Decimals

ICSE Class 6 Mathematics · ML Aggarwal Solutions Understanding ICSE Mathematics Class 6 · 87 solved questions

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1Exercise 7.1

Write each of the following decimal numbers in words:

(i) 30.5

(ii) 0.03

(iii) 108.56

(iv) 47.20

(v) 5.008

(vi) 26.039

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(i) 30.5 is read as thirty point five.

(ii) 0.03 is read as zero point zero three.

(iii) 108.56 is read as one hundred eight point five six.

(iv) 47.20 is read as forty seven point two zero.

(v) 5.008 is read as five point zero zero eight.

(vi) 26.039 is read as twenty six point zero three nine.

2Exercise 7.1

Write each of the following decimal numbers in the place value table:

(i) 4.2

(ii) 0.3

(iii) 205.9

(iv) 0.29

(v) 2.08

(vi) 7200.812

(vii) 38.007

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The given decimal numbers are written in the place value table as under :

PlacesThousandsHundredsTensOnesTenthsHundredthsThousandths
Values10001001011/101/1001/1000
(i) 4.242
(ii) 0.303
(iii) 205.92059
(iv) 0.29029
(v) 2.08208
(vi) 7200.8127200812
(vii) 38.00738007
3Exercise 7.1

Write the following decimal numbers in the expanded form:

(i) 123.7

(ii) 43.06

(iii) 509.306

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(i) We have,

$$\begin{aligned}123.7 = 1 \times 100 + 2 \times 10 + 3 \times 1 + 7 \times \frac{1}{10} \\[1em] = 100 + 20 + 3 + \frac{7}{10}\end{aligned}$$

Hence, 123.7 = 100 + 20 + 3 + $\frac{7}{10}$

(ii) We have,

$$\begin{aligned}43.06 = 4 \times 10 + 3 \times 1 + 0 \times \frac{1}{10} + 6 \times \frac{1}{100} \\[1em] = 40 + 3 + \frac{6}{100}\end{aligned}$$

Hence, 43.06 = 40 + 3 + $\frac{6}{100}$

(iii) We have,

$$\begin{aligned}509.306 = 5 \times 100 + 0 \times 10 + 9 \times 1 + 3 \times \frac{1}{10} + 0 \times \frac{1}{100} + 6 \times \frac{1}{1000} \\[1em] = 500 + 9 + \frac{3}{10} + \frac{6}{1000}\end{aligned}$$

Hence, 509.306 = 500 + 9 + $\frac{3}{10} + \frac{6}{1000}$

4Exercise 7.1

Write each of the following as a decimal number:

(i) $200 + 60 + 5 + \frac{3}{10}$

(ii) $50 + \frac{1}{10} + \frac{6}{100}$

(iii) $70 + 6 + \frac{7}{10} + \frac{9}{1000}$

(iv) $600 + 7 + \frac{3}{100} + \frac{6}{1000}$

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(i) We have,

$$\begin{aligned}200 + 60 + 5 + \frac{3}{10} \\[1em] = 2 \times 100 + 6 \times 10 + 5 \times 1 + 3 \times \frac{1}{10} \\[1em] = 265.3\end{aligned}$$

Hence, 265.3 is the decimal number

(ii) We have,

$$\begin{aligned}50 + \frac{1}{10} + \frac{6}{100} \\[1em] = 5 \times 10 + 0 \times 1 + 1 \times \frac{1}{10} + 6 \times \frac{1}{100} \\[1em] = 50.16\end{aligned}$$

Hence, 50.16 is the decimal number

(iii) We have,

$$\begin{aligned}70 + 6 + \frac{7}{10} + \frac{9}{1000} \\[1em] = 7 \times 10 + 6 \times 1 + 7 \times \frac{1}{10} + 0 \times \frac{1}{100} + 9 \times \frac{1}{1000} \\[1em] = 76.709\end{aligned}$$

Hence, 76.709 is the decimal number

(iv) We have,

$$\begin{aligned}600 + 7 + \frac{3}{100} + \frac{6}{1000} \\[1em] = 6 \times 100 + 0 \times 10 + 7 \times 1 + 0 \times \frac{1}{10} + 3 \times \frac{1}{100} + 6 \times \frac{1}{1000} \\[1em] = 607.036\end{aligned}$$

Hence, 607.036 is the decimal number

5Exercise 7.1

Write each of the following as decimals:

(i) Two ones and five tenths

(ii) Two tens and nine tenths

(iii) Six hundred point eight

(iv) Two hundred five and five hundredths

(v) Seven and fifteen thousandths

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(i) Two ones and five tenths

$= 2 + \frac{5}{10} = 2 + 0.5 = 2.5$

Hence, the decimal form is 2.5.

(ii) Two tens and nine tenths

$= 20 + \frac{9}{10} = 20 + 0.9 = 20.9$

Hence, the decimal form is 20.9.

(iii) Six hundred point eight

$= 600 + \frac{8}{10} = 600 + 0.8 = 600.8$

Hence, the decimal form is 600.8.

(iv) Two hundred five and five hundredths

$= 205 + \frac{5}{100} = 205 + 0.05 = 205.05$

Hence, the decimal form is 205.05.

(v) Seven and fifteen thousandths

$= 7 + \frac{15}{1000} = 7 + 0.015 = 7.015$

Hence, the decimal form is 7.015.

6Exercise 7.1

Write the number given in the following place value table in decimal form:

Thousands (1000)Hundreds(100)Tens(10)Ones(1)Tenths(1/10)Hundredths(1/100)Thousandths(1/1000)
(i)7102306
(ii)211902
(iii)3053015
(iv)7003
(v)540
(vi)719028
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Reading each row of the place value table and writing the digits in the correct decimal places, we get:

(i) 7102.306

(ii) 211.902

(iii) 3053.015

(iv) 70.03

(v) 5.40

(vi) 719.028

7Exercise 7.1

Show the following decimal numbers on the number line:

(i) 0.4

(ii) 1.9

(iii) 1.1

(iv) 2.5

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The unit lengths between successive whole numbers on the number line are divided into 10 equal parts. Where each part represents 0.1

In the given figure,

Show the following decimal numbers on the number line: Decimals, ML Aggarwal Understanding Mathematics Solutions ICSE Class 6.

A = 0.4, B = 1.1, C = 1.9 and D = 2.5

8Exercise 7.1

Write the decimal numbers represented by the points A, B, C and D on the given number line:

Write the decimal numbers represented by the points A, B, C and D on the given number line: Decimals, ML Aggarwal Understanding Mathematics Solutions ICSE Class 6.
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The unit lengths between successive whole numbers on the number line are divided into 10 equal parts. Reading the position of each point, we get:

A = 0.8,   B = 1.3,   C = 2.2,   D = 2.9

9Exercise 7.1

Between which two numbers in tenths place on the number line does each of the given number lie?

(i) 0.06

(ii) 0.45

(iii) 0.66

(iv) 0.92

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(i) The number 0.06 lies between 0.0 and 0.1 on the number line.

(ii) The number 0.45 lies between 0.4 and 0.5 on the number line.

(iii) The number 0.66 lies between 0.6 and 0.7 on the number line.

(iv) The number 0.92 lies between 0.9 and 1.0 on the number line.

1Exercise 7.2

Write the following decimal fractions as decimal numbers:

(i) $\frac{531}{10}$

(ii) $\frac{422}{100}$

(iii) $\frac{58301}{1000}$

(iv) $\frac{7}{10}$

(v) $\frac{3}{100}$

(vi) $\frac{37}{1000}$

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(i) $\frac{531}{10} = 53.1$

Hence, the decimal form is 53.1.

(ii) $\frac{422}{100} = 4.22$

Hence, the decimal form is 4.22.

(iii) $\frac{58301}{1000} = 58.301$

Hence, the decimal form is 58.301.

(iv) $\frac{7}{10} = 0.7$

Hence, the decimal form is 0.7.

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

Hence, the decimal form is 0.03.

(vi) $\frac{37}{1000} = 0.037$

Hence, the decimal form is 0.037.

2Exercise 7.2

Write the following decimal numbers as decimal fractions:

(i) 54.01

(ii) 318.105

(iii) 0.37

(iv) 0.047

(v) 0.03

(vi) 34.5

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(i) 54.01 has 2 decimal places, so the denominator is 100.

$54.01 = \frac{5401}{100}$

Hence, the decimal fraction is $\frac{5401}{100}$.

(ii) 318.105 has 3 decimal places, so the denominator is 1000.

$318.105 = \frac{318105}{1000}$

Hence, the decimal fraction is $\frac{318105}{1000}$.

