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Chapter 5

Decimal Fractions

Class - 6 RS Aggarwal Mathematics Solutions



Exercise 5(A)

Question 1

Write each of the following decimals in figures:

(i) Seventy three point eight four.

(ii) Three hundred forty six point zero nine five.

(iii) Five hundred eight point two six.

(iv) Nineteen point six one eight.

(v) Zero point seven nine three.

(vi) Ten point zero zero seven.

Answer

(i) 73.84

(ii) 346.095

(iii) 508.26

(iv) 19.618

(v) 0.793

(vi) 10.007

Question 2

Write the place value of each digit in each of the following decimals:

(i) 46.38

(ii) 57.095

(iii) 132.405

(iv) 19.006

Answer

(i) We can arrange the digits of 46.38 in the place value chart, as shown below:

TensOnesDecimal pointTenthsHundredths
46.38

The place value of each digit is given below:

Digit4638
Place Value406310\dfrac{3}{10}8100\dfrac{8}{100}

(ii) We can arrange the digits of 57.095 in the place value chart, as shown below:

TensOnesDecimal pointTenthsHundredthsThousandths
57.095

The place value of each digit is given below:

Digit57095
Place Value50709100\dfrac{9}{100}51000\dfrac{5}{1000}

(iii) We can arrange the digits of 132.405 in the place value chart, as shown below:

HundredsTensOnesDecimal pointTenthsHundredthsThousandths
132.405

The place value of each digit is given below:

Digit132405
Place Value100302410\dfrac{4}{10}051000\dfrac{5}{1000}

(iv) We can arrange the digits of 19.006 in the place value chart, as shown below:

TensOnesDecimal pointTenthsHundredthsThousandths
19.006

The place value of each digit is given below:

Digit19006
Place Value1090061000\dfrac{6}{1000}

Question 3

Write each of the following decimals in expanded form:

(i) 53.48

(ii) 6.927

(iii) 10.084

(iv) 8.005

(v) 274.503

(vi) 31.02

(vii) 1635.72

Answer

(i) We have,

53.48=5×10+3×1+4×110+8×1100=50+3+410+810053.48 = 5 \times 10 + 3 \times 1 + 4 \times \dfrac{1}{10} + 8 \times \dfrac{1}{100} \\[1em] = 50 + 3 + \dfrac{4}{10} + \dfrac{8}{100}

(ii) We have,

6.927=6×1+9×110+2×1100+7×11000=6+910+2100+710006.927 = 6 \times 1 + 9 \times \dfrac{1}{10} + 2 \times \dfrac{1}{100} + 7 \times \dfrac{1}{1000} \\[1em] = 6 + \dfrac{9}{10} + \dfrac{2}{100} + \dfrac{7}{1000}

(iii) We have,

10.084=1×10+0×1+0×110+8×1100+4×11000=10+8100+4100010.084 = 1 \times 10 + 0 \times 1 + 0 \times \dfrac{1}{10} + 8 \times \dfrac{1}{100} + 4 \times \dfrac{1}{1000} \\[1em] = 10 + \dfrac{8}{100} + \dfrac{4}{1000}

(iv) We have,

8.005=8×1+0×110+0×1100+5×11000=8+510008.005 = 8 \times 1 + 0 \times \dfrac{1}{10} + 0 \times \dfrac{1}{100} + 5 \times \dfrac{1}{1000} \\[1em] = 8 + \dfrac{5}{1000}

(v) We have,

274.503=2×100+7×10+4×1+5×110+0×1100+3×11000=200+70+4+510+31000274.503 = 2 \times 100 + 7 \times 10 + 4 \times 1 + 5 \times \dfrac{1}{10} + 0 \times \dfrac{1}{100} + 3 \times \dfrac{1}{1000} \\[1em] = 200 + 70 + 4 + \dfrac{5}{10} + \dfrac{3}{1000}

(vi) We have,

31.02=3×10+1×1+0×110+2×1100=30+1+210031.02 = 3 \times 10 + 1 \times 1 + 0 \times \dfrac{1}{10} + 2 \times \dfrac{1}{100} \\[1em] = 30 + 1 + \dfrac{2}{100}

(vii) We have,

1635.72=1×1000+6×100+3×10+5×1+7×110+2×1100=1000+600+30+5+710+21001635.72 = 1 \times 1000 + 6 \times 100 + 3 \times 10 + 5 \times 1 + 7 \times \dfrac{1}{10} + 2 \times \dfrac{1}{100} \\[1em] = 1000 + 600 + 30 + 5 + \dfrac{7}{10} + \dfrac{2}{100}

Question 4

Write each of the following as a decimal number:

(i) 60 + 8 + 410+5100\dfrac{4}{10} + \dfrac{5}{100}

(ii) 500 + 4 + 610+8100+31000\dfrac{6}{10} + \dfrac{8}{100} + \dfrac{3}{1000}

(iii) 300 + 20 + 110+61000\dfrac{1}{10} + \dfrac{6}{1000}

(iv) 2000 + 80 + 4100+81000\dfrac{4}{100} + \dfrac{8}{1000}

(v) 6000 + 3 + 210+71000\dfrac{2}{10} + \dfrac{7}{1000}

Answer

(i) We have,

60+8+410+5100=6×10+8×1+4×110+5×1100=68.4560 + 8 + \dfrac{4}{10} + \dfrac{5}{100} \\[1em] = 6 \times 10 + 8 \times 1 + 4 \times \dfrac{1}{10} + 5 \times \dfrac{1}{100} \\[1em] = 68.45

(ii) We have,

500+4+610+8100+31000=5×100+0×10+4×1+6×110+8×1100+3×11000=504.683500 + 4 + \dfrac{6}{10} + \dfrac{8}{100} + \dfrac{3}{1000} \\[1em] = 5 \times 100 + 0 \times 10 + 4 \times 1 + 6 \times \dfrac{1}{10} + 8 \times \dfrac{1}{100} + 3 \times \dfrac{1}{1000} \\[1em] = 504.683

(iii) We have,

300+20+110+61000=3×100+2×10+0×1+1×110+0×1100+6×11000=320.106300 + 20 + \dfrac{1}{10} + \dfrac{6}{1000} \\[1em] = 3 \times 100 + 2 \times 10 + 0 \times 1 + 1 \times \dfrac{1}{10} + 0 \times \dfrac{1}{100} + 6 \times \dfrac{1}{1000} \\[1em] = 320.106

(iv) We have,

2000+80+4100+81000=2×1000+0×100+8×10+0×1+0×110+4×1100+8×11000=2080.0482000 + 80 + \dfrac{4}{100} + \dfrac{8}{1000} \\[1em] = 2 \times 1000 + 0 \times 100 + 8 \times 10 + 0 \times 1 + 0 \times \dfrac{1}{10} + 4 \times \dfrac{1}{100} + 8 \times \dfrac{1}{1000} \\[1em] = 2080.048

(v) We have,

6000+3+210+71000=6×1000+0×100+0×10+3×1+2×110+0×1100+7×11000=6003.2076000 + 3 + \dfrac{2}{10} + \dfrac{7}{1000} \\[1em] = 6 \times 1000 + 0 \times 100 + 0 \times 10 + 3 \times 1 + 2 \times \dfrac{1}{10} + 0 \times \dfrac{1}{100} + 7 \times \dfrac{1}{1000} \\[1em] = 6003.207

Question 5

Write each of the following numerals as decimals:

(i) 27102\dfrac{7}{10}

(ii) 4331043\dfrac{3}{10}

(iii) 6151006\dfrac{15}{100}

(iv) 18910018\dfrac{9}{100}

(v) 512310005\dfrac{123}{1000}

(vi) 43710004\dfrac{37}{1000}

(vii) 108100010\dfrac{8}{1000}

Answer

(i) We have,

2710=2+710=2+0.7=2.72\dfrac{7}{10}=2+\dfrac{7}{10}=2+0.7=2.7

Hence, the decimal form is 2.7.

(ii) We have,

43310=43+310=43+0.3=43.343\dfrac{3}{10}=43+\dfrac{3}{10}=43+0.3=43.3

Hence, the decimal form is 43.3.

(iii) We have,

615100=6+15100=6+0.15=6.156\dfrac{15}{100}=6+\dfrac{15}{100}=6+0.15=6.15

Hence, the decimal form is 6.15.

