NCERT Solutions For Class 6 Maths Chapter 8 : Decimals
NCERT Solutions For Class 6 Maths Chapter 8 Decimals are intended to help students of Class 6 to get accustomed to the topic of decimals and to practise the concepts thoroughly. Our experts at BYJU’S designed these NCERT Solutions with the aim of providing all possible methods to solve the questions, as per the CBSE syllabus of Class 6 Maths Chapter 8. These PDFs can be downloaded by the students for future reference as well as to practise the question answers of the chapter. Comparing the answers written by the students with the NCERT Solutions for Class 6 prepared by the students will help them in understanding where they went wrong or the steps they can take to improve.Chapter 8 DecimalsExercise 8.1
NCERT Solutions for Class 6 Chapter 8: Decimals Download PDF
Sometimes students find it challenging to solve the exercise problems given at the end of the NCERT book for the practice of students. Therefore, we have introduced NCERT Solutions which will help students to resolve their doubts and difficulties with the help of solved questions. Apart from these, we have notes, question papers, which consist of sample papers and previous year question papers, and preparation tips study materials, which will be helpful for students to score good marks in the final exam.
Access NCERT Solutions for Class 6 Chapter 8: Decimals
Exercise 8.1 page no: 167
1. Write the following numbers in the given table.
Hundreds | Tens | Ones | Tenths |
(100) | (10) | (1) | (1 / 10) |
Solutions:
Rows | Hundreds | Tens | Ones | Tenths |
a | 0 | 3 | 1 | 2 |
b | 1 | 1 | 0 | 4 |
2. Write the following decimals in the place value table.
(a) 19.4
(b) 0.3
(c) 10.6
(d) 205.9
Solutions:
Hundreds | Tens | Ones | Tenths | |
19.4 | 0 | 1 | 9 | 4 |
0.3 | 0 | 0 | 0 | 3 |
10.6 | 0 | 1 | 0 | 6 |
205.9 | 2 | 0 | 5 | 9 |
3. Write each of the following as decimals:
(a) Seven-tenths
(b) Two tens and nine-tenths
(c) Fourteen point six
(d) One hundred and two ones
(e) Six hundred point eight
Solutions:
(a) The decimal form of Seven-tenths is 7 / 10 = 0.7
(b) The decimal form of two tens and nine tenths is 20 + 9 / 10 = 20.9
(c) The decimal form of fourteen point six is 14.6
(d) The decimal form of one hundred and two ones is 100 + 2 = 102.0
(e) The decimal form of six hundred point eight is 600.8
4. Write each of the following as decimals:
(a) 5 / 10
(b) 3 + 7 / 10
(c) 200 + 60 + 5 + 1 / 10
(d) 70 + 8 / 10
(e) 88 / 10
(f)
(g) 3 / 2
(h) 2 / 5
(i) 12 / 5
(j)
(k)
Solutions:
(a) 5 / 10 = 0.5
(b) 3 + 7 / 10 = 3 + 0.7
= 3.7
(c) 200 + 60 + 5 + 1 / 10 = 265 + 0.1
= 265.1
(d) 70 + 8 / 10 = 70 + 0.8
= 70.8
(e) 88 / 10 = 80 / 10 + 8 / 10
= 8 + 0.8
= 8.8
(f)
(g) 3 / 2 = (2 + 1) / 2
= 2 / 2 + 1 / 2
= 1 + 0.5
= 1.5
(h) 2 / 5 = 0.4
(i) 12 / 5 = (10 + 2) / 5
= 10 / 5 + 2 / 5
= 2 + 0.4
= 2.4
(j)
(k)
5. Write the following decimals as fractions. Reduce the fraction to lowest form.
(a) 0.6
(b) 2.5
(c) 1.0
(d) 3.8
(e) 13.7
(f) 21.2
(g) 6.4
Solutions:
(a) 0.6 = 6 / 10
= 3 / 5
(b) 2.5 = 25 / 10
= 5 / 2
(c) 1.0 = 1
= 1
(d) 3.8 = 38 / 10
= 19 / 5
(e) 13. 7 = 137 / 10
(f) 21.2 = 212 / 10
= 106 / 5
(g) 6.4 = 64 / 10
= 32 / 5
6. Express the following as cm using decimals.
(a) 2 mm
(b) 30 mm
(c) 116 mm
(d) 4 cm 2 mm
(e) 162 mm
(f) 83 mm
Solutions:
We know that
1 cm = 10 mm
1 mm = 1 / 10 cm
(a) 2 mm = 2 / 10 cm
= 0.2 cm
(b) 30 mm = 30 / 10 cm
= 3.0 cm
(c) 116 mm = 116 / 10 cm
= 11.6 cm
(d) 4 cm 2 mm = [(4 + 2 / 10)] cm
= 4.2 cm
(e) 162 mm = 162 / 10 cm
= 16.2 cm
(f) 83 mm = 83 / 10 cm
= 8.3 cm
7. Between which two whole numbers on the number line are the given numbers lie?
Which of these whole numbers is nearer the number?
