NCERT Solutions for Class 6 Maths Chapter 8 Decimals in Hindi and English Medium updated for CBSE session 2022-2023.
Class 6 Maths Chapter 8 Solutions in Hindi Medium
Class 6 Maths Chapter 8 all Exercises Solution
NCERT Solutions for Class 6 Maths Chapter 8
Class VI Exercise 8.1, Exercise 8.2, Exercise 8.3, Exercise 8.4, Exercise 8.5 and Exercise 8.6 in English Medium and Hindi Medium updated for current session. Solutions of Prashnavali 8.1, Prashnavali 8.2, Prashnavali 8.3, Prashnavali 8.4, Prashnavali 8.5 and Prashnavali 8.6 in Hindi Medium free to use. You can download these solutions in PDF file format also. Video solutions of each exercise are also available along with the PDF solutions. All the NCERT Solutions Apps have been updated for the session 2022-23 according to NCERT (https://ncert.nic.in/) website, the latest textbooks based on new CBSE Syllabus. Download Class 6 NCERT Solutions Offline Apps for offline free use.
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Download Class 6 Maths Chapter 8 Solutions in PDF Format
Download NCERT Solutions for Class 6 Maths Chapter 8 in PDF format updated for academic session 2022-2023 following the latest NCERT Books. All the contents are simplified and easy to understand. Methods adopted are accordance to standard VI students.
About Class 6 Maths Chapter 8
In 6 Maths Chapter 8 Decimals, we will study various conversion into decimals. We will also learn about tenths, hundredths, etc., places in a decimal number. Place value chart using decimal is also important in this chapter. In the place value system as we go from left to right, at every step the multiplying factor becomes 1/10 of the previous factor. Comparison of number will also learnt here in this chapter.
Let us now compare the numbers 92.44 and 92.4.
In this case , we first compare the whole part.
We see that the whole part for both the numbers is 92 and, hence, equal.
We, however, know that the two numbers are not equal. So, we now compare the tenth part.
We find that for 92.44 and 92.4, the tenth part is also equal, then we compare the hundredth part.
Now the hundredth part of the first number is 4 whereas the second number is not given so, it is 0.
Hence, 92.44 is greater than 92.4.
Important Questions on Class 6 Maths Chapter 8
Write each of the following as decimals: (a) One hundred and two-ones (b) Six hundred point eight
(a) One hundred and 2-ones
= 100 + 2 x 1
= 100 + 2
(b) Six hundred point eight = 600.8
The length of Ramesh’s notebook is 9 cm and 5 mm. What will be its length in cm?
9 cm 5 mm = 9 cm + 5 mm = 9 + 5/10 cm = 9.5 cm
Rashid spent ₹35.75 for Maths book and ₹32.60 for Science book. Find the total amount spent by Rashid.
Money spent for Maths book = ₹35.75
Money spent for Science book = ₹32.60
Total money spent = ₹35.75 + ₹32.60 = ₹68.35
Therefore, total money spent by Rashid is ₹68.35.
Radhika’s mother have her ₹10.50 and her father gave her ₹15.80. Find the total amount given to Radhika by the parents.
Money given by mother = ₹10.50
Money given by father = ₹15.80
Total money received by Radhika = ₹10.50 + ₹15.80 = ₹26.30
Therefore, the total money received by Radhika is ₹26.30.
Nasreen bought 3 m 20 cm cloth for her shirt and 2 m 5 cm cloth for her trouser. Find the total length of cloth bought by her.
Cloth bought for shirt = 3 m 20 cm = 3.20 m
Cloth bought for trouser = 2 m 5 cm = 2.05 m
Total length of cloth bought by Nasreen = 3.20 + 2.05 = 5.25 m
Therefore, the total length of cloth bought by Nasreen is 5.25 m
Naresh walked 2 km 35 m in the morning and 1 km 7 m in the evening. How much distance did he walk in all?
Distance travelled in morning = 2 km 35 m = 2.035 km
Distance travelled in evening = 1 km 7 m = 1.007 km
Total distance travelled = 2.035 + 1.007 = 3.042 km
Therefore, the total distance travelled by Naresh is 3.042 km.
Sunita travelled 15 km 268 m by bus, 7 km 7 m by car and 500 m by foot in order to reach her school. How far is her school from her residence?
Distance travelled by bus = 15 km 268 m = 15.268 km
Distance travelled by car = 7 km 7 m = 7.007 km
Distance travelled on foot = 500 m = 0.500 km
Total distance travelled = 15.268 + 7.007 + 0.500 = 22.775 km
Therefore, total distance travelled by Sunita is 22.775 km.
Ravi purchases 5 kg 400 g rice, 2 kg 20 g sugar and 10 kg 850 g flour. Find the total weight of his purchases.
Weight of Rice = 5 kg 400 g = 5.400 kg
Weight of Sugar = 2 kg 20 g = 2.020 kg
Weight of Flour = 10 kg 850 g = 10.850 kg
Total weight = 5.400 + 2.020 + 10.850 = 18.270 kg
Therefore, the total weight of Ravi’s purchase = 18.270 kg.
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Which concepts will students study in chapter 8 of 6th standard Maths?
The concepts that students will study in chapter 8 of 6th standard Maths are:
- Definition of decimal
- Words to decimals
- Decimals in number line
- Decimals to fractions
- Fractions to decimals
- Using decimals- converting to mm, cm, km, kg, rupees
- Comparing decimals
- Addition of decimal numbers
- Addition of decimal numbers- statement questions
- Subtraction of decimal numbers
- Subtraction of decimal number- statement questions.
Which exercise of chapter 8 of 6th standard Maths has the highest number of illustrations?
Chapter 8 of class 6th Maths has 6 exercises.
The first exercise (Ex 8.1) has three examples (examples 1, 2, 3).
The second exercise (Ex 8.2) has seven examples (examples 4, 5, 6, 7, 8, 9, 10).
The third exercise (Ex 8.3) has one example (example 11).
The fourth exercise (Ex 8.4) has no example.
The fifth exercise (Ex 8.5) has three examples (examples 12, 13, 14).
The sixth exercise (Ex 8.6) has three examples (examples 15, 16, 17).
So, the second exercise (Ex 8.2) has the highest number of examples.
Can students easily solve chapter 8 of Class 6th Maths?
No, students can’t easily solve chapter 8 of grade 6th Maths. Chapter 8 is not very easy and not very tough. It is in between easy and tough. Students require the teacher’s help to solve chapter 8 of 6th standard Maths.
What are the everyday uses of decimals (chapter 8 of 6th standard Maths)?
We use decimals in everyday life while dealing with length, weight, money, distance, height, etc. Decimal numbers are used in those situations where more accuracy is required than the whole numbers can provide. For example, when we measure our weight on the weighing machine, we don’t always find the weight equal to a whole number on the scale. To know our exact weight, we must know what the decimal value on the scale means.