Decimals

EXERCISE-8.1

P1. Write the following as numbers in the given table.

(a)

(b)

Hundreds(100)	Tens (10)	Ones (1)	Tenths $\left(\frac{1}{10}\right)$

Sol. It may be observed that 

Row	Hundreds	Tens	Ones	Tenths
a.	0	3	1	2
b.	1	1	0	4

P2. Write the following decimals in the place value table.

(a) 19.4   

(b) 0.3

(c) 10.6

(d) 205.9

Sol.

Decimal	Hundreds	Tens	Ones	Tenths
19.4	0	1	9	4
0.3	0	0	0	3
10.6	0	1	0	6
205.9	2	0	5	9

P3. 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

Sol. (a) Seven-tenths = $\frac{7}{10}$ = 0.7

(b) Two tens and nine-tenths = 20 + $\frac{9}{10}$ = 20.9

(c) Fourteen point six = 14.6

(d) One hundred and two ones = 100 + 2 = 102.0

(e) Six hundred point eight = 600.8

P4. Write each of the following as decimals :

(a) $\frac{5}{10}$                            (b) $3+\frac{7}{10}$

(c) $200+60+5+\frac{1}{10}$    (d) $70+\frac{8}{10}$

(e) $\frac{88}{10}$                             (f) $4\frac{2}{10}$

(g) $\frac{3}{2}$                               (h) $\frac{2}{5}$

(i) $\frac{12}{5}$                              (j) $3\frac{3}{5}$

(k) $4\frac{1}{2}$

Sol. (a) $\frac{5}{10}=0.5$

(b) $3+\frac{7}{10}=3+0.7=3.7$

(c) $200+60+5+\frac{1}{10}=265+0.1=265.1$

(d) $70+\frac{8}{10}=70+0.8=7.0.8$

(e) $\frac{88}{10}=\frac{80}{10}+\frac{8}{10}=8+0.8=8.8$

(f) $4\frac{2}{10}=4+\frac{2}{10}=4+0.2=4.2$

(g) $\frac{3}{2}=\frac{2+1}{2}=\frac{2}{2}+\frac{1}{2}=1+0.5=1.5$

(h) $\frac{2}{5}=0.4$

(i) $\frac{12}{5}=\frac{10+2}{5}=\frac{10}{5}+\frac{2}{5}=2+0.4=2.4$

(j) $3\frac{3}{5}=3+\frac{3}{5}=3+0.6=3.6$

(k) $4\frac{1}{2}=4+\frac{1}{2}=4+0.5=4.5$

P5. 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

Sol. (a) $0.6=\frac{6}{10}=\frac{3}{5}$

(b) $2.5=\frac{25}{10}=\frac{5}{2}$

(c) $1.0=1$

(d) $3.8=\frac{38}{10}=\frac{19}{5}$

(e) $13.7=\frac{137}{10}$

(f) $21.2=\frac{212}{10}=\frac{106}{5}$

(g) $6.4=\frac{64}{10}=\frac{32}{5}$

P6. 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

Sol. It is known that 1cm = 10 mm

(a) $2\mathrm{mm}=\frac{2}{10}\mathrm{cm}=0.2\mathrm{cm}$

(b) $30\mathrm{mm}=\frac{30}{10}\mathrm{cm}=3.0\mathrm{cm}$

(c) $116\mathrm{mm}=\frac{116}{10}\mathrm{cm}=11.6\mathrm{cm}$

(d) $4\mathrm{cm} 2\mathrm{mm}=\left(4+\frac{2}{10}\right)\mathrm{cm}=4.2\mathrm{cm}$

(e) $162\mathrm{mm}=\frac{162}{10}\mathrm{cm}=16.2\mathrm{cm}$

(f) $83\mathrm{mm}=\frac{83}{10}\mathrm{cm}=8.3\mathrm{cm}$

P7. Between which two whole numbers on the number line are the given numbers lie ? Which of these whole numbers is nearer the number ?

