Class – 6 CH-8 DECIMALS

MATHS NCERT SOLUTIONS

Exercise 8.1

Question 1:

Write the following as numbers in the given table:

Tens Ones Tenths Hundreds Tens Tenths

Hundreds (100) Tens

(10) Ones

(1) Tenths

 1 

10  

Solution 1:

Hundreds (100) Tens

(10) Ones

(1) Tenths

 1 

10  

0 3 2 31.2

1 1 4 110.4

Question 2:

Write the following decimals in the place value table:

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

Solution 2:

(a)

Hundreds Tens Ones Tenths

0 1 9 4

(b)

Hundreds Tens Ones Tenths

0 0 0 3

(c)

Hundreds Tens Ones Tenths

0 1 0 6

(d)

Hundreds Tens Ones Tenths

0 0 5 9

Question 3:

Write each of the following as decimals:

(a) seven-tenths

(b) Two tens and nine-tenths

(d) One hundred and two-ones

(e) Six hundred point eight Solution 3:

(a) seven-tenths = 7 tenths = = 0.7

(b) 2 tens and 9-tenths = 2 x 10 + = 20 + 0.9 =20.9

(d) One hundred and 2-ones = 100 + 2 x 1 = 100 + 2 = 102

(e) Six hundred point eight = 600.8

Question 4:

Write each of the following as decimals:

(a) (b) 3

(d) 70 (e)

(g) (h)

(j) 3 (k) 4

Solution 4:

(a) = 0.5

(b) 3 = 3 + 0.7 = 3.7

(d) 70 = 70 + 0.8 = 70.8

88 80 8 80 8 8

(e)     8 = 8 + 0.8 = 8.8

10 10 10 10 10

2 2

(f) 4  4 = 4 + 0.2 = 4.2

10 10

3 3 5 15 10 5 10  5

(g)      = 1 + 0.5 = 1.5

2 2 5 10 10 10 10

2 2 2 4

(h)   = 0.4

5 5 2 10

12 12 2 24 20 4 20

(i)      = 2 + 0.4 = 2.4

5 5 2 10 10 10

3 3 3 2 6

(j) 3    3 3  3 = 3 + 0.6 = 3.6

5 5 5 2 10

1 1 1 5 5

(k) 4    4 4  4 = 4 + 0.5 = 4.5

2 2 2 5 10

(f) 4

(i)

Write the following decimals as fraction. Reduce the fractions to lowest terms:

(a) 0.6 (b) 2.5 (c) 1.0 (d) 3.8

(e) 13.7 (f) 21.2 (g) 6.4

Solution 5:

625

(a) 0.6 =  (b) 2.5 = 

1010

10 38 19

10 10 5

212

(e) 13.7 = (f) 21.2  =

64 (g) 6.4 = 

Question 6:

Express the following as cm using decimals:

(a) 2 mm (b) 30 mm

(e) 162 mm (f) 83 mm (c) 116 mm (d) 4 cm 2 mm

Solution 6:

(a) 10 mm = 1 cm (b) 10 mm = 1 cm

 1 mm = cm  1 mm = cm

 2 mm = x 2 = 0.2 cm  30 mm = x 30 = 3.0 cm

 1 mm = cm 4 + 0.2 = 4.2 cm

 116 mm = x 116 = 11.6 cm

(e) 10 mm = 1 cm (f) 10 mm = 1 cm

 1 mm = cm  1 mm = cm

 162 mm = x 162 = 16.2 cm  83 mm = x 83 = 8.3 cm

Between which two whole numbers on the number line are the given 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

Solution 7:

(a) From 0 to 1, 0.8 is nearest to 1.

(b) From 5 to 6, 5.1 is nearest to 5. (c) From 2 to 3, 2.6 is nearest to 3.

(d) From 6 to 7, 6.4 is nearest to 6.

(e) From 9 to 10, 9.1 is nearest to 9. (f) From 4 to 5, 4.9 is nearest to 5.

Question 8:

Show the following numbers on the number line:

(b) 1.9 (c) 1.1 (d) 2.5

Question 9:

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

Solution 9:

A = 0 + = 0.8 B = 1 + = 1.3

C = 2 + = 2.2 D = 2 + = 2.9

10:

(a) The length of Ramesh’s notebook is 9 cm and 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.