(a) 0.8
(b) 5.1
(c) 2.6
(d) 6.4
(e) 9.1
(f) 4.9
Solutions:
(a) 0.8 lies between 0 and 1
0.8 is nearer to 1
(b) 5.1 lies between 5 and 6
5.1 is nearer to 5
(c) 2.6 lies between 2 and 3
2.6 is nearer to 3
(d) 6.4 lies between 6 and 7
6.4 is nearer to 6
(e) 9.1 lies between 9 and 10
9.1 is nearer to 9
(f) 4.9 lies between 4 and 5
4.9 is nearer to 5
8. Show the following numbers on the number line.
(a) 0.2
(b) 1.9
(c) 1.1
(d) 2.5
Solutions:
(a) 0.2 lies between the points 0 and 1 on the number line. The space between 0 and 1 is divided into 10 equal parts. Therefore each equal part will be equal to one-tenth. 0.2 is the second point between 0 and 1
(b) 1.9 lies between the points 1 and 2 on the number line. The space between 1 and 2 is divided into 10 equal parts. Therefore each equal part will be equal to one-tenth. 1.9 is the ninth point between 1 and 2
(c) 1.1 lies between the points 1 and 2 on the number line such that the space between 1 and 2 is divided into 10 equal parts. Therefore each equal part will be equal to one-tenth. 1.1 is the first point between 1 and 2
(d) 2.5 lies between the points 2 and 3 on the number line such that the space between 2 and 3 is divided into 10 equal parts. Therefore each equal part will be equal to one-tenth. 2.5 is the fifth point between 2 and 3
9. Write the decimal number represented by the points A, B, C, and D on the given number line.
Solutions:
(a) Point A represents 0.8 cm on the given number line.
(b) Point B represents 1.3 cm on the given number line
(c) Point C represents 2.2 cm on the given number line
(d) Point D represents 2.9 cm on the given number line
10. (a) The length of Ramesh’s notebook is 9 cm 5 mm. What will be its length in cm?
(b) The length of a young gram plant is 65 mm. Express its length in cm.
Solutions:
(a) The length of Ramesh notebook is 9 cm 5 mm
The length in cm is [(9 + 5 / 10)] cm
= 9.5 cm
(b) The length of a gram plant is 65 mm
Hence, the length in cm is 65 / 10
= 6.5 cm
Exercise 8.2 page no: 173
1. Complete the table with the help of these boxes and use decimals to write the number.
Solutions:
Rows | Ones | Tenths | Hundreds | Number |
(a) | 0 | 2 | 6 | 0.26 |
(b) | 1 | 3 | 8 | 1.38 |
(c) | 1 | 2 | 8 | 1.28 |
2. Write the numbers given in the following place value table in decimal form.
Rows | Hundreds | Tens | Ones | Tenths | Hundredths | Thousandths |
100 | 10 | 1 | 1 / 10 | 1/ 100 | 1 / 1000 | |
(a) | 0 | 0 | 3 | 2 | 5 | 0 |
(b) | 1 | 0 | 2 | 6 | 3 | 0 |
(c) | 0 | 3 | 0 | 0 | 2 | 5 |
(d) | 2 | 1 | 1 | 9 | 0 | 2 |
(e) | 0 | 1 | 2 | 2 | 4 | 1 |
Solutions:
