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# NCERT Solutions Class 6 Maths Chapter 8 Decimals

Last updated date: 13th Sep 2024
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## NCERT Solutions for Maths Class 6 Chapter 8 Decimals - FREE PDF Download

Decimal Class 6 covers understanding the place value system for decimals, converting fractions to decimals and vice versa, and performing arithmetic operations such as addition, subtraction, multiplication, and division with decimals. Students will also learn how to compare and order decimal numbers, which helps in understanding their relative sizes. Additionally, the chapter includes real-life applications of decimals in scenarios like money and measurements, making the concepts more relatable and practical. Students can download Class 6 Maths NCERT Solutions from our page which is prepared so that you can understand it easily.

Table of Content
1. NCERT Solutions for Maths Class 6 Chapter 8 Decimals - FREE PDF Download
2. Glance on Maths Chapter 8 Class 6 - Decimals
3. Access Exercise Wise NCERT Solutions for Chapter 8 Maths Class 6
4. Exercises Under NCERT Solutions for Class 6 Maths Chapter 8 Decimals
5. Access NCERT Solutions for Class 6 Maths Chapter 8 – Decimals
5.1Exercise-8.1
5.2Exercise-8.2
5.3Exercise-8.3
5.4Overview of Deleted Syllabus for CBSE Class 6 Maths Decimals
5.5Class 6 Maths Chapter 8: Exercises Breakdown
6. Other Study Material for CBSE Class 6 Maths Chapter 8
7. Chapter-Specific NCERT Solutions for Class 6 Maths
FAQs

Class 6 Math Chapter 8 Solutions are aligned with the updated  Class 6 Maths Syllabus guidelines, ensuring students are well-prepared for exams.

## Glance on Maths Chapter 8 Class 6 - Decimals

• NCERT Solution for Class 6 Math Chapter 8 covers the topic of Comparing Decimals, Using Decimals, Addition of Numbers with Decimals and Subtraction of Decimals.

• Comparing Decimals - Decimals allow us to represent quantities more precisely than whole numbers. Comparing them involves reading the digits place by place, starting from the decimal point. The decimal with the greater value in the first differing place is considered larger.

• Decimals are applied in various real-world scenarios. Example: Lengths or distances (e.g., 5.2 cm, 12.75 km)

• The addition of Decimals is similar to adding whole numbers, but we need to ensure the decimal points are lined up vertically. Then, we add each place value column just like with whole numbers, carrying over the decimal point if necessary.

• Subtraction of Decimals also follows similar principles as subtracting whole numbers. We line up the decimals vertically and subtract each place value column, again borrowing from the left side (higher place value) if needed to ensure a successful subtraction.

• This article contains chapter notes, important questions, exemplar solutions, exercises and video links for Chapter 8 - Decimals, which you can download as PDFs.

• There are four exercises (21 fully solved questions) in Decimal Class 6.

## Access Exercise Wise NCERT Solutions for Chapter 8 Maths Class 6

 S.No. Current Syllabus Exercises of Class 6 Maths Chapter 8 1 NCERT Solutions of Class 6 Maths Decimals Exercise 8.1 2 NCERT Solutions of Class 6 Maths Decimals Exercise 8.2 3 NCERT Solutions of Class 6 Maths Decimals Exercise 8.3 4 NCERT Solutions of Class 6 Maths Decimals Exercise 8.4

## Exercises Under NCERT Solutions for Class 6 Maths Chapter 8 Decimals

### Exercise 8.1  – Hundredth

Just as the tenth, the hundredth decimal is also a part of the student’s syllabus as they learn about how this decimal helps in calculation. There are different examples present to ease the student’s process of learning this segment.

### Exercise 8.2 – Comparing Decimals

This part of  Chapter 8 Class 6 Maths focuses on comparing decimals that usually start with the tenth place. The student needs to understand which decimal is greater and how through this segment.

### Exercise 8.3 – Using Decimals

Decimals are quite applicable in normal day calculation, and the inclusion of Class 6 Chapter 8 Decimals is to enable students to learn more about them. This basic knowledge will prove to be useful for students in later years.

### Exercise 8.4 – Addition of Numbers with Decimals

Learning how to perform the addition of decimals without breaking them can ease the calculation process. This part itself is quite important and is usually where most examination questions tend to come from.

