# Ncert Solutions For Class 6 Maths Ex 8.1

## Ncert Solutions For Class 6 Maths Chapter 8 Ex 8.1

1. Using the given diagram write the data in form of numbers in the table given below:

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

 Hundreds (100) Tens (10) Ones (1) Tenths ($$\frac{1}{10}$$) Number 0 3 1 2 31.2 1 1 0 4 110.4

2. Write the given decimals in the place value table.

(a) 20.4

(b) 0.4

(c) 18.5

(d) 302.1

 Hundreds (100) Tens (10) Ones (1) Tenths ($$\frac{1}{10}$$) Number (a) 0 2 0 4 20.4 (b) 0 0 0 4 0.4 (c) 0 1 8 5 18.5 (d) 3 0 2 1 302.1

3. Write the given values in the form of decimals:

(a) five-tenths

(b) Four tens and seven-tenths

(c) Sixteen point three

(d) Six hundred and eight-ones

(e) Nine hundred point one

(a) five-tenths = 5 tenths

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

= 0.5

(b) Four tens and seven-tenths = 4 tens and 7 tenths

= $$(4 \times 10) + \frac{7}{10}$$

= 40 + 0.7

= 40.7

(c) Sixteen point three = 16.3

(d) Six hundred and eight-ones = 6 hundreds + 8 ones

= 600 + 8

= 608

(e) Nine hundred point one = 900.1

4. Write the fractions given below in the form of 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) $$4\frac{1}{2}$$

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

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

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

= 265.1

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

= 4 + 0.2

= 4.2

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

= $$\frac{15}{10}$$

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

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

= 1 + 0.5

= 1.5

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

= $$\frac{4}{10}$$

= 0.4

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

= $$\frac{24}{10}$$

= $$\frac{20 + 4}{10}$$

= $$\frac{20}{10} + \frac{4}{10}$$

= 2 + 0.4

= 2.4

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

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

= 3 + $$\frac{6}{10}$$

= 3 + 0.6

= 3.6

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

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

= 4 + $$\frac{5}{10}$$

= 4 + 0.5

= 4.5

5. Write the given decimals in the form of fraction. Reduce them to the lowest terms:

(a) 0.6

(b) 2.5

(c) 1.0

(d) 3.8

(e) 13.7

(f) 21.2

(g) 6.4

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

= $$\frac{3}{5}$$

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

= $$\frac{5}{2}$$

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

= 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 = $$\frac{64}{10}$$

= $$\frac{32}{5}$$

6. Express the following as cm using decimals:

(a) 30 mm

(b) 2 mm

(c) 116 mm

(d) 4cm 2mm

(e) 162 mm

(f) 83 mm

(a) Since, 10mm = 1cm

Therefore, 1mm = $$\frac{1}{10}$$ cm

Therefore, 30mm = $$\frac{1}{10}$$ x 30 = 3.0cm

(b) Since, 10 mm =1cm

Therefore, 1mm = $$\frac{1}{10}$$ cm

Therefore, 2mm = $$\frac{1}{10}$$ x 2 = 0.2 cm

(c) Therefore, 116 mm = $$\frac{1}{10}$$ x 116 = 11.6 cm

(d) Since, 10mm = 1cm

$$4cm+\frac{2}{10}cm$$

4 + 0.2 = 4.2 cm

(e) 1mm = $$\frac{1}{10}$$ cm

162 mm = $$\frac{1}{10}\times 162 =16.2cm$$

(f) Since, 10 mm = 1cm

Therefore, 1mm = $$\frac{1}{10}$$ cm

Therefore, 83 mm = $$\frac{1}{10}\times 83 =8.3cm$$

7. Between which two whole numbers do the given numbers lie? Then find the closest whole number to the given numbers.

(a) 5.1

(b) 0.8

(c) 6.4

(d) 2.6

(e) 4.9

(f) 9.1

(a) From 5 to 6, 5.1 is nearest to 5.

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

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

(d) From 2 to 3, 2.6 is nearest to 3.

(e) From 4 to 5, 4.9 is nearest to 5.

(f) From 9 to 10, 9.1 is nearest to 9

8. Show the given numbers on number line:

(a) 0.2

(b)1.9

(c) 1.1

(d) 2.5

9. From the given number line, write down decimal number represented by the points P, Q, R and S.

P = 0 + $$\frac{8}{10}$$

= 0.8

Q = 1 + $$\frac{3}{10}$$

= 1 + 0.3

= 1.3

R = 2 + $$\frac{2}{10}$$

= 2 + 0.2

= 2.2

S = 2 + $$\frac{9}{10}$$

= 2 + 0.9

= 2.9

10. (a) The length of Mahesh’s textbook is 10 cm and 6 mm. What will be the length of textbook in cm?

(b) The length of a rose plant is 75 mm. Write the length in cm form.

(a) 10 cm 6 mm = 9 cm + 6 mm

= 9 + $$\frac{6}{10}$$

= 9.6 cm

(b) 75 mm = $$\frac{75}{10}$$ cm

= 7.5 cm