# NCERT Solutions Class 6 Maths Chapter 8

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

For a better understanding of the chapter, Class 6 students are first introduced to the chapter on fractions and then to the concept of decimals. Class 6 Maths Chapter 8 Decimals starts with a brief explanation of what decimals are and how they are calculated.. It then goes on to explain the significance of tenth and hundredth decimals in computations. In this chapter, students will also learn how to use decimals in a variety of mathematical calculations.

To do well in this chapter, students must try and solve the sums given at the end  of the chapter. For this, students need a reliable guide. On Extramarks, NCERT Solutions are offered to make the study simple and enjoyable. NCERT Solutions For Class 6 Maths Chapter 8 Decimals enables students in Class 6 to practice the questions in a comprehensive manner. These NCERT Solutions were written by subject matter experts as per the CBSE guidelines for Class 6. Students can access these NCERT solutions for future reference for competitive exams as well. They can also be used by students to cross-check and clarify any doubts about difficult questions in the exercises.

## NCERT Solutions for Class 6 Maths Chapter 8

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

If you’re having trouble answering the questions in this chapter, Extramarks is here to help. NCERT Solutions Class 6 Maths Chapter 8 prepares students by providing them with specific, to-the-point questions and answers. The solutions will help students perform well in the exams.

### NCERT Solutions for Class 6 Maths Chapter 8

Students will learn about decimals in NCERT Class 6 Maths Chapter 8. The concept of representing decimals on a number line is covered in this chapter. Students will have a thorough understanding of how to express decimal numbers on a number line.

This chapter covers tenths, hundredths, and decimal comparisons. Following that, the concepts of addition and subtraction with decimals are covered. By the end of this chapter, you should be able to explain how money, length, and weight are represented.

### NCERT Solutions Class 6 Maths Chapter 8 Exercise

#### Exercise 8.1 – Introduction

With the help of this chapter’s section, students will learn what decimals are and how they’re used in calculations. It also teaches the students how to convert a fraction to a decimal.

#### Exercise 8.2 – Tenth

In this subsection, students will learn how to use the tenth decimal. They learn how decimals are formed and how to use them in computations.

#### Exercise 8.3 – Hundredth

The hundredth decimal, like the tenth, is part of the student’s syllabus as they study how this decimal aids in calculating. Various examples are provided to aid the student’s understanding of this segment.

#### Exercise 8.4 – Comparing Decimals

This section of Chapter 8 Class 6 Maths is about comparing decimals that start in the tenth place. Through this segment, the student must comprehend which decimal is greater and how.

#### Exercise 8.5 – Using decimals

Decimals are widely used in everyday calculations, and the inclusion of Class 6 Chapter 8 Decimals allows students to have a better understanding of them. This foundational knowledge will be useful to students in the higher classes..

#### Exercise 8.5.1 – Money

This section covers how decimals are used in the context of money and how they aid in the calculation process.  It also provides students with examples from experiential learning from everyday life.

#### Exercise 8.5.2 – Length

Students will learn about their decimal requirements in numerous elements of calculations in this section of decimals.

#### Exercise 8.5.3 – Weight

This section of the NCERT Maths book Class 6 Chapter 8 Solutions deals with using decimals to calculate weight distribution. Students will comprehend how the calculation process works.

#### Exercise 8.6 – Addition of Numbers with Decimals

Adding decimals is all about calculating the sum of two or more decimal numbers. Although adding decimals is similar to adding whole numbers, there are some rules that must be followed when doing so.

#### Exercise 8.7 – Subtraction of Decimals

This section explains how to subtract with decimals. The second number can be either a whole number, a decimal number, or a natural number. When subtracting decimals, certain rules and procedures must be followed. Students will be able to understand this with the help of appropriate illustrations.

NCERT Solutions for Class 6 Maths Chapter 1 Exercises are given in the table below:

 Chapter 8 – Decimals Exercises Exercise 8.1 10 Questions & Solutions Exercise 8.2 7 Questions & Solutions Exercise 8.3 2 Questions & Solutions Exercise 8.4 5 Questions & Solutions Exercise 8.5 7 Questions & Solutions Exercise 8.6 7 Questions & Solutions

### Benefits of NCERT Solutions Class 6 Maths Chapter 8

There are several benefits to using Extramarks Chapter 8 Decimals answer, some of which are stated below:

• The answers in these solutions are accurate and to the point .
• The answers are written in simple  and easy language.
• They adhere to the guidelines provided by CBSE.
• Illustrations and examples are provided to aid the student’s understanding effectively.
• NCERT Solutions Class 6 Maths Chapter 8 are presented in a well-structured form which  makes it convenient for students to remember everything clearly.

Q.1 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

Ans-

(a) Seven-tenths : 0.7
(b) Two tens and nine-tenths : 20.9
(c Fourteen point six : 14.6
(d) One hundred and two ones : 102.0
(e) Six hundred point eight : 600.8

Q.2 Write each of the following as decimals:

$\begin{array}{l}\left(\text{a}\right)\frac{5}{10}\left(\text{b}\right)3+\frac{7}{10}\left(\text{c}\right)200+60+5+\frac{1}{10}\\ \left(\text{d}\right)70+\frac{8}{10}\left(\text{e}\right)\frac{88}{10}\left(\text{f}\right)4\frac{2}{10}\\ \left(\text{g}\right)\frac{3}{2}\left(\text{h}\right)\frac{2}{5}\left(\text{i}\right)\frac{12}{5}\\ \left(\text{j}\right)3\frac{3}{5}\left(\text{k}\right)4\frac{1}{2}\end{array}$

Ans

$\begin{array}{l}\left(\text{a}\right)\frac{5}{10}=\text{}0.\text{5}\\ \left(\text{b}\right)3+\frac{7}{10}=\text{3}+\text{}0.\text{7}=\text{3}.\text{7}\\ \left(\text{c}\right)200+60+5+\frac{1}{10}=\text{265}+\text{}0.\text{1}=\text{265}.\text{1}\\ \left(\text{d}\right)70+\frac{8}{10}=\text{7}0\text{}+\text{}0.\text{8}=\text{7}0.\text{8}\\ \left(\text{e}\right)\frac{88}{10}=\text{8}.\text{8}\\ \left(\text{f}\right)4\frac{2}{10}=4+\frac{2}{10}=4+0.2=4.2\\ \left(\text{g}\right)\frac{3}{2}=\frac{2+1}{2}=\frac{2}{2}+\frac{1}{2}=1+0.5=1.5\\ \left(\text{h}\right)\frac{2}{5}=\text{}0.\text{4}\\ \left(\text{i}\right)\frac{12}{5}=\frac{10+2}{5}=\frac{10}{5}+\frac{2}{5}=2+\frac{2}{5}=2+0.4=2.4\\ \left(\text{j}\right)3\frac{3}{5}=3+\frac{3}{5}=3+0.6=3.6\\ \left(\text{k}\right)4\frac{1}{2}=4+\frac{1}{2}=4+0.5=4.5\end{array}$