## Access NCERT Solutions for Class 6 Maths Chapter 8 – Decimals

### Exercise-8.1

1. Which is greater?

1. $$0.3$$ or $$0.4$$

2. $$0.07$$ or $$0.02$$

3. $$3$$ or $$0.8$$

4. $$0.5$$ or $$0.05$$

5. $$1.23$$ or $$1.2$$

6. $$0.099$$ or $$0.19$$

7. $$1.5$$ or $$1.50$$

8. $$1.431$$ or $$1.490$$

9. $$3.3$$ or $$3.300$$

10. $$5.64$$ or $$5.603$$

Ans:

1. $$0.3$$ or $$0.4$$

Whole parts for the two numbers are the same but it can be seen that the tenth part of $$0.4$$ is greater than that of $$0.3$$.

$$\therefore \,\,\,\,0.4 > 0.3$$

1. $$0.07$$ or $$0.02$$

Here, given two numbers have the same parts up to the tenth place but the hundredth part of $$0.07$$ is greater than that of $$0.02$$.

$$\therefore \,\,\,\,0.07 > 0.02$$

1. $$3$$ or $$0.8$$

Here, it can be seen that the whole part of $$3$$ is greater than that of $$0.8$$.

$$\therefore \,\,\,\,3 > 0.8$$

1. $$0.5$$ or $$0.05$$

Here, the whole parts of these given two numbers are same but It can be seen that the tenth part of $$0.5$$ is greater than that of $$0.05$$.

$$\therefore \,\,\,\,0.5 > 0.05$$

1. $$1.23$$ or $$1.2$$

Here, the whole and tenth parts of these given two numbers are same but It can be seen that the hundredth part of $$1.23$$ is greater than that of $$1.2$$.

$$\therefore \,\,\,\,1.23 > 1.2$$

1. $$0.099$$ or $$0.19$$

Here, the whole parts of these given two numbers are same but It can be seen that the tenth part of $$0.19$$ is greater than that of $$0.099$$.

$$\therefore \,\,\,\,0.19 > 0.099$$

1. $$1.5$$ or $$1.50$$

Here, the whole and tenth parts of these given two numbers are the same. Also, there is no digit at the hundredth place of $$1.5$$, this implies that this digit will be $$0$$, which is the same as the digit at the hundredth place of $$1.50$$. Therefore, both the numbers are equal.

$$\therefore \,\,\,\,1.5 = 1.50$$

1. $$1.431$$ or $$1.490$$

Here, the whole and tenth parts of these given two numbers are same but It can be seen that the hundredth part of $$1.490$$ is greater than that of $$1.431$$.

$$\therefore \,\,\,\,1.490 > 1.431$$

1. $$3.3$$ or $$3.300$$

Here, the whole and tenth parts of these given two numbers are the same. Also, there is no digit at the hundredth and thousandth place of $$3.3$$, this implies that this digit will be $$0$$, which is the same as the digit at the hundredth and thousandth place of $$3.300$$. Therefore, both the numbers are equal.

$$\therefore \,\,\,\,3.3 = 3.300$$

1. $$5.64$$ or $$5.603$$

Here, the whole and tenth parts of these given two numbers are same but It can be seen that the hundredth part of $$5.64$$ is greater than that of $$5.603$$.

$$\therefore \,\,\,\,5.64 > 5.603$$

Q2. Make five more examples and find the greater:

1. $$1.8$$ or $$1.82$$

2. $$1.0009$$ or $$1.09$$

3. $$10.01$$ or $$100.1$$

4. $$5.100$$ or $$5.0100$$

5.  $$04.213$$ or $$0421.3$$

Ans:

1. $$1.8$$ or $$1.82$$

$$\therefore \,\,\,\,1.82 > 1.8$$

1. $$1.0009$$ or $$1.09$$

$$\therefore \,\,\,\,1.09 > 1.0009$$

1. $$10.01$$ or $$100.1$$

$$\therefore \,\,\,\,100.1 > 10.01$$

1. $$5.100$$ or $$5.0100$$

$$\therefore \,\,\,\,5.100 > 5.0100$$

1.  $$04.213$$ or $$0421.3$$

$$\therefore \,\,\,\,0421.3 > 04.213$$

### Exercise-8.2

Q1. Express as rupees using decimals:

(a) $$5$$ paise

(b) $$75$$ paise

(c) $$20$$ paise

(d) $$50$$ rupees $$90$$ paise

(e) $$725$$ paise

Ans:

We know that:

$$1$$ rupee $$=$$ $$100$$ paise

$$1$$ paise $$=$$ $$\dfrac{1}{{100}}$$ rupee.

1. $$5$$ paise

$$\Rightarrow \,\,5 \times \dfrac{1}{{100}}$$

$$\Rightarrow \,\,\dfrac{5}{{100}} = 0.05$$

$$\therefore \,\,\,5$$ paise $$=$$$$0.05$$ Rs.

