Assam SCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Exercise 2.3

Sudev Chandra Das

Heading with Bright Purple Background
Free Solutions for The Assam State School Education Board (ASSEB) Class 7 Maths Chapter 2 Exercise 2.3 in English Medium

Assam SCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Exercise 2.3

Digital Pipal Academy provides Assam SCERT Solutions for Class 7 Maths Exercise 2.3 in a downloadable PDF format. This exercise focuses on the multiplication of a fraction by a fraction, covering the multiplication of proper, improper, and mixed fractions. By practicing the solutions for Class 7 Maths Chapter 2 Fractions and Decimals, students can strengthen their understanding and improve their performance in Mathematics. It is highly recommended that students attempt the questions independently before referring to the solutions.

Get Free NCERT PDFs

If you want to download free PDFs of any chapter from any subject, click the link below and join our WhatsApp.

Join Our WhatsApp

 

Assam SCERT Solutions for Class 7 Maths Chapter 2: Fractions and Decimals (Exercise 2.3)

SCERT Solution for Class 7 Maths Chapter 2 Fractions and Decimals

SCERT Solution for Class 7 Maths Chapter 2 Fractions and Decimals

Chapter Name Solution Link
Exercise 2.1 Click Here
Exercise 2.2 Click Here
Exercise 2.3 Click Here
Exercise 2.4 Click Here
Exercise 2.5 Click Here

 

What You’ll Learn in Exercise 2.3?

✔️ Multiplication of fractions
✔️ Division of fractions
✔️ Word problems involving fractions

Get step-by-step solutions to master fractions and decimals effortlessly! 🚀📖

 

Chapter – 2: Fractions and Decimals

Chapter – 2: Fractions and Decimals Exercise 2.3


Exercise 2.3

Question 1. Write the decimal in expanded form:

(i) 3.05
(ii) 30.5
(iii) 235.005
(iv) 23005.005

Solution:-

(i) 3.05

 `3.05=3\times1+0\times\frac{1}{10}+5 \times\frac{1}{100}`

(ii) 30.5

`30.5= 3 \times 10 + 0 \times 1 + 5 \times \frac{1}{10}` 

(iii) 235.005 

`235.005= 2 \times 100 + 3 \times 10 + 5 \times 1 + 0 \times \frac{1}{10}`

` + 0 \times \frac{1}{100} + 5 \times \frac{1}{1000}`


(iv) 23005.005

` 23005.005= 2 \times 10000 + 3 \times 1000 + 0 \times 100 + `

`0 \times 10 + 5 \times 1 + 0 \times \frac{1}{10} + 0 \times \frac{1}{100} + 5 \times \frac{1}{1000}` 

 

 Question 2. Use decimals to express in meters and kilometers:

(i) 20 cm
(ii) 267 cm
(iii) 25732 mm
(iv) 3540 cm

 Solution:

(i) 20 cm = 20100\frac{20}{100} = 0.2 m
Again, 20100×1000\frac{20}{100 \times 1000} km = 0.0002 km

(ii) 2.67 m and 0.00267 km

(iii) 25.732 m and 0.025732 km

(iv) 35.4 m and 0.0354 km


 Question 3. Use decimals to express in kilograms:

(i) 520 g
(ii) 4273 g
(iii) 692050 cg
(iv) 2 kg 5 g

Solution:

(i) 520 gm

=5201000 kg= \frac{520}{1000} \text{ kg} =0.52 kg= 0.52 \text{ kg}

(ii) 4.273 kg

(iii) 6.9205 kg

(iv) 2.005 kg

Question 4. Express in Rupees:

(i) 5 paise
(ii) 5 rupees 5 paise
(iii) 55 rupees 55 paise
(iv) 50 rupees 50 paise

 Solution:

(i) 5 paise

=5100 Rupees= \frac{5}{100} \text{ Rupees} =0.05 Rs= 0.05 \text{ Rs}

(ii) 5 rupees 5 paise

=5+5100= 5 + \frac{5}{100} =5+0.05= 5 + 0.05 =5.05 Rs= 5.05 \text{ Rs}

(iii) 55 rupees 55 paise

=55+55100= 55 + \frac{55}{100} =55+0.55= 55 + 0.55 =55.55 Rs= 55.55 \text{ Rs}

(iv) 50 rupees 50 paise

=50+50100= 50 + \frac{50}{100} =50+0.50= 50 + 0.50 =50.50 Rs= 50.50 \text{ Rs}

Question 5. Which one is greater? Answer orally:

(i) 0.2 and 0.02
(ii) 3.03 and 3.30
(iii) 5 and 0.5
(iv) 0.4 and 0.44

Solution:

(i) 0.2 is greater than 0.02
(0.20 > 0.02)

(ii) 3.30 is greater than 3.03
(3.30 > 3.03)

(iii) 5 is greater than 0.5
(5.00 > 0.50)

(iv) 0.44 is greater than 0.4
(0.44 > 0.40)

 

Question 6. In an annual sports competition, in the boys' long jump event, Roktim jumped 3.3 m and Pranjal jumped 333 cm. Who jumped more and by how much?

Solution:

Roktim's jump = 3.3 m
Pranjal's jump = 333 cm = 333 ÷ 100 = 3.33 m
Difference = 3.33 - 3.3 = 0.03 m

Thus, Pranjal jumped 0.03 m more than Roktim.