(iv) We have,

189100=18+9100=18+0.09=18.0918\dfrac{9}{100}=18+\dfrac{9}{100}=18+0.09=18.09

Hence, the decimal form is 18.09.

(v) We have,

51231000=5+1231000=5+0.123=5.1235\dfrac{123}{1000}=5+\dfrac{123}{1000}=5+0.123=5.123

Hence, the decimal form is 5.123.

(vi) We have,

4371000=4+371000=4+0.037=4.0374\dfrac{37}{1000}=4+\dfrac{37}{1000}=4+0.037=4.037

Hence, the decimal form is 4.037.

(vii) We have,

1081000=10+81000=10+0.008=10.00810\dfrac{8}{1000}=10+\dfrac{8}{1000}=10+0.008=10.008

Hence, the decimal form is 10.008.

Exercise 5(B)

Question 1

Convert each of the sets of unlike decimals into like decimals:

(i) 9.4, 65.83, 0.065

(ii) 0.5, 7.573, 4.26, 8.4

(iii) 1.2, 0.06, 3.5, 6.38

(iv) 2.8, 16.85, 64.056, 143.625

Answer

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

9.4 = 9.400

65.83 = 65.830

0.065 = 0.065

As 9.400, 65.830, 0.065 have equal number of decimal places, they are like decimals.

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

0.5 = 0.500

7.573 = 7.573

4.26 = 4.260

8.4 = 8.400

As 0.500, 7.573, 4.260, 8.400 have equal number of decimal places, they are like decimals.

(iii) Maximum number of decimal places in the given numbers is 2. So, we can convert each of the decimals into one having 2 decimal places by annexing zeros.

1.2 = 1.20

0.06 = 0.06

3.5 = 3.50

6.38 = 6.38

As 1.20, 0.06, 3.50, 6.38 have equal number of decimal places, they are like decimals.

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

2.8 = 2.800

16.85 = 16.850

64.056 = 64.056

143.625 = 143.625

As 2.800, 16.850, 64.056, 143.625 have equal number of decimal places, they are like decimals.

Question 2

Fill in the placeholders with > or <:

(i) 63.4 63{\boxed{\phantom{63}}} 36.85

(ii) 4.805 63{\boxed{\phantom{63}}} 4.85

(iii) 8.23 63{\boxed{\phantom{63}}} 8.32

(iv) 5.06 63{\boxed{\phantom{63}}} 5.006

(v) 7.86 63{\boxed{\phantom{63}}} 7.806

(vi) 0.98 63{\boxed{\phantom{63}}} 1

Answer

(i) The given decimals are 63.4 and 36.85. Comparing their whole number parts clearly shows that, 63 > 36.

Therefore, 63.4 > 36.85.

(ii) The given decimals are 4.805 and 4.85.

Writing them as like decimals: 4.805 and 4.850.

Comparing the decimal parts, we find that 805 < 850.

Therefore, 4.805 < 4.85.

(iii) The given decimals are 8.23 and 8.32. Their whole number parts are equal. Comparing their tenths digits, 2 < 3.

Therefore, 8.23 < 8.32.

(iv) The given decimals are 5.06 and 5.006.

Writing them as like decimals: 5.060 and 5.006.

Comparing the decimal parts, 60 > 6.

Therefore, 5.06 > 5.006.

(v) The given decimals are 7.86 and 7.806.

Writing them as like decimals: 7.860 and 7.806.

Comparing the decimal parts, 860 > 806.

Therefore, 7.86 > 7.806.

(vi) The given decimals are 0.98 and 1.

Comparing their whole number parts, 0 < 1.

Therefore, 0.98 < 1.

Question 3

Arrange the following decimals in ascending order:

(i) 3.6, 4.2, 3.57, 4.18, 3.06, 4.08

(ii) 5.37, 3.57, 3.5, 5.3, 5.7, 3.07

(iii) 0.7, 0.007, 0.07, 7.7, 7.07, 0.77

(iv) 2.2, 2.02, 0.22, 0.02, 0.202, 2.002

(v) 76.8, 68.7, 86.7, 87.6, 78.6, 7.86

Answer

(i) The maximum number of decimal places in the given decimals is 2. So, we convert each of the given decimals into like decimals having two decimal places by annexing zeros.

The decimals are: 3.60, 4.20, 3.57, 4.18, 3.06, 4.08

Clearly, 3.06 < 3.57 < 3.60 < 4.08 < 4.18 < 4.20.

Hence, the given decimals in ascending order are:

3.06 < 3.57 < 3.6 < 4.08 < 4.18 < 4.2.

(ii) The maximum number of decimal places in the given decimals is 2. So, we convert each of the given decimals into like decimals having two decimal places by annexing zeros.

The decimals are: 5.37, 3.57, 3.50, 5.30, 5.70, 3.07

Clearly, 3.07 < 3.50 < 3.57 < 5.30 < 5.37 < 5.70.

Hence, the given decimals in ascending order are:

3.07 < 3.5 < 3.57 < 5.3 < 5.37 < 5.7.

(iii) 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: 0.700, 0.007, 0.070, 7.700, 7.070, 0.770

Clearly, 0.007 < 0.070 < 0.700 < 0.770 < 7.070 < 7.700.

Hence, the given decimals in ascending order are:

0.007 < 0.07 < 0.7 < 0.77 < 7.07 < 7.7

(iv) 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: 2.200, 2.020, 0.220, 0.020, 0.202, 2.002

Clearly, 0.020 < 0.202 < 0.220 < 2.002 < 2.020 < 2.200.

Hence, the given decimals in ascending order are:

0.02 < 0.202 < 0.22 < 2.002 < 2.02 < 2.2

(v) The maximum number of decimal places in the given decimals is 2. So, we convert each of the given decimals into like decimals having two decimal places by annexing zeros.

The decimals are: 76.80, 68.70, 86.70, 87.60, 78.60, 7.86

Clearly, 7.86 < 68.70 < 76.80 < 78.60 < 86.70 < 87.60.

Hence, the given decimals in ascending order are:

7.86 < 68.7 < 76.8 < 78.6 < 86.7 < 87.6

Question 4

Arrange the following decimals in descending order:

(i) 6.6, 0.06, 0.6, 0.006, 6.66

(ii) 3.3, 3.03, 0.33, 33.3, 0.303, 30.3

(iii) 2.9, 3.6, 9.2, 6.3, 3.06, 6.03

(iv) 77.7, 0.7, 7.7, 0.07, 7.07, 7.77

(v) 7.35, 3.75, 5.3, 3.57, 5.37, 5.75

Answer

(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: 6.600, 0.060, 0.600, 0.006, 6.660

Clearly, 6.660 > 6.600 > 0.600 > 0.060 > 0.006.

Hence, the given decimals in descending order are:

6.66 > 6.6 > 0.6 > 0.06 > 0.006

(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: 3.300, 3.030, 0.330, 33.300, 0.303, 30.300

Clearly, 33.300 > 30.300 > 3.300 > 3.030 > 0.330 > 0.303.

Hence, the given decimals in descending order are:

33.3 > 30.3 > 3.3 > 3.03 > 0.33 > 0.303

(iii) The maximum number of decimal places in the given decimals is 2. So, we convert each of the given decimals into like decimals having two decimal places by annexing zeros.

The decimals are: 2.90, 3.60, 9.20, 6.30, 3.06, 6.03

Clearly, 9.20 > 6.30 > 6.03 > 3.60 > 3.06 > 2.90.

Hence, the given decimals in descending order are:

9.2 > 6.3 > 6.03 > 3.6 > 3.06 > 2.9

(iv) The maximum number of decimal places in the given decimals is 2. So, we have to convert each of the given decimals to have two places of decimal, by annexing zeros.

The decimals are: 77.70, 0.70, 7.70, 0.07, 7.07, 7.77

Clearly, 77.70 > 7.77 > 7.70 > 7.07 > 0.70 > 0.07.

Hence, the given decimals in descending order are:

77.7 > 7.77 > 7.7 > 7.07 > 0.7 > 0.07

(v) The maximum number of decimal places in the given decimals is 2. So, we have to convert each of the given decimals to have two places of decimal, by annexing zeros.