Q.3 Write the following decimals as fractions. Reduce the fractions to lowest form.
(a) 0.6 (b) 2.5 (c) 1.0 (d) 3.8 (e) 13.7 (f) 21.2 (g) 6.4

Ans-

$\begin{array}{l}\text{(a) 0}\text{.6}=\frac{6}{10}=\frac{3}{5}\\ \text{(b) 2}\text{.5}=\frac{25}{10}=\frac{5}{2}\\ \text{(c) 1}\text{.0}=\frac{10}{10}=1\\ \text{(d) 3}\text{.8}=\frac{38}{10}=\frac{19}{5}\\ \text{(e) 13}\text{.7}=\frac{137}{10}=\frac{137}{10}\\ \text{(f) 21}\text{.2}=\frac{212}{10}=\frac{106}{5}\\ \text{(g) 6}\text{.4}=\frac{64}{10}=\frac{32}{5}\end{array}$

Q.4 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

Ans-

$\begin{array}{l}\left(\text{a}\right)\text{2mm}=\frac{2}{10}\mathrm{cm}=0.2\text{\hspace{0.17em}}\mathrm{cm}\\ \left(\text{b}\right)\text{3}0\text{mm}=\frac{30}{10}\mathrm{cm}=3.0\text{\hspace{0.17em}}\mathrm{cm}\\ \left(\text{c}\right)\text{116mm}=\frac{116}{10}\mathrm{cm}=11.6\text{\hspace{0.17em}}\mathrm{cm}\\ \left(\text{d}\right)\text{4 cm 2mm}=4+\frac{2}{10}\mathrm{cm}=4+\frac{1}{5}=4+0.2=4.2\text{\hspace{0.17em}}\mathrm{cm}\\ \left(\text{e}\right)\text{162mm}=\frac{162}{10}\mathrm{cm}=16.2\text{\hspace{0.17em}}\mathrm{cm}\\ \left(\text{f}\right)\text{83mm}\mathrm{}=\frac{83}{10}\mathrm{cm}=8.3\text{\hspace{0.17em}}\mathrm{cm}\end{array}$

Q.5 Between which two whole numbers on the number line are the given numbers lie?
Which of these whole numbers is nearer the number? Ans-

(a) 0.8 lies between 0 and 1, and is nearer to 1.

(b) 5.1 lies between 5 and 6, and is nearer to 5.

(c) 2.6 lies between 2 and 3, and is nearer to 3.

(d) 6.4 lies between 6 and 7, and is nearer to 6.

(e) 9.1 lies between 9 and 10, and is nearer to 9.

(f) 4.9 lies between 4 and 5, and is nearer to 5.

Q.6 Show the following numbers on the number line:
(a) 0.2
(b) 1.9
(c) 1.1
(d) 2.5

Ans-

(a) 0.2 (b) 1.9 (c) 1.1 (d) 2.5 Q.7 Write the decimal number represented by the points A, B, C, D on the given number line. Ans-

Point A represents 0.8.

Point B represents 1.3.

Point C represents 2.2.

Point D represents 2.9.

Q.8 (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.

Ans-

$\begin{array}{l}\left(\text{a}\right)\text{The length of Ramesh}’\text{s notebook is 9 cm 5 mm}.\text{Its length in cm is}9+\frac{5}{10}=9+0.5=9.5\text{\hspace{0.17em}}\mathrm{cm}\end{array}$

$\begin{array}{l}\left(\text{b}\right)\text{The length of young gram plant is 65 mmIts length in cm is}\frac{65}{10}=6.5\text{\hspace{0.17em}}\mathrm{cm}\end{array}$

Q.9 Write the numbers given in the following place value table in decimal form. Ans-

$\begin{array}{l}\text{(a) 3}+\frac{2}{10}+\frac{5}{100}=3+0.2+0.05=3.25\\ \text{(b) 100}+2+\frac{6}{10}+\frac{3}{100}=102.63\\ \text{(c) 30+}\frac{2}{100}+\frac{5}{1000}=30.025\\ \text{(d) 200}+10+1+\frac{9}{10}+\frac{2}{1000}=211.902\\ \text{(e) 10}+2+\frac{2}{10}+\frac{4}{100}+\frac{1}{1000}=12.241\end{array}$

Q.10 Write the numbers given in the following place value table in decimal form. Ans-

$\begin{array}{l}\text{(a) 3}+\frac{2}{10}+\frac{5}{100}=3+0.2+0.05=3.25\\ \text{(b) 100}+2+\frac{6}{10}+\frac{3}{100}=102.63\\ \text{(c) 30+}\frac{2}{100}+\frac{5}{1000}=30.025\\ \text{(d) 200}+10+1+\frac{9}{10}+\frac{2}{1000}=211.902\\ \text{(e) 10}+2+\frac{2}{10}+\frac{4}{100}+\frac{1}{1000}=12.241\end{array}$

Q.11 Write each of the following as decimals:

$\begin{array}{l}\left(\mathrm{a}\right)\text{ }20+9+\frac{4}{10}+\frac{1}{100}\text{\hspace{0.17em} }\left(\mathrm{b}\right)\text{ }137+\frac{5}{100}\text{\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{c}\right)\text{\hspace{0.17em} }\frac{7}{10}+\frac{6}{100}+\frac{4}{1000}\\ \left(\mathrm{d}\right)\text{ }23+\frac{2}{10}+\frac{6}{1000}\text{ }\left(\mathrm{e}\right)\text{ }700+20+5+\frac{9}{100}\end{array}$

Ans-

$\begin{array}{l}\left(\mathrm{a}\right)\text{\hspace{0.17em}}20+9+\frac{4}{10}+\frac{1}{100}=29.41\\ \left(\mathrm{b}\right)\text{\hspace{0.17em}}137+\frac{5}{100}=137.05\\ \left(\mathrm{c}\right)\text{\hspace{0.17em}}\frac{7}{10}+\frac{6}{100}+\frac{4}{1000}=0.764\\ \left(\mathrm{d}\right)\text{\hspace{0.17em}}23+\frac{2}{10}+\frac{6}{1000}=23.206\\ \left(\mathrm{e}\right)\text{\hspace{0.17em}}700+20+5+\frac{9}{100}=725.09\end{array}$