1. $$75$$ paise

$$\Rightarrow \,\,75 \times \dfrac{1}{{100}}$$

$$\Rightarrow \,\,\dfrac{{75}}{{100}} = 0.75$$

$$\therefore \,\,\,75$$ paise $$=$$$$0.75$$ Rs.

1. $$20$$ paise

$$\Rightarrow \,\,20 \times \dfrac{1}{{100}}$$

$$\Rightarrow \,\,\dfrac{{20}}{{100}} = 0.2$$

$$\therefore \,\,\,20$$ paise $$=$$$$0.2$$ Rs.

1. $$50$$ rupees $$90$$ paise

$$\Rightarrow \,\,5 + 90 \times \dfrac{1}{{100}}$$

$$\Rightarrow \,\,5 + \dfrac{{90}}{{100}} = 5 + 0.9 = 5.9$$

$$\therefore \,\,\,5$$ rupees $$90$$ paise $$=$$$$5.9$$ Rs.

1. $$20$$ paise

$$\Rightarrow \,\,20 \times \dfrac{1}{{100}}$$

$$\Rightarrow \,\,\dfrac{{20}}{{100}} = 0.2$$

$$\therefore \,\,\,20$$ paise $$=$$$$0.2$$ Rs.

Q2. Express as meters using decimals:

(a) $$15$$cm

(b) $$6$$cm

(c) $$2$$m $$45$$cm

(d) $$9$$m $$7$$cm

(e) $$419$$cm

Ans:

We know that:

$$1$$m $$=$$ $$100$$cm

$$1$$cm $$=$$ $$\dfrac{1}{{100}}$$ m.

1. $$15$$ cm

$$\Rightarrow 15 \times \dfrac{1}{{100}}$$

$$\Rightarrow \dfrac{{15}}{{100}} = 0.15$$

$$\therefore 15$$cm $$=$$$$0.15$$m.

1. $$6$$ cm

$$\Rightarrow \,\,6 \times \dfrac{1}{{100}}$$

$$\Rightarrow \,\,\dfrac{6}{{100}} = 0.06$$

$$\therefore \,\,\,6$$cm $$=$$$$0.06$$m.

1. $$2$$ m $$45$$ cm

$$\Rightarrow \,\,2 + 45 \times \dfrac{1}{{100}}$$

$$\Rightarrow \,\,2 + \dfrac{{45}}{{100}} = 2 + 0.45 = 2.45$$

$$\therefore \,\,\,2$$ m $$45$$cm $$=$$$$2.45$$m.

1. $$9$$ m $$7$$ cm

$$\Rightarrow \,\,9 + 7 \times \dfrac{1}{{100}}$$

$$\Rightarrow \,\,9 + \dfrac{7}{{100}} = 9 + 0.07 = 9.07$$

$$\therefore \,\,\,9$$ m $$7$$cm $$=$$$$9.07$$m.

1. $$419$$ cm

$$\Rightarrow \,\,419 \times \dfrac{1}{{100}}$$

$$\Rightarrow \,\,\dfrac{{419}}{{100}} = 4.19$$

$$\therefore \,\,\,419$$cm $$=$$$$4.19$$m.

Q3. Express as cm using decimals:

(a) $$5$$mm

(b) $$60$$mm

(c) $$164$$mm

(d) $$9$$cm $$8$$mm

(e) $$93$$mm

Ans:

We know that:

$$1$$cm $$=$$ $$10$$mm

$$1$$mm $$=$$ $$\dfrac{1}{{10}}$$cm.

1. $$5$$mm

$$\Rightarrow \,\,5 \times \dfrac{1}{{10}}$$

$$\Rightarrow \,\,\dfrac{5}{{10}} = 0.5$$

$$\therefore \,\,\,5$$mm $$=$$ $$0.5$$cm.

1. $$60$$ mm

$$\Rightarrow \,\,60 \times \dfrac{1}{{10}}$$

$$\Rightarrow \,\,\dfrac{{60}}{{10}} = 6.0$$

$$\therefore \,\,\,60$$mm $$=$$ $$6.0$$cm.

1. $$164$$mm

$$\Rightarrow \,\,164 \times \dfrac{1}{{10}}$$

$$\Rightarrow \,\,\dfrac{{164}}{{10}} = 16.4$$

$$\therefore \,\,\,164$$mm $$=$$ $$16.4$$cm.

1. $$9$$ cm $$8$$ mm

$$\Rightarrow \,\,9 + 8 \times \dfrac{1}{{10}}$$

$$\Rightarrow \,\,9 + \dfrac{8}{{10}} = 9 + 0.8 = 9.8$$

$$\therefore \,\,\,9$$ cm $$8$$ mm $$=$$ $$9.8$$ cm.