(iii) 0.37 has 2 decimal places, so the denominator is 100.

$0.37 = \frac{37}{100}$

Hence, the decimal fraction is $\frac{37}{100}$.

(iv) 0.047 has 3 decimal places, so the denominator is 1000.

$0.047 = \frac{47}{1000}$

Hence, the decimal fraction is $\frac{47}{1000}$.

(v) 0.03 has 2 decimal places, so the denominator is 100.

$0.03 = \frac{3}{100}$

Hence, the decimal fraction is $\frac{3}{100}$.

(vi) 34.5 has 1 decimal place, so the denominator is 10.

$34.5 = \frac{345}{10}$

Hence, the decimal fraction is $\frac{345}{10}$.

3Exercise 7.2

Write the following decimal numbers as fractions in lowest terms:

(i) 0.8

(ii) 0.04

(iii) 0.125

(iv) 0.225

(v) 0.066

(vi) 0.092

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(i) Writing the decimal 0.8 as a fraction,

$0.8 = \frac{8}{10} = \frac{4}{5}$

Therefore, the required fraction is $\frac{4}{5}$.

(ii) Writing the decimal 0.04 as a fraction,

$0.04 = \frac{4}{100} = \frac{1}{25}$

Therefore, the required fraction is $\frac{1}{25}$.

(iii) Writing the decimal 0.125 as a fraction,

$0.125 = \frac{125}{1000} = \frac{1}{8}$

Therefore, the required fraction is $\frac{1}{8}$.

(iv) Writing the decimal 0.225 as a fraction,

$0.225 = \frac{225}{1000} = \frac{9}{40}$

Therefore, the required fraction is $\frac{9}{40}$.

(v) Writing the decimal 0.066 as a fraction,

$0.066 = \frac{66}{1000} = \frac{33}{500}$

Therefore, the required fraction is $\frac{33}{500}$.

(vi) Writing the decimal 0.092 as a fraction,

$0.092 = \frac{92}{1000} = \frac{23}{250}$

Therefore, the required fraction is $\frac{23}{250}$.

4Exercise 7.2

Convert the following decimal numbers into mixed fractions:

(i) 31.6

(ii) 3.25

(iii) 7.025

(iv) 95.95

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(i) Writing the decimal 31.6 as a fraction,

$31.6 = \frac{316}{10} = \frac{158}{5} = 31\frac{3}{5}$

Therefore, the required mixed fraction is $31\frac{3}{5}$.

(ii) Writing the decimal 3.25 as a fraction,

$3.25 = \frac{325}{100} = \frac{13}{4} = 3\frac{1}{4}$

Therefore, the required mixed fraction is $3\frac{1}{4}$.

(iii) Writing the decimal 7.025 as a fraction,

$7.025 = \frac{7025}{1000} = \frac{281}{40} = 7\frac{1}{40}$

Therefore, the required mixed fraction is $7\frac{1}{40}$.

(iv) Writing the decimal 95.95 as a fraction,

$95.95 = \frac{9595}{100} = \frac{1919}{20} = 95\frac{19}{20}$

Therefore, the required mixed fraction is $95\frac{19}{20}$.

5Exercise 7.2

Convert the following fractions into decimal numbers:

(i) $\frac{4}{5}$

(ii) $\frac{6}{25}$

(iii) $\frac{112}{125}$

(iv) $\frac{3}{4}$

(v) $\frac{3}{8}$

(vi) $7\frac{3}{40}$

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(i) Converting the fraction $\frac{4}{5}$ into a decimal,

$\frac{4}{5} = \frac{4 \times 2}{5 \times 2} = \frac{8}{10} = 0.8$

Therefore, the decimal form is 0.8.

(ii) Converting the fraction $\frac{6}{25}$ into a decimal,

$\frac{6}{25} = \frac{6 \times 4}{25 \times 4} = \frac{24}{100} = 0.24$

Therefore, the decimal form is 0.24.

(iii) Converting the fraction $\frac{112}{125}$ into a decimal,

$\frac{112}{125} = \frac{112 \times 8}{125 \times 8} = \frac{896}{1000} = 0.896$

Therefore, the decimal form is 0.896.

(iv) Converting the fraction $\frac{3}{4}$ into a decimal,

$\frac{3}{4} = \frac{3 \times 25}{4 \times 25} = \frac{75}{100} = 0.75$

Therefore, the decimal form is 0.75.

(v) Converting the fraction $\frac{3}{8}$ into a decimal,

$\frac{3}{8} = \frac{3 \times 125}{8 \times 125} = \frac{375}{1000} = 0.375$

Therefore, the decimal form is 0.375.

(vi) Converting the fraction $7\frac{3}{40}$ into a decimal,

$7\frac{3}{40} = 7 + \frac{3}{40} = 7 + \frac{3 \times 25}{40 \times 25} = 7 + \frac{75}{1000} = 7 + 0.075 = 7.075$

Therefore, the decimal form is 7.075.

6Exercise 7.2

Convert the following unlike decimal numbers to like decimal numbers:

(i) 17.5, 3.912

(ii) 5.04, 13.1902

(iii) 2.451, 3.7, 28.34

(iv) 3.1, 2.678, 27.0103

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(i) The maximum number of decimal places in the given numbers is 3. So, we convert each of the decimals into one having 3 decimal places by annexing zeros.

17.5 = 17.500

3.912 = 3.912

As 17.500 and 3.912 have equal number of decimal places, they are like decimals.

(ii) The maximum number of decimal places in the given numbers is 4. So, we convert each of the decimals into one having 4 decimal places by annexing zeros.

5.04 = 5.0400

13.1902 = 13.1902

As 5.0400 and 13.1902 have equal number of decimal places, they are like decimals.

(iii) The maximum number of decimal places in the given numbers is 3. So, we convert each of the decimals into one having 3 decimal places by annexing zeros.

2.451 = 2.451

3.7 = 3.700

28.34 = 28.340

As 2.451, 3.700 and 28.340 have equal number of decimal places, they are like decimals.

(iv) The maximum number of decimal places in the given numbers is 4. So, we convert each of the decimals into one having 4 decimal places by annexing zeros.

3.1 = 3.1000

2.678 = 2.6780

27.0103 = 27.0103

As 3.1000, 2.6780 and 27.0103 have equal number of decimal places, they are like decimals.

7Exercise 7.2

In each of the following pairs of decimal numbers, state which number is greater?

(i) 0.3, 0.4

(ii) 1, 0.99

(iii) 1.09, 1.093

(iv) 0.5, 0.05

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(i) The given decimals are 0.3 and 0.4. Their whole number parts are equal. Comparing their tenths digits, 3 < 4.

Therefore, 0.4 is greater.

(ii) The given decimals are 1 and 0.99. Comparing their whole number parts, 1 > 0.

Therefore, 1 is greater.

(iii) The given decimals are 1.09 and 1.093.

Writing them as like decimals: 1.090 and 1.093.

Comparing the decimal parts, 090 < 093.

Therefore, 1.093 is greater.

(iv) The given decimals are 0.5 and 0.05.

Writing them as like decimals: 0.50 and 0.05.

Comparing the decimal parts, 50 > 05.

Therefore, 0.5 is greater.

8Exercise 7.2

In each of the following pairs of decimal numbers, state which number is smaller:

(i) 45.78, 345.8

(ii) 37.701, 37.71

(iii) 5.907, 5.903

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(i) The given decimals are 45.78 and 345.8. Comparing their whole number parts, 45 < 345.

Therefore, 45.78 is smaller.

(ii) The given decimals are 37.701 and 37.71.

Writing them as like decimals: 37.701 and 37.710.

Comparing the decimal parts, 701 < 710.

Therefore, 37.701 is smaller.

(iii) The given decimals are 5.907 and 5.903. Their whole number parts are equal. Comparing the decimal parts, 907 > 903.

Therefore, 5.903 is smaller.

9Exercise 7.2

Arrange the following decimal numbers in ascending order:

(i) 27.35, 27.305, 2.7, 2.543

(ii) 4.53, 4.07, 29.1, 0.9, 0.709

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(i) The maximum number of decimal places in the given decimals is 3. So, we convert each of the given decimals into like decimals having three decimal places by annexing zeros.

The decimals are: 27.350, 27.305, 2.700, 2.543

Clearly, 2.543 < 2.700 < 27.305 < 27.350.

Hence, the given decimals in ascending order are:

2.543 < 2.7 < 27.305 < 27.35

(ii) The maximum number of decimal places in the given decimals is 3. So, we convert each of the given decimals into like decimals having three decimal places by annexing zeros.