Q.12 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

Ans-

(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

Q.13 Between which two numbers in tenths place on the number line does each of the given number lie?
(a) 0.06 (b) 0.45
(c) 0.19 (d) 0.66
(e)0.92 (f)0.57

Ans-

(a) 0.06 lies between 0 and 0.1

(b) 0.45 lies between 0.4 and 0.5

(c) 0.19 lies between 0.1 and 0.2

(d) 0.66 lies between 0.6 and 0.7

(e) 0.92 lies between 0.9 and 1.0

(f) 0.57 lies between 0.5 and 0.6

Q.14 Between which two numbers in tenths place on the number line does each of the given number lie?
(a) 0.06 (b) 0.45
(c) 0.19 (d) 0.66
(e)0.92 (f)0.57

Ans-

(a) 0.06 lies between 0 and 0.1

(b) 0.45 lies between 0.4 and 0.5

(c) 0.19 lies between 0.1 and 0.2

(d) 0.66 lies between 0.6 and 0.7

(e) 0.92 lies between 0.9 and 1.0

(f) 0.57 lies between 0.5 and 0.6

Q.15 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

Ans-

$\begin{array}{l}\left(\text{a}\right)\text{}0.\text{6}0\mathrm{}=\frac{60}{100}=\frac{3}{5}\\ \left(\text{b}\right)\text{}0.0\text{5}\mathrm{}=\frac{5}{100}=\frac{1}{20}\\ \left(\text{c}\right)\text{}0.\text{75}\mathrm{}=\frac{75}{100}=\frac{3}{4}\\ \left(\text{d}\right)\text{}0.\text{18}\mathrm{}=\frac{18}{100}=\frac{9}{50}\\ \left(\text{e}\right)\text{}0.\text{25}\mathrm{}=\frac{25}{100}=\frac{1}{4}\\ \left(\text{f}\right)\text{}0.\text{125}=\frac{125}{1000}=\frac{1}{8}\\ \left(\text{g}\right)\text{}0.0\text{66}=\frac{66}{1000}=\frac{33}{500}\end{array}$

Q.16 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

Ans-

(a) 0.3 or 0.4

The whole parts of these numbers are same.

Now, the tenth part of 0.4 is greater than that of 0.3.

Therefore, 0.4 > 0.3

(b) 0.07 and 0.02

Here, the whole part and the tenth place of both the numbers are same parts.

But the hundredth part of 0.07 is greater than that of 0.02.

Therefore, 0.07 > 0.02

(c) 3 or 0.8

The whole part of 3 is greater than that of 0.8.

Therefore, 3 > 0.8

(d) 0.5 or 0.05

The whole parts of both the numbers are same.

The tenth part of 0.5 is greater than that of 0.05.

Therefore, 0.5 > 0.05

(e) 1.23 or 1.20

Both the numbers have same parts up to the tenth place. But the hundredth part of 1.23 is greater than that of 1.20.

Therefore, 1.23 > 1.20

(f) 0.099 or 0.19

The whole parts of both the numbers are same.

But the tenth part of 0.19 is greater than that of 0.099.

Therefore, 0.19 > 0.099

(g) 1.5 or 1.50

The whole part and the tenth place of both the numbers are same.

Also, there is no digit at hundredth place of 1.5 which implies that the digit at hundredths place will be 0.

Since the hundredths place of both the numbers is also same, therefore, both the numbers are equal.

(h) 1.431 or 1.490

Here, both numbers have the same parts up to the tenth place.

But the hundredth part of 1.490 is greater than that of 1.431.

Therefore, 1.490 > 1.431

(i) 3.3 or 3.300

Here, both numbers have the same parts up to the tenth place.

Also, there is no digit at hundredth and thousandth place of 3.3 which means that the digits at hundredths and thousandths are 0.

Therefore, both the numbers are the same as the digits at all the places are same.

Hence, both these numbers are equal.

(j) 5.64 or 5.603

Here, the whole part and the tenth place of both the numbers is same.

But the hundredth part of 5.64 is greater than that of 5.603.

Therefore, 5.640 > 5.603

Q.17 Make five more examples and find the greater number from them.

1. 4.7 or 4.4
2. 7.34 or 7.304
3. 0.043 or 0.00985
4. 10.598 or 10.61
5. 3.23 or 3.25

Ans-

1. 4.7 or 4.4

Here, the whole part of both the numbers is same. But the tenth part of 4.7 is greater than that of 4.4.

Therefore, 4.7 > 4.4

1. 3.23 or 3.25

Here, the whole parts as well as the tenth part of both the numbers are same.

But the hundredth part of 3.25 is greater than that of a 3.23.

Therefore, 3.25 > 3.23.

1. 7.34 or 7.304

Here, the whole parts and the tenth part of both the numbers are same.

But the hundredth part of 7.34 is greater than that of a 7.304.

Therefore, 7.34 > 7.304

1. 0.043 or 0.00985

Here, the whole parts and the tenth part of both the numbers are same.

But the hundredth part of 0.043 is greater than that of a 0.00985.

Therefore, 0.043 > 0.00985

1. 10.598 or 10.61

Here, the whole parts of both the numbers are same.

But the tenth part of 10.61 is greater than that of a 10.598.

Therefore, 10.61 > 10.598

Q.18 Express as metres using decimals.
(a) 15 cm
(b) 6 cm
(c) 2 m 45 cm
(d) 9 m 7 cm
(e) 419 cm