1. $$93$$ mm

$$\Rightarrow \,\,93 \times \dfrac{1}{{10}}$$

$$26$$

$$\therefore \,\,\,93$$mm $$=$$ $$9.3$$cm.

Q4. Express as km using decimals:

(a) $$8$$ m

(b) $$88$$ m

(c) $$8888$$ m

(d) $$70$$ km $$5$$ m

Ans:

We know that:

$$1$$km $$=$$ $$1000$$m

$$1$$m $$=$$ $$\dfrac{1}{{1000}}$$ km.

1. $$8$$

$$\Rightarrow \,\,8 \times \dfrac{1}{{1000}}$$

$$\Rightarrow \,\,\dfrac{8}{{1000}} = 0.008$$

$$\therefore \,\,\,8$$ m $$=$$ $$0.008$$km.

1. $$88$$

$$\Rightarrow \,\,88 \times \dfrac{1}{{1000}}$$

$$\Rightarrow \,\,\dfrac{{88}}{{1000}} = 0.088$$

$$\therefore \,\,\,88$$ m $$=$$ $$0.088$$km.

1. $$8888$$

$$\Rightarrow \,\,8888 \times \dfrac{1}{{1000}}$$

$$\Rightarrow \,\,\dfrac{{8888}}{{1000}} = 8.888$$

$$\therefore \,\,\,8888$$ m $$=$$ $$8.888$$km.

1. $$70$$ km $$5$$

$$\Rightarrow \,70 + 5 \times \dfrac{1}{{1000}}$$

$$\Rightarrow \,\,70 + \dfrac{5}{{1000}} = 70 + 0.005 = 70.005$$

$$\therefore \,\,\,70$$ km $$5$$ m $$=$$ $$70.005$$ 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

Ans:

We know that:

$$1$$kg $$=$$ $$1000$$g

$$1$$g $$=$$ $$\dfrac{1}{{1000}}$$ kg.

1. $$2$$

$$\Rightarrow \,\,2 \times \dfrac{1}{{1000}}$$

$$\Rightarrow \,\,\dfrac{2}{{1000}} = 0.002$$

$$\therefore \,\,\,2$$ g $$=$$ $$0.002$$ kg.

1. $$100$$

$$\Rightarrow \,\,100 \times \dfrac{1}{{1000}}$$

$$\Rightarrow \,\,\dfrac{{100}}{{1000}} = 0.1$$

$$\therefore \,\,\,100$$ g $$=$$ $$0.1$$ kg.

1. $$3750$$

$$\Rightarrow \,\,3750 \times \dfrac{1}{{1000}}$$

$$\Rightarrow \,\,\dfrac{{3750}}{{1000}} = 3.750$$

$$\therefore \,\,\,3750$$ g $$=$$ $$3.750$$ kg.

1. $$5$$ kg $$8$$

$$\Rightarrow \,\,5 + 8 \times \dfrac{1}{{1000}}$$

$$\Rightarrow \,\,5 + \dfrac{8}{{1000}} = 5 + 0.008 = 5.008$$

$$\therefore \,\,\,5$$ kg $$8$$ g $$=$$ $$5.008$$ kg.

1. $$26$$ kg $$50$$

$$\Rightarrow \,\,26 + 50 \times \dfrac{1}{{1000}}$$

$$\Rightarrow \,\,26 + \dfrac{{50}}{{1000}} = 26 + 0.050 = 26.05$$

$$\therefore \,\,\,26$$ kg $$50$$ g $$=$$ $$26.05$$ kg.

### Exercise-8.3

1. Find the sum in each of the following:

(a) $$0.007{\text{ }} + {\text{ }}8.5{\text{ }} + {\text{ }}30.08$$

(b) $$15{\text{ }} + {\text{ }}0.632{\text{ }} + {\text{ }}13.8$$

(c) $$27.076{\text{ }} + {\text{ }}0.55{\text{ }} + {\text{ }}0.004$$

(d) $$25.65{\text{ }} + {\text{ }}9.005{\text{ }} + {\text{ }}3.7$$

(e) $$0.75{\text{ }} + {\text{ }}10.425{\text{ }} + {\text{ }}2$$

(f) $$280.69{\text{ }} + {\text{ }}25.2{\text{ }} + {\text{ }}38$$

Ans:

(a) $$0.007{\text{ }} + {\text{ }}8.5{\text{ }} + {\text{ }}30.08$$

$\ \ \ \ 0.007$

$+ \ 8.5$

$\underline{+ \ 30.08}$

$\ \ \ \ 38.587$

$$\therefore \,\,\,0.007{\text{ }} + {\text{ }}8.5{\text{ }} + {\text{ }}30.08 = 38.587$$

