}{45}\times\frac{9}{39}`
`=\frac{1\times1}{5\times39}`
`=\frac{1}{195}`
(iii) `\frac{4}{15}\times\frac{9}{10}`
`=\frac{2\times3}{5\times5}`
`=\frac{6}{25}`
(iv) `\frac{51}{40}\times\frac{64}{34}`
`=\frac{3\times8}{5\times2}`
`=\frac{12}{5}`
`=2\frac{2}{5}`
(v) `\frac{4}{5}\times\frac{12}{7}`
`=\frac{4\times12}{5\times7}`
`=\frac{48}{35}`
C. (i) `4\frac{2}{7}\times11\frac{2}{3}`
`=\frac{30}{7}\times\frac{35}{3}`
`=10\times5`
`=50`
(ii) `9\frac{2}{3}\times4\frac{4}{5}`
`=\frac{29}{3}\times\frac{24}{5}`
`=\frac{29\times8}{5}`
`=\frac{232}{5}`
`=46\frac{2}{5}`
(iii) `5\frac{5}{6}\times6\frac{3}{7}`
`=\frac{35}{6}\times\frac{45}{7}`
`=\frac{5\times15}{2\times1}`
`=\frac{75}{2}`
`=37\frac{1}{2}`
(iv) `4\frac{1}{8}\times2\frac{10}{11}`
`=\frac{33}{8}\times\frac{32}{11}`
`=3\times4`
`=12`
(v) `2\frac{2}{17}\times7\frac{2}{9}\times1\frac{33}{52}`
`=\frac{34}{17}\times\frac{65}{9}\times\frac{85}{52}`
`=25`
2. Evaluate –
(i) `\frac{1}{5}of\frac{1}{7}` (ii) `\frac{4}{5}of\frac{2}{3}` (iii)`\frac{15}{14}of\frac{7}{5}` (iv) `\frac{3}{22}of2\frac{3}{4}` (v) `\frac{7}{30}of15`
Solution:
(i) `\frac{1}{5}of\frac{1}{7}`
`=\frac{1}{5}\times\frac{1}{7}`
`=\frac{1}{35}`
(ii) `\frac{4}{5}of\frac{2}{3}`
`=\frac{4}{5}\times\frac{2}{3}`
`=\frac{8}{15}`
(iii) `\frac{15}{14}of\frac{7}{5}`
`=\frac{15}{14}\times\frac{7}{5}`
(iv) `\frac{3}{22}of2\frac{3}{4}`
(v) `\frac{7}{30}of15`
3. What is 
Solution:
= 
= 
= 
4. (i) Evaluating 
of 
and (ii) 
of 
, determine which one is smaller.
Solution:
(i)
of 
= 
= 
(ii)
of 
= 
= 
Now, we compare 
and
and 
Since 0.1667 is smaller than 0.3, we can conclude
Thus, of (which
simplifies to ) is smaller
than of (which simplifies to )
5. Bijit’s has spent 
part on purchasing copy and pen and 
parts for geometry instrument box from the
amount of money his mother gave him as pocket money and kept the rest in hand.
Find the part of money left with Bijit and arrange the three parts in
descending order.
Solution:
Given
Part spent on copy and pen =
Part spent on geometry instrument box
=
To add these fractions, we need a common denominator. The least common multiple (LCM) of 5 and 7 is 35
Total part spent =
= 
= 
The part of money left with Bijit, = 1
= 
= 
So, the part of money left with Bijit is
Now, let's arrange the three parts (money spent on copy and pen, money spent on geometry instrument box, and money left) in descending order,
6. Shyamalee reads 
hours everyday. Out of it she spends 
part in studying Mathematics and English, 
part for in Science and rest in studying other
subjects. Find out her different study times and arrange them in ascending
order.