Ans-

$\begin{array}{l}\left(\mathrm{a}\right)\text{}15\text{}\mathrm{cm}\\ \text{We know that 1}00\text{cm}=\text{1}\mathrm{m}\\ ⇒\text{1 cm}=\frac{1}{100}\text{}\mathrm{m}\\ ⇒15\text{\hspace{0.17em}cm}=\frac{15}{100}\text{\hspace{0.17em}}\mathrm{m}\\ =\text{​}0.15\text{m}\\ \\ \left(\mathrm{b}\right)\text{}6\text{}\mathrm{cm}\text{}\\ \text{We know that 1}00\text{cm}=\text{1}\mathrm{m}\\ ⇒\text{1 cm}=\frac{1}{100}\text{}\mathrm{m}\\ ⇒6\text{\hspace{0.17em}cm}=\frac{6}{100}\text{\hspace{0.17em}}\mathrm{m}\\ =\text{​}0.06\text{m}\\ \\ \left(\mathrm{c}\right)\text{}2\text{}\mathrm{m}\text{}45\text{}\mathrm{cm}\\ \text{We know that 1}00\text{cm}=\text{1}\mathrm{m}\\ ⇒\text{1 cm}=\frac{1}{100}\text{}\mathrm{m}\\ ⇒45\text{\hspace{0.17em}cm}=\frac{45}{100}\text{\hspace{0.17em}}\mathrm{m}\\ =\text{​}0.45\text{m}\\ \\ \text{Now,}2\text{}\mathrm{m}\text{}45\text{}\mathrm{cm}=2\text{\hspace{0.17em}}\mathrm{m}+0.45\text{m}\\ =2.45\text{\hspace{0.17em}}\mathrm{m}\\ \text{}\\ \left(\mathrm{d}\right)\text{}9\text{}\mathrm{m}\text{}7\text{}\mathrm{cm}\\ \text{We know that 1}00\text{cm}=\text{1}\mathrm{m}\\ ⇒\text{1 cm}=\frac{1}{100}\text{}\mathrm{m}\\ ⇒7\text{\hspace{0.17em}cm}=\frac{7}{100}\text{\hspace{0.17em}}\mathrm{m}\\ =\text{​}0.07\text{m}\\ \\ \text{Now,}9\text{}\mathrm{m}\text{}7\text{}\mathrm{cm}=9\text{\hspace{0.17em}}\mathrm{m}+0.07\text{m}\\ =9.07\text{\hspace{0.17em}}\mathrm{m}\text{}\\ \\ \left(\mathrm{e}\right)\text{}419\text{}\mathrm{cm}\\ \text{We know that 1}00\text{cm}=\text{1}\mathrm{m}\\ ⇒\text{1 cm}=\frac{1}{100}\text{}\mathrm{m}\\ ⇒419\text{\hspace{0.17em}cm}=\frac{419}{100}\text{\hspace{0.17em}}\mathrm{m}\\ =\text{​}4.19\text{m}\end{array}$

Q.19 Express as cm using decimals.
(a) 5 mm
(b) 60 mm
(c) 164 mm
(d) 9 cm 8 mm
(e) 93 mm

Ans-

$\begin{array}{l}\left(\mathrm{a}\right)\text{}5\text{m}\mathrm{m}\\ \text{We know that 1}0\text{mm}=\text{1}\mathrm{cm}\\ ⇒\text{1 mm}=\frac{1}{10}\text{c}\mathrm{m}\\ ⇒5\text{mm}=\frac{5}{10}\text{\hspace{0.17em}}\mathrm{cm}\\ =\text{​}0.5\text{cm}\\ \\ \left(\text{b}\right)\text{60 mm}\\ \text{We know that 1}0\text{mm}=\text{1}\mathrm{cm}\\ ⇒\text{1 mm}=\frac{1}{10}\text{c}\mathrm{m}\\ ⇒60\text{mm}=\frac{60}{10}\text{\hspace{0.17em}}\mathrm{cm}\\ =\text{​}6\text{cm or 6.0 cm}\\ \\ \left(\text{c}\right)\text{164 mm}\\ \text{We know that 1}0\text{mm}=\text{1}\mathrm{cm}\\ ⇒\text{1 mm}=\frac{1}{10}\text{c}\mathrm{m}\\ ⇒164\text{mm}=\frac{164}{10}\text{\hspace{0.17em}}\mathrm{cm}\\ =\text{​}16.4\text{cm}\\ \\ \left(\text{d}\right)\text{9 cm 8 mm}\\ \text{We know that 1}0\text{mm}=\text{1}\mathrm{cm}\\ ⇒\text{1 mm}=\frac{1}{10}\text{c}\mathrm{m}\\ ⇒8\text{mm}=\frac{8}{10}\text{\hspace{0.17em}}\mathrm{cm}\\ =\text{​}0.8\text{cm}\\ \mathrm{Now},\text{9 cm 8 mm =}\left(\text{9 + 0.8}\right)\text{cm}\\ \text{= 9.8 cm}\\ \\ \left(\text{e}\right)\text{93 mm}\\ \text{We know that 1}0\text{mm}=\text{1}\mathrm{cm}\\ ⇒\text{1 mm}=\frac{1}{10}\text{c}\mathrm{m}\\ ⇒93\text{mm}=\frac{93}{10}\text{\hspace{0.17em}}\mathrm{cm}\\ =\text{​}9.3\text{cm}\\ \text{}\end{array}$

Q.20 Express as km using decimals.
(a) 8 m
(b) 88 m
(c) 8888 m
(d) 70 km 5 m

Ans-

$\begin{array}{l}\left(\text{a}\right)\text{8 m}\\ \text{We know that 1}000\text{m}=\text{1}\mathrm{km}\\ ⇒\text{1 m}=\frac{1}{1000}\text{k}\mathrm{m}\\ ⇒8\text{m}=\frac{8}{1000}\text{\hspace{0.17em}}\mathrm{km}\\ =\text{​}0.008\text{km}\\ \\ \left(\text{b}\right)\text{88 m}\\ \text{We know that 1}000\text{m}=\text{1}\mathrm{km}\\ ⇒\text{1 m}=\frac{1}{1000}\text{k}\mathrm{m}\\ ⇒88\text{m}=\frac{88}{1000}\text{\hspace{0.17em}}\mathrm{km}\\ =\text{​}0.088\text{km}\\ \text{}\\ \left(\text{c}\right)\text{8888 m}\\ \text{We know that 1}000\text{m}=\text{1}\mathrm{km}\\ ⇒\text{1 m}=\frac{1}{1000}\text{k}\mathrm{m}\\ ⇒8888\text{m}=\frac{8888}{1000}\mathrm{km}\\ =\text{​ 8.8}88\text{km}\\ \\ \left(\text{d}\right)\text{70 km 5 m}\\ \text{We know that 1}000\text{m}=\text{1}\mathrm{km}\\ ⇒\text{1 m}=\frac{1}{1000}\text{k}\mathrm{m}\\ ⇒5\text{m}=\frac{5}{1000}\text{\hspace{0.17em}}\mathrm{km}\\ =\text{​}0.005\text{km}\\ \text{Now, 70 km 5 m=}\left(\text{70 + 0.005}\right)\mathrm{km}\\ =70.005\mathrm{km}\end{array}$