(a) 0.8      (b) 5.1

(c) 2.6      (d) 6.4

(e) 9.1       (f) 4.9

Sol. (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.

P8. Show the following numbers on the number line.

(a) 0.2

(b) 1.9

(c) 1.1

(d) 2.5

Sol. (a) 0.2 represents a point between 0 and 1 on number line, such that the space between 0 and 1 is divided into 10 equal parts. Hence, each equal part will be equal to one-tenth. Now, 0.2 is the second point between 0 and 1.

(b) 1.9 represents a point between 1 and 2 on number line, such that the space between 1 and 2 is divided into 10 equal parts. Hence, each equal part will be equal to one-tenth. Now, 1.9 is the ninth point between 1 and 2.

(c) 1.1 represents a point between 1 and 2 on number line, such that the space between 1 and 2 is divided into 10 equal parts. Hence, each equal part will be equal to one-tenth. Now, 1.1 is the first point between 1 and 2.

(d) 2.5 represents a point between 2 and 3 on number line, such that the space between 2 and 3 is divided into 10 equal parts. Hence, each equal part will be equal to one-tenth. Now, 2.5 is the fifth point between 2 and 3.

P9. Write the decimal number represented by the points A, B, C, D on the given number line ?

Sol. Point A represents 0.8.

Point B represents 1.3.

Point C represents 2.2.

Point D represents 2.9.

P10. (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.

Sol. (a) The length of Ramesh’s notebook is 9 cm 5 mm.

Therefore, the length in cm is $\left(9+\frac{5}{10}\right)\mathrm{cm}=9.5\mathrm{cm}$

(b) The length of a gram plant is 65 mm.

Therefore, the length in cm is $\frac{65}{10}=6.5 \mathrm{cm}$

EXERCISE-8.2

P1. Complete the table with the help of these boxes and use decimals to write the number.

(a)

(b)

(c)

	Ones	Tenths	Hundredths	Number
(a)	–	–	–	–
(b)	–	–	–	–
(c)	–	–	–	–

Sol.

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

P2. Write the numbers given in the following place value table in decimal form.

	Hundreds 100	Tens 10	Ones 1	Tenths $\left(\frac{1}{10}\right)$	Hundredths $\left(\frac{1}{100}\right)$	Thousandths $\left(\frac{1}{1000}\right)$
(a)	0	0	3	2	5	0
(b)	1	0	2	6	3	0
(c)	0	3	0	0	2	5
(d)	2	1	1	9	0	2
(e)	0	1	2	2	4	1

Sol. (a) $3+\frac{21}{10}+\frac{5}{100}=3+0.2+0.05=3.25$

(b) $100+2+\frac{6}{10}+\frac{3}{100}=102+0.6+0.03=102.63$

(c) $30+\frac{2}{100}+\frac{5}{1000}=30+0.02+0.005=30.025$

(d) $200+10+1+\frac{9}{10}+\frac{2}{1000}=211+0.9+0.002=211.902$

(e) $10+2+\frac{2}{10}+\frac{4}{100}+\frac{1}{1000}=12+0.2+0.04+0.001=12.241$

P3. Write the following decimals in the place value table.

(a) 0.29

(b) 2.08

(c) 19.60

(d) 148.32

(e) 200.812

Sol. (a) $0.29=0.2+0.09=\frac{2}{10}+\frac{9}{100}$

(b) $2.08=2+0.08=2+\frac{8}{100}$

(c) $19.60=19+0.60=10+9+\frac{6}{10}$

(d) $148.32=148+0.3+0.02=100+40+8+\frac{3}{10}+\frac{2}{100}$

(e) $200.812=200+0.8+0.01+0.002=200+\frac{8}{10}+\frac{1}{100}+\frac{2}{1000}$

Row	Hundreds	Tens	Ones	Tenths	Hundredths	Thousandths
(a)	0	0	0	2	9	0
(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

P4. Write each of the following decimals.