The decimals are: 7.35, 3.75, 5.30, 3.57, 5.37, 5.75

Clearly, 7.35 > 5.75 > 5.37 > 5.30 > 3.75 > 3.57.

Hence, the given decimals in descending order are:

7.35 > 5.75 > 5.37 > 5.3 > 3.75 > 3.57

Exercise 5(C)

Question 1

Convert each of the following decimals into a fraction with lowest terms:

(i) 0.8

(ii) 0.06

(iii) 0.45

(iv) 0.64

(v) 5.4

(vi) 6.25

(vii) 8.125

(viii) 3.475

Answer

(i) Writing the decimal 0.8 as a fraction,

0.8 = 810=45\dfrac{8}{10}=\dfrac{4}{5}

Therefore, the required fraction is 45\dfrac{4}{5} .

(ii) Writing the decimal 0.06 as a fraction,

0.06 = 6100=350\dfrac{6}{100}=\dfrac{3}{50}

Therefore, the required fraction is 350\dfrac{3}{50} .

(iii) Writing the decimal 0.45 as a fraction,

0.45 = 45100=920\dfrac{45}{100}=\dfrac{9}{20}

Therefore, the required fraction is 920\dfrac{9}{20} .

(iv) Writing the decimal 0.64 as a fraction,

0.64 = 64100=1625\dfrac{64}{100}=\dfrac{16}{25}

Therefore, the required fraction is 1625\dfrac{16}{25} .

(v) Writing the decimal 5.4 as a fraction,

5.4 = 5410=275=525\dfrac{54}{10}=\dfrac{27}{5}=5\dfrac{2}{5}

Therefore, the required fraction is 5255\dfrac{2}{5} .

(vi) Writing the decimal 6.25 as a fraction,

6.25 = 625100=254=614\dfrac{625}{100}=\dfrac{25}{4}=6\dfrac{1}{4}

Therefore, the required fraction is 6146\dfrac{1}{4} .

(vii) Writing the decimal 8.125 as a fraction,

8.125 = 81251000=658=818\dfrac{8125}{1000}=\dfrac{65}{8}=8\dfrac{1}{8}

Therefore, the required fraction is 8188\dfrac{1}{8} .

(viii) Writing the decimal 3.475 as a fraction,

3.475 = 34751000=13940=31940\dfrac{3475}{1000}=\dfrac{139}{40}=3\dfrac{19}{40}

Therefore, the required fraction is 319403\dfrac{19}{40} .

Question 2

Convert each of the following fractions into decimals:

(i) 910\dfrac{9}{10}

(ii) 2310\dfrac{23}{10}

(iii) 42710\dfrac{427}{10}

(iv) 37100\dfrac{37}{100}

(v) 107100\dfrac{107}{100}

(vi) 1423100\dfrac{1423}{100}

(vii) 1231000\dfrac{123}{1000}

(viii) 20451000\dfrac{2045}{1000}

Answer

(i) Converting the fraction 910\dfrac{9}{10} into a decimal,

910=0.9\dfrac{9}{10}=0.9

Therefore, the decimal form is 0.9 .

(ii) Converting the fraction 2310\dfrac{23}{10} into a decimal,

2310=2310=2+310=2+0.3=2.3\dfrac{23}{10}=2\dfrac{3}{10}=2+\dfrac{3}{10}=2+0.3=2.3

Therefore, the decimal form is 2.3 .

(iii) Converting the fraction 42710\dfrac{427}{10} into a decimal,

42710=42710=42+710=42+0.7=42.7\dfrac{427}{10}=42\dfrac{7}{10}=42+\dfrac{7}{10}=42+0.7=42.7

Therefore, the decimal form is 42.7 .

(iv) Converting the fraction 37100\dfrac{37}{100} into a decimal,

37100=0.37\dfrac{37}{100}=0.37

Therefore, the decimal form is 0.37.

(v) Converting the fraction 107100\dfrac{107}{100} into a decimal,

107100=17100=1+7100=1+0.07=1.07\dfrac{107}{100}=1\dfrac{7}{100}=1+\dfrac{7}{100}=1+0.07=1.07

Therefore, the decimal form is 1.07 .

(vi) Converting the fraction 1423100\dfrac{1423}{100} into a decimal,

1423100=1423100=14+23100=14+0.23=14.23\dfrac{1423}{100}=14\dfrac{23}{100}=14+\dfrac{23}{100}=14+0.23=14.23

Therefore, the decimal form is 14.23.

(vii) Converting the fraction 1231000\dfrac{123}{1000} into a decimal,

1231000=0.123\dfrac{123}{1000}=0.123

Therefore, the decimal form is 0.123 .

(viii) Converting the fraction 20451000\dfrac{2045}{1000} into a decimal,

20451000=2451000=2+451000=2+0.045=2.045\dfrac{2045}{1000}=2\dfrac{45}{1000}=2+\dfrac{45}{1000}=2+0.045=2.045

Therefore, the decimal form is 2.045 .

Question 3

Convert each of the following fractions into decimals:

(i) 320\dfrac{3}{20}

(ii) 725\dfrac{7}{25}

(iii) 950\dfrac{9}{50}

(iv) 14\dfrac{1}{4}

(v) 16125\dfrac{16}{125}

(vi) 231250\dfrac{231}{250}

(vii) 17500\dfrac{17}{500}

(viii) 8250\dfrac{8}{250}

Answer

(i) Converting the fraction 320\dfrac{3}{20} into a decimal,

320=3×520×5=15100=0.15\dfrac{3}{20}=\dfrac{3\times5}{20\times5}=\dfrac{15}{100}=0.15

Therefore, the decimal form is 0.15.

(ii) Converting the fraction 725\dfrac{7}{25} into a decimal,

725=7×425×4=28100=0.28\dfrac{7}{25}=\dfrac{7\times4}{25\times4}=\dfrac{28}{100}=0.28

Therefore, the decimal form is 0.28.

(iii) Converting the fraction 950\dfrac{9}{50} into a decimal,

950=9×250×2=18100=0.18\dfrac{9}{50}=\dfrac{9\times2}{50\times2}=\dfrac{18}{100}=0.18

Therefore, the decimal form is 0.18.

(iv) Converting the fraction 14\dfrac{1}{4} into a decimal,

14=1×254×25=25100=0.25\dfrac{1}{4}=\dfrac{1\times25}{4\times25}=\dfrac{25}{100}=0.25

Therefore, the decimal form is 0.25.

(v) Converting the fraction 16125\dfrac{16}{125} into a decimal,

16125=16×8125×8=1281000=0.128\dfrac{16}{125}=\dfrac{16\times8}{125\times8}=\dfrac{128}{1000}=0.128

Therefore, the decimal form is 0.128.

(vi) Converting the fraction 231250\dfrac{231}{250} into a decimal,

231250=231×4250×4=9241000=0.924\dfrac{231}{250}=\dfrac{231\times4}{250\times4}=\dfrac{924}{1000}=0.924

Therefore, the decimal form is 0.924.

(vii) Converting the fraction 17500\dfrac{17}{500} into a decimal,

17500=17×2500×2=341000=0.034\dfrac{17}{500}=\dfrac{17\times2}{500\times2}=\dfrac{34}{1000}=0.034

Therefore, the decimal form is 0.034.

(viii) Converting the fraction 8250\dfrac{8}{250} into a decimal,

8250=8×4250×4=321000=0.032\dfrac{8}{250}=\dfrac{8\times4}{250\times4}=\dfrac{32}{1000}=0.032

Therefore, the decimal form is 0.032.

Question 4

Convert each of the following fractions into decimals:

(i) 5345\dfrac{3}{4}

(ii) 3143\dfrac{1}{4}

(iii) 6156\dfrac{1}{5}

(iv) 124512\dfrac{4}{5}

(v) 7387\dfrac{3}{8}

(vi) 24252\dfrac{4}{25}

(vii) 37203\dfrac{7}{20}

(viii) 411204\dfrac{11}{20}

Answer

(i) 534=5×4+34=2345\dfrac{3}{4} = \dfrac{5 \times 4 + 3}{4} = \dfrac{23}{4}

By actual division, we have:

4) 25.754) 23.004))204) 2304) 2284) 28204) 28204) 28×\begin{array}{l} \phantom{4)}{\space \phantom{2} 5.75} \\ 4\overline{\smash{\big)}\space 23.00} \\ \phantom{4)}\phantom{)}\underline{20} \\ \phantom{4)\space 2}30 \\ \phantom{4)\space 2}\underline{28} \\ \phantom{4)\space 28}20 \\ \phantom{4)\space 28}\underline{20} \\ \phantom{4)\space 28}\times \end{array}

534=234=5.75\therefore 5\dfrac{3}{4} = \dfrac{23}{4} = 5.75

Therefore, the decimal form is 5.75.