Q.21 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-

$\begin{array}{l}\left(\text{a}\right)\text{2 g}\\ \\ \text{We know that 1}000\text{g}=\text{1}kg\\ ⇒\text{1 g}=\frac{1}{1000}\text{k}g\\ ⇒2\text{g}=\frac{2}{1000}\text{\hspace{0.17em}}kg\\ =\text{​}\text{}0.002\text{kg}\\ \\ \left(\text{b}\right)\text{100 g}\\ \text{We know that 1}000\text{g}=\text{1}kg\\ ⇒\text{1 g}=\frac{1}{1000}\text{k}g\\ ⇒100\text{g}=\frac{100}{1000}\text{\hspace{0.17em}}kg\\ =\text{​}\text{}0.1\text{kg}\\ \text{}\\ \left(\text{c}\right)\text{3750 g}\\ \text{We know that 1}000\text{g}=\text{1}kg\\ ⇒\text{1 g}=\frac{1}{1000}\text{k}g\\ ⇒3750\text{g}=\frac{3750}{1000}\text{\hspace{0.17em}}kg\\ =\text{​}\text{3}\text{.750 kg}\\ \\ \left(\text{d}\right)\text{5 kg 8 g}\\ \text{We know that 1}000\text{g}=\text{1}kg\\ ⇒\text{1 g}=\frac{1}{1000}\text{k}g\\ ⇒8\text{g}=\frac{8}{1000}\text{\hspace{0.17em}}kg\\ =\text{​}\text{0}\text{.008 kg}\\ \text{Now, 5 kg 8 g=}\left(\text{5 + 0}\text{.008}\right)kg\\ =5.008\text{}kg\\ \\ \left(\text{e}\right)\text{26 kg 50 g}\\ \text{We know that 1}000\text{g}=\text{1}kg\\ ⇒\text{1 g}=\frac{1}{1000}\text{k}g\\ ⇒50\text{g}=\frac{50}{1000}\text{\hspace{0.17em}}kg\\ =\text{​}\text{}0.05\text{kg}\\ \text{Now, 26 kg 50 g =}\left(\text{26 + 0}\text{.05}\right)kg\\ =26.05\text{}kg\end{array}$

Q.22 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

Ans-

$\begin{array}{l}\left(\text{a}\right)\text{0}\text{.007 + 8}\text{.5 + 30}\text{.08}\\ \text{}\\ \text{0}\text{.007}\\ \text{8}\text{.500}\\ \text{}\underset{¯}{+\text{30}\text{.080}}\\ \text{\hspace{0.17em}}\text{}\underset{¯}{\text{}38.587\text{}}\\ \\ \left(\text{b}\right)\text{15 + 0}\text{.632 + 13}\text{.8}\\ \text{}\\ \text{15}\text{.000}\\ \text{0}\text{.632}\\ \text{}\underset{¯}{+\text{13}\text{.800}}\\ \text{\hspace{0.17em}}\text{}\underset{¯}{\text{}29.432\text{}}\\ \\ \left(\text{c}\right)\text{27}\text{.076 + 0}\text{.55 + 0}\text{.004}\\ \text{}\\ \text{27}\text{.076}\\ \text{0}\text{.550}\\ \text{}\underset{¯}{+\text{0}\text{.004}}\\ \text{\hspace{0.17em}}\text{}\underset{¯}{\text{}27.630\text{}}\\ \\ \left(\text{d}\right)\text{25}\text{.65 + 9}\text{.005 + 3}\text{.7}\\ \text{}\\ \text{25}\text{.650}\\ \text{9}\text{.005}\\ \text{}\underset{¯}{+\text{3}\text{.700}}\\ \text{\hspace{0.17em}}\text{}\underset{¯}{\text{38}\text{.355}}\\ \\ \left(\text{e}\right)\text{0}\text{.75 + 10}\text{.425 + 2}\\ \text{}\\ \text{0}\text{.750}\\ \text{10}\text{.425}\\ \text{}\underset{¯}{+\text{2}\text{.000}}\\ \text{\hspace{0.17em}}\text{}\underset{¯}{\text{13}\text{.175}}\\ \\ \left(\text{f}\right)\text{280}\text{.69 + 25}\text{.2 + 38}\\ \text{}\\ \text{280}\text{.69}\\ \text{25}\text{.20}\\ \text{}\underset{¯}{+\text{38}\text{.00}}\\ \text{\hspace{0.17em}}\text{}\underset{¯}{\text{343}\text{.89}}\end{array}$

Q.23 Nasreen bought 3 m 20 cm cloth for her shirt and 2 m cm cloth for her trouser. Find the total length of cloth bught by her.

Ans-

$\begin{array}{l}\text{Material bought by Nasreen for shirt}\mathrm{}=\text{3 m 2}0\text{cm}\\ =\text{3 m}+\frac{20}{100}\mathrm{}\mathrm{m}=3\text{}\mathrm{m}+\mathrm{}0.2\text{}\mathrm{m}=3.2\text{}\mathrm{m}\mathrm{}\\ \mathrm{}\text{Material bought by Nasreen for trouser}=\text{2 m 5 cm}\\ =\text{2 m}+\frac{5}{100}\mathrm{}\mathrm{m}=2\text{}\mathrm{m}+\mathrm{}0.05\text{}\mathrm{m}=2.05\text{}\mathrm{m}\\ \text{Total length of cloth bought by Nasreen}:\text{}\\ \text{}\\ \text{3.20}\\ \text{}\underset{¯}{\text{+ 2.05}}\\ \text{\hspace{0.17em}}\underset{¯}{\text{5.25}}\\ \text{Hence,the total length of cloth bought by Nasreen is 5.25 m.}\end{array}$

Q.24 Naresh walked 2 km 35 m in the morning and 1 km 7 in the evening. How much distance did he walk in all?

Ans-

$\begin{array}{l}\text{Distance covered by Naresh in the morning}\mathrm{}=\text{2 km 35 m}\\ =\text{2 km}+\frac{35}{1000}\mathrm{}\text{\hspace{0.17em}}\mathrm{km}=2\text{}\mathrm{km}+\text{}\mathrm{}0.035\text{k}\mathrm{m}=2.035\text{km}\\ \text{Distance covered by Naresh in the evening}\mathrm{}=\text{1 km 7 m}\\ =\text{1 km}+\frac{7}{1000}\mathrm{}\text{\hspace{0.17em}}\mathrm{km}=1\text{k}\mathrm{m}+\mathrm{}0.007\text{k}\mathrm{m}=1.007\text{k}\mathrm{m}\\ \text{Total distance walked by Naresh}:\text{}\\ \text{}\\ \text{2.035}\\ \text{}\underset{¯}{\text{+ 1.007}}\\ \text{\hspace{0.17em}}\underset{¯}{\text{3.042}}\\ \text{Hence,the total distance walked by Naresh is 3.042 km.}\end{array}$

Q.25 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?

