Fractions Class 6 Selina Solutions

Concise Mathematics Selina · Class 6

Fractions: Complete Solved Exercises

Interactive step-by-step solutions for Exercises 6(A), 6(B), 6(C), 6(D), 6(E) and Multiple Choice Questions.

Proper and improper fractionsMixed fractionsComparisonFour operationsWord problems

How to study: Open each question, try it first, and then compare your method with the solution.

Exercise 6(A)Exercise 6(B)Exercise 6(C)Exercise 6(D)Exercise 6(E)Multiple Choice QuestionsFAQs

Proper FractionNumerator is smaller than denominator.

Improper FractionNumerator is greater than or equal to denominator.

Like FractionsFractions have the same denominator.

Division RuleMultiply by the reciprocal of the second fraction.

Class 6 Fractions

Exercise 6(A)

Types, conversions and like fractions

25 solved items

Question 1View solution

For each expression, given below, write a fraction:

(i) 2 out of 7 = ........

(ii) 5 out of 17 = ........

(iii) three-fifths = ........

Solution

Method: Use the definition of fractions, identify numerator and denominator, and apply the conversion rule carefully.

As we know, 'out of' means division, so it is written as a fraction partwhole

Hence, 2 out of 7 = 27.

Hence, 5 out of 17 = 517.

Hence, three-fifths = 35.

Question 2(i)View solution

Fill in the blanks:

58 is ........ fraction.

Solution

Method: Use the definition of fractions, identify numerator and denominator, and apply the conversion rule carefully.

Hence, 58 is a proper fraction.

Question 2(ii)View solution

Fill in the blanks:

85 is ........ fraction.

Solution

Method: Use the definition of fractions, identify numerator and denominator, and apply the conversion rule carefully.

Hence, 85 is an improper fraction.

Question 2(iii)View solution

Fill in the blanks:

1515 is ........ fraction.

Solution

Method: Use the definition of fractions, identify numerator and denominator, and apply the conversion rule carefully.

In1515, numerator (15) = denominator (15), i.e. numerator is not less than denominator.

Hence, 1515 is an improper fraction.

Question 2(iv)View solution

Fill in the blanks:

The value of 2323 = ........

Solution

Method: Use the definition of fractions, identify numerator and denominator, and apply the conversion rule carefully.

Hence, the value of 2323 is 1.

Question 2(v)View solution

Fill in the blanks:

The value of 55 = ........

Solution

Method: Use the definition of fractions, identify numerator and denominator, and apply the conversion rule carefully.

Hence, the value of 55 is 1.

Question 2(vi)View solution

Fill in the blanks:

3310 is ........ fraction.

Solution

Method: Use the definition of fractions, identify numerator and denominator, and apply the conversion rule carefully.

3310 consists of a natural number 3 and a proper fraction 310.

Hence, 3310 is a mixed fraction.

Question 2(vii)View solution

Fill in the blanks:

215 and 715 are ........ fractions.

Solution

Method: Use the definition of fractions, identify numerator and denominator, and apply the conversion rule carefully.

215 and 715 have the same denominator 15.

Hence, 215 and 715 are like fractions.

Question 2(viii)View solution

Fill in the blanks:

2312 and 2315 are ........ fractions.

Solution

Method: Use the definition of fractions, identify numerator and denominator, and apply the conversion rule carefully.

2312 and 2315 have different denominators (12 and 15).

Hence, 2312 and 2315 are unlike fractions.

Question 2(ix)View solution

Fill in the blanks:

615 and 2870 are ........ fractions.

Solution

Method: Use the definition of fractions, identify numerator and denominator, and apply the conversion rule carefully.

Reducing to lowest terms, 615 = 25 and 2870 = 25.

Both are equal to 25.

Hence, 615 and 2870 are equal (equivalent) fractions.

Question 2(x)View solution

Fill in the blanks:

824 and 832 are not ........ fractions.

Solution

Method: Use the definition of fractions, identify numerator and denominator, and apply the conversion rule carefully.

824 and 832 have different denominators (24 and 32).

