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## Kerala State Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms

### Kerala Syllabus 9th Standard Maths Decimal Forms Text Book Questions and Answers

Textbook Page No. 26

**Decimal Forms Class 9 Kerala Syllabus Question 1.**

Write the fractions below in decimal form:

- \(\frac{3}{20}\)
- \(\frac{3}{40}\)
- \(\frac{13}{40}\)
- \(\frac{7}{80}\)
- \(\frac{5}{16}\)

Answer:

1. \(\frac{3}{20}\)

= \(\frac{3×5}{20×5}\) = \(\frac{15}{100}\) = 0.15

2. 40 = 2 x 2 x (2 x 5)

Multiplied by 5 x 5

(2 x 5) x (2 x 5) x (2 x 5) = 1000

it is a multiple of 10.

\(\frac{3}{10}\) = \(\frac{3x5x5}{40x4x5x}\) = \(\frac{75}{1000}\) = 0.75

3. 40 = 2 x 2 x (2 x 5)

Multiplied by 5 x 5

(2 x 5) x (2 x 5) x (2 x 5) = 1000

It is a multiple of 10.

\(\frac{13}{40}\) = \(\frac{13x5x5}{20}\) = \(\frac{325}{1000}\) = 0.325

4. 80 = 2 x 2 x 2 x (2 x 5)

Multiplied by 5 x 5 x 5

It is a multiple of 10.

(2 x 5) x (2 x 5) x (2 x 5) x (2 x 5) = 1000

\(\frac{7}{80}\) = \(\frac{7x5x5x5}{80x5x5x5}\) = \(\frac{875}{10000}\) = 0.0875

5. 16 = 2 x 2 x 2 x 2

Multiplied by 5 x 5 x 5 x 5 x 5

(2 x 5) x (2 x 5) x (2 x 5) x (2 x 5) = 1000

It is a multiple of 10.

\(\frac{5}{16}\) = \(\frac{5x5x5x5x5}{16x5x5x5x5}\) = \(\frac{3125}{10000}\) = 0.3125

**Kerala Syllabus 9th Standard Maths Chapter 2 Question 2.**

Find the decimal form of the sums below:

- \(\frac{1}{80}\) + \(\frac{1}{25}\) + \(\frac{1}{125}\)
- \(\frac{1}{5}\) + \(\frac { 1 }{ { 5 }^{ 2 } } \) + \(\frac { 1 }{ { 5 }^{ 3 } } \) + \(\frac { 1 }{ { 5 }^{ 4 } } \)
- \(\frac{1}{2}\) + \(\frac { 1 }{ { 2 }^{ 2 } } \) + \(\frac { 1 }{ { 2 }^{ 3 } } \)

Answer:

1. \(\frac{1}{80}\) + \(\frac{1}{25}\) + \(\frac{1}{125}\)

Denominators of all the fractions should be 125

\(\frac{1×25}{5×25}\) + \(\frac{1×5}{25×5}\) + \(\frac{1×1}{125×1}\)

= \(\frac{25}{125}\) + \(\frac{5}{125}\) + \(\frac{1}{125}\) = \(\frac{31}{125}\)

125 = 5 x 5 x 5

Multiplied by 2 x 2 x 2

(5 x 2) x (5 x 2) x (5 x 2) = 1000

It is a power of 10

\(\frac{31}{125}\) = \(\frac{31x2x2x2}{125x2x2x2}\) = \(\frac{248}{1000}\) = 0.248

2. \(\frac{1}{5}\) + \(\frac { 1 }{ { 5 }^{ 2 } } \) + \(\frac { 1 }{ { 5 }^{ 3 } } \) + \(\frac { 1 }{ { 5 }^{ 4 } } \)

Denominators of all the fractions should be 5^{4}

3. \(\frac{1}{2}\) + \(\frac { 1 }{ { 2 }^{ 2 } } \) + \(\frac { 1 }{ { 2 }^{ 3 } } \)

Denominators of all the fractions should be 2^{3}

**Hss Live Guru 9th Maths Kerala Syllabus Question 3.**

A two – digit number divided by an other two-digit number gives 5.875. What are the numbers?

Answer:

5875 = \(\frac{5875}{1000}\)

5875 = 5 x 5 x 5 x 47

1000 = 5 x 5 x 5 x 8

5.875 = \(\frac{5875}{1000}\) = \(\frac{5x5x5x47}{5x5x5x8}\) = \(\frac{47}{8}\)

But 8 is not a two digit number. So we multiplied both numerator and denominator by 2, we get

\(\frac{47}{8}\) = \(\frac{94}{16}\)

5.875 = \(\frac{94}{16}\)

Textbook Page No. 30

**Kerala Syllabus 9th Standard Maths Notes Question 1.**

For each of the fractions below, find fractions with denominators powers of 10 gefting chaser and closer to it and hence write its decimal form:

- \(\frac{2}{3}\)
- \(\frac{5}{6}\)
- \(\frac{1}{9}\)

Answer:

1. \(\frac{2}{3}\) = \(\frac{2×10}{3×10}\) = \(\frac{2}{10}\) x \(\frac{10}{3}\) = \(\frac{2}{10}\) x (3 + \(\frac{1}{3}\))

The fraction \(\frac{6}{10}\), \(\frac{66}{100}\), \(\frac{666}{1000}\), ……… and so on get closer and closer to \(\frac{2}{3}\).

i.e., 0.6, 0.66, 0.666, …………..