MCQs – Assam SCERT Class 7 Maths Chapter 2 (Exercise 2.3)

  1. What is the product of 35\frac{3}{5} and 23\frac{2}{3}?
    a) 615\frac{6}{15}
    b) 25\frac{2}{5}
    c) 56\frac{5}{6}
    d) 15\frac{1}{5}
    Answer: b) 25\frac{2}{5}

  2. Which of the following fractions is the reciprocal of 47\frac{4}{7}?
    a) 74\frac{7}{4}
    b) 47\frac{4}{7}
    c) 37\frac{3}{7}
    d) 73\frac{7}{3}
    Answer: a) 74\frac{7}{4}

  3. If 58\frac{5}{8} is divided by 23\frac{2}{3}, what is the result?
    a) 512\frac{5}{12}
    b) 1516\frac{15}{16}
    c) 108\frac{10}{8}
    d) 1024\frac{10}{24}
    Answer: b) 1516\frac{15}{16}

  4. Which property is used in a×1a=1a \times \frac{1}{a} = 1?
    a) Commutative property
    b) Associative property
    c) Distributive property
    d) Reciprocal property
    Answer: d) Reciprocal property

  5. What is the result of 6÷346 \div \frac{3}{4}?
    a) 8
    b) 9
    c) 92\frac{9}{2}
    d) 7
    Answer: b) 9

  6. What is the product of 79\frac{7}{9} and 35\frac{3}{5}?
    a) 2145\frac{21}{45}
    b) 715\frac{7}{15}
    c) 1027\frac{10}{27}
    d) 49\frac{4}{9}
    Answer: b) 715\frac{7}{15}

  7. If a fraction is multiplied by its reciprocal, what is the result?
    a) The same fraction
    b) 0
    c) 1
    d) The numerator of the fraction
    Answer: c) 1

  8. Which of the following represents division of 89\frac{8}{9} by 23\frac{2}{3}?
    a) 89×32\frac{8}{9} \times \frac{3}{2}
    b) 89×23\frac{8}{9} \times \frac{2}{3}
    c) 89÷32\frac{8}{9} \div \frac{3}{2}
    d) 89+23\frac{8}{9} + \frac{2}{3}
    Answer: a) 89×32\frac{8}{9} \times \frac{3}{2}

  9. If 12÷4512 \div \frac{4}{5}, then the answer is:
    a) 10
    b) 15
    c) 604\frac{60}{4}
    d) 9
    Answer: b) 15

  10. A cake is divided into 6 equal parts, and Rina ate 2 parts. What fraction of the cake did she eat?
    a) 26\frac{2}{6}
    b) 13\frac{1}{3}
    c) 12\frac{1}{2}
    d) 25\frac{2}{5}
    Answer: a) 26\frac{2}{6} or 13\frac{1}{3}


Why Choose Digital Pipal Academy?

Q: Is Digital Pipal Academy the No. 1 Educational Website for Assam Education for Free Learning?

Answer: Yes! Digital Pipal Academy provides high-quality, free educational resources for Assam Board students, making learning easy and accessible.

Q: What is the Best Website for Learning Class 7 Maths in Assam?

Answer: Digital Pipal Academy is the best website for learning Class 7 Maths in Assam. Visit www.pipalacademy.com for free study materials, NCERT & SCERT solutions, and expert guidance.

At Digital Pipal Academy, we offer:
✔️ Detailed SCERT solutions with clear explanations 📖
✔️ Easy-to-understand concepts and real-life applications
✔️ Exam-focused content for SEBA Board students 🎯
✔️ Free and accessible learning materials 💡
✔️ Regular updates and the latest syllabus coverage 📅


FAQs – Assam SCERT Class 7 Maths Chapter 2 (Exercise 2.3)

Q1: What topics are covered in Exercise 2.3 of Chapter 2?

Answer: Exercise 2.3 covers multiplication and division of fractions, including real-life word problems.

Q2: How will these solutions help in exam preparation?

Answer: The step-by-step solutions ensure a clear understanding of concepts, helping students perform better in exams.

Q3: Are the solutions provided here as per the SCERT Assam syllabus?

Answer: Yes, all solutions strictly follow the SCERT Assam syllabus and are written in an exam-friendly format.

Q4: Can I download the solutions for Exercise 2.3?

Answer: Currently, the solutions are available online for free at Digital Pipal Academy. Stay tuned for downloadable PDF versions in the future!

Q5: Where can I find solutions for other Class 7 Maths chapters?

Answer: You can explore Digital Pipal Academy for free solutions to all SCERT Class 7 Maths chapters, as well as other subjects like Science and English.


📌 Related Posts:

📖 Chapter 2 (Exercise 2.2): Fractions and Decimals - SCERT Solutions
📖 Chapter 3: Data Handling - SCERT Solutions

For more Assam SCERT Class 7 Maths Solutions, visit Digital Pipal Academy and stay updated with our latest educational content! 🚀

Sudev Chandra Das

About the Publisher

Hi! I'm Sudev Chandra Das (B.Sc. Mathematics), the Founder of Digital Pipal Academy. I've dedicated myself to guiding students toward better education. I believe, 'Success comes from preparation, hard work, and learning from failure.' Let’s embark on a journey of growth and digital excellence together!

 

Note for Users

If you find any incorrect answers, please notify us via Instagram at @pipalacademy or email us at info@pipalacademy.com. For content that may infringe copyright, kindly refrain from copying our content. Thank you for supporting Digital Pipal Academy!

Join Our WhatsApp

 

Our website uses cookies to enhance your experience. Learn More
Accept !