(b) $$15{\text{ }} + {\text{ }}0.632{\text{ }} + {\text{ }}13.8$$

$\ \ \ \ 15.000$

$+ \ \ 0.632$

$\underline{+ \ 13.8}$

$\ \ \ \ 29.132$

$$\therefore \,\,\,15{\text{ }} + {\text{ }}0.632{\text{ }} + {\text{ }}13.8 = 29.132$$

(c) $$27.076{\text{ }} + {\text{ }}0.55{\text{ }} + {\text{ }}0.004$$

$\ \ \ \ 27.076$

$+ \ \ 0.55$

$\underline{+ \ \ 0.0004}$

$\ \ \ \ 27.6304$

$$\therefore \,\,\,\,27.076{\text{ }} + {\text{ }}0.55{\text{ }} + {\text{ }}0.004 = 27.630$$

(d) $$25.65{\text{ }} + {\text{ }}9.005{\text{ }} + {\text{ }}3.7$$

$\ \ \ \ 25.65$

$+ \ \ 9.005$

$\underline{+ \ \ 3.7}$

$\ \ \ \ 38.355$

$\therefore \,\,\,\,25.65{\text{ }} + {\text{ }}9.005{\text{ }} + {\text{ }}3.7 = 38.355$$(e)$$0.75{\text{ }} + {\text{ }}10.425{\text{ }} + {\text{ }}2$$\ \ \ \ \ 0.75 + \ 10.425 \underline{+ \ \ 2.0} \ \ \ \ 13.175$$\therefore \,\,\,\,0.75{\text{ }} + {\text{ }}10.425{\text{ }} + {\text{ }}2 = 13.175$$(f)$$280.69{\text{ }} + {\text{ }}25.2{\text{ }} + {\text{ }}38$$\ \ \ \ 280.69 + \ \ 25.2 \underline{+ \ \ 38.0} \ \ \ \ 343.89$$\therefore \,\,\,\,280.69{\text{ }} + {\text{ }}25.2{\text{ }} + {\text{ }}38 = 343.89$$Q2. Rashid spent$$Rs.{\text{ }}35.75$$for Maths book and$$Rs.{\text{ }}32.60$$for Science book. Find the total amount spent by Rashid. Ans: The money spent on Maths books =$$Rs.{\text{ }}35.75$$The money spent for Science books =$$Rs.{\text{ }}32.60$$Total money spent =$$Rs.{\text{ }}35.75{\text{ }} + {\text{ }}Rs.{\text{ }}32.60{\text{ }} = {\text{ }}Rs.{\text{ }}68.35$$Therefore, the total amount spent by Rashid is$$Rs.{\text{ }}68.35$$. Q3. Radhika’s mother gave her$$Rs.{\text{ }}10.50$$and her father gave her$$Rs.{\text{ }}15.80$$. Find the total amount given to Radhika by the parents. Ans: Money given by Radhika’s mother =$$Rs.{\text{ }}10.50$$Money given by Radhika’s father =$$Rs.{\text{ }}15.80$$Total money received by Radha =$$Rs.{\text{ }}10.50{\text{ }} + {\text{ }}Rs.{\text{ }}15.80{\text{ }} = {\text{ }}Rs.{\text{ }}26.30$$Therefore, the total amount given to Radhika by her parents is$$Rs.{\text{ }}26.30$$. Q4. Nasreen bought$$3m{\text{ }}20cm$$cloth for her shirt and$$2m{\text{ }}5cm$$cloth for her trouser. Find the total length of cloth bought by her. Ans: The cloth bought for shirt =$$3m{\text{ }}20cm{\text{ }} = 3.20m$$The cloth bought for trouser =$$2m{\text{ }}5cm{\text{ }} = {\text{ }}2.05m$$Total length of cloth bought by Nasreen =$$3.20+2.05 = 5.25m$$Therefore, the total length of cloth bought by Nasreen is$$5.25{\text{ }}m\,\, = \,5m\,25cm$$. Q5. Naresh walked$$2km\,35m$$in the morning and$$1km\,7m$$in the evening. How much distance did he walk in all? Ans: The distance travelled in morning =$$2{\text{ }}km{\text{ }}35{\text{ }}m{\text{ }} = {\text{ }}2.035{\text{ }}km$$The