Solution 10:

(a) 9 cm 5 mm = 9 cm + 5 mm = 9 + = 9.5 cm

(b) 65 mm = cm = 6.5 cm

Exercise 8.2

Question 1:

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

Ones Tenths Hundredths Numbers

a

b

c

Solution 1:

Ones Tenths Hundredths Numbers

a 0 2 6 0.26

b 1 3 8 1.38

c 1 2 8 1.28

Question 2:

Write the numbers given in the following place value table in decimal form:

Hundreds 100 Tens 10 Ones 1 Tenths

Hundredths

Thousandths

a

b

c

d

e 0

0 0

1 3

2 2

2 5

4 0

Solution 2:

(a) 0 x 100 + 0 x 10 + 3 x 1 + 2 x + 5 x + 0 x

= 0 + 0 + 3 + 0.2 + 0.05 + 0 = 3.25

(b) 1 x 100 + 0 x 10 + 2 x 1 + 6 x + 3 x + 0 x

= 1 + 0 + 2 + 0.6 + 0.03 + 0 = 102.63

= 0 + 30 + 0 + 0 + 0.02 + 0.005 = 30.025

(d) 2 x 100 + 1 x 10 + 1 x 1 + 9 x + 0 x + 2 x

= 200 + 10 + 1 + 0.9 + 0 + 0.002 = 211.902

(e) 0 x 100 + 1 x 10 + 2 x 1 + 2 x + 4 x + 1 x

=0 + 10 + 2 + 0.2 + 0.04 + 0.001 = 12.241

Question 3:

Write the following decimals in the place value table:

(a) 0.29 (b) 2.08 (c) 19.60 (d) 148.32 (e) 200.812 Solution 3:

Numbers Hundreds Tens Ones Tenths Hundredths Thousandths

100 10 1

(a) 0.29 0 0 0 2 9 0

(b) 2.08 0 0 2 0 8 0

(d) 148.32 1 4 8 3 2 0

(e) 200.812 2 0 0 8 1 2

Question 4:

Write each of the following as decimals:

4 17 6 4

(a) 20 9   (b) 137 (c)  

10 10010 100 1000

2 6

(d) 23  (e) 700 20 5  

10 1000

Solution 4:

(a) 20 + 9 + 0.4 + 0.01 = 29.41

(b) 137 + 0.05 = 137.05

(d) 23 + 0.2 + 0.006 = 23.206

(e) 700 + 20 + 5 + 0.09 = 725.09

Question 5:

Write each of the following decimals in words:

(a) 0.03 (b) 1.20

(e) 0.032 (f) 5.008

Solution 5:

(a) Zero point zero three

(b) One point two zero

(d) Ten point zero seven

(e) Zero point zero three two

(f) Five point zero zero eight

Question 6:

Between which two numbers in tenths place on the number line does each of the given number lie?

(a) 0.06

(b) 0.45

(d) 0.66

(e) 0.92

(f) 0.57

Solution 6:

All the numbers lie between 0 and 1.

(a) 0.06 is nearer to 0.1.

(b) 0.45 is nearer to 0.5.

(d) 0.66 is nearer to 0.7.

(e) 0.92 is nearer to 0.9.

(f) 0.57 is nearer to 0.6.

Question 7:

Write as fractions in lowest terms:

(a) 0.60

(b) 0.05

(d) 0.18

(e) 0.25

(f) 0.125

(g) 0.066

Solution 7:

(a) 0.60 =

(b) 0.05 =

(d) 0.18 =

(e) 0.25 =

(f) 0.125 = 

(f) 0.066 = 

Exercise 8.3

Question 1:

Which is greater:

(a) 0.3 or 0.4

(b) 0.07 or 0.02

(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 Solution 1:

Before comparing, we write both terms in like decimals:

(a) 0.3 < 0.4

(b) 0.07 > 0.02

(d) 0.50 or 0.05  0.50 > 0.05

(e) 1.23 or 1.20  1.23 > 1.20

(f) 0.099 or 0.190  0.099 < 0.190

(g) 1.50 or 1.50  1.50 = 1.50

(h) 1.431 < 1.490

(i) 3.300 or 3.300  3.300 = 3.300

(j) 5.640 or 5.603  5.640 > 5.603

Question 2:

Make five more examples and find the greater:

(a) 1.8 or 1.82

(b) 1.0009 or 1.09

(d) 5.100 or 5.0100 (e) 04.213 or 0421.3 Solution 2:

Before comparing, we write both terms in like decimals

(a) 1.80 or 1.82  1.82 is greater than 1.8

(b) 1.0009 or 1.0900  1.09 is greater than 1.0009

(d) 5.1000 or 5.0100  5.100 is greater than 5.0100

(e) 04.213 or 0421.300

 0421.3 is greater than 04.213

Exercise 8.4

Question 1:

Express as rupees using decimals:

(a) 5 paise (b) 75 paise

(e) 725 paise (d) 50 rupees 90 paise

Solution 1:

(a) 1 paisa = ₹ (b) 1 paisa = ₹

 5 paise = x 5 = ₹ 0.05  75 paise = x 5 = ₹ 0.75

 20 paise = x 5 = ₹ 0.05 ₹ 50+90paise=50+ x90 =₹50.90

(e) 1 paisa = ₹

 725 paise = x 725  = ₹ 7.25

Question 2:

Express as meters using decimals:

(a) 15 cm (b) 6 cm

(e) 419 cm Solution 2:

(a) 1 cm = m (b) 1 cm = m

 15 cm = x 15 = 0.15 m  6 cm = x 6 = 0.06 m

 2 m 45 cm = 2 + x 45 = 2.45 m  9 m 7 cm = 9 + x 7 = 9.07 m

(e) 1 cm = m

 419 cm = x 419 =  4.19 m

Question 3:

Express as cm using decimals:

(a) 5 mm (b) 60 mm

(e) 93 mm (d) 9 cm 8 mm

Solution 3:

(a) 1 mm = cm (b) 1 mm = cm

 5 mm = x 5 = 0.5 cm  60 mm = x 60 = 6 cm

 164 mm = x 164 = 16.4 cm  9cm 8mm = 9+ x 8 =9+0.8=9.8 cm

(e) 1 mm = cm

 93 mm = x 93 = 9.3 cm

Question 4:

Express as km using decimals:

(a) 8 m (b) 88 m

Solution 4:

(a) 1 m = km (b) 1 m = km

 8 m = x 8 = 0.008 km  88 m = x 88 = 0.088 km

 8888 m = x 8888 = 8.888 km  70 km 5m =70+ x 5 =70.005 km

Question 5:

Express as kg using decimals:

(a) 2 g (b) 100 g

(e) 26 kg 50 g

Solution 5:

(a) 1 g = kg (b) 1 g = kg

 2 g = x 2 = 0.002 kg  100 g = x 100 = 0.1 kg

 3750 g = x 3750 = 3.750 kg  5 kg 8 g = 5 + x 8 = 5.008 kg

(e) 1 g = kg

 26 kg 50 g = 26 + x 50 = 26.050 kg

Exercise 8.5

Question 1:

Find the sum in each of the following:

(a) 0.007 + 8.5 + 30.08 (b) 15 + 0.632 + 13.8

(e) 0.75 + 10.425 + 2 (f) 280.69 + 25.2 + 38 Solution 1:

(a) H T O . Tenth Hund. Thou. 0 . 0 0 7

8 . 5

+ 3 0 . 0 8

3 8 . 5 8 7

H T O . Tenth Hund. Thou.

0 1 5 . 0 0 0

. 6 3 2

1 3 . 8

2 9 . 4 3 2

H T O . Te nth Hund. Thou.

2 7 . 0 7 6

. 5 5

. 0 0 4

2 7 . 6 3 0

H T O . Te nth Hu nd. Th ou.

2 5 . 6 5

9 . 0 0 5

3 . 7

3 8 . 3 5 5

. Tenth Hund. Thou.

. 7 5

1 0 . 4 2 5

2 .

1 3 . 1 7 5

= 38.587

(b)

= 29.432

(c)

= 27.630

(d)

= 38.355

(e)

= 13.175

(f) H T O . Tenth Hund. Thou.

2 8 0 . 6 9 2 5 . 2 + 3 8 .

4 3 . 8 = 343.89

Question 2:

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

Solution 2:

Money spent for Maths book = ₹35.75

Money spent for Science book = ₹32.60 Total money spent = ₹35.75 + ₹32.60 = ₹68.35 Therefore, total money spent by Rashid is ₹68.35.

Question 3:

Radhika’s mother have her ₹10.50 and her father gave her ₹15.80. Find the total amount given to Radhika by the parents.

Solution 3:

Money given by mother = ₹10.50

Money given by father = ₹15.80

Total money received by Radha = ₹10.50 + ₹15.80 = ₹26.30 Therefore, the total money received by Radha is ₹26.30.