(a) $20+9+\frac{4}{10}+\frac{1}{100}$

(b) $137+\frac{5}{100}$

(c) $\frac{7}{10}+\frac{6}{100}+\frac{4}{1000}$

(d) $23+\frac{2}{10}+\frac{6}{1000}$

(e) $700+20+5+\frac{9}{100}$

Sol. (a) $20+9+\frac{4}{10}+\frac{1}{100}$$=29+0.4+0.01=29.41$

(b) $137+\frac{5}{100}=137+0.05=137.05$

(c) $\frac{7}{10}+\frac{6}{100}+\frac{4}{1000}$$=0.7+0.06+0.004=0.764$

(d) $23+\frac{2}{10}+\frac{6}{1000}$$=23+0.2+0.006=23$

(e) $700+20+5+\frac{9}{100}$$=725+0.09=725.09$

P5. 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

Sol. (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

P6. 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

Sol. (a) 0.06 $\to$ 0 and 0.1

(b) 0.45 $\to$ 0.4 and 0.5

(c) 0.19 $\to$ 0.1 and 0.2

(d) 0.66 $\to$ 0.6 and 0.7

(e) 0.92 $\to$ 0.9 and 1.0

(f) 0.57 $\to$ 0.5 and 0.6

P7. 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

Sol. (a) $0.60=\frac{60}{100}=\frac{6}{10}=\frac{3}{5}$

(b) $0.05=\frac{5}{100}=\frac{1}{20}$

(c) $0.75=\frac{75}{100}=\frac{3}{4}$

(d) $0.18=\frac{18}{100}=\frac{9}{50}$

(e) $0.25=\frac{25}{100}=\frac{1}{4}$

(f) $0.125=\frac{125}{1000}=\frac{1}{8}$

(g) $0.066=\frac{66}{1000}=\frac{33}{500}$

EXERCISE-8.3 

P1. 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

Sol. (a) 0.3 or 0.4

The whole parts of these numbers are same. It can be seen that the tenth part of 0.4 is greater than that of 0.3.

Hence, 0.4 > 0.3

(b) 0.07 and 0.02

Here, both numbers have same parts up to the tenth place. However, the hundredth part of 0.07 is greater than that of 0.02.

Hence, 0.07 > 0.02

(c) 3 or 0.8

It can be seen that the whole part of 3 is greater than that of 0.8.

Hence, 3 > 0.8

(d) 0.5 or 0.05

The whole parts of these numbers are same. It can be seen that the tenth part of 0.5 is greater than that of 0.05.

Hence, 0.5 > 0.05

(e) 1.23 or 1.20

Here, both numbers have same parts up to the tenth place. However, the hundredth part of 1.23 is greater than that of 1.20.

Hence, 1.23 > 1.20

(f) 0.099 or 0.19

The whole parts of these numbers are same. It can be seen that the tenth part of 0.19 is greater than that of 0.099.

Hence, 0.099 < 0.19

(g) 1.5 or 1.50

Here, both numbers have the same parts up to the tenth place. Also, there is no digit at hundredth place of 1.5. This implies that this digit will be 0, which is same as the digit at the hundredth place of 1.50. Therefore, both these numbers are equal.

(h) 1.431 or 1.490

Here, both numbers have the same parts up to the tenth place. However, the hundredth part of 1.490 is greater than that of 1.431.

Hence, 1.431 < 1.490

(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. This implies that these digits are 0, which are the same as the digits at the hundredth and thousandth place of 3.300. Therefore, both these numbers are equal.

(j) 5.64 or 5.603

Here, both numbers have the same parts up to the tenth place. However, the hundredth part of 5.64 is greater than that of 5.603.

Hence, 5.640 > 5.603

EXERCISE-8.4

P1. Express as rupees using decimals.

(a) 5 paise       

(b) 75 paise

(c) 20 paise

(d) 50 rupees 90 paise

(e) 725 paise

Sol. It is known that there are 100 paise in 1 rupee.

