SCERT Solutions for Class 7 Maths Chapter 2 Fraction and Decimal Exercise 2.1

Sudev Chandra Das

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.


 


SCERT Assam Solutions for Class 7 Maths Chapter 2 Fraction and Decimal Exercise 2.1

 

Class: 7

Mathematics

Chapter: 2

Exercise2.1

Session:

SEBA 2024 - 25

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)          

 


Digital Pipal Academy Verified By

Sudev Chandra Das(B.Sc. Mathematics)

 

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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.

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