(a) 3 + 2 / 10 + 5 / 100
= 3 + 0.2 + 0.05
= 3.25
(b) 100 + 2 + 6 / 10 + 3 / 100
= 102 + 0.6 + 0.03
= 102.63
(c) 30 + 2 / 100 + 5 / 1000
= 30 + 0.02 + 0.005
= 30.025
(d) 200 + 10 + 1 + 9 / 10 + 2 / 1000
= 211 + 0.9 + 0.002
= 211.902
(e) 10 + 2 + 2 / 10 + 4 / 100 + 1 / 1000
= 12 + 0.2 + 0.04 + 0.001
= 12.241
3. Write the following decimals in the place value table.
(a) 0.29
(b) 2.08
(c) 19.60
(d) 148.32
(e) 200.812
Solutions:
(a) 0.29
= 0.2 + 0.09
= 2 / 10 + 9 / 100
(b) 2.08
= 2 + 0.08
= 2 + 8 / 100
(c) 19.60
= 19 + 0.60
= 10 + 9 + 6 / 10
(d) 148.32
= 148 + 0.3 + 0.02
= 100 + 40 + 8 + 3 / 10 + 2 / 100
(e) 200.812
= 200 + 0.8 + 0.01 + 0.002
=200 + 8 / 10 + 1 / 100 + 2 / 1000
Hundreds | Tens | Ones | Tenths | Hundredths | Thousandths |
0 | 0 | 0 | 2 | 9 | 0 |
0 | 0 | 2 | 0 | 8 | 0 |
0 | 1 | 9 | 6 | 0 | 0 |
1 | 4 | 8 | 3 | 2 | 0 |
2 | 0 | 0 | 8 | 1 | 2 |
4. Write each of the following as decimals.
(a) 20 + 9 + 4 / 10 + 1 / 100
(b) 137 + 5 / 100
(c) 7 / 10 + 6 / 100 + 4 / 1000
(d) 23 + 2 / 10 + 6 / 1000
(e) 700 + 20 + 5 + 9 / 100
Solutions:
(a) 20 + 9 + 4 / 10 + 1 / 100
= 29 + 0.4 + 0.01
= 29.41
(b) 137 + 5 / 100
= 137 + 0.05
= 137.05
(c) 7 / 10 + 6 / 100 + 4 / 1000
= 0.7 + 0.06 + 0.004
=0.764
(d) 23 + 2 / 10 + 6 / 1000
= 23 + 0.2 + 0.006
= 23.206
(e) 700 + 20 + 5 + 9 / 100
= 725 + 0.09
= 725.09
5. Write each of the following decimals in words.
(a) 0.03
(b) 1.20
(c) 108.56
(d) 10. 07
(e) 0.032
(f) 5.008
Solutions:
The following are the decimals in words
(a) 0.03 = zero point zero three
(b) 1.20 = one point two zero
(c) 108.56 = one hundred eight point five six
(d) 10.07 = ten point zero seven
(e) 0.032 = zero point zero three two
(f) 5.008 = five point zero zero eight
6. Between which two numbers in tenths place on the number line does each of the given number lie?
(a) 0.60
(b) 0.45
(c) 0.19
(d) 0.66
(e) 0.92
(f) 0.57
Solutions:
(a) 0.60 lies between 0 and 0.1 in tenths place
(b) 0.45 lies between 0.4 and 0.5 in tenths place
(c) 0.19 lies between 0.1 and 0.2 in tenths place
(d) 0.66 lies between 0.6 and 0.7 in tenths place
(e) 0.92 lies between 0.9 and 1.0 in tenths place
(f) 0.57 lies between 0.5 and 0.6 in tenths place
7. Write as fractions in lowest terms.
(a) 0.60
(b) 0.05
(c) 0.75
(d) 0.18
(e) 0.25
(f) 0.125
(g) 0.066
Solutions:
(a) 0.60 = 60 / 100
= 6 / 10
= 3 / 5
(b) 0.05 = 5 / 100
= 1 / 20
(c) 0.75 = 75 / 100
= 3 / 4
(d) 0.18 = 18 / 100
= 9 / 50
(e) 0.25 = 25 / 100
= 1 / 4
(f) 0.125 = 125 / 1000
= 1 / 8
(g) 0.066 = 66 / 1000
= 33 / 500
Exercise 8.3 page no: 175
1. Which is greater?
(a) 0.3 or 0.4
(b) 0.07 or 0.02
(c) 3 or 0.8
(d) 0.5 or 0.05
(e) 1.23 or 1.2
(f) 0.099 or 0.19
(g) 1.5 or 1.50
(h) 1.431 or 1.490
(i) 3.3 or 3.300
(j) 5.64 or 5.603
Solutions:
(a) 0.3 or 0.4
Whole parts for both the numbers are same. We know that the tenth part of 0.4 is greater than that of 0.3
∴ 0.4 > 0.3
(b) 0.07 or 0.02
Both the numbers have same parts up to the tenth place but the hundredth part of 0.07 is greater than that of 0.02
∴ 0.07 > 0.02
(c) 3 or 0.8
The whole part of 3 is greater than that of 0.8
∴ 3 > 0.8
(d) 0.5 or 0.05
Whole parts for both the numbers are same. Here the tenth part of 0.5 is greater than that of 0.05
∴ 0.5 > 0.05
(e) 1.23 or 1.20
Here both the numbers have same parts up to the tenth place. The hundredth part of 1.23 is greater than that of 1.20
∴ 1.23 > 1.20
(f) 0.099 or 0.19
Whole parts for both the numbers are same. Here the tenth part of 0.19 is greater than that of 0.099
∴ 0.099 < 0.19
(g) 1.5 or 1.50
We may find that both numbers have same parts up to the tenth place. Here 1.5 have no digit at hundredth place. It represents that this digit is 0, which is equal to the digit at hundredth place of 1.50.