Hence, 824 and 832 are not like fractions.

Question 2(xi)View solution

Fill in the blanks:

3213 = 3 × 13 + ........ 13 = ........

Solution

Method: Use the definition of fractions, identify numerator and denominator, and apply the conversion rule carefully.

Hence, 3213 = (3×13+2)13 = 4113.

Question 2(xii)View solution

Fill in the blanks:

435 = ........ = ........

Solution

Method: Use the definition of fractions, identify numerator and denominator, and apply the conversion rule carefully.

Hence, 435 = (4×5+3)5 = 235.

Question 3View solution

From the following fractions, separate (i) proper fractions and (ii) improper fractions:

29, 43, 715, 1120, 2011, 1823, 2735.

Solution

Method: Use the definition of fractions, identify numerator and denominator, and apply the conversion rule carefully.

A fraction is proper if numerator < denominator, and improper if numerator ≥ denominator.

(i) Proper fractions (numerator < denominator) are:

(ii) Improper fractions (numerator > denominator) are:

Hence, proper fractions are 29, 715, 1120, 1823, 2735 and improper fractions are 43, 2011.

Question 4(i)View solution

Change the following mixed fractions to improper fractions:

215

Solution

Method: Use the definition of fractions, identify numerator and denominator, and apply the conversion rule carefully.

Hence, 215 = 115.

Question 4(ii)View solution

Change the following mixed fractions to improper fractions:

314

Solution

Method: Use the definition of fractions, identify numerator and denominator, and apply the conversion rule carefully.

Hence, 314 = 134.

Question 4(iii)View solution

Change the following mixed fractions to improper fractions:

718

Solution

Method: Use the definition of fractions, identify numerator and denominator, and apply the conversion rule carefully.

Hence, 718 = 578.

Question 4(iv)View solution

Change the following mixed fractions to improper fractions:

2111

Solution

Method: Use the definition of fractions, identify numerator and denominator, and apply the conversion rule carefully.

Hence, 2111 = 2311.

Question 5(i)View solution

Change the following improper fractions to mixed fractions:

10017

Solution

Method: Use the definition of fractions, identify numerator and denominator, and apply the conversion rule carefully.

Dividing 100 by 17, we get quotient = 5 and remainder = 15.

Hence, 10017 = 51517.

Question 5(ii)View solution

Change the following improper fractions to mixed fractions:

8111

Solution

Method: Use the definition of fractions, identify numerator and denominator, and apply the conversion rule carefully.

Dividing 81 by 11, we get quotient = 7 and remainder = 4.

Hence, 8111 = 7411.

Question 5(iii)View solution

Change the following improper fractions to mixed fractions:

2097

Solution

Method: Use the definition of fractions, identify numerator and denominator, and apply the conversion rule carefully.

Dividing 209 by 7, we get quotient = 29 and remainder = 6.

Hence, 2097 = 2967.

Question 5(iv)View solution

Change the following improper fractions to mixed fractions:

11315

Solution

Method: Use the definition of fractions, identify numerator and denominator, and apply the conversion rule carefully.

Dividing 113 by 15, we get quotient = 7 and remainder = 8.

Hence, 11315 = 7815.

Question 6(i)View solution

Change the following groups of fractions to like fractions:

13, 25, 34, 16

Solution

Method: Use the definition of fractions, identify numerator and denominator, and apply the conversion rule carefully.

Hence, the required like fractions are 2060, 2460, 4560 and 1060.

Question 6(ii)View solution

Change the following groups of fractions to like fractions:

56, 78, 1112, 310

Solution

Method: Use the definition of fractions, identify numerator and denominator, and apply the conversion rule carefully.

Hence, the required like fractions are 100120, 105120, 110120 and 36120.

Question 6(iii)View solution

Change the following groups of fractions to like fractions:

27, 78, 514, 916

Solution

Method: Use the definition of fractions, identify numerator and denominator, and apply the conversion rule carefully.

Hence, the required like fractions are 32112, 98112, 40112 and 63112.