The decimals are: 4.530, 4.070, 29.100, 0.900, 0.709

Clearly, 0.709 < 0.900 < 4.070 < 4.530 < 29.100.

Hence, the given decimals in ascending order are:

0.709 < 0.9 < 4.07 < 4.53 < 29.1

10Exercise 7.2

Arrange the following decimal numbers in descending order:

(i) 3.303, 33.03, 3.3, 30.33

(ii) 72.5, 2.75, 27.505, 0.275, 2.507

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(i) The maximum number of decimal places in the given decimals is 3. So, we convert each of the given decimals into like decimals having three decimal places by annexing zeros.

The decimals are: 3.303, 33.030, 3.300, 30.330

Clearly, 33.030 > 30.330 > 3.303 > 3.300.

Hence, the given decimals in descending order are:

33.03 > 30.33 > 3.303 > 3.3

(ii) The maximum number of decimal places in the given decimals is 3. So, we convert each of the given decimals into like decimals having three decimal places by annexing zeros.

The decimals are: 72.500, 2.750, 27.505, 0.275, 2.507

Clearly, 72.500 > 27.505 > 2.750 > 2.507 > 0.275.

Hence, the given decimals in descending order are:

72.5 > 27.505 > 2.75 > 2.507 > 0.275

1Exercise 7.3

Add:

(i) 17.5, 8.8

(ii) 9.999, 0.03

(iii) 5.87, 1.03, 0.1

(iv) 23.71, 9.9, 4.023

(v) 4.5, 16.024, 7.99

(vi) 8.79, 23.001, 5.41, 0.875

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(i) Writing the decimals in column form and adding, we get:

$$\begin{array}{r} \phantom{+}17.4500 \\ -\space 7.9702 \\ \hline 9.4798 \end{array}$$

Therefore, the sum is 26.3.

(ii) Converting the given decimals into like decimals, each having 3 places of decimal, we get:

9.999, 0.030

Writing these decimals in column form and adding, we get:

$$\begin{array}{r} \phantom{+}9.999 \\ +\space 0.030 \\ \hline 10.029 \end{array}$$

Therefore, the sum is 10.029.

(iii) Converting the given decimals into like decimals, each having 2 places of decimal, we get:

5.87, 1.03, 0.10

Writing these decimals in column form and adding, we get:

$$\begin{array}{r} \phantom{+}5.87 \\ 1.03 \\ +\space 0.10 \\ \hline 7.00 \end{array}$$

Therefore, the sum is 7.

(iv) Converting the given decimals into like decimals, each having 3 places of decimal, we get:

23.710, 9.900, 4.023

Writing these decimals in column form and adding, we get:

$$\begin{array}{r} \phantom{+}23.710 \\ 9.900 \\ +\space 4.023 \\ \hline 37.633 \end{array}$$

Therefore, the sum is 37.633.

(v) Converting the given decimals into like decimals, each having 3 places of decimal, we get:

4.500, 16.024, 7.990

Writing these decimals in column form and adding, we get:

$$\begin{array}{r} \phantom{+}4.500 \\ 16.024 \\ +\space 7.990 \\ \hline 28.514 \end{array}$$

Therefore, the sum is 28.514.

(vi) Converting the given decimals into like decimals, each having 3 places of decimal, we get:

8.790, 23.001, 5.410, 0.875

Writing these decimals in column form and adding, we get:

$$\begin{array}{r} \phantom{+}8.790 \\ 23.001 \\ 5.410 \\ +\space 0.875 \\ \hline 38.076 \end{array}$$

Therefore, the sum is 38.076.

2Exercise 7.3

Calculate:

(i) 5.82 - 2.65

(ii) 19.01 - 12.234

(iii) 15.4 + 3.015 - 14.237

(iv) 7.4 - 2.19 - 0.456 - 3.5

(v) 19.27 - 3.6 - 8.812 + 0.84

(vi) 6.4 - 2.351 - 1.45 - 0.999

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(i) Writing the decimals in column form and subtracting, we get:

$$\begin{array}{r} \phantom{+}5.82 \\ -\space 2.65 \\ \hline 3.17 \end{array}$$

Therefore, 5.82 - 2.65 = 3.17.

(ii) Converting the given decimals into like decimals, each having 3 places of decimal, we get 19.010 and 12.234.

Writing in column form and subtracting, we get:

$$\begin{array}{r} \phantom{+}19.010 \\ -\space 12.234 \\ \hline 6.776 \end{array}$$

Therefore, 19.01 - 12.234 = 6.776.

(iii) Writing each of the given numbers with three decimal places, we get:

Given expression = 15.400 + 3.015 - 14.237

Now,

$$\begin{array}{r} \phantom{+}15.400 \\ +\space 3.015 \\ \hline 18.415 \end{array} \qquad \text{and} \qquad \begin{array}{r} \phantom{+}18.415 \\ -\space 14.237 \\ \hline 4.178 \end{array}$$

Therefore, 15.4 + 3.015 - 14.237 = 4.178.

(iv) Writing each of the given numbers with three decimal places, we get:

Given expression = 7.400 - 2.190 - 0.456 - 3.500

$\phantom{Given expression}$ = 7.400 - (2.190 + 0.456 + 3.500)

Now,

$$\begin{array}{r} \phantom{+}2.190 \\ 0.456 \\ +\space 3.500 \\ \hline 6.146 \end{array}$$

So,

$$\begin{array}{r} \phantom{+}7.400 \\ -\space 6.146 \\ \hline 1.254 \end{array}$$

Therefore, 7.4 - 2.19 - 0.456 - 3.5 = 1.254.

(v) Writing each of the given numbers with three decimal places, we get:

Given expression = 19.270 - 3.600 - 8.812 + 0.840

$\phantom{Given expression}$ = (19.270 + 0.840) - (3.600 + 8.812)

Now,

$$\begin{array}{r} \phantom{+}19.270 \\ +\space 0.840 \\ \hline 20.110 \end{array} \qquad \text{and} \qquad \begin{array}{r} \phantom{+}3.600 \\ +\space 8.812 \\ \hline 12.412 \end{array}$$

So,

$$\begin{array}{r} \phantom{+}20.110 \\ -\space 12.412 \\ \hline 7.698 \end{array}$$

Therefore, 19.27 - 3.6 - 8.812 + 0.84 = 7.698.

(vi) Writing each of the given numbers with three decimal places, we get:

Given expression = 6.400 - 2.351 - 1.450 - 0.999 = 6.400 - (2.351 + 1.450 + 0.999)

Now,

$$\begin{array}{r} \phantom{+}2.351 \\ 1.450 \\ +\space 0.999 \\ \hline 4.800 \end{array}$$

So,

$$\begin{array}{r} \phantom{+}6.400 \\ -\space 4.800 \\ \hline 1.600 \end{array}$$

Therefore, 6.4 - 2.351 - 1.45 - 0.999 = 1.6.

3Exercise 7.3

What number added to 0.756 gives 1?

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Let the required number be x. Then,

0.756 + x = 1

Solving for x,

x = 1 - 0.756

Rewriting the numbers with three decimal places,

x = 1.000 - 0.756

Writing in column form and subtracting, we get:

$$\begin{array}{r} \phantom{+}1.000 \\ -\space 0.756 \\ \hline 0.244 \end{array}$$

Therefore, the required number is 0.244.

4Exercise 7.3

By how much should 17.45 be decreased to get 7.9702?

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Let the required decrease be x. Then,

17.45 - x = 7.9702

Solving for x,

x = 17.45 - 7.9702

Rewriting the numbers with four decimal places,

x = 17.4500 - 7.9702

Writing in column form and subtracting, we get:

$$\begin{array}{r} \phantom{+}17.4500 \\ -\space 7.9702 \\ \hline 9.4798 \end{array}$$

Therefore, 17.45 should be decreased by 9.4798.

1Exercise 7.4

Evaluate the following:

(i) 3.7 × 4.5

(ii) 12.08 × 9.3

(iii) 238.06 × 7.5

(iv) 0.79 × 32.4

(v) 3.6 × 1.4 × 0.7

(vi) 9.01 × 2.5 × 1.6

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(i) Ignoring the decimal points, we multiply 37 and 45.

37 × 45 = 1665

The two given numbers together have 1 + 1 = 2 decimal places.

So, placing the decimal point such that the product has 2 decimal places,

3.7 × 4.5 = 16.65

Therefore, 3.7 × 4.5 = 16.65.

(ii) Ignoring the decimal points, we multiply 1208 and 93.

1208 × 93 = 112344

The two given numbers together have 2 + 1 = 3 decimal places.