(ii) 314=3×4+14=1343\dfrac{1}{4}=\dfrac{3\times4 + 1}{4}=\dfrac{13}{4}

By actual division, we have:

4) 23.254) 13.004))124) 2104) 2284) 28204) 28204) 28×\begin{array}{l} \phantom{4)}{\space \phantom{2} 3.25} \\ 4\overline{\smash{\big)}\space 13.00} \\ \phantom{4)}\phantom{)}\underline{12} \\ \phantom{4)\space 2}10 \\ \phantom{4)\space 22}\underline{8} \\ \phantom{4)\space 28}20 \\ \phantom{4)\space 28}\underline{20} \\ \phantom{4)\space 28}\times \end{array}

Therefore, the decimal form is 3.25.

(iii) 615=6×5+15=3156\dfrac{1}{5}=\dfrac{6\times5 + 1}{5}=\dfrac{31}{5}

5) 26.25) 31.05))305) 2.105) 2.105) 28×\begin{array}{l} \phantom{5)}{\space \phantom{2} 6.2} \\ 5\overline{\smash{\big)}\space 31.0} \\ \phantom{5)}\phantom{)}\underline{30} \\ \phantom{5)\space 2.}10 \\ \phantom{5)\space 2.}\underline{10} \\ \phantom{5)\space 28}\times \end{array}

Therefore, the decimal form is 6.2.

(iv) 1245=12×5+45=64512\dfrac{4}{5}=\dfrac{12 \times 5 + 4}{5}=\dfrac{64}{5}

5) 12.85) 64.05))55) .145) .105) 21405) 55405) 28.×\begin{array}{l} \phantom{5)}{\space 12.8} \\ 5\overline{\smash{\big)}\space 64.0} \\ \phantom{5)}\phantom{)}\underline{5} \\ \phantom{5)\space .}14 \\ \phantom{5)\space .}\underline{10} \\ \phantom{5)\space 21}40 \\ \phantom{5)\space 55}\underline{40} \\ \phantom{5)\space 28.}\times \end{array}

Therefore, the decimal form is 12.8.

(v) 738=7×8+38=5987\dfrac{3}{8}=\dfrac{7 \times 8 + 3}{8}=\dfrac{59}{8}

8) )7.3758) 59.0008))568) )0308) )2248) 22)608) 55)568) 222)408) 552)408) 2831×\begin{array}{l} \phantom{8)}{\space \phantom{)} 7.375} \\ 8\overline{\smash{\big)}\space 59.000} \\ \phantom{8)}\phantom{)}\underline{56} \\ \phantom{8)\space )}030 \\ \phantom{8)\space )}\underline{\phantom{2}24} \\ \phantom{8)\space 22)}60 \\ \phantom{8)\space 55}\underline{\phantom{)}56} \\ \phantom{8)\space 222)}40 \\ \phantom{8)\space 55}\underline{\phantom{2)}40} \\ \phantom{8)\space 2831}\times \end{array}

Therefore, the decimal form is 7.375.

(vi) 2425=25×2+425=54252\dfrac{4}{25}=\dfrac{25 \times 2 + 4 }{25}=\dfrac{54}{25}

25) )2.1625) 54.0025))5025) ))4025) ))2525) ))15025) ))15025) 28)×\begin{array}{l} \phantom{25)}{\space \phantom{)} 2.16} \\ 25\overline{\smash{\big)}\space 54.00} \\ \phantom{25)}\phantom{)}\underline{50} \\ \phantom{25)\space ))}40 \\ \phantom{25)\space ))}\underline{25} \\ \phantom{25)\space ))}150 \\ \phantom{25)\space ))}\underline{150} \\ \phantom{25)\space 28)}\times \end{array}

Therefore, the decimal form is 2.16.

(vii) 3720=3×20+720=67203\dfrac{7}{20}=\dfrac{3 \times 20 + 7}{20}=\dfrac{67}{20}

20) )3.3520) 67.0020))6020) ))7020) ))6020) ))10020) ))10020) 28)×\begin{array}{l} \phantom{20)}{\space \phantom{)} 3.35} \\ 20\overline{\smash{\big)}\space 67.00} \\ \phantom{20)}\phantom{)}\underline{60} \\ \phantom{20)\space ))}70 \\ \phantom{20)\space ))}\underline{60} \\ \phantom{20)\space ))}100 \\ \phantom{20)\space ))}\underline{100} \\ \phantom{20)\space 28)}\times \end{array}

Therefore, the decimal form is 3.35.

(viii) 41120=4×20+1120=91204\dfrac{11}{20}=\dfrac{4 \times 20 + 11}{20}=\dfrac{91}{20}

20) )4.5520) 91.0020))8020) )11020) )10020) )))10020) )))10020) 28))×\begin{array}{l} \phantom{20)}{\space \phantom{)} 4.55} \\ 20\overline{\smash{\big)}\space 91.00} \\ \phantom{20)}\phantom{)}\underline{80} \\ \phantom{20)\space )}110 \\ \phantom{20)\space )}\underline{100} \\ \phantom{20)\space )))}100 \\ \phantom{20)\space )))}\underline{100} \\ \phantom{20)\space 28))}\times \end{array}

Therefore, the decimal form is 4.55.

Exercise 5(D)

Question 1

Add:

(i) 27.5, 8.93, 0.759, 9.6

(ii) 0.83, 8.07, 17.9, 294.678

(iii) 0.53, 7.07, 77.7, 0.835

(iv) 9.38, 3.895, 109.5, 72.87, 0.005

(v) 9.008, 27.689, 247.96, 39.8

(vi) 325.7, 98.079, 0.87, 236.68

Answer

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

27.500, 8.930, 0.759, 9.600

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

+27.5008.9300.759+ 9.60046.789\begin{array}{r} \phantom{+}27.500 \\ 8.930 \\ 0.759 \\ +\space 9.600 \\ \hline 46.789 \end{array}

Therefore, the sum is 46.789.

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

0.830, 8.070, 17.900, 294.678

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

+0.8308.07017.900+ 294.678321.478\begin{array}{r} \phantom{+}0.830 \\ 8.070 \\ 17.900 \\ +\space 294.678 \\ \hline 321.478 \end{array}

Therefore, the sum is 321.478.

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

0.530, 7.070, 77.700, 0.835

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

+0.5307.07077.700+ 0.83586.135\begin{array}{r} \phantom{+}0.530 \\ 7.070 \\ 77.700 \\ +\space 0.835 \\ \hline 86.135 \end{array}

Therefore, the sum is 86.135.

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

9.380, 3.895, 109.500, 72.870, 0.005

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

+9.3803.895109.50072.870+ 0.005195.650\begin{array}{r} \phantom{+}9.380 \\ 3.895 \\ 109.500 \\ 72.870 \\ +\space 0.005 \\ \hline 195.650 \end{array}

Therefore, the sum is 195.65.

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

9.008, 27.689, 247.960, 39.800

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

+9.00827.689247.960+ 39.800324.457\begin{array}{r} \phantom{+}9.008 \\ 27.689 \\ 247.960 \\ +\space 39.800 \\ \hline 324.457 \end{array}

Therefore, the sum is 324.457.

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

325.700, 98.079, 0.870, 236.680

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

+325.70098.0790.870+ 236.680661.329\begin{array}{r} \phantom{+}325.700 \\ 98.079 \\ 0.870 \\ +\space 236.680 \\ \hline 661.329 \end{array}

Therefore, the sum is 661.329.

Question 2

Find the difference:

(i) 17.05 - 9.78

(ii) 30.23 - 17.495

(iii) 107.5 - 58.69

(iv) 60 - 18.35

(v) 48.3 - 29.571

(vi) 100 - 72.745

(vii) 215.4 - 69.845

(viii) 1.5 - 0.703

(ix) 42 - 0.764

Answer

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

+17.05 9.787.27\begin{array}{r} \phantom{+}17.05 \\ -\space 9.78 \\ \hline 7.27 \end{array}

Therefore, the difference is 7.27.