Solution:
Given,
The total study time of Shyamalee = hours
= 
hours
Out of it she spends for Mathematics and English = 
part
Out of it she spends for Science = 
part
Out of it she spends for other subjects is
= 
= 
= 
= 
= 
= 
= 
Now,
= 
= 
In ascending
order
or <
7. In school competition of ‘Model Reading’
Utpal completed 
parts of the allotted portion in 3 minutes and
Runjun completed 
parts of the allotted page in the same time.
Who read more?
Solution:
Given,
In school competition of ‘Model Reading’ Utpal
completed = 
part
and Runjun completed = 
part
Convert each fraction to have a denominator of 66
Since 
Therefore, Runjun read more.
8. Rita read part of a short story book of 75 pages. How
many pages of the book are left to read.
Solution:
Given,
Rita read 
part of a
75-page book.
The number of pages Rita read = 
= 15×3
= 45
Now,
subtract the pages read from the total number of pages
Pages left = 75−45
= 30 pages
So, Rita has 30 pages left to read.
9. A man has ₹ 200 within. He paid one fifth of the amount as bus fare. How much amount was left with him?
Solution:
Given,
The total amount of money the man had = ₹ 200
The bus fare from the total amount = 
= 1
= ₹ 40
Therefore, the man has ₹160 left.
10. Each of the two water tanks in Ranjus
house, contain 500 litres of water in the morning. parts of water of one tank was used for
bathing and washing and 
parts of water of the other tank was used for
cooking. What was the total water remaining in the two tanks?
Solution:
Given,
Ranjus first tank contained water = 500 litres
The amount of water used from the first tank for bathing and washing
= 
= 3
= 300 litres
Ranjus second tank contained water = 500 litres
The amount of water used from the second tank for cooking
= 
= 1
= 125 litres
Now,
Water remaining in the first tank =500−300
= 200 litres
Water remaining in the second tank=500−125
=375 litres
The total water remaining =200+375
= 575 litres
Therefore, the total water remaining in the two tanks is 575 litres.
11. Bobita took
part of a cake for eating. In the mean time her brother came and
snatched away from her part. what part of the cake colud
Bobita eat ?
Solution:
Given,
Bobita took 
part of a cake
to eat
The amount of cake her brother snatched away from her part
= 
= 
=
=
Part Bobita could eat = 
= 
= 
= 
Therefore, Bobita could eat of the cake.
12. Out of ready made shirt in a shop 
part are white. part are blue and the rest are yellow.
If the number of yellow shirts is 72. What are the numbers shirts of other
varities. What is the total number of shirts in shop ?
Solution:
Given,
White shirts = 
part
Blue shirts = part
Let 𝑥 be the total number of shirts.
The part of yellow shirts = 
=
=
=
=
=
Since the number of yellow shirts is 72
According to Question 
Now,
Number of white shirts = 
= 
= 24
Number of blue shirts = 
= 
= 
= 120
Therefore, the shop has 24 white shirts, 120 blue shirts, and 72 yellow shirts, with a total of 216 shirts.
13. Fill in the blanks
(i)
=
(ii) 
(iii) 
(iv) 
Solution:
(i)
=
(ii)
(iii)
(iv)
SEBA SCERT Solutions for Class 7 Maths Chapter 2: Fractions and Decimals
Exercise 2.1: Introduction to Fractions
In this exercise, students will delve into the fundamental concepts of fractions. The exercise is designed to reinforce their understanding of how fractions represent parts of a whole and how different fractions can express the same value in various ways.
Topics Covered:
Understanding Fractions: Students will learn the basics of fractions, including the numerator and denominator, and how they are used to represent parts of a whole.
Equivalent Fractions: This section will help students understand how to determine if two fractions are equivalent by cross-multiplying and simplifying fractions to their lowest terms.
Simplifying Fractions: Students will practice reducing fractions to their simplest form by dividing the numerator and the denominator by their greatest common divisor (GCD).
Comparing Fractions: This topic will guide students through comparing fractions by finding a common denominator or by converting them to decimals.