distance travelled in evening =$$1{\text{ }}km{\text{ }}7{\text{ }}m{\text{ }} = {\text{ }}1.007{\text{ }}km$$Total distance travelled =$$2.035+1.007 = 3.042km$$Therefore, the total distance covered by Naresh is$$3.042km\,\, = \,3km\,42m$$. Q6. Sunita travelled$$15{\text{ }}km{\text{ }}268{\text{ }}m$$by bus,$$7{\text{ }}km{\text{ }}7{\text{ }}m$$by car and$$500{\text{ }}m$$by foot in order to reach her school. How far is her school from her residence? Ans: The distance travelled by bus =$$15{\text{ }}km{\text{ }}268{\text{ }}m{\text{ }} = {\text{ }}15.268{\text{ }}km$$The distance travelled by car =$$7km{\text{ }}7m = 7.007km$$The distance travelled on foot =$$500m= 0.500km$$Total distance travelled =$$15.268+7.007+0.500 = 22.775km$$Therefore, the distance between Sunita’s residence and her school is$$22.775{\text{ }}km\,\, = \,\,22km\,\,775m$$. Q7. Ravi purchases$$5{\text{ }}kg{\text{ }}400{\text{ }}g$$rice,$$2{\text{ }}kg{\text{ }}20{\text{ }}g$$sugar and$$10{\text{ }}kg{\text{ }}850{\text{ }}g$$flour. Find the total weight of his purchases. Ans: The weight of Rice$$ = {\text{ }}5{\text{ }}kg{\text{ }}400{\text{ }}g{\text{ }} = {\text{ }}5.400{\text{ }}kg$$The weight of Sugar =$$2{\text{ }}kg{\text{ }}20{\text{ }}g{\text{ }} = {\text{ }}2.020{\text{ }}kg$$The weight of Flour =$$10{\text{ }}kg{\text{ }}850{\text{ }}g{\text{ }} = {\text{ }}10.850{\text{ }}kg$$Total weight =$$5.400{\text{ }} + {\text{ }}2.020{\text{ }} + {\text{ }}10.850{\text{ }} = {\text{ }}18.270{\text{ }}kg$$Therefore, the total weight of Ravi purchases is$$18.270{\text{ }}kg\,\, = \,\,18kg\,\,270g$$. Exercise-8.4 Q1. Subtract: (a)$$Rs.{\text{ }}18.25$$from$$Rs.{\text{ }}20.75$$(b)$$202.54{\text{ }}m$$from$$250{\text{ }}m$$(c)$$Rs.{\text{ }}5.36$$from$$Rs.{\text{ }}8.40$$(d)$$2.051{\text{ }}km$$from$$5.206{\text{ }}km$$(e)$$0.314{\text{ }}kg$$from$$2.107{\text{ }}kg$$Ans: Given that both quantities have the same quantity, we can subtract. 1.$$Rs.{\text{ }}18.25$$from$$Rs.{\text{ }}20.75$$20,.75 \underline { - \,18\,\,.\,\,25} 2.50 Therefore,$$Rs.\,\,20.75\,\, - \,\,Rs.{\text{ }}18.25 = Rs.\,2.50$$1.$$202.54{\text{ }}m$$from$$250{\text{ }}m$$250.00 \underline{ -202.54} 47.46 Therefore,$$250m - 202.54m = 47.46m$$1.$$Rs.{\text{ }}5.36$$from$$Rs.{\text{ }}8.40$$8\,\,.\,\,40 \underline {-5.36} 3.04 Therefore,$$Rs.\,\,8.40 - Rs.{\text{ }}5.36 = Rs.\,\,3.04$$1.$$2.051{\text{ }}km$$from$$5.206{\text{ }}km$$5.206 \underline { -2.051} 3.155 Therefore,$$5.206\,km\, - 2.051{\text{ }}km = 3.155\,km$$1.$$0.314{\text{ }}kg$$from$$2.107{\text{ }}kg$$2.107 \underline {-0.314} 1.793 Therefore,$$2.107\,kg\, - 0.314{\text{ }}kg = 1.793\,kg$$2. Find the value of: (a)$$9.756{\text{ }}-{\text{ }}6.28$$(b)$$21.05{\text{ }}-{\text{ }}15.27$$(c)$$18.5{\text{ }}-{\text{ }}6.79$$(d)$$11.6{\text{ }}-{\text{ }}9.847$$Ans: 1.