So, placing the decimal point such that the product has 3 decimal places,

12.08 × 9.3 = 112.344

Therefore, 12.08 × 9.3 = 112.344.

(iii) Ignoring the decimal points, we multiply 23806 and 75.

23806 × 75 = 1785450

The two given numbers together have 2 + 1 = 3 decimal places.

So, placing the decimal point such that the product has 3 decimal places,

238.06 × 7.5 = 1785.450 = 1785.45

Therefore, 238.06 × 7.5 = 1785.45.

(iv) Ignoring the decimal points, we multiply 79 and 324.

79 × 324 = 25596

The two given numbers together have 2 + 1 = 3 decimal places.

So, placing the decimal point such that the product has 3 decimal places,

0.79 × 32.4 = 25.596

Therefore, 0.79 × 32.4 = 25.596.

(v) Ignoring the decimal points, we multiply 36, 14 and 7.

36 × 14 × 7 = 504 × 7 = 3528

The three given numbers together have 1 + 1 + 1 = 3 decimal places.

So, placing the decimal point such that the product has 3 decimal places,

3.6 × 1.4 × 0.7 = 3.528

Therefore, 3.6 × 1.4 × 0.7 = 3.528.

(vi) Ignoring the decimal points, we multiply 901, 25 and 16.

901 × 25 × 16 = 22525 × 16 = 360400

The three given numbers together have 2 + 1 + 1 = 4 decimal places.

So, placing the decimal point such that the product has 4 decimal places,

9.01 × 2.5 × 1.6 = 36.0400 = 36.04

Therefore, 9.01 × 2.5 × 1.6 = 36.04.

2Exercise 7.4

Calculate the following:

(i) 70.756 ÷ 4

(ii) 2.46 ÷ 6

(iii) 3.016 ÷ 8

(iv) 8.64 ÷ 3.6

(v) 72.8 ÷ 0.04

(vi) 0.144 ÷ 0.02

Show SolutionHide Solution↓

(i) By actual division, we have:

$$\begin{array}{l} \phantom{4}{\space \phantom{} 17.689} \\ 4\overline{\smash{\big)}\space 70.756} \\ \phantom{4}\phantom{}\underline{-4} \\ \phantom{4)\space)}30 \\ \phantom{4\space}\underline{-28} \\ \phantom{4)\space 28}27 \\ \phantom{4)\space)}\underline{-24} \\ \phantom{4)\space 28))}35 \\ \phantom{4)\space 28}\underline{-32} \\ \phantom{4)\space 28))))}36 \\ \phantom{4)\space 28))}\underline{-36} \\ \phantom{4)\space 28))))}\times \end{array}$$

Therefore, 70.756 ÷ 4 = 17.689.

(ii) By actual division, we have:

$$\begin{array}{l} \phantom{6)\space }0.41 \\ 6\overline{\smash{\big)}\space 2.46} \\ \phantom{(}\underline{-0\phantom{.}} \\ \phantom{6(\space) )}24 \\ \phantom{\space)))}\underline{-24} \\ \phantom{6())\space 2.}6 \\ \phantom{6\space 2.)}\underline{-6} \\ \phantom{6)\space 2.4}\times \end{array}$$

Therefore, 2.46 ÷ 6 = 0.41.

(iii) By actual division, we have:

$$\begin{array}{l} \phantom{8)\space }0.377 \\ 8\overline{\smash{\big)}\space 3.016} \\ \phantom{8}\underline{-0\phantom{.}} \\ \phantom{8)\space }30 \\ \phantom{\space }\underline{-24} \\ \phantom{8)\space 1}61 \\ \phantom{\space 1}\underline{-56} \\ \phantom{8)\space 1\phantom{1}}56 \\ \phantom{\space ()\phantom{1}}\underline{-56} \\ \phantom{8)\space 3.1}\times \end{array}$$

Therefore, 3.016 ÷ 8 = 0.377.

(iv) To divide 8.64 by 3.6, multiply both the dividend and divisor by 10 so that the divisor becomes a whole number:

$8.64 \div 3.6 = \frac{8.64 \times 10}{3.6 \times 10} = \frac{86.4}{36}$

By actual division, we have:

$$\begin{array}{l} \phantom{36)\space }2.4 \\ 36\overline{\smash{\big)}\space 86.4} \\ \phantom{36}\underline{-72} \\ \phantom{36(\space)}144 \\ \phantom{+\space)}\underline{-144} \\ \phantom{36\space 86.}\times \end{array}$$

Therefore, 8.64 ÷ 3.6 = 2.4.

(v) To divide 72.8 by 0.04, multiply both the dividend and divisor by 100 so that the divisor becomes a whole number:

$72.8 \div 0.04 = \frac{72.8 \times 100}{0.04 \times 100} = \frac{7280}{4} = 1820$

Therefore, 72.8 ÷ 0.04 = 1820.

(vi) To divide 0.144 by 0.02, multiply both the dividend and divisor by 100 so that the divisor becomes a whole number:

$0.144 \div 0.02 = \frac{0.144 \times 100}{0.02 \times 100} = \frac{14.4}{2} = 7.2$

Therefore, 0.144 ÷ 0.02 = 7.2.

3Exercise 7.4

Multiply each of the following numbers by 10, 100 and 1000 (orally):

(i) 4.7

(ii) 3.45

(iii) 0.234

Show SolutionHide Solution↓

When we multiply a decimal number by 10, 100 or 1000, the decimal point shifts one, two or three places to the right respectively.

(i) For 4.7:

4.7 × 10 = 47

4.7 × 100 = 470

4.7 × 1000 = 4700

(ii) For 3.45:

3.45 × 10 = 34.5

3.45 × 100 = 345

3.45 × 1000 = 3450

(iii) For 0.234:

0.234 × 10 = 2.34

0.234 × 100 = 23.4

0.234 × 1000 = 234

4Exercise 7.4

Divide each of the following numbers by 10, 100 and 1000 (orally):

(i) 4.7

(ii) 3.45

(iii) 23.01

Show SolutionHide Solution↓

When we divide a decimal number by 10, 100 or 1000, the decimal point shifts one, two or three places to the left respectively.

(i) For 4.7:

4.7 ÷ 10 = 0.47

4.7 ÷ 100 = 0.047

4.7 ÷ 1000 = 0.0047

(ii) For 3.45:

3.45 ÷ 10 = 0.345

3.45 ÷ 100 = 0.0345

3.45 ÷ 1000 = 0.00345

(iii) For 23.01:

23.01 ÷ 10 = 2.301

23.01 ÷ 100 = 0.2301

23.01 ÷ 1000 = 0.02301

5Exercise 7.4

Find the value of the following:

(i) (3.5)2

(ii) (0.4)3

Show SolutionHide Solution↓

(i) We have,

(3.5)2 = 3.5 × 3.5

Ignoring the decimal points, 35 × 35 = 1225.

The two factors together have 1 + 1 = 2 decimal places, so:

(3.5)2 = 12.25

Therefore, (3.5)2 = 12.25.

(ii) We have,

(0.4)3 = 0.4 × 0.4 × 0.4

Ignoring the decimal points, 4 × 4 × 4 = 64.

The three factors together have 1 + 1 + 1 = 3 decimal places, so:

(0.4)3 = 0.064

Therefore, (0.4)3 = 0.064.

1Exercise 7.5

Express as rupees using decimals:

(i) 75 paise

(ii) 1025 paise

(iii) 63 rupees 9 paise

Show SolutionHide Solution↓

We know that 1 paisa = $\frac{1}{100}$ rupee = ₹0.01.

(i) 75 paise = ₹$\frac{75}{100}$ = ₹0.75.

(ii) 1025 paise = ₹$\frac{1025}{100}$ = ₹10.25.

(iii) 63 rupees 9 paise = ₹63 + ₹$\frac{9}{100}$ = ₹63 + ₹0.09 = ₹63.09.

2Exercise 7.5

Express as cm using decimals:

(i) 8 mm

(ii) 263 mm

(iii) 13 cm 3 mm

Show SolutionHide Solution↓

We know that 1 mm = $\frac{1}{10}$ cm = 0.1 cm.

(i) 8 mm = $\frac{8}{10}$ cm = 0.8 cm.

(ii) 263 mm = $\frac{263}{10}$ cm = 26.3 cm.

(iii) 13 cm 3 mm = 13 cm + $\frac{3}{10}$ cm = 13 cm + 0.3 cm = 13.3 cm.