∴ Both these numbers are equal
(h) 1.431 or 1.490
Here, both the numbers have same parts up to the tenth place but the hundredth part of 1.490 is greater than that of 1.431
∴ 1.431 < 1.490
(i) 3.3 or 3.300
Here, both numbers have same parts up to the tenth place. There are no digits at hundredth and thousandth place of 3.3. It represents that these numbers are 0, which is equal to the digits at hundredth and thousandth place of 3.300.
∴ Both these numbers are equal
(j) 5.64 or 5.603
Here both numbers have same parts up to the tenth place but the hundredth part of 5.64 is greater than that of 5.603
∴ 5.64 > 5.603
2. Make five more examples and find the greater number from them.
Solutions:
Five more examples are
(a) 32.55 or 32.5
Whole parts for both the numbers are same. The tenth part are also equal, but the hundredth part of 32.55 is greater than that of 32.5
Hence, 32.55 > 32.5
(b) 1 or 0.99
The whole part of 1 is greater than that of 0.99
∴ 1 > 0.99
(c) 1.09 or 1.093
Here both the numbers have same parts up to the hundredth. But the thousandth part of 1.093 is greater than that of 1.09
∴ 1.093 > 1.09
(d) 2 or 1.99
The whole part of 2 is greater than that of 1.99
∴ 2 > 1.99
(e) 2.08 or 2.085
Here both the numbers have same parts up to the hundredth. But the thousandth part of 2.085 is greater than that of 2.08
∴ 2.085 > 2.08
Exercise 8.4 page no: 177
1. Express as rupees using decimals.
(a) 5 paise
(b) 75 paise
(c) 20 paise
(d) 50 rupees 90 paise
(e) 725 paise
Solutions:
We know that there are 100 paise in 1 rupees
(a) 5 paise = 5 / 100 rupees
= Rupess 0.05
(b) 75 paise = 75 / 100 rupees
= Rupees 0.75
(c) 20 paise = 20 / 100 rupees
= Rupees 0.20
(d) 50 rupees 90 paise = [(50 + 90 / 100)] rupees
= Rupees 50.90
(e) 725 paise = 725 / 100 rupees
= Rupees 7.25
2. Express as metres using decimals.
(a) 15 cm
(b) 6 cm
(c) 2 m 45 cm
(d) 9 m 7 cm
(e) 419 cm
Solutions:
We know that there are 100 cm in 1 metre
(a) 15 cm = 15 / 100 m
= 0.15 m
(b) 6 cm = 6 / 100 m
= 0.06 m
(c) 2 m 45 cm = [(2 + 45 / 100)] m
= 2.45 m
(d) 9 m 7 cm = [(9 + 7 / 100)] m
= 9.07 m
(e) 419 cm = 419 / 100 m
= 4.19 m
3. Express as cm using decimals
(a) 5 mm
(b) 60 mm
(c) 164 mm
(d) 9 cm 8 mm
(e) 93 mm
Solutions:
We know that there are 10 mm in 1 cm
(a) 5 mm = 5 / 10 cm
= 0.5 cm
(b) 60 mm = 60 / 10 cm
= 6.0 cm
(c) 164 mm = 164 / 10 cm
= 16.4 cm
(d) 9 cm 8 mm = [(9 + 8 / 10)] cm
= 9.8 cm
(e) 93 mm = 93 / 10 cm
= 9.3 cm
4. Express as km using decimals.
(a) 8 m
(b) 88 m
(c) 8888 m
(d) 70 km 5 m
Solutions:
We know that there are 1000 metres in 1 km
(a) 8 m = 8 / 1000 km
= 0.008 km
(b) 88 m = 88 / 1000 km
= 0.088 km
(c) 8888 m = 8888 / 1000 km
= 8.888 km
(d) 70 km 5 m = [(70 + 5 / 1000)] km
= 70.005 km
5. Express as kg using decimals.
(a) 2 g
(b) 100 g
(c) 3750 g
(d) 5 kg 8 g
(e) 26 kg 50 g
Solutions:
We know that there are 1000 grams in 1 kg
(a) 2 g = 2 / 1000 kg
= 0.002 kg
(b) 100 g = 100 / 1000 kg
= 0.1 kg
(c) 3750 g = 3750 / 1000 kg
= 3.750 kg
(d) 5 kg 8 g = [(5 + 8 / 1000)] kg
= 5.008 kg
(e) 26 kg 50 g = [(26 + 50 / 1000)] kg
= 26.050 kg
Exercise 8.5 page no: 179
1. Find the sum in each of the following:
(a) 0.007 + 8.5 + 30.08
(b) 15 + 0.632 + 13.8
(c) 27.076 + 0.55 + 0.004
(d) 25.65 + 9.005 + 3.7
(e) 0.75 + 10.425 + 2
(f) 280.69 + 25.2 + 38
Solutions:
(a) Sum of 0.007 + 8.5 + 30.08
0.007
8.500
+ 30.080
__________
38.587
__________
(b) Sum of 15 + 0.632 + 13.8
15.000
0.632
+ 13.800
_________
29.432
__________
(c) Sum of 27.076 + 0.55 + 0.004
27.076
0.550
+ 0.004
_____________
27.630
______________
(d) Sum of 25.65 + 9.005 + 3.7
25.650
9.005
+ 3.700
__________
38.355
___________
(e) Sum of 0.75 + 10.425 + 2
0.750
10.425
+ 2.000
_________
13.175
__________
(f) Sum of 280.69 + 25.2 + 38
280.69
25.20
+ 38.00
__________
343.89
___________
2. Rashid spent ₹ 35.75 for Maths book and ₹ 32.60 for Science book. Find the total amount spent by Rashid.
Solutions:
Cost of Maths book = ₹ 35.75
Cost of Science book = ₹ 32.60
Total amount spent by Rashid is
35.75
+ 32.60
__________
68.35
___________
∴ Total amount of money spent by Rashid is ₹ 68.35
3. Radhika’s mother gave her ₹ 10.50 and her father gave her ₹ 15.80, find the total amount given to Radhika by the parents.
Solutions:
Amount given by Radhika’s mother = ₹ 10.50
Amount given by Radhika’s father = ₹ 15.80
Total amount given by her parents
10.50
+ 15.80
__________
26.30
___________
∴ Total amount of money given by Radhika’s parents is ₹ 26.30
4. 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.
Solutions:
Cloth of shirt = 3 m 20 cm
Cloth of trouser = 2 m 5 cm
Total length of cloth is
3.20
+ 2.05
________
5.25
_________
∴ Total length of cloth bought by Nasreen is 5.25 m
5. 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?
Solutions:
Distance walked by Naresh in the morning = 2 km 35 m
= [(2 + 35 /1000)] km
= 2.035 km
Distance walked by him in the evening = 1 km 7 m
= [(1 + 7 / 1000)] km
= 1.007 km
Total distance walked by Naresh is
2.035
+ 1.007
_______
3.042
________
∴ Total distance walked by Naresh is 3.042 km
6. Sunita travelled 15 km 268 m by bus, 7 km 7 m by car and 500 m on foot in order to reach her school. How far is her school from her residence?
Solutions:
