NCERT Solutions for Class 6 Maths Chapter 8 Decimals (Ex 8.6) Exercise 8.6

NCERT Solutions for Class 6 Maths Chapter 8 Decimals (Ex 8.6) Exercise 8.6

NCERT textbooks, which can help students get ready for their future experiences, are the main mandatory readings for CBSE students. NCERT books are an excellent resource since they help students in a variety of ways to improve their base knowledge. NCERT textbooks make it simple to thoroughly investigate any topic’s foundations. Students can gain a thorough understanding of the subjects by carefully reading the NCERT texts. Students who have the necessary conceptual clarity are able to solve problems quickly and successfully, which has long-term advantages. A learner won’t need to study an idea if they can understand it. The most crucial source for students’ basic knowledge is NCERT texts.

When students take their annual examinations at the end of the year, this might be a significant advantage. Students will be better equipped to respond to more questions on the exam if they are aware of the many question types that may be asked on exams. They could be able to answer more questions truthfully as a result, which could raise their test grade. Success in Class 6 has a tremendous impact on a student’s academic career. Class 6 lays the groundwork for learning in Class 8. Higher education colleges commonly use the results of the Class 8 board exams to determine who gets admitted. Passing the board examinations is typically required for enrollment in various courses. However, even studying for the yearly Class 6 exam might be difficult. It will require effort and careful planning. Students should be required to practise answering questions frequently, especially in a subject like Maths. The CBSE recommends using the NCERT textbook as a part of its curriculum. Students must complete NCERT lessons in order to learn and do well on examinations.

If students want to ace their board examinations, they must have access to the best study materials. Additionally, they need a plan for efficient planning. Completing the NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6 is one of the best ways to increase exam performance. Due to the wide variety of questions and answers in the NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6, students can practise answering a variety of questions. On the tests, they were able to answer more questions. For each subject, there are a number of important questions that students should consider. They can get accustomed to answering numerous questions, which will eventually enable them to write more rapidly. Access to the Class 6 Maths Chapter 8 Exercise 8.6 may thereby improve a student’s time management skills. Before testing, pupils could become accustomed to the format of the questions and create a successful strategy for answering them.

Before the annual examinations, students need to have access to a lot of questions and in-depth explanations. Illustrations from NCERT textbooks are used in the majority of question banks. The pupils must respond to each of these inquiries. Numerous highly important questions can be found in the NCERT practice examinations. Extramarks provides NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6 ready to assist students with the questions and answers. When pupils have access to NCERT Solutions Class 6 Maths Chapter 8 Exercise 8.6, answering these questions is less difficult.

Access Other Exercises of Class 6 Maths Chapter 8

Chapter 8 – Decimals Exercises
Exercise 8.1
10 Questions & Solutions
Exercise 8.2
7 Questions & Solutions
Exercise 8.3
2 Questions & Solutions
Exercise 8.4
5 Questions & Solutions
Exercise 8.5
7 Questions & Solutions

NCERT Solutions for Class 6 Maths, Chapter 8: Decimals (Ex 8.6) Exercise 8.6

The NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6 can be downloaded from the Extramarks’ website or mobile application so that CBSE Class 6 students can access and study the solutions even when they are not online. To better understand the issues, they must complete NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6.

The NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6 is beneficial for students preparing for yearly exams since they are written in simple language that every student can understand. The NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6 was created step-by-step to help students succeed in their exams. The Class 6th Math Exercise 8.6 was developed using the most recent CBSE syllabus and recommendations. Students can avoid wasting time and effort by getting the information they need regarding the courses covered in the CBSE curriculum and how exam grades are calculated without having to look elsewhere. In the NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6, all of the problems from the lesson are fully answered.

Students may react to questions like these quickly and accurately in examinations if they consistently practice the NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6. Students do not require any additional assistance when utilizing the NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6, because they were written in an extremely simple style. Subject experts prepared the NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6 in a kind and instructive style. Maths is regarded as one of the more challenging classes for sixth graders. However, students may easily solve the problems if they use the in-depth solutions provided in the NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6. The questions on the upcoming yearly examinations will be easier for students who put a lot of work into these exercises.

The NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6 includes application examples and explanations that simplify the exercises in the lesson. As a result, it enhances students’ understanding of the subject, which enhances their exam performance. Giving pupils enough examples to practice can help them better understand each subject. In order to help students achieve the maximum possible grades on their exams, the NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6, is written in great depth. In Class 6, concepts and lessons covered in Class 5 are further developed. It also prepares students for further study in maths and related disciplines. Online learning tools are readily available for Class 6 students. Some of these resources might not be required or useful for Class 6 students to adequately prepare for their exams. Because NCERT textbooks are required for all students by the CBSE, they must read the NCERT book for Maths.

Access NERT Solutions for Class 6 Maths, Chapter 8 – Decimals Exercise 8.6 

Students can access the  NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6 on the Extramarks website and mobile app. Students can quite easily use this curriculum to pass competitive tests like the Joint Entrance Exam (JEE) if they choose to do so. Students who desire to apply to the Indian Institutes of Technology (IITs) and National Institutes of Technology (NITs) can use this to gain basic knowledge. Through competitive exams like the Central Universities Entrance Tests (CUET) and others, students can pursue careers outside of engineering. Studying NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6 would help students understand the material because MCQs may be present in these tests.