3Exercise 7.5

Express as metres using decimals:

(i) 6 cm

(ii) 528 cm

(iii) 7 m 55 cm

Show SolutionHide Solution↓

We know that 1 cm = $\frac{1}{100}$ m = 0.01 m.

(i) 6 cm = $\frac{6}{100}$ m = 0.06 m.

(ii) 528 cm = $\frac{528}{100}$ m = 5.28 m.

(iii) 7 m 55 cm = 7 m + $\frac{55}{100}$ m = 7 m + 0.55 m = 7.55 m.

4Exercise 7.5

Express as km using decimals:

(i) 5 m

(ii) 888 m

(iii) 15 km 88 m

Show SolutionHide Solution↓

We know that 1 m = $\frac{1}{1000}$ km = 0.001 km.

(i) 5 m = $\frac{5}{1000}$ km = 0.005 km.

(ii) 888 m = $\frac{888}{1000}$ km = 0.888 km.

(iii) 15 km 88 m = 15 km + $\frac{88}{1000}$ km = 15 km + 0.088 km = 15.088 km.

5Exercise 7.5

Express as kg using decimals:

(i) 37 g

(ii) 100 g

(iii) 5 kg 8 g

Show SolutionHide Solution↓

We know that 1 g = $\frac{1}{1000}$ kg = 0.001 kg.

(i) 37 g = $\frac{37}{1000}$ kg = 0.037 kg.

(ii) 100 g = $\frac{100}{1000}$ kg = 0.1 kg.

(iii) 5 kg 8 g = 5 kg + $\frac{8}{1000}$ kg = 5 kg + 0.008 kg = 5.008 kg.

6Exercise 7.5

Express as kL using decimals:

(i) 6 L

(ii) 555 L

(iii) 3 kL 95 L

Show SolutionHide Solution↓

We know that 1 L = $\frac{1}{1000}$ kL = 0.001 kL.

(i) 6 L = $\frac{6}{1000}$ kL = 0.006 kL.

(ii) 555 L = $\frac{555}{1000}$ kL = 0.555 kL.

(iii) 3 kL 95 L = 3 kL + $\frac{95}{1000}$ kL = 3 kL + 0.095 kL = 3.095 kL.

7Exercise 7.5

Anita bought 2 m 70 cm cloth for her shirt and 2 m 85 cm cloth for her trouser. Find the total length of the cloth bought by her.

Show SolutionHide Solution↓

Length of cloth for shirt = 2 m 70 cm

= 2 m + $\frac{70}{100}$ m = 2 m + 0.70 m = 2.70 m

Length of cloth for trouser = 2 m 85 cm

= 2 m + $\frac{85}{100}$ m = 2 m + 0.85 m = 2.85 m

Total length of the cloth = 2.70 + 2.85

Writing in column form and adding:

$$\begin{array}{r} \phantom{+}2.70 \\ +\space 2.85 \\ \hline 5.55 \end{array}$$

Hence, the total length of the cloth bought by Anita is 5.55 m or 5 m 55 cm.

8Exercise 7.5

Sunita travelled 15 km 268 m by bus, 7 km 7 m by car and 500 m on foot in order to reach her school. How far is her school from her residence?

Show SolutionHide Solution↓

Distance travelled by bus = 15 km 268 m

= 15 km + $\frac{268}{1000}$ km = 15 km + 0.268 km = 15.268 km

Distance travelled by car = 7 km 7 m

= 7 km + $\frac{7}{1000}$ km = 7 km + 0.007 km = 7.007 km

Distance travelled on foot = 500 m

= $\frac{500}{1000}$ km = 0.500 km

Total distance = 15.268 + 7.007 + 0.500

Writing in column form and adding:

$$\begin{array}{r} \phantom{+}15.268 \\ 7.007 \\ +\space 0.500 \\ \hline 22.775 \end{array}$$

Hence, Sunita's school is 22.775 km or 22 km 775 m from her residence.

9Exercise 7.5

Rahul bought 4 kg 90 g apples, 2 kg 60 g grapes and 5 kg 300 g mangoes. Find the total weight of all the fruits he bought.

Show SolutionHide Solution↓

Weight of apples = 4 kg 90 g

= 4 kg + $\frac{90}{1000}$ kg = 4 kg + 0.090 kg = 4.090 kg

Weight of grapes = 2 kg 60 g

= 2 kg + $\frac{60}{1000}$ kg = 2 kg + 0.060 kg = 2.060 kg

Weight of mangoes = 5 kg 300 g

= 5 kg + $\frac{300}{1000}$ kg = 5 kg + 0.300 kg = 5.300 kg

Total weight = 4.090 + 2.060 + 5.300

Writing in column form and adding:

$$\begin{array}{r} \phantom{+}4.090 \\ 2.060 \\ +\space 5.300 \\ \hline 11.450 \end{array}$$

Hence, the total weight of the fruits bought is 11.450 kg or 11 kg 450 g.

10Exercise 7.5

Rani has ₹18.50. She bought one ice cream for ₹11.75. How much money does she have now?

Show SolutionHide Solution↓

Money Rani had = ₹18.50

Cost of ice cream = ₹11.75

Money left with her = 18.50 - 11.75

Writing in column form and subtracting:

$$\begin{array}{r} \phantom{+}18.50 \\ -\space 11.75 \\ \hline 6.75 \end{array}$$

Hence, Rani now has ₹6.75.

11Exercise 7.5

Tina had 20 m 5 cm long cloth. She cuts 4 m 50 cm length of cloth from this for making a curtain. How much length of cloth is left with her?

Show SolutionHide Solution↓

Total length of cloth = 20 m 5 cm

= 20 m + $\frac{5}{100}$ m = 20 m + 0.05 m = 20.05 m

Length of cloth cut for curtain = 4 m 50 cm

= 4 m + $\frac{50}{100}$ m = 4 m + 0.50 m = 4.50 m

Length of cloth left = 20.05 - 4.50

Writing in column form and subtracting:

$$\begin{array}{r} \phantom{+}20.05 \\ -\space 4.50 \\ \hline 15.55 \end{array}$$

Hence, the length of cloth left with Tina is 15.55 m or 15 m 55 cm.

12Exercise 7.5

Ruby bought a watermelon weighing 5 kg 300 g. Out of which she gave 2 kg 680 g to her neighbour. What is the weight of the watermelon left with Ruby?

Show SolutionHide Solution↓

Weight of watermelon = 5 kg 300 g

= 5 kg + $\frac{300}{1000}$ kg = 5 kg + 0.300 kg = 5.300 kg

Weight given to neighbour = 2 kg 680 g

= 2 kg + $\frac{680}{1000}$ kg = 2 kg + 0.680 kg = 2.680 kg

Weight left with Ruby = 5.300 - 2.680

Writing in column form and subtracting:

$$\begin{array}{r} \phantom{+}5.300 \\ -\space 2.680 \\ \hline 2.620 \end{array}$$

Hence, the weight of the watermelon left with Ruby is 2.620 kg or 2 kg 620 g.

13Exercise 7.5

The cost of 1 metre of cloth is ₹35.80. What will be cost of 9.8 metres of cloth?

Show SolutionHide Solution↓

Cost of 1 metre of cloth = ₹35.80

Cost of 9.8 metres of cloth = ₹35.80 × 9.8

Ignoring the decimal points, 3580 × 98 = 350840.

The two factors together have 2 + 1 = 3 decimal places, so:

35.80 × 9.8 = 350.840 = 350.84

Hence, the cost of 9.8 metres of cloth is ₹350.84.

14Exercise 7.5

Farida bought some bags of cement, each weighing 49.8 kg. If the total weight of all the bags is 1792.8 kg, how many bags did she buy?

Show SolutionHide Solution↓

Weight of each bag = 49.8 kg

Total weight of all the bags = 1792.8 kg

Number of bags = $\frac{1792.8}{49.8}$

Multiplying both the numerator and denominator by 10, we get:

Number of bags = $\frac{1792.8 \times 10}{49.8 \times 10} = \frac{17928}{498}$

By actual division, we have:

$$\begin{array}{l} \phantom{498)\space 179}36 \\ 498\overline{\smash{\big)}\space 17928} \\ \phantom{498}\underline{-1494} \\ \phantom{498)\space )1}2988 \\ \phantom{498)\space)}\underline{-2988} \\ \phantom{498)\space 1792}\times \end{array}$$

Hence, Farida bought 36 bags of cement.

1Objective Type Questions - Mental Maths

Fill in the following blanks:

(i) The decimal point in a decimal number is placed between ones digit and .... digit.

(ii) The place value of the digit 3 in the decimal number 15.437 is .....