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Class 6 Fractions

Exercise 6(B)

Lowest terms, comparison and ordering

26 solved items

Question 1(i)View solution

Reduce the given fractions to their lowest terms:

810

Solution

Method: Use lowest terms, equivalent fractions, LCM or cross-multiplication to compare and arrange fractions.

Hence, 810 in lowest terms is 45.

Question 1(ii)View solution

Reduce the given fractions to their lowest terms:

5075

Solution

Method: Use lowest terms, equivalent fractions, LCM or cross-multiplication to compare and arrange fractions.

Hence, 5075 in lowest terms is 23.

Question 1(iii)View solution

Reduce the given fractions to their lowest terms:

1881

Solution

Method: Use lowest terms, equivalent fractions, LCM or cross-multiplication to compare and arrange fractions.

Hence, 1881 in lowest terms is 29.

Question 1(iv)View solution

Reduce the given fractions to their lowest terms:

40120

Solution

Method: Use lowest terms, equivalent fractions, LCM or cross-multiplication to compare and arrange fractions.

Hence, 40120 in lowest terms is 13.

Question 1(v)View solution

Reduce the given fractions to their lowest terms:

10570

Solution

Method: Use lowest terms, equivalent fractions, LCM or cross-multiplication to compare and arrange fractions.

Hence, 10570 in lowest terms is 32 or 112.

Question 2(i)View solution

State whether true or false:

25 = 1015

Solution

Method: Use lowest terms, equivalent fractions, LCM or cross-multiplication to compare and arrange fractions.

Since 30 ≠ 50, the cross products are not equal.

Hence, the statement is False.

Question 2(ii)View solution

State whether true or false:

3542 = 56

Solution

Method: Use lowest terms, equivalent fractions, LCM or cross-multiplication to compare and arrange fractions.

Since the cross products are equal, the fractions are equal.

Hence, the statement is True.

Question 2(iii)View solution

State whether true or false:

54 = 45

Solution

Method: Use lowest terms, equivalent fractions, LCM or cross-multiplication to compare and arrange fractions.

Since 25 ≠ 16, the cross products are not equal.

Hence, the statement is False.

Question 2(iv)View solution

State whether true or false:

79 = 117

Solution

Method: Use lowest terms, equivalent fractions, LCM or cross-multiplication to compare and arrange fractions.

79 is a proper fraction (numerator < denominator), while 117 = 87 is an improper fraction (denominator < numerator).

Hence, the statement is False.

Question 2(v)View solution

State whether true or false:

97 = 117

Solution

Method: Use lowest terms, equivalent fractions, LCM or cross-multiplication to compare and arrange fractions.

Converting, improper fraction into mixed fraction:

97 = 127, which is not equal to 117.

Hence, the statement is False.

Question 3(i)View solution

Which fraction is greater?

35 or 23

Solution

Method: Use lowest terms, equivalent fractions, LCM or cross-multiplication to compare and arrange fractions.

Hence, 23 is greater.

Question 3(ii)View solution

Which fraction is greater?

59 or 34

Solution

Method: Use lowest terms, equivalent fractions, LCM or cross-multiplication to compare and arrange fractions.

Hence, 34 is greater.

Question 3(iii)View solution

Which fraction is greater?

1114 or 2635

Solution

Method: Use lowest terms, equivalent fractions, LCM or cross-multiplication to compare and arrange fractions.

Hence, 1114 is greater.

Question 4(i)View solution

Which fraction is smaller?

38 or 45

Solution

Method: Use lowest terms, equivalent fractions, LCM or cross-multiplication to compare and arrange fractions.

Hence, 38 is smaller.

Question 4(ii)View solution

Which fraction is smaller?

815 or 47

Solution

Method: Use lowest terms, equivalent fractions, LCM or cross-multiplication to compare and arrange fractions.

Hence, 815 is smaller.

Question 4(iii)View solution

Which fraction is smaller?

726 or 1039

Solution

Method: Use lowest terms, equivalent fractions, LCM or cross-multiplication to compare and arrange fractions.