Students may easily access NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6, PDF format because of a learning program called Extramarks that makes them usable even when they are offline. NCERT Solutions is the best source for instructing students on Class 6 Maths Chapter 8 Exercise 8.6. Most students find Class 6 challenging because the material gets more challenging after that grade. However, some students encounter decision-making challenges occasionally as they move from a basic to an intermediate level. Because it is identified as NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6, finding it on the Extramarks website and mobile application is simple.

Everyone who intends to continue their education throughout the year should have a PDF copy of the NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6. Additionally, it could be difficult to get Hindi NCERT solutions. NCERT solutions and other educational materials are rarely available on websites in Hindi, as opposed to the vast majority of websites that do so in English. Although NCERT Solutions are accessible online, ongoing offline access is preferred. Due to offline access, students can now choose to learn whenever and wherever they choose. For thorough learning, students can use the NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6 from Extramarks at any time.

Unexpected network issues and poor network access won’t be a barrier to learning anymore. The skillfully crafted NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6 provides a chance for study and exam practice by including every problem from the textbook. Students can download the NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6 in PDF format from the Extramarks website and mobile application to use as part of their ongoing test preparation

Exercise 8.6

The CBSE-recommended syllabus for the annual examinations is fully covered in the NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6 from Extramarks. To assist students in preparing for the test questions, qualified experts have developed The NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6. The NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6 can be used by students to prepare for the JEE and NEET, among other college entrance exams. Students should read the NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6 at least twice in order to fully comprehend the problems.

Decimal subtraction is covered in NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6. Before the practice problems, a collection of examples are solved to help the students become proficient in these ideas. By using the PDF of NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6 created by Extramarks educators, students can promptly get their questions answered. The finest explanations of the solutions are provided in order to assist students in passing the test.

NCERT Solutions for Class 6 Maths Chapter 8 Decimals Exercise 8.6

Students who have access to the NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6 will benefit from having a firm grasp of both basic and advanced concepts while responding to problems from the NCERT school textbooks. Students will have a strong conceptual understanding of the chapter once they have finished it. They can assess themselves and make corrections by comparing their solutions to the NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6. The Extramarks’ NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6 is a study aid that focuses on a specific chapter. Students who want to do better in their exams can choose from a variety of learning modules and study tools, including chapter-by-chapter worksheets and practice examinations.

How to subtract two decimal values is demonstrated in the NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6. As with addition, decimal subtraction can be accomplished by subtracting hundredths from hundredths, tenths from tenths, ones from ones, and so forth. When subtracting decimal numbers, one must additionally regroup the numbers. The thorough manual in NCERT Solutions for Class 6 Maths, Chapter 8, Exercise 8.6, aids students in staying ahead.

There are a total of 7 questions in Exercise 8.6 of chapter 8 of the NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6. Two of these seven questions are in the form of equations, while the other five are word problems. The NCERT Solutions for Class 6 Maths, Chapter 8, Exercise 8.6, are provided below:

Q.1 Subtract :
(a) 18.25 from20.75
(b) 202.54 m from 250 m
(c) 5.36 from 8.40
(d) 2.051 km from 5.206 km
(e) 0.314 kg from 2.107 kg

Ans.

(a) 18.25 from20.75

20.75 − 18.25 ¯ 2.50 ¯   MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiGacaGaaeqabaWaaqaafaaakqaabeqaaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGYaGaaeimaiaab6cacaqG3aGaaeynaaqaaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiabgkHiTiaabccadaadaaqaaiaabccacaqGGaGaaeymaiaabIdacaqGUaGaaeOmaiaabwdacaqGGaaaaaqaaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaWaaWaaaeaacaqGGaGaaeiiaiaabccacaaIYaGaaiOlaiaaiwdacaaIWaGaaeiiaiaabccaaaGaaGjbVlaabccaaaaa@5F27@

Therefore, 20.75 − 18.25 = 2.50
(b) 202.54 m from 250 m

250.00 − 202.54 ¯ 47.46 ¯ MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiGacaGaaeqabaWaaqaafaaakqaabeqaaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGYaGaaeynaiaabcdacaqGUaGaaeimaiaabcdaaeaacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacqGHsislcaqGGaWaaWaaaeaacaqGGaGaaeiiaiaabkdacaqGWaGaaeOmaiaab6cacaqG1aGaaeinaiaabccaaaaabaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccadaadaaqaaiaabccacaqGGaGaaeiiaiaabsdacaqG3aGaaeOlaiaabsdacaaI2aGaaeiiaiaabccaaaaaaaa@5F05@

Therefore, 250 m – 202.54 m = 47.46 m
(c) 5.36 from 8.40

8.40 − 5.36 ¯ 3.04  ¯ MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@6460@