$$9.756{\text{ }}-{\text{ }}6.28$$9.756 \underline {-6.280} 3.476$$\therefore 9.756{\text{ }}-{\text{ }}6.28 = 3.476$$1.$$21.05{\text{ }}-{\text{ }}15.27$$21.05 \underline {-15.27} 5.78$$\therefore \,\,\,21.05{\text{ }}-{\text{ }}15.27 = 5.78$$1.$$18.5{\text{ }}-{\text{ }}6.79$$18.50 \underline {-6.79} 11.71$$\therefore \,\,\,18.5{\text{ }}-{\text{ }}6.79 = 11.71$$1.$$18.5{\text{ }}-{\text{ }}6.79$$11.600 \underline {-9.847} 1.753$$\therefore \,\,\,11.6{\text{ }}-{\text{ }}9.847 = 1.753$$Q3. Raju bought a book of$$Rs.{\text{ }}35.65.$$He gave$$Rs.{\text{ }}50$$to the shopkeeper. How much money did he get back from the shopkeeper? Ans: The total amount given to shopkeeper =$$Rs.{\text{ }}50$$The cost of book =$$Rs.{\text{ }}35.65$$Amount left =$$Rs.{\text{ }}50.00{\text{ - }}Rs.{\text{ }}35.65{\text{ }} = {\text{ }}Rs.{\text{ }}14.35$$Therefore,$$Rs.{\text{ }}14.35$$will get back from the shopkeeper to Raju. Q4. Rani had$$Rs.{\text{ }}18.50.$$She bought one ice-cream for$$Rs.{\text{ }}11.75.$$How much money does she have now? Ans: The total money =$$Rs.{\text{ }}18.50$$The cost of Ice-cream =$$Rs.{\text{ }}11.75$$Amount left =$$Rs.{\text{ }}18.50{\text{ }}-{\text{ }}Rs.{\text{ }}11.75{\text{ }} = {\text{ }}Rs.{\text{ }}6.75$$Therefore, Rani has$$Rs.{\text{ }}6.75$$. Q5. Tina had$$20{\text{ }}m{\text{ }}5{\text{ }}cm$$long cloth. She cuts$$4m{\text{ }}50cm$$length of cloth from this for making a curtain. How much cloth is left with her? Ans: The total length of cloth$$ = {\text{ }}20{\text{ }}m{\text{ }}5{\text{ }}cm$$=$$20.05{\text{ }}m$$The length of cloth used =$$4{\text{ }}m{\text{ }}50{\text{ }}cm{\text{ }} = {\text{ }}4.50{\text{ }}m$$Remaining cloth =$$20.05{\text{ }}m{\text{ }}-{\text{ }}4.50{\text{ }}m{\text{ }} = {\text{ }}15.55{\text{ }}m$$Therefore$$15.55{\text{ }}m$$cloth is left with her. Q6. Namita travels$$20{\text{ }}km{\text{ }}50m$$every day. Out of this she travels$$10km{\text{ }}200m$$by bus and the rest by auto. How much distance does she travel by auto? Ans: The total distance travel =$$20km{\text{ }}50m = 20.050km$$Distance travelled by bus =$$10km{\text{ }}200m = 10.200km$$Distance travelled by auto =$$20.050-10.200 = 9.850km$$Therefore,$$11.6-9.847$$distance covered by Namita through auto. Q7. Aakash bought vegetables weighing$$10kg.$$Out of this$$3kg{\text{ }}500g$$in onions,$$2kg{\text{ }}75g$$is tomatoes and the rest is potatoes. What is the weight of the potatoes? Ans: The weight of onions =$$3kg{\text{ }}500g = 3.500kg$$The weight of tomatoes =$$2kg{\text{ }}75g = 2.075kg$$Total weight of onions and tomatoes =$$3.500+2.075 = 5.575kg$$The weight of potatoes =$$10kg - 5.575kg = 4.425kg$$Therefore, the weight of potatoes is$$4kg\,425g$\$.

