# RBSE Solutions for Class 6 MATHS Chapter 8: Decimals

Access RBSE Solutions for Class 6 Chapter 8: Decimals Navigating the intricate realm of decimals is a crucial aspect of Class 6 Mathematics in the Rajasthan Board of Secondary Education (RBSE) curriculum. Chapter 8, aptly titled “Decimals,” serves as a gateway for students to grasp the nuances of this numerical concept. To facilitate a thorough understanding, RBSE solutions have been meticulously crafted by subject-matter experts at BYJU’S. Aimed at aligning with the RBSE syllabus for Class 6 Maths Chapter 8, these solutions provide diverse methods for problem-solving. In this article, we delve into the key features of these RBSE solutions and their importance in ensuring a strong grasp of decimal concepts.

1. Unpacking the Significance of Decimals:

Before delving into the RBSE solutions, it is imperative to acknowledge the importance of decimals in mathematical understanding. Decimals, a fundamental aspect of numerical representation, find applications in various real-life scenarios, laying the groundwork for more advanced mathematical concepts.

2. RBSE Class 6 Maths Chapter 8 Overview:

Chapter 8 focuses on introducing students to the world of decimals. Key concepts covered in this chapter include:

• Understanding decimal notation.
• Comparing and ordering decimals.
• Addition and subtraction of decimals.
• Practical applications of decimals.

3. RBSE Solutions: A Comprehensive Approach:

Designed with the goal of aligning with the RBSE syllabus, these solutions offer comprehensive assistance to students. Key features of the RBSE Class 6 Maths Chapter 8 solutions include:

• In-depth explanations for each concept.
• Multiple problem-solving methods to cater to diverse learning styles.
• Step-by-step solutions for better comprehension.

• Thorough Practice: Students can engage in extensive practice of decimal-related problems, enhancing their proficiency in the subject.
• Error Identification: Comparing their answers with the RBSE solutions allows students to identify errors, understand misconceptions, and take corrective steps.
• Conceptual Clarity: The step-by-step solutions aid in developing a deep understanding of decimal concepts, ensuring a solid foundation for future mathematical learning.

## RBSE Solutions for Class 6 MATHS Chapter 8 Exercise 8.1 Page No. 167

1. Write the following numbers in the given table.

Solutions:

2. Write the following decimals in the place value table.

(a) 19.4

(b) 0.3

(c) 10.6

(d) 205.9

Solutions:

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 the 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 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. Thus, 0.2 is the second point between 0 and 1.

(b) 1.9 lies between 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. Thus, 1.9 is the ninth point between 1 and 2.

(c) 1.1 lies between 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. Thus, 1.1 is the first point between 1 and 2.

(d) 2.5 lies between 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. Thus, 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’s notebook is 9 cm 5 mm

The length in cm is [(9 + 5 / 10)] cm

= 9.5 cm

(b) The length of the gram plant is 65 mm

Hence, the length in cm is 65 / 10

= 6.5 cm

## RBSE Solutions for Class 6 MATHS Chapter 8 Exercise 8.2 Page No. 173

1. Complete the table with the help of these boxes and use decimals to write the number.

Solutions:

2. Write the numbers given in the following place value table in decimal form.

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

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 the 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

## RBSE Solutions for Class 6 MATHS Chapter 8 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 1.20

(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

The whole parts for both numbers are the 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 the 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

The whole parts for both numbers are the 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 the 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

The whole parts for both numbers are the 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 the same parts up to the tenth place. Here 1.5 have no digit at the hundredth place. It represents that this digit is 0, which is equal to the digit at the hundredth place of 1.50.

∴ Both these numbers are equal.

(h) 1.431 or 1.490

Here, both the numbers have the 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 the same parts up to the tenth place. There are no digits at the hundredth and thousandth place of 3.3. It represents that these numbers are 0, which is equal to the digits at the hundredth and thousandth place of 3.300.

∴ Both these numbers are equal.

(j) 5.64 or 5.603

Here, both numbers have the 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 given below:

(a) 32.55 or 32.5

The whole parts for both numbers are the same. The tenth part is 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 numbers have the 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 numbers have the same parts up to the hundredth. But the thousandth part of 2.085 is greater than that of 2.08,

∴ 2.085 > 2.08,

## RBSE Solutions for Class 6 MATHS Chapter 8 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 rupee.

(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

## RBSE Solutions for Class 6 MATHS Chapter 8 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 the Maths book and ₹ 32.60 for the Science book. Find the total amount spent by Rashid.

Solutions:

Cost of Maths book = ₹ 35.75

Cost of Science book = ₹ 32.60

The total amount spent by Rashid is

35.75

+ 32.60

__________

68.35

___________

∴ The 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:

The amount given by Radhika’s mother = ₹ 10.50

The amount given by Radhika’s father = ₹ 15.80

The total amount given by her parents is

10.50

+ 15.80

__________

26.30

___________

∴ The 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

The total length of the cloth is

3.20

+ 2.05

________

5.25

_________

∴ The 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

The total distance walked by Naresh is

2.035

+ 1.007

_______

3.042

________

∴ The 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

The total distance of the school from her residence is

15.268

7.007

+ 0.500

________

22.775

________

∴ The 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

The total weight of his purchases is

5.400

2.020

+ 10.850

___________

18.270

____________

∴ The total weight of his purchases is 18.270 kg

## RBSE Solutions for Class 6 MATHS Chapter 8 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 the following:

(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

The money Raju will get back from the shopkeeper will be the difference between these two.

∴ The money left with Raju is

50.00

– 35.65

___________

14.35

___________

Hence, the 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, the money left with Rani will be the difference between these two.

Hence, the money left with her is

18.50

– 11.75

__________

6.75

___________

∴ The 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

The length of cloth left with Tina will be the difference between these two.

Thus, the 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.

In conclusion, as Class 6 students in the RBSE curriculum explore the intricacies of decimals in Chapter 8, the RBSE solutions provided by BYJU’S become indispensable. These solutions offer a comprehensive approach, enabling students to not only solve problems but also gain a profound understanding of decimal concepts. Utilizing the downloadable PDFs and engaging in thorough practice will empower students to navigate the world of decimals with confidence, laying a robust foundation for their mathematical journey.