NCERT Solutions Class 6 Maths Chapter 8

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

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

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-

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

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-

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

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

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:

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

Ans-

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

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-

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

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

2. 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.

3. 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

4. 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

5. 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-

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

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-

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

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

Ans-

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

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\\ \Rightarrow \text{1 g}=\frac{1}{1000}\text{k}g\\ \Rightarrow 2\text{g}=\frac{2}{1000}kg\\ =\text{}0.002\text{kg}\\ \\ \left(\text{b}\right)\text{100 g}\\ \text{We know that 1}000\text{g}=\text{1}kg\\ \Rightarrow \text{1 g}=\frac{1}{1000}\text{k}g\\ \Rightarrow 100\text{g}=\frac{100}{1000}kg\\ =\text{}0.1\text{kg}\\ \\ \left(\text{c}\right)\text{3750 g}\\ \text{We know that 1}000\text{g}=\text{1}kg\\ \Rightarrow \text{1 g}=\frac{1}{1000}\text{k}g\\ \Rightarrow 3750\text{g}=\frac{3750}{1000}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\\ \Rightarrow \text{1 g}=\frac{1}{1000}\text{k}g\\ \Rightarrow 8\text{g}=\frac{8}{1000}kg\\ =\text{}\text{0}\text{.008 kg}\\ \text{Now, 5 kg 8 g=}\left(\text{5 + 0}\text{.008}\right)kg\\ =5.008kg\\ \\ \left(\text{e}\right)\text{26 kg 50 g}\\ \text{We know that 1}000\text{g}=\text{1}kg\\ \Rightarrow \text{1 g}=\frac{1}{1000}\text{k}g\\ \Rightarrow 50\text{g}=\frac{50}{1000}kg\\ =\text{}0.05\text{kg}\\ \text{Now, 26 kg 50 g =}\left(\text{26 + 0}\text{.05}\right)kg\\ =26.05kg\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{0}\text{.007}\\ \text{8}\text{.500}\\ \underset{¯}{+\text{30}\text{.080}}\\ \underset{¯}{38.587}\\ \\ \left(\text{b}\right)\text{15 + 0}\text{.632 + 13}\text{.8}\\ \\ \text{15}\text{.000}\\ \text{0}\text{.632}\\ \underset{¯}{+\text{13}\text{.800}}\\ \underset{¯}{29.432}\\ \\ \left(\text{c}\right)\text{27}\text{.076 + 0}\text{.55 + 0}\text{.004}\\ \\ \text{27}\text{.076}\\ \text{0}\text{.550}\\ \underset{¯}{+\text{0}\text{.004}}\\ \underset{¯}{27.630}\\ \\ \left(\text{d}\right)\text{25}\text{.65 + 9}\text{.005 + 3}\text{.7}\\ \\ \text{25}\text{.650}\\ \text{9}\text{.005}\\ \underset{¯}{+\text{3}\text{.700}}\\ \underset{¯}{\text{38}\text{.355}}\\ \\ \left(\text{e}\right)\text{0}\text{.75 + 10}\text{.425 + 2}\\ \\ \text{0}\text{.750}\\ \text{10}\text{.425}\\ \underset{¯}{+\text{2}\text{.000}}\\ \underset{¯}{\text{13}\text{.175}}\\ \\ \left(\text{f}\right)\text{280}\text{.69 + 25}\text{.2 + 38}\\ \\ \text{280}\text{.69}\\ \text{25}\text{.20}\\ \underset{¯}{+\text{38}\text{.00}}\\ \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-

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

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-

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

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-

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

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-

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

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

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

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-

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

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-

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

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-

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

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

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

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

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

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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

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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-

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

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

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

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

Ans-

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

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.

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

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

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- 

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

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

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

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

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

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

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

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

\begin{array}{l}\\ \text{2}\text{.107}\\ -\underset{¯}{\text{0}\text{.314}}\\ \underset{¯}{1.793}\\ \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

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

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

Therefore, the money left with Rani is ₹ 6.75.