Question 4:

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

Solution 4:

Cloth bought for shirt = 3 m 20 cm = 3.20 m

Cloth bought for trouser = 2 m 5 cm = 2.05 m

Total length of cloth bought by Nasreen = 3.20 + 2.05 = 5.25 m

Therefore, the total length of cloth bought by Nasreen is 5.25 m

Question 5:

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

Solution 5:

Distance travelled in morning = 2 km 35 m = 2.035 km

Distance travelled in evening = 1 km 7 m = 1.007 km

Total distance travelled = 2.035 + 1.007 = 3.042 km Therefore, the total distance travelled by Naresh is 3.042 km.

Question 6:

Sunita travelled 15 km 268 m by bus, 7 km 7 m by car and 500 m by foot in order to reach her school. How far is her school from her residence?

Solution 6:

Distance travelled by bus = 15 km 268 m = 15.268 km

Distance travelled by car = 7 km 7 m = 7.007 km

Distance travelled on foot = 500 m = 0.500 km Total distance travelled = 15.268 + 7.007 + 0.500 = 22.775 km Therefore, total distance travelled by Sunita is 22.775 km.

Question 7:

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

Solution 7:

Weight of Rice = 5 kg 400 g = 5.400 kg

Weight of Sugar = 2 kg 20 g = 2.020 kg

Weight of Flour = 10 kg 850 g = 10.850 kg

Total weight = 5.400 + 2.020 + 10.850 = 18.270 kg

Therefore, the total weight of Ravi’s purchase = 18.270 kg.

Question 1:

Subtract:

(a) ₹18.25 from ₹20.75

(a) 2 0 . 7 5

– 1 8 . 2 5

0 2 . 5 0

= ₹2.50

– 5 . 3 6

3 . 0 4

= ₹3.04

(e) 2 . 1 0 7

– 0 . 3 1 4

1 . 7 9 3

= 1.793 kg

Question 2:

Find the value of:

(a) 9.756 – 6.28 (c) 18.5 – 6.79 Solution 2:

(a) 9 . 7 5 6

– 6 . 2 8

3 . 4 7 6

= 3.476

Exercise 8.6

(b) 202.54 m from 250

(d) 2.051 km from 5.206 km

(b) 2 5 0 . 0 0

– 2 0 2 . 5 4

4 7 . 4 6

= 47.46 m

(d) 5 . 2 0 6

– 2 . 0 5 1 3 . 1 5 5

= 3.155 km

(b) 21.05 – 15.27

(d) 11.6 – 9.847

(b) 2 1 . 0 5

– 1 5 . 2 7

0 5 . 7 8

= 5.78

– 6 . 7 9 – 9 . 8 4 7

1 1 . 7 1 1 . 7 5 3

= 11.71 = 1.753

Question 3:

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

Solution 3:

Total amount given to shopkeeper = ₹50

Cost of book = ₹35.65

Amount left = ₹50.00 – ₹35.65

= ₹14.35

Therefore, Raju got back ₹14.35 from the shopkeeper.

Question 4:

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

Solution 4:

Total money = ₹18.50

Cost of Ice-cream = ₹11.75

Amount left = ₹18.50 – ₹11.75

= ₹6.75

Therefore, Rani has ₹6.75 now.

Question 5:

Tina had 20 m 5 cm long cloth. She cuts 4 m 50 cm length of cloth from this for making a curtain. How much cloth is left with her?

Solution 5:

Total length of cloth = 20 m 5 cm = 20.05 m

Length of cloth used = 4 m 50 cm = 4.50 m Remaining cloth = 20.05 m – 4.50 m = 15.55 m Therefore, 15.55 m of cloth is left with Tina.

Question 6:

Namita travels 20 km 50 m every day. Out of this she travels 10 km 200 m by bus and the rest by auto. How much distance does she travel by auto?

Solution 6:

Total distance travel = 20 km 50 m = 20.050 km

Distance travelled by bus = 10 km 200 m = 10.200 km Distance travelled by auto = 20.050 – 10.200 = 9.850 km Therefore, 9.850 km distance travels by auto.

Question 7:

Aakash bought vegetables weighing 10 kg. Out of this 3 kg 500 g in onions, 2 kg 75 g is tomatoes and the rest is potatoes. What is the weight of the potatoes?

Solution 7:

Weight of onions = 3 kg 500 g = 3.500 kg

Weight of tomatoes = 2 kg 75 g = 2.075 kg

Total weight of onions and tomatoes = 3.500 + 2.075 = 5.575 kg Therefore, weight of potatoes = 10.000 – 5.575 = 4.425 kg Thus, the weight of potatoes is 4.425 kg.

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Sunday, December 17, 2023