(a) $5\text{ paise }=\frac{5}{100}\text{ rupees }=\mathrm{Rs} 0.05$

(b) $75\text{ paise }=\frac{75}{100}\text{ rupees }=\mathrm{Rs} 0.75$

(c) $20\text{ paise }=\frac{20}{100}\text{ rupees }=\mathrm{Rs} 0.20$

(d) $50\text{ rupees }90\text{ paise }=\left(50+\frac{90}{100}\right)\text{ rupee }=Rs50.90$

(e) $725\text{ paise }=\frac{725}{100}\text{ rupees }=\mathrm{Rs} 7.25$

P2. Express as metres using decimals.

(a) 15 cm

(b) 6 cm

(c) 2 m 45 cm

(d) 9 m 7 cm

(e) 419 cm

Sol. It is known that there are 100 cm in 1 metre.

(a) $15\mathrm{cm}=\frac{15}{100}\mathrm{m}=0.15\mathrm{m}$

(b) $6\mathrm{cm}=\frac{6}{100}\mathrm{m}=0.06\mathrm{m}$

(c) $2\mathrm{m} 45\mathrm{cm}=\left(2+\frac{45}{100}\right)\mathrm{m}=2.45\mathrm{m}$

(d) $9\mathrm{m} 7\mathrm{cm}=\left(9+\frac{7}{100}\right)\mathrm{m}=9.07\mathrm{m}$

(e) $419\mathrm{m}=\frac{419}{100}\mathrm{m}=4.19\mathrm{m}$

P3. Express as cm using decimals.

(a) 5 mm

(b) 60 mm

(c) 164 mm

(d) 9 cm 8 mm

(e) 93 mm

Sol. It is known that there are 10 mm in 1 cm.

(a) $5\mathrm{mm}=\frac{5}{10}\mathrm{cm}=0.5\mathrm{cm}$

(b) $60\mathrm{mm}=\frac{60}{10}\mathrm{cm}=6.0\mathrm{cm}$

(c) $164\mathrm{mm}=\frac{164}{10}\mathrm{cm}=16.4\mathrm{cm}$

(d) $9\mathrm{cm} 8\mathrm{mm}=\left(9+\frac{8}{10}\right)\mathrm{cm}=9.8\mathrm{cm}$

(e) $93\mathrm{mm}=\frac{93}{10}\mathrm{cm}=9.3\mathrm{cm}$

P4. Express as km using decimals.

(a) 8 m

(b) 88 m

(c) 8888 m

(d) 70 km 5 m

Sol. It is known that there are 1000 metres in 1 km.

(a) $8\mathrm{m}=\frac{8}{1000}\mathrm{km}=0.008\mathrm{km}$

(b) $88\mathrm{m}=\frac{88}{1000}\mathrm{km}=0.088\mathrm{km}$

(c) $8888\mathrm{m}=\frac{8888}{1000}\mathrm{km}=8.888\mathrm{km}$

(d) $70\mathrm{km} 5\mathrm{m}=\left(70+\frac{5}{1000}\right)\mathrm{km}=70.005\mathrm{km}$

P5. Express as kg using decimals.

(a) 2 g

(b) 100 g

(c) 3750 g

(d) 5 kg 8 g

(e) 26 kg 50 g

Sol. It is known that there are 1000 grams in 1 kg.

(a) $2\mathrm{g}=\frac{2}{1000}\mathrm{kg}=0.002\mathrm{kg}$

(b) $100\mathrm{g}=\frac{100}{1000}\mathrm{kg}=0.1\mathrm{kg}$

(c) $3750\mathrm{g}=\frac{3750}{1000}\mathrm{kg}=3.75\mathrm{kg}$

(d) $5\mathrm{kg} 8\mathrm{g}=\left(5+\frac{8}{1000}\right)\mathrm{kg}=5.008\mathrm{kg}$

(e) $26\mathrm{kg} 50\mathrm{g}=\left(26+\frac{50}{1000}\right)\mathrm{kg}=26.050\mathrm{kg}$

