ICSE Class 8 Maths Numbers Chapter 04 Decimals

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ICSE Class 8 Mathematics Numbers Chapter 4 Decimals Digital Edition

For Class 8 Mathematics, this chapter in ICSE Class 8 Maths Numbers Chapter 04 Decimals provides a detailed overview of important concepts. We highly recommend using this text alongside the ICSE Solutions for Class 8 Mathematics to learn the exercise questions provided at the end of the chapter.

Numbers Chapter 4 Decimals ICSE Book Class Class 8 PDF (2026-27)

4 Decimals

A fraction with some power of 10 as the denominator is called a decimal fraction.

\(\frac{3}{10}\), \(\frac{17}{100}\) and \(\frac{317}{1000}\) are some decimal fractions. \(\frac{3}{10}\) can be written as .3 or 0.3.

We call 0.3 a decimal or decimal number. Similarly, \(\frac{17}{100}\) and \(\frac{317}{1000}\) can be written as the decimal numbers 0.17 and 0.317 respectively.

For writing decimals, the positional system of writing numbers can be extended beyond the units place, to include places for tenths, hundredths, thousandths, ten thousandths, etc.

The place-value chart for decimal numbers is as follows.

Place...Ten thousandsThousandsHundredsTensUnits.TenthsHundredthsThousandthsTen thousandths...
Value...100001000100101.\(\frac{1}{10}\)\(\frac{1}{100}\)\(\frac{1}{1000}\)\(\frac{1}{10000}\)...

Consider the decimal number 17352.9876. It would be expanded as

\[1 \times 10000 + 7 \times 1000 + 3 \times 100 + 5 \times 10 + 2 \times 1 + 9 \times \frac{1}{10} + 8 \times \frac{1}{100} + 7 \times \frac{1}{1000} + 6 \times \frac{1}{10000}\]

We read 17352.9876 as "seventeen thousand three hundred and fifty-two point nine eight seven six". 17352 is called its integral part and .9876 is called its decimal part.

Conversion Of Fractions Into Decimals

Method - First convert the fraction into a fraction with a power of 10 as the denominator. Count the number of zeros in the denominator. Then counting from the right of the numerator, place the decimal point after as many digits as the number of zeros in the denominator.

Examples

(i) \(\frac{47}{100} = 0.47\) (also written as .47).

(ii) \(\frac{13}{125} = \frac{13 \times 8}{125 \times 8} = \frac{104}{1000} = 0.104\).

(iii) \(1\frac{4}{5} = \frac{9}{5} = \frac{9 \times 2}{5 \times 2} = \frac{18}{10} = 1.8\).

Teacher's Note

Decimals are used in currency, measurements, and sports statistics in everyday life, making this concept practically relevant for students.

Conversion Of Decimals Into Ordinary Fractions

Method - In the numerator, write the given number leaving out the decimal point. In the denominator, write 1 followed by as many zeros as the number of digits in the decimal part. Reduce the fraction to its simplest form.

Examples

(i) \(13.4 = \frac{134}{10} = \frac{67}{5}\).

(ii) \(23.87 = \frac{2387}{100}\).

(iii) \(352.9171 = \frac{3529171}{10000}\).

(iv) \(0.78929 = \frac{78929}{100000}\).

Teacher's Note

Understanding decimal to fraction conversion helps students in cooking recipes and calculating discounts during shopping.

Fundamental Operations On Decimals

You are already familiar with the four fundamental operations (addition, subtraction, multiplication and division) on decimals.

Example

(i) Find 1.735 + 2.9872 + 12.3 - 3.95.

(ii) Divide the product of 1.21 and 0.056 by 0.28.

Solution

(i)

1.7350

2.9872

+ 12.3000

17.0222

- 3.9500

13.0722

(ii) First, we find the product of 1.21 and 0.056.

121

\(\times\) 56

726

605

6776

\(\therefore\) 1.21 \(\times\) 0.056 = 0.06776.

Now, we divide 0.06776 by 0.28.

28) 6.776 (0.242

- 56

117

- 112

56

- 56

\(\times\)

\(\therefore\) 0.06776 \(\div\) 0.28 = 0.242.

Hence, the required quotient = 0.242.