Therefore, 8.40 – 18.25 = 3.04
(d) 2.051 km from 5.206 km

5.206 − 2.051 ¯ 3.155 ¯ MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiGacaGaaeqabaWaaqaafaaakqaabeqaaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabwdacaqGUaGaaeOmaiaabcdacaqG2aaabaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaeyOeI0IaaeiiamaamaaabaGaaeiiaiaabccacaqGYaGaaeOlaiaabcdacaqG1aGaaeymaiaabccaaaaabaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccadaadaaqaaiaabccacaqGGaGaaG4maiaac6cacaaIXaGaaGynaiaaiwdacaqGGaaaaaaaaa@5BC6@

Therefore, 5.206 km – 2.051 km = 3.155 km
(e) 0.314 kg from 2.107 kg

2.107 − 0.314 ¯ 1.793  ¯ ​ MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@715A@

Therefore, 2.107 kg – 0.314 kg = 1.793 kg

Q.2 Find the value of:
(a) 9.756 – 6.28
(b) 21.05 – 15.27
(c) 18.5 – 6.79
(d) 11.6 – 9.847

Ans.

(a) 9.756 – 6.28                9.756 − 6.280 ¯ 3.476 ¯ (b) 21.05 – 15.27 21.05 − 15.27 ¯ 5.78 ¯ (c) 18.5 – 6.79 18.50 − 6.79 ¯ 11.71  ¯ (d) 11.6 – 9.847 9.756 11.600 − 9.847 ¯ 1.753 ¯

Q.3 Raju bought a book for 35.65. He gave 50 to the shopkeeper. How much money did he get back from the shopkeeper?

Ans.

Cost price of the book = 35.65
Money given to the shopkeeper = 50
So, the money that Raju will get back is

50.00 − 35.65¯ 14.35 ¯

Therefore, the money that Raju will get back is 14.35.

Q.4 Rani had 18.50. She bought one ice-cream for 11.75. How much money does she have now?

Ans.

Total money with Rani = 18.50

The money spent for an ice-cream= 11.75

So, the money left with Rani =

18.50 − 11.75 ¯ 6.75 ¯ MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@6360@

Therefore, the money left with Rani is 6.75.

Q.5 Tina had 20 m 5 cm long cloth. She cuts 4 m 50 cm length of cloth from this for making a curtain. How much cloth is left with her?

Ans.

Total length of the cloth =20 m 5 cm=20 m+5100  m=20 m+ 0.05 m =20.05 mLength of the cloth cut to make a curtain=4 m 50 cm=4 m+50100  m=4 m+ 0.5 m =4.5 mSo, the cloth left with her: 20.05 − 4.50 ¯ 15.55 ¯ Hence,the length of the cloth left with Tina is 15.55 m.

Q.6 Namita travels 20 km 50 m every day. Out of this she travels 10 km 200 m by bus and the rest by auto. How much distance does she travel by auto?

Ans.

Distance travelled by Namita everyday=20 km 50 m =20 km+501000  km=20 km+ 0.05 km =20.05 kmDistance travelled by bus=10 km 200 m=10 km+2001000  km=10 km+ 0.2 km =10.2 kmHence, the distance travelled by auto : 20.05 − 10.20 ¯ 9.85 ¯ Therefore,the distance travelled by auto is 9.850 km.

Q.7 Aakash bought vegetables weighing 10 kg. Out of this, 3 kg 500 g is onions, 2 kg 75 g is tomatoes and the rest is potatoes. What is the weight of the potatoes?

Ans.

Total Weight of vegetables=10 kgWeight of onions=3 kg 500 g=3 kg+5001000  kg=3 kg+ 0.5 kg =3.5 kgWeight of tomatoes=2 kg 75 g=2 kg+751000  kg=2 kg+ 0.075 kg =2.075 kgTherefore, the weight of potatoes= Total weight of vegetables−( weight of tomatoes + weight of onions) = 10 −(2.075+3.5) First we will solve the bracket. 2.075 + 3.500 ¯ 5.575 ¯ Now,we will subtract the sum of the weights of tomatoes and onions from the weight of vegetables. 10.000 − 5.575 ¯ 4.425 ¯Therefore,the weight of potatoes is 4.425 kg.

Please register to view this section

FAQs (Frequently Asked Questions)

1. How can students get the NCERT Solutions Class 6 Maths Chapter 8 Exercise 8.6?

Students may only access and download the NCERT Solutions Class 6 Maths Chapter 8 Exercise 8.6 in PDF format from the Extramarks website and mobile app. For children preparing for board examinations and other admissions exams, that makes sense and is helpful.

2. Do the detailed solutions in Extramarks' NCERT Solutions for Class 6 Maths Chapter 5 Exercise 5.2 exist?

Each question is fully answered in the NCERT Solutions Class 6 Maths Chapter 5 Exercise 5.2 by Extramarks. The responses go into great detail on each step. The complete range of theoretical issues has been addressed. Diagrams are included whenever they are needed. Students can start writing their answers with confidence and earn the best result by carefully reading the NCERT Solutions Class 6 Maths Chapter 5 Exercise 5.2.