(iii) The decimal number 27.025 has ..... decimal places.

(iv) The decimal number 5.06 is read as five point .....

(v) If an object is divided into 1000 equal parts, then its 27 parts are represented by .....

(vi) Two decimal numbers having different number of ..... are called unlike decimal numbers.

(vii) 4 tens, 3 ones, 2 tenths, 0 hundredths and 5 thousandths in decimal form is written as .....

(viii) The smallest decimal number upto three decimal places is ......

(ix) The largest four digit decimal number less than 1 using the digits 1, 5, 3 and 8 once is ...

Show SolutionHide Solution↓

(i) The decimal point in a decimal number is placed between ones digit and tenths digit.

(ii) The place value of the digit 3 in the decimal number 15.437 is $\frac{3}{100}$.

(iii) The decimal number 27.025 has 3 decimal places.

(iv) The decimal number 5.06 is read as five point zero six.

(v) If an object is divided into 1000 equal parts, then its 27 parts are represented by 0.027.

(vi) Two decimal numbers having different number of decimal places are called unlike decimal numbers.

(vii) 4 tens, 3 ones, 2 tenths, 0 hundredths and 5 thousandths in decimal form is written as 43.205.

(viii) The smallest decimal number upto three decimal places is 0.001.

(ix) The largest four digit decimal number less than 1 using the digits 1, 5, 3 and 8 once is 0.8531.

2Objective Type Questions - Mental Maths

State whether the following statements are true (T) or false (F):

(i) Every decimal number can be represented by a point on a number line.

(ii) Fractions with denominator 10, 100, 1000, ..... are called decimal fractions.

(iii) A decimal number having 3 decimal places can be written as a fraction with denominator 1000.

(iv) The value of a decimal number remains the same if any number of extra zeros are written at the end of a decimal number.

(v) If a decimal number is multiplied by 10, then the decimal point moves by one place to the left.

Show SolutionHide Solution↓

(i) True. Every decimal number corresponds to a unique point on a number line.

(ii) True. By definition, fractions with denominator 10, 100, 1000, ... are called decimal fractions.

(iii) True. For a decimal number with 3 decimal places, the denominator of its decimal fraction is 1000.

(iv) True. For example, 3.7 = 3.70 = 3.700.

(v) False. When a decimal number is multiplied by 10, the decimal point moves by one place to the right, not to the left.

31 Mark · Multiple Choice Questions

Five and seven hundredths is equal to

Show SolutionHide Solution↓

Five and seven hundredths = $5 + \frac{7}{100} = 5 + 0.07 = 5.07$

Hence, option 2 is the correct option.

41 Mark · Multiple Choice Questions

Sixty three thousandths is equal to

Show SolutionHide Solution↓

Sixty three thousandths = $\frac{63}{1000} = 0.063$

Hence, option 3 is the correct option.

51 Mark · Multiple Choice Questions

$3\frac{7}{100}$ is equal to

Show SolutionHide Solution↓

$3\frac{7}{100} = 3 + \frac{7}{100} = 3 + 0.07 = 3.07$

Hence, option 1 is the correct option.

61 Mark · Multiple Choice Questions

$5\frac{3}{1000}$ is equal to

Show SolutionHide Solution↓

$5\frac{3}{1000} = 5 + \frac{3}{1000} = 5 + 0.003 = 5.003$

Hence, option 3 is the correct option.

71 Mark · Multiple Choice Questions

The place value of the digit 7 in the decimal number 5.0378 is

Show SolutionHide Solution↓

In the number 5.0378, the digit 7 is at the thousandths place.

So, place value of 7 = $\frac{7}{1000}$.

Hence, option 4 is the correct option.

81 Mark · Multiple Choice Questions

The place value of the digit 0 in the decimal number 13.405 is

Show SolutionHide Solution↓

In the number 13.405, the digit 0 is at the hundredths place. The place value of 0 at any position is 0.

Hence, option 1 is the correct option.

91 Mark · Multiple Choice Questions

The value of $5 + \frac{7}{10} + \frac{3}{1000}$ is

Show SolutionHide Solution↓

$5 + \frac{7}{10} + \frac{3}{1000} = 5 + 0.7 + 0.003 = 5.703$

Hence, option 2 is the correct option.

101 Mark · Multiple Choice Questions

The value of $\frac{3}{25}$ is

Show SolutionHide Solution↓

$\frac{3}{25} = \frac{3 \times 4}{25 \times 4} = \frac{12}{100} = 0.12$

Hence, option 3 is the correct option.

111 Mark · Multiple Choice Questions

The value of $5\frac{1}{25}$ is

Show SolutionHide Solution↓

$5\frac{1}{25} = 5 + \frac{1}{25} = 5 + \frac{1 \times 4}{25 \times 4} = 5 + \frac{4}{100} = 5 + 0.04 = 5.04$

Hence, option 3 is the correct option.

121 Mark · Multiple Choice Questions

The decimal number not equivalent to 5.7 is

Show SolutionHide Solution↓

The value of a decimal number remains the same if extra zeros are written at the end. So, 5.7 = 5.70 = 5.700 = 5.7000. But 5.07 is different from 5.7.

Hence, option 2 is the correct option.

131 Mark · Multiple Choice Questions

When 0.04 is written as a fraction in its simplest form, then the sum of numerator and denominator is

Show SolutionHide Solution↓

$0.04 = \frac{4}{100} = \frac{1}{25}$

Sum of numerator and denominator = 1 + 25 = 26.

Hence, option 3 is the correct option.

141 Mark · Multiple Choice Questions

13.572 correct to the tenth place is

Show SolutionHide Solution↓

To round 13.572 to the tenth place, look at the digit in the hundredths place, which is 7. Since 7 ≥ 5, we round up the tenths digit.

So, 13.572 ≈ 13.6.

Hence, option 4 is the correct option.

151 Mark · Multiple Choice Questions

1 g is equal to

Show SolutionHide Solution↓

$1 \text{ g} = \frac{1}{1000} \text{ kg} = 0.001 \text{ kg}$

Hence, option 3 is the correct option.

161 Mark · Multiple Choice Questions

2 km 7 m is equal to

Show SolutionHide Solution↓

$2 \text{ km } 7 \text{ m} = 2 \text{ km} + \frac{7}{1000} \text{ km} = 2 \text{ km} + 0.007 \text{ km} = 2.007 \text{ km}$

Hence, option 3 is the correct option.

171 Mark · Multiple Choice Questions

Among 2.34, 2.43, 2.344 and 2.4, the greatest number is

Show SolutionHide Solution↓

Converting the given decimals into like decimals (3 decimal places):

2.340, 2.430, 2.344, 2.400

Clearly, 2.430 is the greatest. So, the greatest number is 2.43.

Hence, option 2 is the correct option.

181 Mark · Multiple Choice Questions

5.2 - 3.6 is equal to

Show SolutionHide Solution↓

$$\begin{array}{r} \phantom{+}5.2 \\ -\space 3.6 \\ \hline 1.6 \end{array}$$

Hence, option 4 is the correct option.

191 Mark · Multiple Choice Questions

A decimal number lying between 2.2 and 2.22 is

Show SolutionHide Solution↓

Writing 2.2 and 2.22 as like decimals, we get 2.200 and 2.220.

Among the options, 2.219 lies between 2.200 and 2.220.

Hence, option 3 is the correct option.

201 Mark · Multiple Choice Questions

0.023 lies between

Show SolutionHide Solution↓

Writing 0.023 with 3 decimal places, we see that 0.020 < 0.023 < 0.030. So, 0.023 lies between 0.02 and 0.03.

Hence, option 2 is the correct option.

211 Mark · Multiple Choice Questions

0.7499 lies between

Show SolutionHide Solution↓

Writing all as like decimals, 0.7490 < 0.7499 < 0.7500. So, 0.7499 lies between 0.749 and 0.75.

Hence, option 3 is the correct option.

221 Mark · Multiple Choice Questions

Which of the following decimal numbers is the greatest?

Show SolutionHide Solution↓

Comparing all four decimal numbers, clearly 0.291 > 0.219 > 0.182 > 0.038. So, 0.291 is the greatest.

Hence, option 4 is the correct option.

231 Mark · Multiple Choice Questions

Which of the following decimal numbers is the smallest?

Show SolutionHide Solution↓

Converting to like decimals (3 decimal places): 0.108, 1.080, 0.801, 0.810.

Clearly, 0.108 < 0.801 < 0.810 < 1.080. So, 0.108 is the smallest.

Hence, option 1 is the correct option.