Ans-

$\begin{array}{l}\mathrm{Distance}\text{}\mathrm{travelled}\text{}\mathrm{by}\text{}\mathrm{Sunita}\text{}\mathrm{in}\text{}\mathrm{bus}\text{}\mathrm{}=15\text{}\mathrm{km}\text{}268\text{}\mathrm{m}\\ =\text{15 km}+\frac{268}{1000}\mathrm{}\text{\hspace{0.17em}}\mathrm{km}=15\text{}\mathrm{km}+\text{}\mathrm{}0.268\text{k}\mathrm{m}=15.268\text{km}\\ \mathrm{Distance}\text{}\mathrm{travelled}\text{}\mathrm{by}\text{}\mathrm{Sunita}\text{}\mathrm{in}\text{}\mathrm{car}\text{}\mathrm{}=7\text{}\mathrm{km}\text{}7\text{}\mathrm{m}\\ =\text{7 km}+\frac{7}{1000}\mathrm{}\text{\hspace{0.17em}}\mathrm{km}=7\text{k}\mathrm{m}+\mathrm{}0.007\text{k}\mathrm{m}=7.007\text{k}\mathrm{m}\\ \mathrm{Distance}\mathrm{}\mathrm{travelled}\mathrm{}\mathrm{by}\mathrm{}\mathrm{Sunita}\mathrm{on} \mathrm{foot}=500\text{}\mathrm{m}\\ =\frac{500}{1000}\text{}\mathrm{km}=0.5\mathrm{km}\\ \text{Total distance travelled by Sunita:}\\ \text{}\\ \text{15.268}\\ \text{7.007}\\ \text{\hspace{0.17em}+}\underset{¯}{\text{0.500}}\\ \text{​}\underset{¯}{\text{22.775}}\\ \text{Hence,the total distance travelled by Sunita is 22.775 km.}\end{array}$

Q.26 Ravi purchased 5 kg 400 g rice, 2 kg 20 g sugar and 10 kg 850g flour. Find the total weight of his purchases.

Ans-

$\begin{array}{l}\mathrm{Quantity}\text{}\mathrm{of}\text{}\mathrm{rice}\text{}\mathrm{purchased}\text{}\mathrm{by}\text{}\mathrm{Ravi}\text{}\mathrm{}=5\text{}\mathrm{kg}\text{}400\mathrm{g}\\ =5\text{}\mathrm{kg}+\frac{400}{1000}\mathrm{}\text{\hspace{0.17em}}\mathrm{kg}=5\text{}\mathrm{kg}+\text{}\mathrm{}0.4\text{}\mathrm{kg}=5.4\text{}\mathrm{kg}\\ \\ \mathrm{Quantity}\text{}\mathrm{of}\text{}\mathrm{sugar}\text{}\mathrm{purchased}\text{}\mathrm{by}\text{}\mathrm{Ravi}\text{}\mathrm{}=2\text{}\mathrm{kg}\text{}20\mathrm{g}\\ =2\text{}\mathrm{kg}+\frac{20}{1000}\mathrm{}\text{\hspace{0.17em}}\mathrm{kg}=2\text{}\mathrm{kg}+\text{}\mathrm{}0.02\text{}\mathrm{kg}=2.02\text{}\mathrm{kg}\\ \\ \mathrm{Quantity}\text{}\mathrm{of}\text{}\mathrm{flour}\text{}\mathrm{purchased}\text{}\mathrm{by}\text{}\mathrm{Ravi}\text{}\mathrm{}=10\text{}\mathrm{kg}\text{}850\mathrm{g}\\ 10\text{}\mathrm{kg}+\frac{850}{1000}\mathrm{}\text{\hspace{0.17em}}\mathrm{kg}=10\text{}\mathrm{kg}+\text{}\mathrm{}0.85\text{}\mathrm{kg}=10.85\text{}\mathrm{kg}\\ \text{Total quantity purchased by Ravi:}\\ \text{}\\ \text{5.40}\\ \text{2.02}\\ \text{\hspace{0.17em}+}\underset{¯}{\text{10.85}}\\ \text{​}\underset{¯}{\text{18.27}}\\ \\ \text{Hence,the total quantity purchased by Ravi is 18.27 kg.}\end{array}$

Q.27 Find the value of:
(a) 9.756 – 6.28
(b) 21.05 – 15.27
(c) 18.5 – 6.79
(d) 11.6 – 9.847

Ans-

$\begin{array}{l}\left(\text{a}\right)\text{9}.\text{756}–\text{6}.\text{28}\\ \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{9}.\text{756}\\ \text{}-\text{}\underset{¯}{\text{6}\text{.280}}\\ \text{}\underset{¯}{\text{}3.476\text{}}\\ \\ \left(\text{b}\right)\text{21}.0\text{5}–\text{15}.\text{27}\\ \text{}\\ \text{21}\text{.05}\\ \text{}-\text{}\underset{¯}{\text{15}\text{.27}}\\ \text{}\underset{¯}{\text{}5.78\text{}}\\ \text{​}\text{}\\ \left(\text{c}\right)\text{18}.\text{5}–\text{6}.\text{79}\\ \text{}\\ \text{18}\text{.50}\\ \text{}-\text{}\underset{¯}{\text{6}\text{.79}}\text{}\\ \text{}\underset{¯}{11.71\text{\hspace{0.17em}}\text{}}\text{}\\ \text{​}\text{}\\ \left(\text{d}\right)\text{11}.\text{6}–\text{9}.\text{847 9}\text{.756}\\ \text{}\\ \text{11}\text{.600}\\ \text{}-\underset{¯}{\text{9}\text{.847}}\\ \text{}\underset{¯}{\text{}1.753\text{}}\end{array}$

Q.28 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?

Ans-

$\begin{array}{l}\text{Total length of the cloth}\mathrm{}=\text{20 m 5 cm}\\ =20\text{}\mathrm{m}+\frac{5}{100}\mathrm{}\text{\hspace{0.17em}}\mathrm{m}=20\text{}\mathrm{m}+\text{}\mathrm{}0.05\text{}\mathrm{m}=20.05\text{}\mathrm{m}\\ \\ \text{Length of the cloth cut to make a curtain}=\text{4 m 50 cm}\\ =4\text{}\mathrm{m}+\frac{50}{100}\mathrm{}\text{\hspace{0.17em}}\mathrm{m}=4\text{}\mathrm{m}+\text{}\mathrm{}0.5\text{}\mathrm{m}=4.5\text{}\mathrm{m}\\ \text{So, the cloth left with her:}\\ \text{}\\ \text{20.05}\\ \text{}-\underset{¯}{\text{4.50}}\text{}\\ \text{}\underset{¯}{\text{15.55}}\text{}\\ \\ \text{Hence,the length of the cloth left with Tina is 15.55 m.}\end{array}$

Q.29 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?