Hence, 1039 is smaller.

Question 5(i)View solution

Arrange the given fractions in descending order of magnitude:

516, 1324 and 78

Solution

Method: Use lowest terms, equivalent fractions, LCM or cross-multiplication to compare and arrange fractions.

Hence, the descending order is 78 > 1324 > 516.

Question 5(ii)View solution

Arrange the given fractions in descending order of magnitude:

45, 715, 1120 and 34

Solution

Method: Use lowest terms, equivalent fractions, LCM or cross-multiplication to compare and arrange fractions.

Hence, the descending order is 45 > 34 > 1120 > 715.

Question 5(iii)View solution

Arrange the given fractions in descending order of magnitude:

57, 38 and 911

Solution

Method: Use lowest terms, equivalent fractions, LCM or cross-multiplication to compare and arrange fractions.

Hence, the descending order is 911 > 57 > 38.

Question 6(i)View solution

Arrange the given fractions in ascending order of magnitude:

916, 712 and 14

Solution

Method: Use lowest terms, equivalent fractions, LCM or cross-multiplication to compare and arrange fractions.

Hence, the ascending order is 14 < 916 < 712.

Question 6(ii)View solution

Arrange the given fractions in ascending order of magnitude:

56, 27, 89 and 13

Solution

Method: Use lowest terms, equivalent fractions, LCM or cross-multiplication to compare and arrange fractions.

Hence, the ascending order is 27 < 13 < 56 < 89.

Question 6(iii)View solution

Arrange the given fractions in ascending order of magnitude:

23, 59, 56 and 38

Solution

Method: Use lowest terms, equivalent fractions, LCM or cross-multiplication to compare and arrange fractions.

Hence, the ascending order is 38 < 59 < 23 < 56.

Question 7(i)View solution

Insert the symbol '=' or '>' or '<' between each of the pairs of fractions given below:

6 11 ........ 5 9

Solution

Method: Use lowest terms, equivalent fractions, LCM or cross-multiplication to compare and arrange fractions.

Hence, 611 < 59.

Question 7(ii)View solution

Insert the symbol '=' or '>' or '<' between each of the pairs of fractions given below:

3 7 ........ 9 13

Solution

Method: Use lowest terms, equivalent fractions, LCM or cross-multiplication to compare and arrange fractions.

Hence, 37 < 913.

Question 7(iii)View solution

Insert the symbol '=' or '>' or '<' between each of the pairs of fractions given below:

56 64 ........ 7 8

Solution

Method: Use lowest terms, equivalent fractions, LCM or cross-multiplication to compare and arrange fractions.

Reducing5664 to lowest terms, 5664 = 78.

Hence, 5664 = 78.

Question 7(iv)View solution

Insert the symbol '=' or '>' or '<' between each of the pairs of fractions given below:

5 12 ........ 8 33

Solution

Method: Use lowest terms, equivalent fractions, LCM or cross-multiplication to compare and arrange fractions.

Hence, 512 > 833.

Class 6 Fractions

Exercise 6(C)

Addition and subtraction

16 solved items

Question 1(i)View solution

Add the following fractions:

134 and 38

Solution

Method: Convert mixed numbers when needed, take the LCM of denominators, and then add or subtract the numerators.

Hence, the required sum = 218.

Question 1(ii)View solution

Add the following fractions:

25, 2315 and 710

Solution

Method: Convert mixed numbers when needed, take the LCM of denominators, and then add or subtract the numerators.

Hence, the required sum = 3310.

Question 1(iii)View solution

Add the following fractions:

178, 112 and 134

Solution

Method: Convert mixed numbers when needed, take the LCM of denominators, and then add or subtract the numerators.

Hence, the required sum = 518.

Question 1(iv)View solution

Add the following fractions:

334, 216 and 158

Solution

Method: Convert mixed numbers when needed, take the LCM of denominators, and then add or subtract the numerators.

Hence, the required sum = 71324.

Question 1(v)View solution

Add the following fractions:

289, 1118 and 356

Solution

Method: Convert mixed numbers when needed, take the LCM of denominators, and then add or subtract the numerators.