Distance travelled by bus = 15 km 268 m
= [(15 + 268 / 1000)] km
= 15.268 km
Distance travelled by car = 7 km 7 m
= [(7 + 7 / 1000)] km
= 7.007 km
Distance walked by Sunita = 500 m
= 500 / 1000
= 0.500 km
Total distance of school from her residence is
15.268
7.007
+ 0.500
________
22.775
________
∴ Total distance of the school from her residence is 22.775 km
7. Ravi purchased 5 kg 400 g rice, 2 kg 20 g sugar and 10 kg 850 g flour. Find the total weight of his purchases.
Solutions:
Weight of rice = 5 kg 400 g
= [(5 + 400 / 1000)] kg
= 5.400 kg
Weight of sugar = 2 kg 20 g
= [(2 + 20 / 1000)] kg
= 2.020 kg
Weight of flour = 10 kg 850 g
= [(10 + 850 / 1000)] kg
= 10.850 kg
Total weight of his purchases is
5.400
2.020
+ 10.850
___________
18.270
____________
∴ Total weight of his purchases is 18.270 kg
Exercise 8.6 page no: 181
1. Subtract:
(a) ₹ 18.25 from ₹ 20.75
(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
Solutions:
(a) ₹ 20.75 – ₹ 18.75
20.75
– 18.25
__________
2.50
___________
₹ 2.50
(b) 250 m – 202.54 m
250.00
– 202.54
___________
47.46
____________
47.46 m
(c) ₹ 8.40 – ₹ 5.36
8.40
– 5.36
_________
3.04
_________
₹ 3.04
(d) 5.206 km – 2.051 km
5.206
– 2.051
__________
3.155
__________
3.155 km
(e) 2.107 kg – 0.314 kg
2.107
– 0.314
_________
1.793
__________
1.793 kg
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
Solutions:
(a) 9.756
– 6.280
_________
3.476
_________
(b) 21.05
– 15.27
___________
5.78
____________
(c) 18.50
– 6.79
___________
11.71
___________
(d) 11.600
– 9.847
____________
1.753
____________
3. Raju bought a book for ₹ 35.65. He gave ₹ 50 to the shopkeeper. How much money did he get back from the shopkeeper?
Solutions:
Money given to shopkeeper = ₹ 50.00
Price of the book = ₹ 35.65
Money that Raju will get back from the shopkeeper will be the difference of these two
∴ Money left with Raju is
50.00
– 35.65
___________
14.35
___________
Hence, money left with Raju is ₹ 14.35
4. Rani had ₹ 18.50. She bought one ice cream for ₹ 11.75. How much money does she have now?
Solutions:
Money with Rani = ₹ 18.50
Price of an ice cream = ₹ 11.75
Now money left with Rani will be the difference of these two
Hence, money left with her is
18.50
– 11.75
__________
6.75
___________
∴ Money left with Rani is ₹ 6.75
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?
Solutions:
Length of cloth = 20 m 5 cm
= 20.05 m
Length of cloth to make a curtain = 4 m 50 cm
= 4.50 m
Length of cloth left with Tina will be the difference of these two
Thus length of cloth left with her is
20.05
– 4.50
________
15.55
________
∴ The length of the remaining cloth left with Tina is 15.55 m
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?
Solutions:
Total distance travelled by Namita = 20 km 50 m
= 20.050 km
Distance travelled by bus = 10 km 200 m
= 10.200 km
Distance travelled by auto = Total distance travelled – Distance travelled by bus
∴ Distance to be travelled by auto is
20.050
– 10.200
________
9.850
________
∴ Namita travelled 9.850 km by auto
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?
Solutions:
Total weight of vegetables Aakash bought = 10.000 kg
Weight of onions = 3 kg 500 g
= 3.500 kg
Weight of tomatoes = 2 kg 75 g
= 2.075 kg
Weight of potatoes = Total weight of vegetables bought – (weight of onions + weight of tomatoes)
= 10.000 – (3.500 + 2.075)
3.500
+ 2.075
________
5.575
________
10.000
– 5.575
_________
4.425
_________
∴ 4.425 kg is the weight of the potatoes
Frequently Asked Questions on NCERT Solutions for Class 6 Maths Chapter 8
Explain the concept of decimal numbers covered in Chapter 8 of NCERT Solutions for Class 6 Maths.
The numbers used to represent numbers smaller than unit 1 are called decimal numbers. The decimal point or the period plays a significant part in a Decimal Number. This period separates the fractional part and whole number part in a decimal number. Place value of a digit can be defined as the value of a digit as per the place of that digit in a number. The introduction of this chapter has definitions of terms which are important for the exams. Students can now study and be updated about the latest syllabus of the CBSE board using the NCERT Solutions which are available in PDF format.
Will the NCERT Solutions for Class 6 Maths Chapter 8 help students to understand the concepts which are important from the exam perspective?
In order to understand the expansion procedure, students can refer to the examples which are present in the NCERT textbook before solving the exercise wise problems. Each problem in the solutions are solved in a stepwise manner to help students in understanding the concepts in a better way. By using the solutions PDF, students will be well versed with the method of solving these equations and score well in the exam.
In decimals, explain recurring and terminating decimals covered in the NCERT Solutions for Class 6 Maths Chapter 8?
A terminating decimal is a decimal number which has a finite number of digits after the decimal. A recurring decimal is one where its decimal representation becomes periodic or the same sequence of digits keeps repeating indefinitely. Question paper in the annual exam would target the chapters which are simple for the students but tricky to solve. For this purpose, students should go through the NCERT Solutions if they aspire to score good marks.