Ans-

$\begin{array}{l}\text{Distance travelled by Namita everyday}=\text{20 km 50 m}\mathrm{}\\ =20\text{k}\mathrm{m}+\frac{50}{1000}\mathrm{}\text{\hspace{0.17em}}\mathrm{km}=20\text{k}\mathrm{m}+\text{}\mathrm{}0.05\text{k}\mathrm{m}=20.05\text{k}\mathrm{m}\\ \\ \text{Distance travelled by bus}=\text{10 km 200 m}\\ =10\text{k}\mathrm{m}+\frac{200}{1000}\mathrm{}\text{\hspace{0.17em}}\mathrm{km}=10\text{k}\mathrm{m}+\text{}\mathrm{}0.2\text{k}\mathrm{m}=10.2\text{k}\mathrm{m}\\ \text{Hence, the distance travelled by auto :}\\ \text{}\\ \text{20.05}\\ \text{}-\underset{¯}{\text{10.20}}\text{}\\ \text{}\underset{¯}{\text{9.85}}\text{}\\ \\ \text{Therefore,the distance travelled by auto is 9.850 km.}\end{array}$

Q.30 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?

Ans-

$\begin{array}{l}\text{Total Weight of vegetables}=10\text{\hspace{0.17em}kg}\\ \\ \text{Weight of onions}=\text{3 kg 500 g}\\ =3\text{k}\mathrm{g}+\frac{500}{1000}\mathrm{}\text{\hspace{0.17em}}\mathrm{kg}=3\text{k}\mathrm{g}+\text{}\mathrm{}0.5\text{k}\mathrm{g}=3.5\text{k}\mathrm{g}\\ \text{Weight of tomatoes}=\text{2 kg 75 g}\\ =2\text{k}\mathrm{g}+\frac{75}{1000}\mathrm{}\text{\hspace{0.17em}}\mathrm{kg}=2\text{k}\mathrm{g}+\text{}\mathrm{}0.075\text{k}\mathrm{g}=2.075\text{k}\mathrm{g}\\ \\ \text{Therefore,}\\ \text{the weight of potatoes}=\text{Total weight of vegetables}-\text{( weight of}\\ \text{tomatoes + weight of onions)}\\ =\text{}10\text{}-\left(2.075+3.5\right)\text{}\\ \text{}\\ \text{First we will solve the bracket.}\\ \text{2.075}\\ \text{}+\underset{¯}{\text{3.500}}\text{}\\ \text{}\underset{¯}{\text{5.575}}\text{}\\ \text{Now,we will subtract the sum of the weights of tomatoes and}\\ \text{onions from the weight of vegetables.}\\ \\ \text{10.000}\\ \text{}-\underset{¯}{\text{5.575}}\text{}\\ \text{}\underset{¯}{\text{4.425}}\\ \\ \text{Therefore,the weight of potatoes is 4.425 kg.}\end{array}$

Q.31 Write the following as numbers in the given table: Ans-

 Hundreds(100) Tens(10) Ones(1) Tenths(1/10) (a) 0 3 1 2 (b) 1 1 0 4

Q.32 Write the following decimals in the place value table.

(a) 19.4 (b) 0.3 (c) 10.6 (d) 205.9

Ans-

 Hundreds(100) Tens(10) Ones(1) Tenths(1/10) 19.4 0 1 9 4 0.3 0 0 0 3 10.6 0 1 0 6 205.9 2 0 5 9

Q.33 Complete the table with the help of these boxes and use decimals to write the number. Ans-

 Ones Tenths Hundredths Number (a) 0 2 6 0.26 (b) 1 3 8 1.38 (c) 1 2 8 1.28

Q.34 Write the following decimals in the place value table.
(a) 0.29 (b) 2.08 (c) 19.60 (d) 148.32 (e) 200.812

Ans-

 Hundreds Tens Ones Tenths Hundredths Thousandths a 0 0 0 2 9 9 b 0 0 2 0 8 0 c 0 1 9 6 0 0 d 1 4 8 3 2 0 e 2 0 0 8 1 2

Q.35 Express as rupees using decimals.
(a) 5 paise (b) 75 paise (c) 20 paise
(d) 50 rupees 90 paise (e) 725 paise

Ans-

$\begin{array}{l}\left(\mathbf{a}\right)\text{}\mathbf{5}\text{}\mathbf{paise}\\ \text{We know that 1}00\text{paise}=\text{1}\\ ⇒\text{1 paise}=₹\frac{1}{100}\text{}\\ ⇒5\text{}\mathrm{paise}=₹\frac{5}{100}\text{\hspace{0.17em}}\\ =\text{​}₹\text{}0.05\text{}\\ \end{array}$

$\begin{array}{l}\left(\mathbf{b}\right)\text{}\mathbf{75}\text{}\mathbf{paise}\\ \text{We know that 1}00\text{paise}=₹\text{1}\\ ⇒\text{1 paise}=₹\frac{1}{100}\text{}\\ ⇒75\text{}\mathrm{paise}=₹\frac{75}{100}\text{\hspace{0.17em}}\\ =\text{​}₹\text{}0.75\text{}\\ \\ \left(\mathbf{c}\right)\text{}\mathbf{20}\text{}\mathbf{paise}\\ \text{We know that 1}00\text{paise}=₹\text{1}\\ ⇒\text{1 paise}=₹\frac{1}{100}\text{}\\ ⇒20\text{}\mathrm{paise}=₹\frac{20}{100}\text{\hspace{0.17em}}\\ =\text{​}₹\text{}0.20\\ \text{}\\ \left(\mathbf{d}\right)\text{}\mathbf{50}\text{}₹\text{}\mathbf{90}\text{}\mathbf{paise}\\ \text{We know that 1}00\text{paise}=₹\text{1}\\ ⇒\text{1 paise}=₹\frac{1}{100}\text{}\\ ⇒90\text{}\mathrm{paise}=₹\frac{90}{100}\text{\hspace{0.17em}}\\ =\text{​}₹\text{}0.90\\ \\ \text{}\mathrm{}\therefore 50₹90\mathrm{paise}=50₹+\text{}₹\text{}0.90\\ =\text{\hspace{0.17em}}₹\text{}50.90\end{array}$

$\begin{array}{l}\left(\mathbf{e}\right)\text{}\mathbf{725}\text{}\mathbf{paise}\\ \text{We know that 1}00\text{paise}=₹\text{1}\\ ⇒\text{1 paise}=₹\frac{1}{100}\text{}\\ ⇒725\text{}\mathrm{paise}=₹\frac{725}{100}\text{\hspace{0.17em}}\\ =\text{​}₹\text{}7.25\\ \end{array}$

Q.36 Rashid spent ₹ 35.75 for Maths book and ₹ 32.60 for Science book. Find the total amount spent by Rashid.

