RBSE Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.1

Rajasthan Board RBSE Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.1 Textbook Exercise Questions and Answers.

RBSE Class 6 Maths Solutions Chapter 8 Decimals Ex 8.1

Question 1.
Write the following as numbers in the given table:

Answer:

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

 

Question 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
Answer:
(a) Seven-tenths = 7 × \(\frac{1}{10}=\frac{7}{10}\) = 0.7

(b) Two-tens and nine-tenths
= 2 tens + 0 ones + 9 tenths
= 2 × 10 + 0 × 1 + 9 × \(\frac{1}{10}\)
= 20 + 0 + \(\frac{9}{10}\) = 20.9

(c) Fourteen point six = 14.6

(d) One hundred and two ones
= 1 hundred + 0 tens + 2 ones
= 1 × 100 + 0 × 10 + 2 × 1 
= 100 + 0 + 2 = 102

(e) Six hundred point eight = 600.8 

Question 4.
Write each of the following as decimals:
(a) \(\frac{5}{10} \)
Answer:
\(\frac{5}{10} \) = 0.5

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

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

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

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

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

(g) \(\frac{3}{2}\)
Answer:
\(\frac{3}{2}=\frac{3 \times 5}{2 \times 5}=\frac{15}{10}=\frac{10+5}{10}=\frac{10}{10}+\frac{5}{10}\)
= 1 + 0.5 = 1.5

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

(i) \(\frac{12}{5}\)
Answer:
\(\frac{12}{5}=\frac{12 \times 2}{5 \times 2}=\frac{24}{10}=\frac{20+4}{10}\)
= \(\frac{20}{10}+\frac{4}{10}\) = 2 + 0.4 = 2.4

(j) 3\(\frac{3}{5}\)
Answer:
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\(\frac{1}{2}\)
Answer:
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

Question 5.
Write the following decimals as fractions. Reduce the fractions to lowest form:
(a) 0.6
Answer:
0.6 = \(\frac{6}{10}=\frac{6 \div 2}{10 \div 2}=\frac{3}{5}\)
{∵ HCF of 6, 10 = 2}

(b) 2.5
Answer:
2.5 = \(\frac{25}{10}=\frac{25 \div 5}{10 \div 5}=\frac{5}{2}\)
{∵ H.C.F of25, 10 = 5}

(c) 1.0
Answer:
1.0 = \(\frac{10}{10}=\frac{10 \div 10}{10 \div 10}=\frac{1}{1}\) = 1
{∵ HCF of 10, 10 = 10}

(d) 3.8
Answer:
3.8 = \(\frac{38}{10}=\frac{38 \div 2}{10 \div 2}=\frac{19}{5}\)
{∵HCF of 38, 10 = 2}

(e) 13.7
Answer:
13.7 = \(\frac{137}{10}\)

(f) 21.2
Answer:
21.2 = \(\frac{212}{10}=\frac{212 \div 2}{10 \div 2}=\frac{106}{5}\)
{∵ HCF of 212, 10 = 2}

(g) 6.4
Answer:
6.4 = \(\frac{64}{10}=\frac{64 \div 2}{10 \div 2}=\frac{32}{5}\)
{∵ HCF of 64, 10 = 2}

Question 6.
Express the following as cm using decimals:
(a) 2 mm
Answer:

(b) 30 mm
Answer:

(c) 116 mm
Answer:

(d) 4 cm 2 mm
Answer:

= 4 cm + 0.2 cm = 4.2 cm

(e) 162 mm
Answer:
162 mm = 160 mm + 2 mm

(f) 83 mm
Answer:

Question 7.
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
Answer:
(a) 0.8, the number 0.8 lies between 0 and 1. The whole number 1 is nearer to 0.8.
(b) 5.1, the number 5.1 lies between 5 to 6. The whole number 5 is nearer to 5.1.
(c) 2.6, the number 2.6 lies between 2 and 3. The whole number 3 is nearer to 2.6.
(d) 6.4, the number 6.4 lies between 6 and 7. The whole number 6 is nearer to 6.4.
(e) 9.1, the number 9.1 lies between 9 and 10. The whole number 9 is nearer to 9.1.
(f) 4.9, the number 4.9 lies between 4 and 5. The whole number 5 is nearer to 4.9.

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

Question 9.
Write the decimal number represented by the points A, B, C, D on the given number line :

Answer:
(a) Since point A lies between 0 and 1, when 0 and 1 is divided into 10 equal parts, and 8 parts have been taken, so A represent 0.8.
(b) As the unit length between 1 and 2 have been divided into 10 equal parts and 3 parts have been taken, so B represent 1.3.
(c) Since C lies between 2 and 3, where unit length between 2 and 3 is divided into 10 equal parts and 2 parts have been taken, so C represent 2.2.
(d) Since D lies between 2 and 3 and unit length between 2 and 3 is divided into 10 equal parts and 9 parts have been taken, so D represent 2.9.

Question 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.
Answer:
(a) Length of Ramesh’s notebook = 9 cm 5 mm 
= 9 cm + 5 mm
= 9 cm + \(\frac{5}{10}\) cm
= 9 cm + 0.5 cm = 9.5 cm

(b) Length of young gram plant = 65 mm
= 60 mm + 5 mm 60 5
= \(\frac{60}{10}\) cm + \(\frac{5}{10}\) cm
= 6 cm + 0.5 cm = 6.5 cm