241 Mark · Multiple Choice Questions

0.003 × 0.2 is equal to

Show SolutionHide Solution↓

Ignoring the decimal points, 3 × 2 = 6.

The two factors together have 3 + 1 = 4 decimal places.

So, 0.003 × 0.2 = 0.0006.

Hence, option 4 is the correct option.

251 Mark · Multiple Choice Questions

0.45 ÷ 0.9 is equal to

Show SolutionHide Solution↓

$0.45 \div 0.9 = \frac{0.45 \times 10}{0.9 \times 10} = \frac{4.5}{9} = 0.5$

Hence, option 3 is the correct option.

26Statement I-II Type Questions

Statement I: In the number 34.17, the whole number part is 34 and the decimal part is 0.17.

Statement II: $34.17 = 3 \times 10 + 4 \times 1 + 1 \times \frac{1}{10} + 7 \times \frac{1}{100}$

Show SolutionHide Solution↓

Statement I: In the number 34.17, the part to the left of the decimal point is 34 (whole number part) and the part to the right of the decimal point is 0.17 (decimal part). So, Statement I is true.

Statement II: Writing 34.17 in expanded form,

$34.17 = 3 \times 10 + 4 \times 1 + 1 \times \frac{1}{10} + 7 \times \frac{1}{100}$

So, Statement II is true.

Hence, option 3 is the correct option.

27Statement I-II Type Questions

Statement I: $\frac{1}{20}$ is a terminating fraction.

Statement II: The denominator of $\frac{1}{20}$ is the 20th multiple of its numerator.

Show SolutionHide Solution↓

Statement I: We have $\frac{1}{20} = \frac{1 \times 5}{20 \times 5} = \frac{5}{100} = 0.05$, which is a terminating decimal. So, Statement I is true.

Statement II: The denominator of $\frac{1}{20}$ is 20, and $20 = 20 \times 1$, which is the 20th multiple of the numerator 1. So, Statement II is true.

Hence, option 3 is the correct option.

28Statement I-II Type Questions

Statement I: According to a report from 2021, about 24.62% of our country's total land area is covered by forests. When we round this percentage to the nearest whole number, it represents $\frac{1}{4}$ th of the total land area.

Statement II: 24.62 rounded off to the nearest whole number is 25. Also, 25% is $\frac{1}{4}$.

Show SolutionHide Solution↓

Statement II: To round 24.62 to the nearest whole number, look at the tenths digit, which is 6 ≥ 5, so we round up. So 24.62 ≈ 25.

Also, $25% = \frac{25}{100} = \frac{1}{4}$. So Statement II is true.

Statement I: 24.62% ≈ 25% = $\frac{1}{4}$. So Statement I is also true.

Hence, option 3 is the correct option.

29Statement I-II Type Questions

Statement I: 42.500 and 42.123 two unlike decimal numbers.

Statement II: 42.500 = 42.5

Show SolutionHide Solution↓

Statement I: Both 42.500 and 42.123 have 3 decimal places, so they are like decimals, not unlike. So, Statement I is false.

Statement II: The value of a decimal number does not change if extra zeros are written at the end of its decimal part. So $42.500 = 42.5$. So, Statement II is true.

Hence, option 2 is the correct option.

30Statement I-II Type Questions

Statement I: The cost of 68.45 m of cloth at the rate of ₹400 per metre is ₹273.80 × 100

Statement II: 40.0 kL = 40,000.00 L

Show SolutionHide Solution↓

Statement I: Cost of 68.45 m of cloth = ₹$(68.45 \times 400)$ = ₹27380.

Also, ₹273.80 × 100 = ₹27380. So, Statement I is true.

Statement II: $1 \text{ kL} = 1000 \text{ L}$, so $40.0 \text{ kL} = 40 \times 1000 = 40000 \text{ L} = 40,000.00 \text{ L}$. So, Statement II is true.

Hence, option 3 is the correct option.

1Check Your Progress

Convert the following decimal numbers into fractions (in lowest terms):

(i) 6.015

(ii) 0.876

(iii) 23.375

Show SolutionHide Solution↓

(i) Writing the decimal 6.015 as a fraction,

$6.015 = \frac{6015}{1000} = \frac{1203}{200} = 6\frac{3}{200}$

Therefore, the required fraction is $6\frac{3}{200}$.

(ii) Writing the decimal 0.876 as a fraction,

$0.876 = \frac{876}{1000} = \frac{219}{250}$

Therefore, the required fraction is $\frac{219}{250}$.

(iii) Writing the decimal 23.375 as a fraction,

$23.375 = \frac{23375}{1000} = \frac{187}{8}= 23\frac{3}{8}$

Therefore, the required fraction is $23\frac{3}{8}$.

2Check Your Progress

Write the following fractions as decimal numbers:

(i) $\frac{5}{8}$

(ii) $2\frac{31}{125}$

(iii) $13\frac{7}{40}$

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(i) Converting the fraction $\frac{5}{8}$ into a decimal,

$\frac{5}{8} = \frac{5 \times 125}{8 \times 125} = \frac{625}{1000} = 0.625$

Therefore, the decimal form is 0.625.

(ii) Converting the fraction $2\frac{31}{125}$ into a decimal,

$2\frac{31}{125} = 2 + \frac{31}{125} = 2 + \frac{31 \times 8}{125 \times 8} = 2 + \frac{248}{1000} = 2 + 0.248 = 2.248$

Therefore, the decimal form is 2.248.

(iii) Converting the fraction $13\frac{7}{40}$ into a decimal,

$13\frac{7}{40} = 13 + \frac{7}{40} = 13 + \frac{7 \times 25}{40 \times 25} = 13 + \frac{175}{1000} = 13 + 0.175 = 13.175$

Therefore, the decimal form is 13.175.

3Check Your Progress

Arrange the following decimal numbers in ascending order:

(i) 123.8, 74.205, 74.209, 7.4209

(ii) 85.01, 85.1, 85.001, 85.103

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(i) The maximum number of decimal places in the given decimals is 4. So, we convert each of the given decimals into like decimals having four decimal places by annexing zeros.

The decimals are: 123.8000, 74.2050, 74.2090, 7.4209

Clearly, 7.4209 < 74.2050 < 74.2090 < 123.8000.

Hence, the given decimals in ascending order are:

7.4209 < 74.205 < 74.209 < 123.8

(ii) The maximum number of decimal places in the given decimals is 3. So, we convert each of the given decimals into like decimals having three decimal places by annexing zeros.

The decimals are: 85.010, 85.100, 85.001, 85.103

Clearly, 85.001 < 85.010 < 85.100 < 85.103.

Hence, the given decimals in ascending order are:

85.001 < 85.01 < 85.1 < 85.103

4Check Your Progress

Arrange the following decimal numbers in descending order:

(i) 6.45, 4.65, 6.405, 64.5

(ii) 73.5, 35.7, 7.35, 7.035

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(i) The maximum number of decimal places in the given decimals is 3. So, we convert each of the given decimals into like decimals having three decimal places by annexing zeros.

The decimals are: 6.450, 4.650, 6.405, 64.500

Clearly, 64.500 > 6.450 > 6.405 > 4.650.

Hence, the given decimals in descending order are:

64.5 > 6.45 > 6.405 > 4.65

(ii) The maximum number of decimal places in the given decimals is 3. So, we convert each of the given decimals into like decimals having three decimal places by annexing zeros.

The decimals are: 73.500, 35.700, 7.350, 7.035

Clearly, 73.500 > 35.700 > 7.350 > 7.035.

Hence, the given decimals in descending order are:

73.5 > 35.7 > 7.35 > 7.035

5Check Your Progress

If the school bags of Garima and Nakul weigh 5.2 kg and 4.832 kg respectively, find

(i) the total weight

(ii) the difference in weight of the bags

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Converting the given decimals into like decimals having three decimal places, we get 5.200 kg and 4.832 kg.

(i) The total weight = 5.200 + 4.832

Writing in column form and adding:

$$\begin{array}{r} \phantom{+}5.200 \\ +\space 4.832 \\ \hline 10.032 \end{array}$$

Hence, the total weight of the bags is 10.032 kg.

(ii) The difference in weight = 5.200 - 4.832

Writing in column form and subtracting:

$$\begin{array}{r} \phantom{+}5.200 \\ -\space 4.832 \\ \hline 0.368 \end{array}$$

Hence, the difference in weight of the bags is 0.368 kg.