Simplification

An expression involving decimals is simplified by following the rule of BODMAS.

Example

Simplify the following.

(i) (4.001 + 24.9 + 93.93 + 0.682 - 3) \(\div\) 0.03

(ii) \(\left(\frac{14.28 \times 3.5}{17} + \frac{3.42}{0.018}\right)\) of 0.3

Solution

(i)

4.001

24.900

93.930

+ 0.682

123.513

- 3.000

120.513

\(\frac{120.513 \div 0.03}{} = \frac{12051.3}{3} = 4017.1\)

\(\therefore\) 120.513 \(\div\) 0.03 = 4017.1.

Hence, the given expression = 4017.1.

(ii)

1428

\(\times\) 35

7140

4284

49980

\(\therefore\) 14.28 \(\times\) 3.5 = 49.98.

\(\therefore\) \(\frac{14.28 \times 3.5}{17} = \frac{49.98}{17} = 2.94\)

17) 49.98 (2.94

- 34

159

- 153

68

- 68

\(\times\)

\(\therefore\) \(\frac{3.42}{0.018} = \frac{3420}{18} = 190\).

18) 3420 (190

- 18

162

- 162

\(\times\) 0

0

\(\times\)

Hence, the given expression = (2.94 + 190) \(\times\) 0.3 = 192.94 \(\times\) 0.3 = 57.882.

Teacher's Note

These simplification techniques are essential for solving real-world problems in finance, engineering, and everyday calculations.

Recurring Decimals

\(\frac{1}{2} = 0.5\), \(\frac{1}{4} = 0.25\) and \(\frac{3}{8} = 0.375\). Such decimals are called terminating decimals as the dividend is exactly divisible by the divisor.

Now, consider the fraction \(\frac{5}{9}\).

9) 5.0 (0.555...

- 45

50

- 45

50

- 45

5...

The process of division is endless and we get a succession of 5s in the quotient.

The number 0.555... is called a recurring decimal and is written as 0.\(\overline{5}\) or 0.5 in short. In general, a decimal in which a digit or a finite group of digits recurs endlessly (that is, the division never gets completed) is called a recurring decimal. In such decimals, either a bar is placed over the recurring group of digits or a dot is placed over each of the first and last digits of the group.

Examples

(i) \(\frac{4}{11} = 0.363636...\) is written as 0.\(\overline{36}\) or 0.36 as the digits 3 and 6 are repeated endlessly.

(ii) \(\frac{41}{333} = 0.123123...\) is written as 0.\(\overline{123}\) or 0.123.

Here, the group of digits 123 recurs repeatedly in the quotient. So, we place a dot each above the first and last digits of the group or a bar over the group.

(iii) \(\frac{2}{7} = 0.285714285714...\) is written as 0.\(\overline{285714}\) or 0.285714.

Conversion Of Recurring Decimals Into Fractions

Example

Convert the following recurring decimals into fractions.

(i) 0.\(\overline{3}\)

(ii) 0.2\(\overline{5}\)

(iii) 0.\(\overline{25}\)

Solution

(i) Let x = 0.\(\overline{3}\) = 0.333...

Multiplying both sides by 10, we get

10x = 3.333...

Subtracting (1) from (2), we get

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

Hence, 0.\(\overline{3}\) = \(\frac{3}{9} = \frac{1}{3}\)

(ii) Let x = 0.2\(\overline{5}\) = 0.2555...

Multiplying both sides by 10, we get

10x = 2.555...

Subtracting (1) from (2), we get

10x - x = 2.3 or 9x = 2.3 or \(x = \frac{2.3}{9} = \frac{23}{90}\).

Hence, 0.2\(\overline{5}\) = \(\frac{23}{90}\).

(iii) Let x = 0.\(\overline{25}\) = 0.252525...

Multiplying both sides by 100, we get

100x = 25.252525...

Subtracting (1) from (2), we get

99x = 25 \(\Rightarrow\) \(x = \frac{25}{99}\)

Hence, 0.\(\overline{25}\) = \(\frac{25}{99}\).

Teacher's Note

Understanding recurring decimals is important for scientific calculations and programming, where precision and pattern recognition are crucial.

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ICSE Book Class 8 Mathematics Numbers Chapter 4 Decimals

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