Hence, the required sum = 713.

Question 2(i)View solution

Simplify:

11112 − 1316

Solution

Method: Convert mixed numbers when needed, take the LCM of denominators, and then add or subtract the numerators.

Hence, 11112 − 1316 = 1548.

Question 2(ii)View solution

Simplify:

234 − 156

Solution

Method: Convert mixed numbers when needed, take the LCM of denominators, and then add or subtract the numerators.

Hence, 234 − 156 = 1112.

Question 2(iii)View solution

Simplify:

257 + 314 − 1321

Solution

Method: Convert mixed numbers when needed, take the LCM of denominators, and then add or subtract the numerators.

Hence, 257 + 314 − 1321 = 21342.

Question 2(iv)View solution

Simplify:

356 − 16 − 1112

Solution

Method: Convert mixed numbers when needed, take the LCM of denominators, and then add or subtract the numerators.

Hence, 356 − 16 − 1112 = 2712.

Question 2(v)View solution

Simplify:

6 + 310 − 1 8 15 6+

Solution

Method: Convert mixed numbers when needed, take the LCM of denominators, and then add or subtract the numerators.

Hence, 6 + 310 − 1815 = 4 23 30 6+.

Question 2(vi)View solution

Simplify:

134 + 257 − 1314

Solution

Method: Convert mixed numbers when needed, take the LCM of denominators, and then add or subtract the numerators.

Hence, 134 + 257 − 1314 = 314.

Question 2(vii)View solution

Simplify:

4 + 318 − 3 1 6 4+3

Solution

Method: Convert mixed numbers when needed, take the LCM of denominators, and then add or subtract the numerators.

Hence, 4 + 318 − 316 = 3 23 24 4+3.

Question 2(viii)View solution

Simplify:

6 − 312 − 215 6−3

Solution

Method: Convert mixed numbers when needed, take the LCM of denominators, and then add or subtract the numerators.

Hence, 6 − 312 − 215 = 310 6−3.

Question 2(ix)View solution

Simplify:

158 − 216 + 334

Solution

Method: Convert mixed numbers when needed, take the LCM of denominators, and then add or subtract the numerators.

Hence, 158 − 216 + 334 = 3524.

Question 2(x)View solution

Simplify:

312 + 123 − 214

Solution

Method: Convert mixed numbers when needed, take the LCM of denominators, and then add or subtract the numerators.

Hence, 312 + 123 − 214 = 21112.

Question 2(xi)View solution

Simplify:

435 − 279 − 1215 − 245

Solution

Method: Convert mixed numbers when needed, take the LCM of denominators, and then add or subtract the numerators.

Hence, 435 − 279 − 1215 − 245 = 2945.

Class 6 Fractions

Exercise 6(D)

Multiplication, division and BODMAS

32 solved items

Question 1(i)View solution

Simplify:

37 × 25

Solution

Method: Replace “of” with multiplication, change division into multiplication by the reciprocal, and follow BODMAS.

Hence, 37 × 25 = 635.

Question 1(ii)View solution

Simplify:

49 × 35

Solution

Method: Replace “of” with multiplication, change division into multiplication by the reciprocal, and follow BODMAS.

Hence, 49 × 35 = 415.

Question 1(iii)View solution

Simplify:

512 × 8

Solution

Method: Replace “of” with multiplication, change division into multiplication by the reciprocal, and follow BODMAS.

Hence, 512 × 8 = 313.

Question 1(iv)View solution

Simplify:

76 of 314

Solution

Method: Replace “of” with multiplication, change division into multiplication by the reciprocal, and follow BODMAS.

Hence, 76 of 314 = 14.

Question 1(v)View solution

Simplify:

338 × 367

Solution

Method: Replace “of” with multiplication, change division into multiplication by the reciprocal, and follow BODMAS.

Hence, 338 × 367 = 13156.

Question 1(vi)View solution

Simplify:

12 of 13 × 34

Solution

Method: Replace “of” with multiplication, change division into multiplication by the reciprocal, and follow BODMAS.