(ii) Converting the given decimals into like decimals, each having 3 places of decimal, we get 30.230 and 17.495.

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

+30.230 17.49512.735\begin{array}{r} \phantom{+}30.230 \\ -\space 17.495 \\ \hline 12.735 \end{array}

Therefore, the difference is 12.735.

(iii) Converting the given decimals into like decimals, each having 2 places of decimal, we get 107.50 and 58.69.

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

+107.50 58.6948.81\begin{array}{r} \phantom{+}107.50 \\ -\space 58.69 \\ \hline 48.81 \end{array}

Therefore, the difference is 48.81 .

(iv) Converting the given decimals into like decimals, each having 2 places of decimal, we get 60.00 and 18.35.

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

+60.00 18.3541.65\begin{array}{r} \phantom{+}60.00 \\ - \space 18.35 \\ \hline 41.65 \end{array}

Therefore, the difference is 41.65 .

(v) Converting the given decimals into like decimals, each having 3 places of decimal, we get 48.300 and 29.571.

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

+48.300 29.57118.729\begin{array}{r} \phantom{+}48.300 \\ -\space 29.571 \\ \hline 18.729 \end{array}

Therefore, the difference is 18.729

(vi) Converting the given decimals into like decimals, each having 3 places of decimal, we get 100.000 and 72.745.

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

+100.000 72.74527.255\begin{array}{r} \phantom{+}100.000 \\ -\space 72.745 \\ \hline 27.255 \end{array}

Therefore, the difference is 27.255.

(vii) Converting the given decimals into like decimals, each having 3 places of decimal, we get 215.400 and 69.845.

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

+215.400 69.845145.555\begin{array}{r} \phantom{+}215.400 \\ -\space 69.845 \\ \hline 145.555 \end{array}

Therefore, the difference is 145.555.

(viii) Converting the given decimals into like decimals, each having 3 places of decimal, we get 1.500 and 0.703.

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

+1.500 0.7030.797\begin{array}{r} \phantom{+}1.500 \\ -\space 0.703 \\ \hline 0.797 \end{array}

Therefore, the difference is 0.797.

(ix) Converting the given decimals into like decimals, each having 3 places of decimal, we get 42.000 and 0.764.

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

+42.000 0.76441.236\begin{array}{r} \phantom{+}42.000 \\ -\space 0.764 \\ \hline 41.236 \end{array}

Therefore, the difference is 41.236.

Question 3

Simplify:

(i) 74.8 + 68.79 - 47.609 - 7.08

(ii) 68.3 - 102.735 + 217.96 - 37.7

(iii) 203.5 - 43.76 - 9.89 - 13.086

(iv) 77.7 - 7.555 - 5.77

Answer

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

Given expression = 74.800 + 68.790 - 47.609 - 7.080

74.800+68.790\phantom{74.800 + 68.790-}= (74.800 + 68.790) - (47.609 + 7.080)

Now,

+74.800+ 68.790143.590and+47.609+ 7.08054.689\begin{array}{r} \phantom{+}74.800 \\ +\space 68.790 \\ \hline 143.590 \end{array} \qquad \text{and} \qquad \begin{array}{r} \phantom{+}47.609 \\ +\space 7.080 \\ \hline 54.689 \end{array}

So,

+143.590 54.68988.901\begin{array}{r} \phantom{+}143.590 \\ -\space 54.689 \\ \hline 88.901 \end{array}

Therefore, 74.8 + 68.79 - 47.609 - 7.08 = 88.901.

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

Given expression = 68.300 - 102.735 + 217.960 - 37.700

68.300+217.960\phantom{68.300 + 217.960-}= (68.300 + 217.960) - (102.735 + 37.700)

Now,

+68.300+ 217.960286.260and+102.735+ 37.700140.435\begin{array}{r} \phantom{+}68.300 \\ +\space 217.960 \\ \hline 286.260 \end{array} \qquad \text{and} \qquad \begin{array}{r} \phantom{+}102.735 \\ +\space 37.700 \\ \hline 140.435 \end{array}

So,

+286.260 140.435145.825\begin{array}{r} \phantom{+}286.260 \\ -\space 140.435 \\ \hline 145.825 \end{array}

Therefore, 68.3 - 102.735 + 217.96 - 37.7 = 145.825.

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

Given expression = 203.500 - 43.760 - 9.890 - 13.086

203.500+217.960\phantom{203.500 + 217.960}= 203.500 - ( 43.760 + 9.890 + 13.086)

Now,

+43.7609.890+ 13.08666.736\begin{array}{r} \phantom{+}43.760 \\ 9.890 \\ +\space 13.086 \\ \hline 66.736 \end{array}

So,

+203.500 66.736136.764\begin{array}{r} \phantom{+}203.500 \\ -\space 66.736 \\ \hline 136.764 \end{array}

Therefore, 203.5 - 43.76 - 9.89 - 13.086 = 136.764.

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

Given expression = 77.700 - 7.555 - 5.770

203.500+217.960\phantom{203.500 + 217.960}= 77.700 - ( 7.555 + 5.770)

Now,

+7.555+ 5.77013.325\begin{array}{r} \phantom{+}7.555 \\ +\space 5.770 \\ \hline 13.325 \end{array}

So,

+77.700 13.32564.375\begin{array}{r} \phantom{+}77.700 \\ -\space 13.325 \\ \hline 64.375 \end{array}

Therefore, 77.7 - 7.555 - 5.77 = 64.375.

Question 4

What is to be added to 73.6 to get 91.03 ?

Answer

Let the number be x. Then,

73.6 + x = 91.03

Solving for x,

x = 91.03 - 73.6

Rewriting the numbers with two decimal places,

x = 91.03 - 73.60

Writing in column form and subtracting, we get:

+91.03 73.6017.43\begin{array}{r} \phantom{+}91.03 \\ -\space 73.60 \\ \hline 17.43 \end{array}

Therefore, x = 17.43.

Question 5

What is to be subtracted from 6.1 to get 0.785 ?

Answer

Let the number to be subtracted from 6.1 be x. Then,

6.1 − x = 0.785

Solving for x,

x = 6.1 − 0.785

Rewriting the numbers with three decimal places,

x = 6.100 − 0.785

Writing in column form and subtracting, we get:

+6.100 0.7855.315\begin{array}{r} \phantom{+}6.100 \\ -\space 0.785 \\ \hline 5.315 \end{array}

Therefore, x = 5.315.

Question 6

By how much should 32.987 be increased to get 60 ?

Answer

Let the required increase be x. Then,

32.987 + x = 60

Solving for x,

x = 60 − 32.987

Rewriting the numbers with three decimal places,

x = 60.000 − 32.987

Writing in column form and subtracting, we get:

+60.000 32.98727.013\begin{array}{r} \phantom{+}60.000 \\ -\space 32.987 \\ \hline 27.013 \end{array}

Therefore, x = 27.013.

Question 7

By how much should 72.5 be decreased to get 18.76 ?

Answer

Let the required decrease be x. Then,

72.5 − x = 18.76

Solving for x,

x = 72.5 − 18.76

Rewriting the numbers with two decimal places,

x = 72.50 − 18.76

Writing in column form and subtracting, we get:

+72.50 18.7653.74\begin{array}{r} \phantom{+}72.50 \\ -\space 18.76 \\ \hline 53.74 \end{array}

Therefore, x = 53.74.

Question 8

The sum of two numbers is 34.02. If one of the numbers is 16.268, find the other.

Answer

Let the other number be x. Then,

x + 16.268 = 34.02

Solving for x,

x = 34.02 − 16.268

Rewriting the numbers with three decimal places,

x = 34.020 − 16.268

Writing in column form and subtracting, we get:

+34.020 16.26817.752\begin{array}{r} \phantom{+}34.020 \\ -\space 16.268 \\ \hline 17.752 \end{array}

Therefore, x = 17.752.

Question 9

Sajjal bought a geometry box for ₹ 139.75 and an exercise book for ₹ 59.60 from a book shop. He gave a 500-rupee note to the shopkeeper. What amount did he get back?