\(\frac{1}{2}\) = 0.666

2. \(\frac{5}{6}\) = \(\frac{5×10}{6×10}\) = \(\frac{5}{10}\) x \(\frac{10}{6}\) = \(\frac{5}{10}\) x (1 + \(\frac{4}{6}\)

The fractions \(\frac{80}{100}\), \(\frac{830}{1000}\), \(\frac{8330}{5}\), ………. and so on get closer and closer to \(\frac{5}{6}\).

∴ \(\frac{5}{6}\) = 0.8333………

3. \(\frac{1}{9}\) = \(\frac{1×10}{9×10}\) = \(\frac{1}{10}\) x \(\frac{10}{9}\)

The fractions \(\frac{1}{10}\), \(\frac{11}{100}\), \(\frac{111}{1000}\), ………… and so on get closer and closer to \(\frac{1}{9}\).

∴ \(\frac{1}{9}\) = 0.1111 ……

**Hsslive Maths Class 9 Kerala Syllabus Question 2.**

- Using algebra, explain why \(\frac{1}{10}\), \(\frac{11}{100}\), \(\frac{1}{1000}\), ………… of any number get closer and closer to \(\frac{1}{9}\) of that number.
- Use the general principle got above to single-digit numbers to find the decimal forms of \(\frac{2}{9}\), \(\frac{4}{9}\), \(\frac{5}{9}\), \(\frac{7}{9}\), \(\frac{8}{9}\).
- What can we say in general about decimal forms with a single digit repeating?

Answer:

1. Let x be the number

The numbers \(\frac{x}{10}\), \(\frac{11x}{100}\), \(\frac{111x}{1000}\), ……… comes closer and closer to \(\frac{x}{9}\).

So the fraction \(\frac{1}{10}\), \(\frac{11}{100}\), \(\frac{111}{1000}\), ……. comes closer and closer to \(\frac{1}{9}\).

2. a.

\(\frac{1}{10}\) part of a number comes closer to its \(\frac{1}{9}\)

b. \(\frac{1}{10}\) part of a number comes closer to its \(\frac{1}{9}\)

c. \(\frac{1}{10}\) part of a number comes closer to its \(\frac{1}{9}\)

d. Since \(\frac{1}{9}\) is closes to \(\frac{1}{10}\)

e. Since \(\frac{1}{9}\) is closes to \(\frac{1}{10}\)

3. The denominator of a fraction is 9 or multiple of 9 then its decimal froms is always a single digit repetition of their own numerator.

**Hsslive Guru 9th Maths Kerala Syllabus Question 3.**

- Find the decimal form of \(\frac{1}{11}\).
- Find the decimal forms of \(\frac{2}{11}\), \(\frac{3}{11}\)
- What is the decimal form of \(\frac{10}{11}\)?

Answer:

1. \(\frac{1}{11}\) = \(\frac{1×10}{11×10}\) = \(\frac{1}{10}\) x \(\frac{10}{11}\)

Continuing like this,

= 0.090909

2. a. \(\frac{2}{11}\) = \(\frac{2×10}{11×10}\) + \(\frac{2}{10}\) x \(\frac{10}{11}\)

Continuing like this,

\(\frac{2}{11}\) = 0.18181818 …….

b. \(\frac{3}{11}\) = \(\frac{3×10}{11×10}\) = \(\frac{3}{10}\) x \(\frac{10}{11}\)

Continuing like this

\(\frac{3}{11}\) = 0.272727…..

3. \(\frac{10}{11}\) = \(\frac{10×10}{11×10}\) = \(\frac{10}{10}\) x \(\frac{10}{11}\)

Continuing like this

\(\frac{10}{11}\) = 0.90909……….

**Hss Live Guru 9 Maths Kerala Syllabus Question 4.**

Write the results of the operations below as decimals:

- 0.111…. + 0.222 …….
- 0.333…. + 0.777 …..
- 0.333…. x 0.666…
- (0.333….)
^{2} - \(\sqrt { 0.444…… } \)

Answer:

1. 0.111 = \(\frac{1}{9}\)

0.222…. = \(\frac{2}{9}\)

0.111 …. + 0.2222 …….. = \(\frac{1}{9}\) + \(\frac{2}{9}\) = \(\frac{3}{9}\)

= 0.3333

2. 0.3333 = \(\frac{3}{9}\)

0.7777…. = \(\frac{7}{9}\)

0.3333 ……. + 0.7777…. = \(\frac{1}{9}\) = \(\frac{2}{9}\) = \(\frac{3}{9}\)

= 1.1111……….

