NCERT Solutions Class 6 Maths Chapter 8 Decimals Ex 8.1

NCERT Solutions Class 6 Maths
Chapter – 8 (Decimals)

The NCERT Solutions in English Language for Class 6 Mathematics Chapter – 8 Decimals Exercise 8.1 has been provided here to help the students in solving the questions from this exercise.

Chapter 8: Decimals

Exercise – 8.1

1. Write the following numbers in the given table.
$NCERT Class 6 Math Solution$

Hundreds (100)	Tens (10)	Ones (1)	Tenths $\left ( \frac{1}{2} \right )$

Solutions:

	Hundreds (100)	Tens (10)	Ones (1)	Tenths $\left ( \frac{1}{2} \right )$
a	0	3	1	2
b	1	1	0	4

2. Write the following decimals in the place value table.
(a) 19.4
(b) 0.3
(c) 10.6
(d) 205.9

Solutions:

		Hundreds	Tens	Ones	Tenths
(a)	19.4	0	1	9	4
(b)	0.3	0	0	0	3
(c)	10.6	0	1	0	6
(d)	205.9	2	0	5	9

3. 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

Solutions:

(a) The decimal form of Seven-tenths = $\frac{7}{10}$ = 0.7

(b) The decimal form of two tens and nine tenths = 20 + $\frac{9}{10}$ = 20 + 0.9 = 20.9

(c) The decimal form of fourteen point six = 14.6
(d) The decimal form of one hundred and two ones = 100 + 2 = 102.0
(e) The decimal form of six hundred point eight = 600.8

4. Write each of the following as decimals:
(a) $\frac{5}{10}$

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

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

Solutions:

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

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

(d) 70 + $\frac{8}{10}$ = 70 + 0.8 = 70.8

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

(f) $4\frac{2}{10}$ = $\frac{10 \times 4 + 2}{10} = \frac{40 + 2}{10} = \frac{40}{10} + \frac{2}{10}$

= $4 + 0.2= 4.2$

(g) $\frac{3}{2}$ = $\frac{2+1}{2} = \frac{2}{2} + \frac{1}{2} = 1 + \frac{1\times 5}{2 \times 5} = 1 + \frac{5}{10}$

= 1 + 0.5 = 1.5

(h) $\frac{2}{5}$ = $\frac{2\times 2}{5\times 2} = \frac{4}{10} = 0.4$

(i) $\frac{12}{5}$ = $\frac{10 + 2}{5} = \frac{10}{5} + \frac{2}{5} = 2 + \frac{2\times 2}{5\times 2} = 2 + \frac{4}{10}$

= 2 + 0.4 = 2.4

(j) $3\frac{3}{5}$ = $\frac{3\times 5 + 3}{5} = \frac{15 + 3}{3} = \frac{15}{3} + \frac{3}{5}$

= $3 + \frac{3}{5} = 3 + \frac{3\times 2}{5\times 2} = 3+ \frac{6}{10}$

= 3 + 0.6 = 3.6

(k) $4\frac{1}{2}$ = $\frac{4\times 2 + 1}{2} = \frac{8+1}{2} = \frac{8}{2} + \frac{1}{2}$

= $4 + \frac{1}{2} = 4 + \frac{1\times 5}{2\times 5} = 4+\frac{5}{10}$

= 4 + 0.5 = 4.5

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

Solutions:

(a) 0.6 = $\frac{6}{10}$ = $\frac{6\div2}{10\div2}$

$\because$ HCF (6,10) = 2

= $\frac{3}{5}$

(b) 2.5 = $\frac{25}{10}$ = $\frac{25 \div5}{10\div5}$

$\because$ HCF (25, 10) = 5

= $\frac{5}{2}$

(c) 1.0 = $\frac{10}{10} = 1$

(d) 3.8 = $\frac{38}{10} = \frac{38\div2}{10\div2}$

$\because$ HCF (38, 10) = 2

= $\frac{19}{5}$

(e) 13. 7 = $\frac{137}{10}$

(f) 21.2 = $\frac{212}{10} = \frac{212 \div 2}{10 \div2}$

$\because$ HCF (212, 10) = 2

= $\frac{106}{5}$

(g) 6.4 = $\frac{64}{10} = \frac{64\div2}{10\div2}$

$\because$ HCF (64, 10) = 2

= $\frac{32}{5}$

6. 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

Solutions:

We know that
1 cm = 10 mm
1 mm = $\frac{1}{10}$ cm

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

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

(c) 116 mm = $\frac{116}{10} cm = \frac{110 + 6}{10}cm = 11 cm+ \frac{6}{10}cm$

= 11 cm + 0.6 cm = 11.6 cm

(d) 4 cm 2 mm = 4cm + $\frac{2}{10}$ cm

= 4 cm + 0.2 cm = 4.2 cm

(e) 162 mm = $\frac{162}{10} cm = \frac{160 + 2}{10}cm = \frac{160}{10}cm + \frac{2}{10}cm$

= 16cm + 0.2 cm = 16.2 cm

(f) 83 mm = $\frac{83}{10} cm = \frac{80+3}{10}cm = \frac{80}{10}cm + \frac{3}{10}cm$

= 8 cm + 0.3 cm = 8.3 cm

7. Between which two whole numbers on the number line are the given numbers lie? Which of these whole numbers is nearer the number?
$NCERT Class 6 Math Solution$
(a) 0.8
(b) 5.1
(c) 2.6
(d) 6.4
(e) 9.1
(f) 4.9

Solutions:

(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.

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

Solutions:

(a) 0.2 lies between the points 0 and 1 on the number line. The space between 0 and 1 is divided into 10 equal parts. Therefore each equal part will be equal to one-tenth. 0.2 is the second point between 0 and 1.
$NCERT Class 6 Math Solution$

(b) 1.9 lies between the points 1 and 2 on the number line. The space between 1 and 2 is divided into 10 equal parts. Therefore each equal part will be equal to one-tenth. 1.9 is the ninth point between 1 and 2.
$NCERT Class 6 Math Solution$

(c) 1.1 lies between the points 1 and 2 on the number line such that the space between 1 and 2 is divided into 10 equal parts. Therefore each equal part will be equal to one-tenth. 1.1 is the first point between 1 and 2.
$NCERT Class 6 Math Solution$

(d) 2.5 lies between the points 2 and 3 on the number line such that the space between 2 and 3 is divided into 10 equal parts. Therefore each equal part will be equal to one-tenth. 2.5 is the fifth point between 2 and 3.
$NCERT Class 6 Math Solution$

9. Write the decimal number represented by the points A, B, C, and D on the given number line.
$NCERT Class 6 Math Solution$

Solutions:

(a) Point A represents 0.8 cm on the given number line.
(b) Point B represents 1.3 cm on the given number line.
(c) Point C represents 2.2 cm on the given number line.
(d) Point D represents 2.9 cm on the given number line.

10. (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.

Solutions:

(a) The length of Ramesh notebook is 9 cm 5 mm
The length in cm is = 9 cm + 5 mm
= $9 + \frac{5}{10}$ cm $\because 1 mm = \frac{1}{10} cm$

= 9 + 0.5 cm
= 9.5 cm

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

Hence, the length in cm = $\frac{65}{10}$ $\because 1 mm = \frac{1}{10} cm$

= $\frac{60 + 5}{10} = \frac{60}{10} + \frac{5}{10}$

= 6 + 0.5 cm
= 6.5 cm

NCERT Class 6^thSolution

NCERT Solutions Class 6 English

NCERT Solutions Class 6 Hindi

NCERT Solutions Class 6 Maths

NCERT Solutions Class 6 Sanskrit

NCERT Solutions Class 6 Science

NCERT Solutions Class 6 Social Science

NCERT Solutions Class 6 Maths Chapter 8 Decimals Ex 8.1