Hence, 12 of 13 × 34 = 18.

Question 1(vii)View solution

Simplify:

37 × 59 × 415

Solution

Method: Replace “of” with multiplication, change division into multiplication by the reciprocal, and follow BODMAS.

Hence, 37 × 59 × 415 = 1 =1.

Question 1(viii)View solution

Simplify:

113 × 127 of 114

Solution

Method: Replace “of” with multiplication, change division into multiplication by the reciprocal, and follow BODMAS.

Hence, 113 × 127 of 114 = 217.

Question 2(i)View solution

Simplify:

23 ÷ 115

Solution

Method: Replace “of” with multiplication, change division into multiplication by the reciprocal, and follow BODMAS.

Hence, 23 ÷ 115 = 59.

Question 2(ii)View solution

Simplify:

412 ÷ 49

Solution

Method: Replace “of” with multiplication, change division into multiplication by the reciprocal, and follow BODMAS.

Hence, 412 ÷ 49 = 1018.

Question 2(iii)View solution

Simplify:

1 ÷ 25

Solution

Method: Replace “of” with multiplication, change division into multiplication by the reciprocal, and follow BODMAS.

Hence, 1 ÷ 25 = 2 1 2 1÷.

Question 2(iv)View solution

Simplify:

49 ÷ 49

Solution

Method: Replace “of” with multiplication, change division into multiplication by the reciprocal, and follow BODMAS.

Hence, 49 ÷ 49 = 1 =1.

Question 2(v)View solution

Simplify:

213 ÷ 134

Solution

Method: Replace “of” with multiplication, change division into multiplication by the reciprocal, and follow BODMAS.

Hence, 213 ÷ 134 = 113.

Question 3(i)View solution

Simplify:

14 of 227 ÷ 35

Solution

Method: Replace “of” with multiplication, change division into multiplication by the reciprocal, and follow BODMAS.

Hence, 14 of 227 ÷ 35 = 2021.

Question 3(ii)View solution

Simplify:

114 × 12 ÷ 113

Solution

Method: Replace “of” with multiplication, change division into multiplication by the reciprocal, and follow BODMAS.

Hence, 114 × 12 ÷ 113 = 1532.

Question 3(iii)View solution

Simplify:

617 × 0 × 538

Solution

Method: Replace “of” with multiplication, change division into multiplication by the reciprocal, and follow BODMAS.

Any number multiplied by 0 is 0.

Hence, 617 × 0 × 538 = 0.

Question 3(iv)View solution

Simplify:

34 × 113 ÷ 37 of 258

Solution

Method: Replace “of” with multiplication, change division into multiplication by the reciprocal, and follow BODMAS.

Hence, 34 × 113 ÷ 37 of 258 = 89.

Question 3(v)View solution

Simplify:

214 ÷ 27 of 113 × 23

Solution

Method: Replace “of” with multiplication, change division into multiplication by the reciprocal, and follow BODMAS.

Hence, 214 ÷ 27 of 113 × 23 = 31516.

Question 4(i)View solution

Simplify: 5 − (811 − 3311)

Solution

Method: Replace “of” with multiplication, change division into multiplication by the reciprocal, and follow BODMAS.

Using BODMAS, simplify the bracket first.

Inside the bracket, 811 − 3311 = −2611.

Hence, 5 − (811 − 3311) = 7611.

Question 4(ii)View solution

Simplify: 12 ÷ (78 − 35)

Solution

Method: Replace “of” with multiplication, change division into multiplication by the reciprocal, and follow BODMAS.

Using BODMAS, simplify the bracket first.

78 − 35 = 1140.

Hence, 12 ÷ (78 − 35) = 1911.

Question 4(iii)View solution

Simplify: 213 ÷ (512 + 334)

Solution

Method: Replace “of” with multiplication, change division into multiplication by the reciprocal, and follow BODMAS.

Using BODMAS, simplify the bracket first.

512 + 334 = 374.

Hence, 213 ÷ (512 + 334) = 28111.