EXERCISE-8.5 

P1. 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

Sol.(a) 0.007 + 8.5 + 30.08

$\begin{array}{r}0.007\\ 8.500\\ +30.080\\ 38.587\end{array}$

(b) 15 + 0.632 + 13.8

$\begin{array}{r}15.000\\ 0.632\\ +13.800\\ 29.432\end{array}$

(c) 27.076 + 0.55 + 0.004

$\begin{array}{r}27.076\\ 0.550\\ +0.004\\ 27.630\end{array}$

(d) 25.65 + 9.005 + 3.7

$\begin{array}{r}25.650\\ 9.005\\ +3.700\\ 38.355\end{array}$

(e) 0.75 + 10.425 + 2

$\begin{array}{r}0.750\\ 10.425\\ +2.000\\ 13.175\end{array}$

(f) 280.69 + 25.2 + 38

$\begin{array}{r}280.69\\ 25.20\\ +38.00\\ 343.89\end{array}$

P2. Rashid spent Rs 35.75 for Maths book and Rs 32.60 for Science book. Find the total amount spent by Rashid.

Sol. Price of Maths book = Rs. 35.75

Price of Science book = Rs. 32.60

Total amount spent by Rashid is

$\begin{array}{r}35.75\\ +32.60\\ 68.35\end{array}$

Therefore, the amount spent by Rashid is Rs 68.35.

P3. Radhika’s mother gave her Rs 10.50 and her father gave her Rs 15.80, find the total amount given to Radhika by the parents.

Sol. Amount given by mother = Rs. 10.50

Amount given by mother = Rs. 15.80

Total amount given by parents is

$\begin{array}{r}10.50\\ +15.80\\ 26.30\end{array}$

Therefore, the amount given by her parents is Rs 26.30.

P4. Nasreen bought 3 m 20 cm cloth for her shirt and 2 m 5 cm cloth for her trouser. Find the total length of cloth bought by her.

Sol. Cloth for shirt = 3 m 20 cm = $\left(3+\frac{20}{100}\right)\mathrm{m}=3.20\mathrm{m}$

Cloth for trouser = 2 m 5 cm

$=\left(2+\frac{5}{100}\right)\mathrm{m}=2.05\mathrm{m}$

Total length of cloth is

$\begin{array}{r}3.20\\ +2.05\\ 5.25\end{array}$

Hence, the total length of cloth bought by her is 5.25 m.

P5. Naresh walked 2 km 35 m in the morning and 1 km 7 m in the evening. How much distance did he walk in all ?

Sol. Distance walked in the morning = 2 km 35 m

$=\left(2+\frac{35}{1000}\right)\mathrm{km}$ = 2.035 km

Distance walked in the evening = 1 km 7 m

$=\left(1+\frac{7}{1000}\right)\mathrm{km}$ = 1.007 km

Total distance walked by him is

$\begin{array}{r}2.035\\ +1.007\\ 3.042 km\end{array}$

P6. 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 ?

Sol. Distance travelled by bus = 15 km 268 m

$=\left(15+\frac{268}{1000}\right)\mathrm{km}$ = 15.268 km

Distance travelled by car = 7 km 7 m

$=\left(7+\frac{7}{1000}\right)\mathrm{km}$ = 7.007 km

Distance travelled on foot = 500 m

$=\frac{500}{100} \mathrm{km}$ = 0.500 km

Total distance of school from her residence is

$\begin{array}{r}15.268\\ 7.007\\ + 0.500\\ 22.775\end{array}$

P7. Ravi purchased 5 kg 400 g rice, 2 kg 20 g sugar and 10 kg 850 g flour. Find the total weight of his purchases.