Answer

Cost of geometry box = ₹ 139.75

Cost of exercise book = ₹ 59.60

The total cost is,

+139.75+ 59.60199.35\begin{array}{r} \phantom{+}139.75 \\ +\space 59.60 \\ \hline 199.35 \end{array}

Money given to the shopkeeper = ₹ 500

Amount returned = 500 − 199.35

Rewriting with two decimal places,

Amount returned = 500.00 − 199.35

Writing in column form and subtracting, we get:

+500.00 199.35300.65\begin{array}{r} \phantom{+}500.00 \\ -\space 199.35 \\ \hline 300.65 \end{array}

Therefore, the amount received back is ₹ 300.65.

Exercise 5(E)

Question 1

Rajan purchased a geometry box for ₹ 136.75, a colour box for ₹ 48.80 and a register for ₹ 64.50. What amount did he pay to the shopkeeper?

Answer

Cost of the geometry box = ₹ 136.75

Cost of the colour box = ₹ 48.80

Cost of the register = ₹ 64.50

Total amount = 136.75 + 48.80 + 64.50

Writing in column form and adding:

+136.7548.80+ 64.50250.05\begin{array}{r} \phantom{+}136.75 \\ 48.80 \\ +\space 64.50 \\ \hline 250.05 \end{array}

Hence, the total amount paid is ₹ 250.05.

Question 2

Sakshi bought 3 m 28 cm cloth for her shirt and 2 m 7 cm for her trousers. What is the total length of cloth bought by her?

Answer

Length of cloth for shirt = 3 m 28 cm

= 3 m + 28100\dfrac{28}{100} m = 3 m + 0.28 m = 3.28 m

Length of cloth for trousers = 2 m 7 cm

= 2 m + 7100\dfrac{7}{100} m = 2 m + 0.07 m = 2.07 m

Total length of the cloth = 3.28 + 2.07

Writing in column form and adding:

+3.28+ 2.075.35\begin{array}{r} \phantom{+}3.28 \\ +\space 2.07 \\ \hline 5.35 \end{array}

Hence, the total length of the cloth bought by Sakshi is 5.35 m or 5 m 35 cm.

Question 3

A bag contains 95 kg 80 g of rice and the mass of the empty bag is 860 g. What is the weight of the bag filled with rice?

Answer

Mass of rice = 95 kg 80 g

= 95 kg + 801000\dfrac{80}{1000} kg = 95 kg + 0.080 kg = 95.080 kg

Mass of empty bag = 860 g = 8601000\dfrac{860}{1000} = 0.860 kg

Total weight of the bag filled with rice = 95.080 + 0.860

Writing in column form and adding:

+95.080+ 0.86095.940\begin{array}{r} \phantom{+}95.080 \\ +\space 0.860 \\ \hline 95.940 \end{array}

Hence, the total weight of the bag filled with rice is 95.940 kg or 95 kg 940 g.

Question 4

Distance between Delhi and Meerut is 85 km 850 m and the distance between Meerut and Muzaffarnagar is 56 km 85 m. What is the distance between Delhi and Muzaffarnagar?

Answer

Distance between Delhi and Meerut = 85 km 850 m

= 85 km + 8501000\dfrac{850}{1000} = 85 km + 0.850 = 85.850 km

Distance between Meerut and Muzaffarnagar = 56 km 85 m

= 56 km + 851000\dfrac{85}{1000} = 56 km + 0.085= 56.085 km

The total distance between Delhi and Muzaffarnagar = 85.850 + 56.085

+85.850+ 56.085141.935\begin{array}{r} \phantom{+}85.850 \\ +\space 56.085 \\ \hline 141.935 \end{array}

Hence, the total distance between Delhi and Muzaffarnagar is 141.935 km or 141 km 935 m.

Question 5

Divya purchased a book worth ₹ 256.85 and a set of note-books worth ₹ 546.70. She gave a ₹ 2000 note to the shopkeeper. How much money did she get back?

Answer

Cost of the book = ₹ 256.85

Cost of note-books = ₹ 546.70

Total cost = 256.85 + 546.70

Writing in column form and adding:

+256.85+ 546.70803.55\begin{array}{r} \phantom{+}256.85 \\ +\space 546.70 \\ \hline 803.55 \end{array}

Money given to the shopkeeper = ₹ 2000

Amount returned to Divya = 2000.00 - 803.55

Writing in column form and subtracting:

+2000.00 803.551196.45\begin{array}{r} \phantom{+}2000.00 \\ -\space 803.55 \\ \hline 1196.45 \end{array}

Hence, the total amount returned by the shopkeeper is ₹ 1196.45.

Question 6

The weight of a gas cylinder filled with gas is 32 kg 200 g. If the weight of the gas contained in it is 17 kg 80 g, find the weight of the empty cylinder.

Answer

Weight of the filled cylinder = 32 kg 200 g

= 32 kg + 2001000\dfrac{200}{1000} kg = 32 kg + 0.200 kg = 32.200 kg

Weight of gas contained in the cylinder = 17 kg 80 g

= 17 kg + 801000\dfrac{80}{1000} kg = 17 kg + 0.080 kg = 17.080 kg

Weight of empty cylinder = 32.200 − 17.080

Writing in column form and subtracting:

+32.200 17.08015.120\begin{array}{r} \phantom{+}32.200 \\ -\space 17.080 \\ \hline 15.120 \end{array}

Hence, weight of empty cylinder is 15.120 kg or 15 kg 120 g.

Question 7

Stuti went to the market with ₹ 3000 in cash. Out of this money, she purchased one frock, one toy and one bag costing ₹ 675.85, ₹ 318 and ₹ 1972.75 respectively. How much money is left with her?

Answer

Initial amount with Stuti = ₹ 3000

Total amount spent:

Cost of the frock = ₹ 675.85

Cost of the toy = ₹ 318.00

Cost of the bag = ₹ 1972.75

Total amount spent = 675.85 + 318.00 + 1972.75

Writing in column form and adding:

+675.85318.00+ 1972.752966.60\begin{array}{r} \phantom{+}675.85 \\ 318.00 \\ +\space 1972.75 \\ \hline 2966.60 \end{array}

Money left = 3000.00 - 2966.60

Writing in column form and subtracting:

+3000.00 2966.6033.40\begin{array}{r} \phantom{+}3000.00 \\ -\space 2966.60 \\ \hline 33.40 \end{array}

Hence, Stuti has ₹ 33.40 left with her.

Question 8

The total weight of a bag containing 25 kg 750 g of potatoes and 18 kg 80 g of melon is 44 kg 200 g. How much is the weight of the empty bag?

Answer

Weight of potatoes = 25 kg 750 g

= 25 kg + 7501000\dfrac{750}{1000} kg = 25 kg + 0.750 kg= 25.750 kg

Weight of melon = 18 kg 80 g

= 18 kg + 801000\dfrac{80}{1000} kg = 18 kg + 0.080 kg = 18.080 kg

Total weight of the bag = 44 kg 200 g

= 44 kg + 2001000\dfrac{200}{1000} kg = 44 kg + 0.200 kg = 44.200 kg

Weight of empty bag = 44.200 − (25.750 + 18.080)

+25.750+ 18.08043.830and+44.200 43.8300.370\begin{array}{r} \phantom{+}25.750 \\ +\space 18.080 \\ \hline 43.830 \end{array} \qquad \text{and} \qquad \begin{array}{r} \phantom{+}44.200 \\ -\space 43.830 \\ \hline 0.370 \end{array}

Hence, the weight of the empty bag is 0.370 kg or 370 g.

Question 9

If the school bags of Divya and Rahul weigh 8 kg 60 g and 6 kg 275 g respectively, whose bag is heavier and by how much?

Answer

Weight of Divya’s bag = 8 kg 60 g

= 8 kg + 601000\dfrac{60}{1000} kg = 8 kg + 0.060 kg = 8.060 kg

Weight of Rahul’s bag = 6 kg 275 g

= 6 kg + 2751000\dfrac{275}{1000} kg = 6 kg + 0.275 kg = 6.275 kg

Since 8.060 > 6.275, Divya's bag is heavier than Rahul's bag.

The difference in weight = 8.060 - 6.275

Writing in column form and subtracting:

+8.060 6.2751.785\begin{array}{r} \phantom{+}8.060 \\ -\space 6.275 \\ \hline 1.785 \end{array}

Divya's bag is heavier by 1.785 kg or 1 kg 785 g.

