NCERT Solutions for Class 6 Maths Chapter 8

NCERT Solutions for Class 6 Maths Chapter 8 Decimals 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 session 2020-2021.

Solutions of प्रश्नावली 8.1, प्रश्नावली 8.2, प्रश्नावली 8.3, प्रश्नावली 8.4, प्रश्नावली 8.5 and प्रश्नावली 8.6 in हिंदी मीडियम 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 and Offline Apps have been updated for the session 2020-21 according to latest NCERT Books based on new CBSE Syllabus 2020-21. Download Class 6 NCERT Solutions Offline Apps for offline use.




NCERT Solutions for Class 6 Maths Chapter 8

Class:6
Subject:Maths – गणित
Chapter 8:Decimals

Download Class 6 Maths Chapter 8 Solutions in PDF Form

Download NCERT Solutions for Class 6 Maths Chapter 8 in PDF format updated for academic session 2020-2021 following the latest NCERT Books 2020-21. 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.




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Important Questions on 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
= 102
= 102.0

(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
निम्न को दशमलव रूप में लिखिए: 20 + 9 + 4/10 + 1/100.
20 + 9 + 4/10 + 1/100
=20 + 9 + 0.4 + 0.01
= 29.41
निम्न दशमलव को शब्दों में लिखिए: 108.56
एक सौ आठ दशमलव पाँच छः
The length of a young gram plant is 65 mm. Express its length in cm.
65 mm
= 60 mm + 5 mm
= 6 cm + 5/10 cm
= 6.5 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.

0.45 संख्या रेखा के किन दो बिन्दुओं के बीच स्थित हैं?
0.45 संख्या 0 और 1 के बीच स्थित हैं।

संख्या 0.45 बिंदु 0.5 के पास स्थित है।

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

राजू एक पुस्तक ₹35.65 की खरीदता है। उसने दुकानदार को ₹50 दिए। दुकानदार ने उसे कितने रुपये वापिस दिए?
दुकानदार को दिए गए कुल रूपये = ₹50
पुस्तक की कीमत = ₹35.65

शेष राशि
= ₹50.00 – ₹35.65
= ₹14.35

अतः, दुकानदार ने राजू को ₹14.35 वापिस किए।

रानी के पास ₹18.50 हैं। उसने ₹11.75 की एक आइसक्रीम खरीदी। अब उसके पास कितने रूपये बचे?
रानी के पास कुल धनराशि = ₹18.50
आइसक्रीम की कीमत = ₹11.75

शेष राशि
= ₹18.50 – ₹11.75
= ₹6.75

अतः, रानी के पास अब ₹6.75 बचे हैं।

टीना के पास 20 मी 5 सेमी लंबा कपड़ा है। उसमें से उसने एक पर्दा बनाने के लिए 4 मी 50 सेमी कपड़ा काट लिया। टीना के पास अब कितना लंबा कपड़ा बचा?
कपड़े की कुल लंबाई
= 20 मी 5 सेमी
= 20.05 मी

पर्दे के लिए काटा गया कपड़ा
= 4 मी 50 सेमी
= 4.50 मी

शेष बचा कपड़ा
= 20.05 मी – 4.50 मी
= 15.55 मी

अतः, टीना के पास अब 15.55 मी लंबा कपड़ा शेष है।

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.

नमिता प्रतिदिन 20 किमी 50 मी की दूरी तय करती है। इसमें से 10 किमी 200 मी दूरी वह बस द्वारा तय करती है और शेष ऑटो-रिक्शा द्वारा। नमिता ऑटो-रिक्शा द्वारा कितनी दूरी तय करती है?
कुल तय दूरी
= 20 किमी 50 मी
= 20.050 किमी

बस द्वारा तय दूरी
= 10 किमी 200 मी
= 10.200 किमी

इसलिए, ऑटो-रिक्शा द्वारा तय दूरी
= 20.050 किमी – 10.200 किमी
= 9.850 किमी

अतः, नमिता ऑटो-रिक्शा द्वारा 9.850 किमी की दूरी तय करती है।

आकाश 10 किग्रा सब्जी खरीदता है जिसमें से 3 किग्रा 500 ग्रा प्याज, 2 किग्रा 75 ग्रा टमाटर और शेष आलू हैं। आलू का वजन ज्ञात कीजिए।
सब्जी का कुल भार
= 10.000 किग्रा

प्याज का भार
= 3 किग्रा 500 ग्रा
= 3.500 किग्रा

टमाटर का भार
= 2 किग्रा 75 ग्रा
= 2.075 किग्रा

प्याज और टमाटर का कुल भार
= 3.500 किग्रा + 2.075 किग्रा
= 5.575 किग्रा

इसलिए, आलू का भार
= 10.000 किग्रा – 5.575 किग्रा
= 4.425 किग्रा

अतः, आलू का वजन 4.425 किग्रा है।

NCERT Solutions for Class 6 Maths Chapter 8
NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.1
NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.1 in english medium
NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.1 in pdf form free download
NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.2
NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.2 in pdf form free download
NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.2 in english medium
Class 6 Maths Chapter 8 Exercise 8.3
NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.4
NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.4 in english medium free pdf download
NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.5
NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.5 in free pdf download
NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6
NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.6 pdf all question download
6 Maths Chapter 8 Exercise 8.1
6 Maths 8.1
Class vi maths 8.1
ex 8.1 class vi
6 Maths Chapter 8 Exercise 8.2
6 Maths 8.2
ex 8.2 class vi
6 Maths Chapter 8 Exercise 8.3
6 Maths Chapter 8 Exercise 8.4
Class vi maths 8.4
6 Maths Chapter 8 Exercise 8.5
class vi ex 8.5
ex 8.5 class 6 solutions
6 Maths Chapter 8 Exercise 8.6
Class vi solutions ex 8.6