Sol. Weight of rice = 5 kg 400 g = $\left(5+\frac{400}{1000}\right)\mathrm{kg}=5.400\mathrm{kg}$

Weight of sugar = 2 kg 20 g = $\left(2+\frac{20}{1000}\right)\mathrm{kg}=2.020\mathrm{kg}$

Weight of flour = 10 kg 850 g = $\left(10+\frac{850}{1000}\right)\mathrm{kg}=10.850\mathrm{kg}$

Total weight of his purchases is

$\begin{array}{r}5.400\\ 2.020\\ +10.850\\ 18.270\end{array}$

EXERCISE-8.6

P1. Subtract :

(a) Rs 18.25 from Rs 20.75

(b) 202.54 m from 250 m

(c) Rs 5.36 from Rs 8.40

(d) 2.051 km from 5.206 km

(e) 0.314 kg from 2.107 kg

Sol. (a) Rs 20.75 − Rs 18.25

$\begin{array}{r}20.75\\ – 18.25\\ 2.50\end{array}$

(b) 250 m − 202.54 m

$\begin{array}{r}250.00\\ – 202.54\\ 47.46\end{array}$

(c) Rs 8.40 − Rs 5.36

$\begin{array}{r}8.40\\ – 5.36\\ 3.04\end{array}$

(d) 5.206 km − 2.051 km

$\begin{array}{r}5.206\\ – 2.051\\ 3.155\end{array}$

(e) 2.107 kg − 0.314 kg

$\begin{array}{r}2.107\\ – 0.314\\ 1.793\end{array}$

P2. Find the value of :

(a) 9.756 − 6.28

(b) 21.05 − 15.27

(c) 18.5 − 6.79

(d) 11.6 − 9.847

Sol. (a) $\begin{array}{r}9.756\\ – 6.280\\ 3.476\end{array}$

(b) $\begin{array}{r}21.05\\ – 15.27\\ 5.78\end{array}$

(c) $\begin{array}{r}18.50\\ – 6.79\\ 11.71\end{array}$

(d) $\begin{array}{r}11.600\\ – 9.847\\ 1.753\end{array}$

P3. Raju bought a book for Rs 35.65. He gave Rs 50 to the shopkeeper. How much money did he get back from the shopkeeper ?

Sol. Money given to shopkeeper = Rs 50.00

Cost of book = Rs 35.65

Money that Raju will get back will be the difference of these two.

Hence, money that Raju will get back is

$\begin{array}{r}50.00\\ – 35.65\\ 14.35\end{array}$

Therefore, he will get back Rs 14.35.

P4. Rani had Rs 18.50. She bought one ice-cream for Rs 11.75. How much money does she have now ?

Sol. Money with Rani = Rs 18.50

Money spent for an ice cream = Rs 11.75

The money left with Rani will be the difference of these two.

Hence, the money left is

$\begin{array}{r}18.50\\ – 11.75\\ 6.75\end{array}$

P5. 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 ?

Sol. Length of Cloth = 20 m 5 cm = 20.05 m

Length of cloth cut so as to make a curtain = 4 m 50 m = 4.50 m

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

Hence, the length of the cloth left with her is

$\begin{array}{r}20.05\\ – 4.50\\ 15.55\end{array}$

Therefore, 15.55 m cloth will be remaining.

P6. 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 ?

Sol. Total distance travelled by Namita = 20 km 50m = 20.050 km

Distance travelled by bus = 10 km 200 m = 10.200 km

Distance travelled by auto = Total distance travelled − Distance travelled by bus

Hence, the distance travelled by auto is

$\begin{array}{r}20.050\\ – 10.200\\ 9.850\end{array}$

P7. 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 ?

Sol. Total weight of vegetables bought = 10.000 kg

Weight of onions = 3 kg 500 g = 3.500 kg

Weight of tomatoes = 2 kg 75 g = 2.075 kg

Weight of potatoes = Total weight of vegetables bought − (Weight of onions + Weight of tomatoes)

= 10.000 − (3.500 + 2.075)

$\begin{array}{r}3.500\\ + 2.075\\ 5.575\end{array} \begin{array}{r}10.000\\ – 5.575\\ 4.425\end{array}$

Hence, the weight of the potatoes was 4.425 kg.