Question 10

The distance between Naman's office and his house is 19 km. He covers 14 km 65 m by scooter, 3 km 75 m by bus and the rest on foot. How much distance does he cover by walking?

Answer

Total distance = 19 km

Distance covered by scooter = 14 km 65 m

= 14 km + 651000\dfrac{65}{1000} km = 14 km + 0.065 km = 14.065 km

Distance covered by bus = 3 km 75 m

= 3 km + 751000\dfrac{75}{1000} km = 3 km + 0.075 km= 3.075 km

Total distance covered = 14.065 + 3.075

Writing in column form and adding:

+14.065+ 3.07517.140\begin{array}{r} \phantom{+}14.065 \\ +\space 3.075 \\ \hline 17.140 \end{array}

So, the total distance covered = 17.140 km

Distance covered on foot = 19.000 - 17.140

Writing in column form and subtracting:

+19.000 17.1401.860\begin{array}{r} \phantom{+}19.000 \\ -\space 17.140 \\ \hline 1.860 \end{array}

The distance covered by walking is 1.860 km or 1 km 860 m.

Question 11

There was an electric wire of length 100 m. Out of it, three pieces of length 26 m 85 cm, 3 m 18 cm, and 18 m 70 cm were cut off. What is the length of the remaining wire?

Answer

Total length of the wire = 100 m

Length of the first piece = 26 m 85 cm

= 26 m + 85100\dfrac{85}{100} m = 26 m + 0.85 m = 26.85 m

Length of the second piece = 3 m 18 cm

= 3 m + 18100\dfrac{18}{100} m = 3 m + 0.18 m = 3.18 m

Length of the third piece = 18 m 70 cm

= 18 m + 70100\dfrac{70}{100} m = 18 m + 0.70 m = 18.70 m

Total length cut off = 26.85 + 3.18 + 18.70

Writing in column form and adding:

+26.853.18+ 18.7048.73\begin{array}{r} \phantom{+}26.85 \\ 3.18 \\ +\space 18.70 \\ \hline 48.73 \end{array}

Total length cut off = 48.73 m

Length of remaining wire = 100.00 - 48.73

+100.00 48.7351.27\begin{array}{r} \phantom{+}100.00 \\ -\space 48.73 \\ \hline 51.27 \end{array}

The remaining wire is 51.27 m or 51 m 27 cm long.

Question 12

A bag contained 98 kg 300 g of wheat. There was a consumption of 36 kg 450 g, 28 kg 360 g and 32 kg 880 g during three months. How much wheat remains in the bag?

Answer

Total amount of wheat present in the bag = 98 kg 300 g

= 98 kg + 3001000\dfrac{300}{1000} kg = 98 kg + 0.300 kg = 98.300 kg

Consumption in month 1 = 36 kg 450 g

= 36 kg + 4501000\dfrac{450}{1000} kg = 36 kg + 0.450 kg = 36.450 kg

Consumption in month 2 = 28 kg 360 g

= 28 kg + 3601000\dfrac{360}{1000} kg = 28 kg + 0.360 kg = 28.360 kg

Consumption in month 3 = 32 kg 880 g

= 32 kg + 8801000\dfrac{880}{1000} kg = 32 kg + 0.880 kg = 32.880 kg

Total consumption during the three months = (36.450 + 28.360 + 32.880 ) kg

Writing in column form and adding:

+36.45028.360+ 32.88097.690\begin{array}{r} \phantom{+}36.450 \\ 28.360 \\ +\space 32.880 \\ \hline 97.690 \end{array}

Total wheat consumed in three months = 97.690 kg

Amount of wheat remaining in the bag = 98.300 - 97.690

+98.300 97.6900.610\begin{array}{r} \phantom{+}98.300 \\ -\space 97.690 \\ \hline 0.610 \end{array}

Therefore 0.610 kg or 610 g of wheat is remaining in the bag.

Question 13

Out of 260 litres of milk, a milkman sold 36 l 500 ml, 75 l 750 ml, 68 l 400 ml and 42 l 850 ml. How much milk is left with him?

Answer

Initial amount of milk = 260 l

We know that,

36 l 500 ml = 36 l + 5001000\dfrac{500}{1000} l = 36 l + 0.500 l = 36.500 l

75 l 750 ml = 75 l + 7501000\dfrac{750}{1000} l = 75 l + 0.750 l = 75.750 l

68 l 400 ml = 68 l + 4001000\dfrac{400}{1000} l = 68 l + 0.400 l = 68.400 l

42 l 850 ml = 42 l + 8501000\dfrac{850}{1000} l = 42 l + 0.850 l = 42.850 l

Total amount of milk sold = (36.500 + 75.750 + 68.400 + 42.850) l

Writing in column form and adding:

+36.50075.75068.400+ 42.850223.500\begin{array}{r} \phantom{+}36.500 \\ 75.750 \\ 68.400 \\ +\space 42.850 \\ \hline 223.500 \end{array}

Total amount of milk sold is 223.500 l.

Milk remaining = (260.000 - 223.500) l

Writing in column form and subtracting,

+260.000 223.50036.500\begin{array}{r} \phantom{+}260.000 \\ -\space 223.500 \\ \hline 36.500 \end{array}

The milkman has 36.500 l or 36 l 500 ml of milk left.

Exercise 5(F) — Multiple Choice Questions

Question 1

The place value of 3 in 5.136 is

  1. 3
  2. 310\dfrac{3}{10}
  3. 3100\dfrac{3}{100}
  4. 31000\dfrac{3}{1000}

Answer

The number is 5.136. The digit 3 is in the hundredths place.

So, its place value is 3100\dfrac{3}{100}.

Hence, Option 3 is the correct option.

Question 2

The place value of 4 in 3.046 is

  1. 4
  2. 40
  3. 440\dfrac{4}{40}
  4. 4100\dfrac{4}{100}

Answer

The number is 3.046. The digit 4 is in the hundredths place.

So, its place value is 4100\dfrac{4}{100}.

Hence, Option 4 is the correct option.

Question 3

3710=?3\dfrac{7}{10} = ?

  1. 3.7
  2. 3.07
  3. 7.3
  4. none of these

Answer

We have,

3710=3+710=3+0.7=3.73 \dfrac{7}{10}=3+\dfrac{7}{10}=3+0.7 = 3.7

Hence, Option 1 is the correct option.

Question 4

0.125 when expressed as a fraction in lowest terms is

  1. 14\dfrac{1}{4}

  2. 18\dfrac{1}{8}

  3. 25\dfrac{2}{5}

  4. 38\dfrac{3}{8}

Answer

The given decimal is 0.125.

Writing the decimal as a fraction,

0.125 = 1251000=18\dfrac{125}{1000} = \dfrac{1}{8}

Hence, Option 2 is the correct option.

Question 5

8 rupees 5 paise can be expressed as

  1. ₹ 8.50
  2. ₹ 80.50
  3. ₹ 8.05
  4. ₹ 8.5

Answer

We know that 1 rupee = 100 paise

So, 5 paise = 5100=0.05\dfrac{5}{100} = 0.05 rupees

So 8 rupees 5 paise = 8 + 0.05 = 8.05 rupees

Hence, Option 3 is the correct option.

Question 6

1 + 0.1 + 0.01 = ?

  1. 1.2
  2. 1.11
  3. 1.21
  4. 1.01

Answer

The maximum number of decimal places in the given decimals is 2. So, we have to convert each of the given decimals to have two places of decimal, by annexing zeros.

So the expression is, 1.00 + 0.10 + 0.01

Writing in column form and adding:

+1.000.10+ 0.011.11\begin{array}{r} \phantom{+}1.00 \\ 0.10 \\ +\space 0.01 \\ \hline 1.11 \end{array}

So the sum is 1.11.

Hence, Option 2 is the correct option.

Question 7

6191000=?6\dfrac{19}{1000} = ?

  1. 6.190
  2. 6.19
  3. 6.019
  4. none of these

Answer

We have,

6191000=6+191000=6+0.019=6.0196\dfrac{19}{1000} =6+\dfrac{19}{1000}=6+0.019=6.019

So the decimal form is 6.019.

Hence, Option 3 is the correct option.