3. 0.3333 = \(\frac{3}{9}\)

0.6666…. = \(\frac{6}{9}\)

0.3333….. x 0.6666…… = \(\frac{3}{9}\) x \(\frac{6}{9}\)

= \(\frac{18}{81}\) = \(\frac{2}{9}\) = 0.2222……..

4. 0.3333 …….. = \(\frac{3}{9}\)

(0.3333)^{2} ….. = (\(\frac{3}{9}\))^{2}

= \(\frac{3×3}{9×9}\) = \(\frac{1}{9}\) = 0.1111…………

5. 0.4444 ….. = \(\frac{4}{9}\)

### Kerala Syllabus 9th Standard Maths Decimal Forms Exam Oriented Question and Answers

Question 1.

Write the deciamal form of \(\frac{1}{6}\)

Answer:

\(\frac{1}{6}\) = 0.1666…….

Question 2.

Write in deciamals

- \(\frac{1}{9}\)
- \(\frac{2}{9}\)
- \(\frac{1}{7}\)
- \(\frac{1}{11}\)
- \(\frac{2}{11}\)
- \(\frac{1}{12}\)

Answer:

- 0.111……
- 0.222………
- 0.14285…..
- 0.090909…..
- 0.181818….
- 0.08333……

Question 3.

Find the fraction of denominator is a power of 10 equal to each of the fractions below, and then write their decimal forms:

i. \(\frac{1}{50}\)

ii. \(\frac{3}{40}\)

iii.\(\frac{5}{16}\)

iv. \(\frac{12}{625}\)

Answer:

Question 4.

Find the fraction of denominator is a power of 10 getting closer and closer to each of the fractions be low, and then write their decimal forms.

- \(\frac{5}{6}\)
- \(\frac{3}{11}\)
- \(\frac{23}{11}\)
- \(\frac{1}{13}\)

Answer:

1. \(\frac{5}{6}\) = \(\frac{5}{10}\) x \(\frac{10}{6}\) = \(\frac{1}{10}\)(\(\frac{50}{6}\))

2. \(\frac{3}{11}\) = \(\frac{3}{100}\) x \(\frac{100}{11}\) = \(\frac{3}{100}\)(9 + \(\frac{1}{11}\))

3. \(\frac{23}{11}\) = \(\frac{23}{100}\) x \(\frac{100}{11}\)

4. \(\frac{1}{13}\) = \(\frac{1}{100}\) x \(\frac{100}{13}\)

Question 5.

- Explain using algebra, that the fractions \(\frac{1}{10}\), \(\frac{11}{100}\), \(\frac{111}{1000}\)… gets closer and closer to \(\frac{1}{9}\)
- Using the general principle above on single digit numbers, find the decimal forms of \(\frac{2}{9}\), \(\frac{4}{9}\), \(\frac{5}{9}\), \(\frac{7}{9}\), \(\frac{8}{9}\) (Why \(\frac{3}{9}\) and \(\frac{6}{9}\) left out in this?)
- What can we say in general about those decimal forms In which a single digit repeats?

Answer:

1. Let x be the number

2. \(\frac{2}{9}\) – \(\frac{2}{10}\) = \(\frac{2}{90}\)

(\(\frac{3}{9}\), \(\frac{6}{9}\) These fractions having common factor in the numerator and the denominator)

3. Repeated deciamals.

Question 6.

- Find the decimal form of \(\frac{1}{4}\).
- Write the decimal form of \(\frac{7}{10}\) + \(\frac{3}{100}\) + \(\frac{4}{1000}\).

Answer:

1. \(\frac{1}{4}\) = 0.25

2. \(\frac{7}{10}\) = 0.7

\(\frac{3}{100}\) = 0.03

\(\frac{4}{1000}\) = 0.004

\(\frac{7}{10}\) + \(\frac{3}{100}\) + \(\frac{4}{1000}\)

= 0.7 + 0.03 + 0.004

= 0.734

Question 7.

- Write the decimal forms of \(\frac{1}{3}\) and \(\frac{1}{9}\).
- What is the decimal form of (0.3333…..)
^{2}?

Answer:

1. \(\frac{1}{3}\) = 0.3333…..

\(\frac{1}{9}\) = 0.1111……

2. (0.3333…..)^{2} = (\(\frac{1}{3}\))^{2} = \(\frac{1}{9}\) = 0.1111….

Question 8.

Write the decimal forms of \(\frac{3}{25}\) and \(\frac{1}{8}\).

Answer:

Question 9.

a. Write the decimal form of the fractions and \(\frac{1}{2}\) and \(\frac{2}{5}\).

b. 1f f is a fraction between and \(\frac{2}{5}\) What is k?

c. Write the decimal form of the fraction \(\frac{4}{k}\).

Answer:

If \(\frac{4}{k}\) is between \(\frac{1}{2}\) and \(\frac{2}{5}\), then \(\frac{4}{k}\) is between \(\frac{4}{8}\) and \(\frac{4}{10}\). Then the number is \(\frac{4}{9}\).

∴ k = 9

c. \(\frac{4}{9}\) = 0.4444…….