Question 4(iv)View solution

Simplify: (378 − 335) ÷ 12

Solution

Method: Replace “of” with multiplication, change division into multiplication by the reciprocal, and follow BODMAS.

Using BODMAS, simplify the bracket first.

378 − 335 = 1140.

Hence, (378 − 335) ÷ 12 = 1120.

Question 4(v)View solution

Simplify: 47 ÷ (13 × 245)

Solution

Method: Replace “of” with multiplication, change division into multiplication by the reciprocal, and follow BODMAS.

Using BODMAS, calculate the multiplication inside the bracket first.

13 × 245 = 1415.

Hence, 47 ÷ (13 × 245) = 3049.

Question 5(i)View solution

Simplify: (12 + 13) ÷ (14 − 16)

Solution

Method: Replace “of” with multiplication, change division into multiplication by the reciprocal, and follow BODMAS.

Using BODMAS, simplify both brackets first.

12 + 13 = 56 and 14 − 16 = 112.

Hence, (12 + 13) ÷ (14 − 16) = 10.

Question 5(ii)View solution

Simplify: (2435 ÷ 67 + 59) × 34

Solution

Method: Replace “of” with multiplication, change division into multiplication by the reciprocal, and follow BODMAS.

Using BODMAS, solve the division inside the bracket first.

2435 ÷ 67 = 45.

Hence, (2435 ÷ 67 + 59) × 34 = 1160.

Question 5(iii)View solution

Simplify: 34 of 618 − 23 of 214

Solution

Method: Replace “of” with multiplication, change division into multiplication by the reciprocal, and follow BODMAS.

Replace “of” with multiplication and calculate each product first.

Hence, 34 of 618 − 23 of 214 = 3332.

Question 5(iv)View solution

Simplify: 730 of (13 + 715) ÷ (56 − 35)

Solution

Method: Replace “of” with multiplication, change division into multiplication by the reciprocal, and follow BODMAS.

Using BODMAS, simplify both brackets before multiplication and division.

Hence, 730 of (13 + 715) ÷ (56 − 35) = 45.

Question 5(v)View solution

Simplify: 212 − 312 × 134 + 212

Solution

Method: Replace “of” with multiplication, change division into multiplication by the reciprocal, and follow BODMAS.

Using BODMAS, solve the multiplication first.

Hence, 212 − 312 × 134 + 212 = −118.

Question 5(vi)View solution

Simplify: 457 × (318 ÷ 1112)

Solution

Method: Replace “of” with multiplication, change division into multiplication by the reciprocal, and follow BODMAS.

Using BODMAS, simplify the bracket first.

Hence, 457 × (318 ÷ 1112) = 16114.

Question 5(vii)View solution

Simplify: 25 of (17 − 112) of 125

Solution

Method: Replace “of” with multiplication, change division into multiplication by the reciprocal, and follow BODMAS.

Using BODMAS, simplify the bracket first and replace “of” with multiplication.

Hence, 25 of (17 − 112) of 125 = 130.

Question 5(viii)View solution

Simplify: (12 − 13) × (34 − 45) ÷ (12 − 25 + 17)

Solution

Method: Replace “of” with multiplication, change division into multiplication by the reciprocal, and follow BODMAS.

Using BODMAS, simplify all brackets first.

Hence, (12 − 13) × (34 − 45) ÷ (12 − 25 + 17) = −7204.

Question 5(ix)View solution

Simplify: 56 − 35 × (13 + 211)

Solution

Method: Replace “of” with multiplication, change division into multiplication by the reciprocal, and follow BODMAS.

Using BODMAS, simplify the bracket and then multiply.

Hence, 56 − 35 × (13 + 211) = 173330.

Class 6 Fractions

Exercise 6(E)

Word problems

11 solved items

Question 1View solution

From a rope 1012 m long, 458 m is cut off. Find the length of the remaining rope.

Solution

Method: Translate the word problem into fraction operations, calculate step by step, and include the correct unit.

Total length of the rope = 1012 m.

Length cut off = 458 m.