Question 8

6.9, 5.4, 3.5 and 7.2 are

  1. like decimals
  2. unlike decimals
  3. equal decimals
  4. none of these

Answer

The decimals 6.9, 5.4, 3.5 and 7.2 have one decimal place each. So, they are like decimals.

Hence, Option 1 is the correct option.

Question 9

Which is greater, 5.08 or 5.8?

  1. 5.08
  2. 5.8
  3. both are equal
  4. none of these

Answer

Rewriting 5.08 and 5.8 as like decimals, we have 5.08 and 5.80.

Comparing the tenths place, we know that 8 > 0. So, 5.80 > 5.08.

Hence, Option 2 is the correct option.

Question 10

4.17 + 3.06 + 5.135 = ?

  1. 12.158
  2. 12.365
  3. 12.635
  4. none of these

Answer

Rewriting 4.17 + 3.06 + 5.135 as like decimals,

4.170 + 3.060 + 5.135

Writing in column form and adding:

+4.1703.060+ 5.13512.365\begin{array}{r} \phantom{+}4.170 \\ 3.060 \\ +\space 5.135 \\ \hline 12.365 \end{array}

So the sum is 12.365.

Hence, Option 2 is the correct option.

Question 11

5 tenths, when expressed as a decimal, is

  1. 0.5
  2. 0.05
  3. 5.0
  4. 0.005

Answer

5 tenths means: 510\dfrac{5}{10}

Converting into decimal,

510=0.5\dfrac{5}{10} = 0.5

Hence, 5 tenths as a decimal is 0.5.

Hence, Option 1 is the correct option.

Question 12

0.3 + 0.33 = ?

  1. 0.36
  2. 0.333
  3. 0.63
  4. none of these

Answer

Rewriting 0.3 + 0.33 as like decimals: 0.30 + 0.33

Writing in column form and adding:

+0.30+ 0.330.63\begin{array}{r} \phantom{+}0.30 \\ +\space 0.33 \\ \hline 0.63 \end{array}

So the sum is 0.63.

Hence, Option 3 is the correct option.

Question 13

6 km 225 m + 3 km 75 m = ?

  1. (6.225 + 3.75) km
  2. (6.225 + 3.075) km
  3. (6.25 + 3.750) km
  4. none of these

Answer

We have,

6 km 225 m = 6 km + 2251000\dfrac{225}{1000} km = 6 km + 0.225 km = 6.225 km

3 km 75 m = 3 km + 751000\dfrac{75}{1000} km = 3 km + 0.075 km = 3.075 km

So,

6 km 225 m + 3 km 75 m = (6.225 + 3.075) km

So the correct representation is (6.225 + 3.075) km.

Hence, Option 2 is the correct option.

Exercise 5(F) — Mental Maths

Question 1

Fill in the blanks:

(i) 6 hundredths in decimal form is ............... .

(ii) 625\dfrac{6}{25} in decimal form is ............... .

(iii) 2 m 15 cm + 6 m 5 cm = ............... metres.

(iv) 15P + 5P = ₹ ............... .

(v) 15 km + 75 m = ............... .

(vi) (2 cm - 5 mm) = ............... cm.

Answer

(i) 6 hundredths is represented as 6100\dfrac{6}{100}

So, 6100=0.06\dfrac{6}{100}=0.06

6 hundredths in decimal form is 0.06.

(ii) We have,

625=6×425×4=24100=0.24\dfrac{6}{25} = \dfrac{6 \times 4}{25 \times 4} = \dfrac{24}{100} = 0.24

625\dfrac{6}{25} in decimal form is 0.24.

(iii) We know that,

2 m 15 cm = 2 m + 15100\dfrac{15}{100} = 2 m + 0.15 = 2.15 m

6 m 5 cm = 6 m + 5100\dfrac{5}{100} = 6 m + 0.05 = 6.05 m

Writing in column form and adding:

+2.15+ 6.058.20\begin{array}{r} \phantom{+}2.15 \\ +\space 6.05 \\ \hline 8.20 \end{array}

2 m 15 cm + 6 m 5 cm = 8.20 metres.

(iv) We have,

15 P = 15100=0.15\dfrac{15}{100} = ₹ 0.15

5 P = 5100=0.05\dfrac{5}{100} = ₹ 0.05

Writing in column form and adding:

+0.15+ 0.050.20\begin{array}{r} \phantom{+}0.15 \\ +\space 0.05 \\ \hline 0.20 \end{array}

So, 15 P + 5 P = ₹ ( 0.15 + 0.05 ) = ₹ 0.20.

(v) We know that, 75 m = 751000\dfrac{75}{1000} km = 0.075 km.

15 km + 75 m = ( 15.000 + 0.075 ) km

Writing in column form and adding:

+15.000+ 0.07515.075\begin{array}{r} \phantom{+}15.000 \\ +\space 0.075 \\ \hline 15.075 \end{array}

So, 15 km + 75 m = 15.075 km.

(vi) We have, 5 mm = 510\dfrac{5}{10} cm = 0.5 cm.

(2 cm - 5 mm) = (2.0 - 0.5) cm

Writing in column form and subtracting:

+2.0 0.51.5\begin{array}{r} \phantom{+}2.0 \\ -\space 0.5 \\ \hline 1.5 \end{array}

So, (2 cm - 5 mm) = 1.5 cm.

Question 2

Write (T) for true and (F) for false for each of the statements given below:

(i) 3 m 5 cm = 3.5 m

(ii) 16 km 30 m = 16.30 km

(iii) ₹ 8 + 6 P = ₹ (8.60)

(iv) 0.03 = 3 hundredths

(v) 7 kg 80 g = 7.8 kg

(vi) 0.1 + 0.01 + 0.001 = 0.111

Answer

(i) F

Reason —

3 m 5 cm = 3 m + 5100\dfrac{5}{100} m = 3 m + 0.05 m = 3.05 m, and not 3.5 m.

(ii) F

Reason —

16 km 30 m = 16 km + 301000\dfrac{30}{1000} km = 16 km + 0.030 km = 16.030 km, not 16.30 km.

(iii) F

Reason — Here, 6 P = 6100=0.06\dfrac{6}{100} ₹= ₹ 0.06.

So, ₹ 8 + 6 P = ₹ (8 + 0.06) = ₹ 8.06.

(iv) T

Reason —

0.03 = 3100\dfrac{3}{100}

So, 0.03 is 3 hundredths.

(v) F

Reason — We have, 80 g = 801000\dfrac{80}{1000} kg = 0.08 kg.

So, 7 kg 80 g =(7 + 0.08) kg = 7.08 kg

(vi) T

Reason — Rewriting the decimals to be like decimals,

0.1 + 0.01 + 0.001 = 0.100 + 0.010 + 0.001

Writing in column form and adding:

+0.1000.010+ 0.0010.111\begin{array}{r} \phantom{+}0.100 \\ 0.010 \\ +\space 0.001 \\ \hline 0.111 \end{array}

So the sum is 0.111.

Exercise 5(F) — Assertions and Reasons Questions

Question 1

Assertion. The place value of the digit 4 in 2.147 is 4100\dfrac{4}{100}

Reason. The place value of a digit at hundredths place is 10 times the same digit at the tenths place.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

  2. Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

  3. Assertion (A) is true but Reason (R) is false.

  4. Assertion (A) is false but Reason (R) is true.

Answer

In the number 2.147, the digit 4 is in the hundredths place. So, its place value is 4100\dfrac{4}{100}.

So the Assertion (A) is true.

The place value of a digit at the hundredths place is actually 110\dfrac{1}{10} of the value it would have at the tenths place, not 10 times.

So the Reason (R) is false.

Option 3 is the correct option.

Question 2

Assertion. The decimal numbers 2.5, 2.05, 2.005 and 2.0005 are equivalent.

Reason. Any number of zeros put at the end of a decimal number does not change its value.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

  2. Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

  3. Assertion (A) is true but Reason (R) is false.

  4. Assertion (A) is false but Reason (R) is true.

Answer

The decimal numbers 2.5, 2.05, 2.005 and 2.0005 are not equivalent because the zero is being placed between the decimal point and the digit 55 which changes its place value from tenths to hundredths, and so on.

Therefore the Assertion (A) is false.

Adding zeros at the end of a decimal number does not change its value. For example, 2.5 = 2.50, the value does not change.

So the Reason (R) is true.

Option 4 is the correct option.

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