### Overview of Deleted Syllabus for CBSE Class 6 Maths Decimals

 Chapter Dropped Topics Decimals 8.2 - Tenth

### Class 6 Maths Chapter 8: Exercises Breakdown

 Exercise Number of Questions Exercise 8.1 2 Questions & Solutions Exercise 8.2 5 Questions & Solutions Exercise 8.3 7 Questions & Solutions Exercise 8.4 7 Questions & Solutions

## Conclusion

In conclusion, Class 6 Maths Chapter 8 Decimals Solutions provides a comprehensive understanding of decimals, an important mathematical concept with wide-ranging applications. Through Class 6 Chapter 8 Decimals PDF, students have learned to identify and use the place value system for decimals, convert between fractions and decimals, and perform basic arithmetic operations involving decimals. They have also developed skills to compare and order decimal numbers, enhancing their numerical literacy.

## Other Study Material for CBSE Class 6 Maths Chapter 8

 S.No. Important Links for Chapter 8 Decimals 1 Class 6 Decimals Important Questions 2 Class 6 Decimals Revision Notes 3 Class 6 Decimals RD Sharma Solutions 4 Class 6 Decimals RS Aggarwal Solutions

## Chapter-Specific NCERT Solutions for Class 6 Maths

Given below are the chapter-wise NCERT Solutions for Class 6 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions Class 6 Maths Chapter 8 Decimals

1. What are NCERT Solutions of Class 6 Maths Chapter 8 about?

The basis of decimals is to simplify the student’s process of calculation by breaking down numbers to tenth and hundredth. Decimals comprise of various illustrations and exercises that proceed from simple to advanced to help students achieve a better score. This chapter offers students with primary knowledge enabling them to create a foundation for their mathematical abilities.

Reason for including these exercises is to help a student achieve good grades in this subject matter. It consists of segments including, tenth, hundredth, comparing decimals, and using decimals for the calculation of money, length, weight, addition and subtraction of decimals. Proper understanding of the subject will allow the student to score well in examinations.

2. How to Complete NCERT Solution Class 6 Chapter 8 Faster?

If a student wishes to complete the NCERT class 6 chapter 8 faster, then referring to the NCERT solutions for class 6 Maths chapter decimals in Vedantu is the smarter option. The solutions are well-formatted and written in simple language, making it easy for the student to understand. Students are also provided with a proper question and answers format making it easy for them to grasp the topic.

The solution also follows an examination format; this makes it easy for the student to follow it and be prepared for their examinations. The entire solution covers every segment of the chapter. It includes sections that require emphasis and may appear in examinations such as the use of decimals for the calculation of money, length, and weight.

For students who wish to score well in their examinations and are underprepared or need to finish the chapter faster can refer to this solution on Vedantu.

3. How many exercises are present in NCERT Solutions for Class 6 Maths Chapter 8 Decimals?

In Class 6 Maths Chapter 8 Decimals, there are a total of six exercises. Students can download NCERT Solutions from Vedantu. NCERT Solutions for all exercises are given and explained simply and easily. Students can understand the concepts about decimals and score high marks in their exams. The solutions provided by Vedantu(vedantu.com) are free of cost. They are also available on the Vedantu Mobile app.

4. How can I score full marks in Class 6 Maths Chapter 8 Decimals?

Students can score full marks in Class 6 Maths chapter 8 Decimals by practicing all NCERT solutions. Students should revise all NCERT textbook questions of Class 6 Maths Chapter 8 Decimals. They can find NCERT Solutions for Class 6 Maths Chapter 8 online. They can click here to download the PDF file for NCERT Solutions of Class 6 Maths Chapter 8 Decimals. All the Solutions explain the concepts of decimals in a simple manner for easy understanding of the students.

5. What are Decimals according to Chapter 8 of Class 6 Maths?

Decimals are the numbers in algebra that have a whole number part and a fractional part. For example- 89.678, 7654.7, 879.0 are some of the decimal numbers. The dot present between the numbers is called the decimal point. There are three types of decimal numbers-

· Recurring decimal numbers

• Non-recurring decimal numbers

• Decimal fraction

All the arithmetic operations like addition, subtraction, multiplication, and division can be performed on decimals.

6. How do you introduce decimals to 6th Graders?

In 6th Grade, students will learn about decimals. They will learn about:

• Like and unlike decimals

• How to solve different operations of decimals

• Addition, subtraction, multiplication, and division of decimals. Students can download NCERT Solutions Class 6 Maths Chapter 8 for understanding the different concepts of decimals. NCERT Solutions will help students to learn about the basic ideas and other concepts about decimals.

7. How do you add decimals as taught in Class 6 Maths?

The addition of decimals is very easy. Students can score high marks by understanding the basic concepts of decimals in Class 6. It is similar to the addition of the whole numbers. Students have to write the given decimals in vertical form. Make sure that the decimal point comes under the decimal point. Then add the numbers in the same manner as whole numbers are added. For example, 23.4 + 26.5= 49.9.

8. What are like and unlike decimals?

Like decimals are the decimals that have the same number of digits after the decimal point. Unlike decimals are the decimals that have the different number of digits after the decimal point. Examples of like decimals include 45.23, 23.34, 46.45. Examples of unlike decimals include 23.456, 23.22, 45.1. Students need to understand the concept of like and unlike decimals in Class 6 to score high marks.

9. How do you convert a fraction to a decimal in Class 6 Maths Chapter 8 Decimals Solutions?

To convert a fraction to a decimal, divide the numerator by the denominator. The result is the decimal form of the fraction.

10. What is the place value system for decimals in class 6 chapter 8 maths?

The place value system for decimals includes positions such as tenths, hundredths, thousandths, etc., to the right of the decimal point, each position representing a power of ten.