Ans-

$\begin{array}{l}\text{Amount spent for Maths book}=\text{}₹\text{35}.\text{75}\\ \text{Amount spent for Science book}=\text{}₹\text{32}.\text{6}0\\ \\ \text{Total amount spent by Rashid is:}\\ \text{}\\ \text{35}\text{.75}\\ \text{}\underset{¯}{\text{+ 32}\text{.60}}\\ \text{\hspace{0.17em}}\underset{¯}{\text{68}\text{.35}}\\ \\ \text{Therefore},\text{the amount spent by Rashid is ₹ 68}.\text{35}.\end{array}$

Q.37 Radhika’s mother gave her ₹ 10.50 and her father gave her ₹ 15.80, find the total amount given to Radhika by the parents.

Ans-

$\begin{array}{l}\text{Amount given by mother}=\text{₹ 1}0.\text{5}0\\ \text{Amount given by father}=\text{₹ 15}.\text{8}0\end{array}$

$\begin{array}{l}\text{Total amount given by parents is:}\\ \text{}\\ \text{10}\text{.50}\\ \text{}\underset{¯}{\text{+ 15}\text{.80}}\\ \text{\hspace{0.17em}}\underset{¯}{\text{26}\text{.30}}\\ \\ \text{Therefore},\text{the amount given by her parents is ₹ 26}.\text{3}0.\end{array}$

Q.38 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

Ans-

(a) ₹ 18.25 from ₹20.75

$\begin{array}{l}\text{20}\text{.75}\\ \text{}-\text{}\underset{¯}{\text{18}\text{.25}}\\ \text{}\underset{¯}{\text{}2.50\text{}}\text{\hspace{0.17em}}\text{}\end{array}$

Therefore, ₹ 20.75 − ₹ 18.25 = ₹ 2.50
(b) 202.54 m from 250 m

$\begin{array}{l}\text{250}\text{.00}\\ \text{}-\text{}\underset{¯}{\text{202}\text{.54}}\\ \text{}\underset{¯}{\text{47}\text{.4}6\text{}}\end{array}$

Therefore, 250 m – 202.54 m = 47.46 m
(c) ₹ 5.36 from ₹ 8.40

$\begin{array}{l}\text{}\text{}\\ \text{8}\text{.40}\\ \text{}-\text{}\underset{¯}{\text{5}\text{.36}}\\ \text{}\underset{¯}{\text{}3.04\text{\hspace{0.17em}}}\end{array}$

Therefore, ₹ 8.40 – ₹ 18.25 = ₹ 3.04
(d) 2.051 km from 5.206 km

$\begin{array}{l}\text{5}\text{.206}\\ \text{}-\text{}\underset{¯}{\text{2}\text{.051}}\\ \text{}\underset{¯}{\text{}3.155\text{}}\end{array}$

Therefore, 5.206 km – 2.051 km = 3.155 km
(e) 0.314 kg from 2.107 kg

$\begin{array}{l}\text{}\\ \text{2}\text{.107}\\ \text{}-\text{}\underset{¯}{\text{0}\text{.314}}\\ \text{}\underset{¯}{\text{}1.793\text{\hspace{0.17em}}\text{}}\\ \text{​}\text{}\\ \text{}\end{array}$

Therefore, 2.107 kg – 0.314 kg = 1.793 kg

Q.39 Raju bought a book for ₹ 35.65. He gave ₹ 50 to the shopkeeper. How much money did he get back from the shopkeeper?

Ans-

Cost price of the book = ₹ 35.65
Money given to the shopkeeper = ₹ 50
So, the money that Raju will get back is

$\begin{array}{l}\text{}\\ \text{50}\text{.00}\\ \text{}-\underset{¯}{\text{35}\text{.65}}\\ \text{}\underset{¯}{\text{}14.35\text{}}\text{​}\text{}\\ \text{}\end{array}$

Therefore, the money that Raju will get back is ₹ 14.35.

Q.40 Rani had ₹ 18.50. She bought one ice-cream for ₹ 11.75. How much money does she have now?

Ans-

Total money with Rani = ₹ 18.50

The money spent for an ice-cream= ₹ 11.75

So, the money left with Rani =

$\begin{array}{l}\text{}\\ \text{18}\text{.50}\\ \text{}-\underset{¯}{\text{11}\text{.75}}\\ \text{}\underset{¯}{\text{}6.75\text{}}\text{}\\ \text{}\end{array}$

Therefore, the money left with Rani is ₹ 6.75.

## 1. Explain the concept of decimal numbers covered in Chapter 8 of NCERT Solutions for Class 6 Maths.

Decimal numbers are numbers that are used to represent numbers that are not whole or complete. The decimal point, often known as the period (.), is an important element of a decimal number. In a decimal number, this period divides the fractional and whole number parts. A digit’s place value can be defined as the value of a digit in relation to its position in a number.

## 2. Explain the concepts of recurring and terminating decimals covered in the NCERT Solutions for Class 6 Maths Chapter 8?

A decimal number with a finite number of digits after the decimal is known as a terminating decimal. The decimal representation of a recurring decimal becomes periodic, or the same sequence of digits keeps repeating endlessly.

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

There are a total of six exercises in Class 6 Maths Chapter 8 Decimals. Given below are the exercises in this chapter:

Ex 8.1 – 10 Questions

Ex 8.2 – 7 Questions

Ex 8.3 – 2 Questions

Ex 8.4 – 5 Questions

Ex 8.5 – 7 Questions

Ex 8.6 – 7 Questions

## 4. What are like and unlike decimals?

Decimals with the same number of digits following the decimal point are known as like decimals. The decimals with a varied number of digits after the decimal point are known as unlike decimals. 45.23, 23.34, and 46.45 are examples of like decimals. 23.456, 23.22, and 45.1 are examples of unlike decimals.

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

In algebra, decimals are numbers that have both a whole number and a fractional part. For example, some decimal numbers are 89.678, 7654.7, and 879.0. The decimal point is the dot that appears between the numbers. There are three forms of decimal numbers: one, two, and three.

• Recurring decimal numbers
• Non-recurring decimal numbers
• Decimal fraction

All arithmetic operations, such as addition, subtraction, multiplication, and division, are possible with decimals.