6Check Your Progress

Evaluate the following:

(i) 31.42 - 17.853 - 6.43

(ii) 13.01 - 5.428 - 3.703 + 2.99

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(i) Writing each of the given numbers with three decimal places, we get:

Given expression = 31.420 - 17.853 - 6.430

$\phantom{Given expression}$ = 31.420 - (17.853 + 6.430)

Now,

$$\begin{array}{r} \phantom{+}17.853 \\ +\space 6.430 \\ \hline 24.283 \end{array}$$

So,

$$\begin{array}{r} \phantom{+}31.420 \\ -\space 24.283 \\ \hline 7.137 \end{array}$$

Therefore, 31.42 - 17.853 - 6.43 = 7.137.

(ii) Writing each of the given numbers with three decimal places, we get:

Given expression = 13.010 - 5.428 - 3.703 + 2.990

$\phantom{Given expression}$ = (13.010 + 2.990) - (5.428 + 3.703)

Now,

$$\begin{array}{r} \phantom{+}13.010 \\ +\space 2.990 \\ \hline 16.000 \end{array} \qquad \text{and} \qquad \begin{array}{r} \phantom{+}5.428 \\ +\space 3.703 \\ \hline 9.131 \end{array}$$

So,

$$\begin{array}{r} \phantom{+}16.000 \\ -\space 9.131 \\ \hline 6.869 \end{array}$$

Therefore, 13.01 - 5.428 - 3.703 + 2.99 = 6.869.

7Check Your Progress

By how much does the sum of 15.453 and 31.647 exceed the sum of 18.47 and 19.506?

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First sum = 15.453 + 31.647

$$\begin{array}{r} \phantom{+}15.453 \\ +\space 31.647 \\ \hline 47.100 \end{array}$$

Second sum = 18.47 + 19.506

Converting into like decimals, we get 18.470 and 19.506.

$$\begin{array}{r} \phantom{+}18.470 \\ +\space 19.506 \\ \hline 37.976 \end{array}$$

Required difference = 47.100 - 37.976

$$\begin{array}{r} \phantom{+}47.100 \\ -\space 37.976 \\ \hline 9.124 \end{array}$$

Hence, the sum of 15.453 and 31.647 exceeds the sum of 18.47 and 19.506 by 9.124.

8Check Your Progress

Convert 2435 m to km and express the result as mixed fraction.

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We know that 1 m = $\frac{1}{1000}$ km.

So, 2435 m = $\frac{2435}{1000} = 2.435$ km

Reducing to lowest terms,

$\frac{2435}{1000} = \frac{487}{200}$

Converting to a mixed fraction,

$\frac{487}{200} = 2\frac{87}{200}$

Hence, 2435 m = 2.435 km and 2435 m = $2\frac{87}{200}$ km.

9Check Your Progress

Namita travels 20 km 50 m every day. Out of this she travels 10 km 200 m by bus and the rest by auto. How much distance does she travel by auto?

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Total distance travelled = 20 km 50 m

= 20 km + $\frac{50}{1000}$ km = 20 km + 0.050 km = 20.050 km

Distance travelled by bus = 10 km 200 m

= 10 km + $\frac{200}{1000}$ km = 10 km + 0.200 km = 10.200 km

Distance travelled by auto = 20.050 - 10.200

Writing in column form and subtracting:

$$\begin{array}{r} \phantom{+}20.050 \\ -\space 10.200 \\ \hline 9.850 \end{array}$$

Hence, Namita travels 9.850 km or 9 km 850 m by auto.

10Check Your Progress

Asha purchased 5 kg 400 g rice, 2 kg 20 g sugar and 10 kg 850 g flour. Find the total weight of his purchases.

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Weight of rice = 5 kg 400 g

= 5 kg + $\frac{400}{1000}$ kg = 5 kg + 0.400 kg = 5.400 kg

Weight of sugar = 2 kg 20 g

= 2 kg + $\frac{20}{1000}$ kg = 2 kg + 0.020 kg = 2.020 kg

Weight of flour = 10 kg 850 g

= 10 kg + $\frac{850}{1000}$ kg = 10 kg + 0.850 kg = 10.850 kg

Total weight = 5.400 + 2.020 + 10.850

Writing in column form and adding:

$$\begin{array}{r} \phantom{+}5.400 \\ 2.020 \\ +\space 10.850 \\ \hline 18.270 \end{array}$$

Hence, the total weight of Asha's purchases is 18.270 kg or 18 kg 270 g.

11Check Your Progress

1 kg of pure milk contains 0.263 kg of fat. How much fat is there 15.5 kg of milk?

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Quantity of fat in 1 kg of milk = 0.263 kg

Quantity of fat in 15.5 kg of milk = 0.263 × 15.5

Ignoring the decimal points, 263 × 155 = 40765.

The two factors together have 3 + 1 = 4 decimal places, so:

0.263 × 15.5 = 4.0765

Hence, there is 4.0765 kg of fat in 15.5 kg of milk.

12Check Your Progress

The product of two numbers is 15.275. If one number is 4.7, find the other.

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Product of the two numbers = 15.275

One number = 4.7

The other number = $\frac{15.275}{4.7}$

Multiplying both the numerator and denominator by 10, we get:

The other number = $\frac{15.275 \times 10}{4.7 \times 10} = \frac{152.75}{47}$

By actual division, we have:

$$\begin{array}{l} \phantom{47)\space 1}3.25 \\ 47\overline{\smash{\big)}\space 152.75} \\ \phantom{47}\underline{-141} \\ \phantom{47)))\space }117 \\ \phantom{47)))\space}\underline{-94\phantom{1}} \\ \phantom{47)))\space 1}235 \\ \phantom{47)))\space }\underline{-235} \\ \phantom{47)\space 152.7}\times \end{array}$$

Hence, the other number is 3.25.

13Check Your Progress

On her birthday, Ayushi is taking her 5 friends to a movie and treats them with cold drink. The cost of a ticket is ₹150 and a cold drink costs ₹28.50. How much Ayushi has to spend?

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Total number of persons = Ayushi + 5 friends = 6

Cost of one ticket = ₹150

Cost of 6 tickets = ₹150 × 6 = ₹900

Cost of one cold drink = ₹28.50

Cost of 6 cold drinks = ₹28.50 × 6 = ₹171.00

Total amount Ayushi has to spend = ₹900 + ₹171

Writing in column form and adding:

$$\begin{array}{r} \phantom{+}900.00 \\ +\space 171.00 \\ \hline 1071.00 \end{array}$$

Hence, Ayushi has to spend ₹1,071.

14Check Your Progress

Write digits in the boxes of the number:

$3\boxed{\phantom{6}}6\boxed{\phantom{6}}.8\boxed{\phantom{6}} \boxed{\phantom{6}}$ to obtain (i) greatest possible number (ii) smallest possible number. Repetition of digits in a number is not allowed.

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The number is of the form 3⬜6⬜.8 ⬜⬜ where the digits 3, 6 and 8 are already used. The four boxes have to be filled with four distinct digits chosen from {0, 1, 2, 4, 5, 7, 9}.

(i) Greatest possible number:

To get the greatest number, we fill the boxes with the four largest available digits in decreasing order from the leftmost box: 9, 7, 5, 4.

So the boxes are filled as: 3967.854

Greatest possible number = 3967.854

(ii) Smallest possible number:

To get the smallest number, we fill the boxes with the four smallest available digits in increasing order from the leftmost box: 0, 1, 2, 4.

So the boxes are filled as: 3061.824

Smallest possible number = 3061.824

15Check Your Progress

Arrange the following numbers in descending order:

$5\frac{3}{4}$, 5.721, $5\frac{7}{8}$, $5\frac{17}{25}$, 5.693

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First, convert each mixed fraction into a decimal number.

$5\frac{3}{4} = 5 + \frac{3}{4} = 5 + \frac{3 \times 25}{4 \times 25} = 5 + \frac{75}{100} = 5 + 0.75 = 5.75$

$5\frac{7}{8} = 5 + \frac{7}{8} = 5 + \frac{7 \times 125}{8 \times 125} = 5 + \frac{875}{1000} = 5 + 0.875 = 5.875$

$5\frac{17}{25} = 5 + \frac{17}{25} = 5 + \frac{17 \times 4}{25 \times 4} = 5 + \frac{68}{100} = 5 + 0.68 = 5.68$

So the given numbers in decimal form are:

5.75, 5.721, 5.875, 5.68, 5.693

Converting into like decimals, each having three decimal places:

5.750, 5.721, 5.875, 5.680, 5.693

Clearly, 5.875 > 5.750 > 5.721 > 5.693 > 5.680.

Hence, the given numbers in descending order are:

$5\frac{7}{8}$ > $5\frac{3}{4}$ > 5.721 > 5.693 > $5\frac{17}{25}$

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