Remaining length = 1012 − 458 = 578 m.

Hence, the length of the remaining rope is 578 m.

Question 2View solution

A piece of cloth is 5 m long. After washing, it shrinks by 125 of its length. What is the length of the cloth after washing?

Solution

Method: Translate the word problem into fraction operations, calculate step by step, and include the correct unit.

Shrinkage = 125 of 5 m = 15 m.

Length after washing = 5 − 15 = 445 m.

Hence, the length of the cloth after washing is 445 m.

Question 3View solution

I bought wheat worth ₹ 1212, rice worth ₹ 2534 and vegetables worth ₹ 1014. I gave a hundred-rupee note to the shopkeeper; how much money did he return to me?

Solution

Method: Translate the word problem into fraction operations, calculate step by step, and include the correct unit.

Total money spent = 1212 + 2534 + 1014

So, total money spent = ₹ 4812.

Money returned by the shopkeeper = 100 − 4812

Hence, the shopkeeper returned ₹ 5112.

Question 4View solution

Out of 500 oranges in a box, 325 are rotten and 15 are kept for some guests. How many oranges are left in the box?

Solution

Method: Translate the word problem into fraction operations, calculate step by step, and include the correct unit.

Rotten oranges = 325 × 500 = 60.

Oranges kept for guests = 15 × 500 = 100.

Oranges left = 500 − 60 − 100 = 340.

Hence, the number of oranges left in the box is 340.

Question 5View solution

An ornament piece is made of gold and copper. Its total weight is 96 g. If 112 of the ornament is copper, find the weight of gold in it.

Solution

Method: Translate the word problem into fraction operations, calculate step by step, and include the correct unit.

Total weight of the ornament = 96 g.

Weight of copper = 112 of 96 = 112 × 96 ×96 = 8 g.

Weight of gold = Total weight − Weight of copper

Hence, the weight of gold = 88 g.

Question 6View solution

A girl did half of some work on Monday and one-third of it on Tuesday. How much will she have to do on Wednesday in order to complete the work?

Solution

Method: Translate the word problem into fraction operations, calculate step by step, and include the correct unit.

Work done on Monday = 12.

Work done on Tuesday = 13.

Total work done in two days = 12 + 13

Work left for Wednesday = Total work - Work done in two days

Hence, she will have to do 16 of the work on Wednesday.

Question 7View solution

A man spends 38 of his money and still has ₹ 720 left with him. How much money did he have at first?

Solution

Method: Translate the word problem into fraction operations, calculate step by step, and include the correct unit.

Part of money spent = 38.

Part of money left = 1 − 38 = 88 − 38 = 58.

Given, 58 of the money = ₹ 720.

Hence, the man had ₹ 1,152 at first.

Question 8View solution

In a school, 45 of the students are boys, and the number of girls is 100. Find the number of boys.

Solution

Method: Translate the word problem into fraction operations, calculate step by step, and include the correct unit.

Girls form 1 − 45 = 15 of the school.

If 15 equals 100, the total number of students is 500.

Boys = 45 × 500 = 400.

Hence, the number of boys is 400.

Question 9View solution

After finishing 34 of my journey, I find that 12 km of my journey is covered. How much distance is still left to be covered?

Solution

Method: Translate the word problem into fraction operations, calculate step by step, and include the correct unit.

Given, 34 of the journey = 12 km.

Distance left = Total journey − Distance covered

Hence, the distance still left to be covered = 4 km.

Question 10View solution

When Ajit travelled 15 km, he found that one-fourth of his journey was still left. What was the full length of the journey?

Solution

Method: Translate the word problem into fraction operations, calculate step by step, and include the correct unit.

Part of journey left = 14.

Part of journey travelled = 1 − 14 = 34.

Given, 34 of the journey = 15 km.

Hence, the full length of the journey = 20 km.

Question 11View solution

In a particular month, a man earns ₹ 7,200. Out of this income, he spends 310 on food, 14 on house rent, 110 on insurance and 225 on holidays. How much did he save in that month?

Solution

Method: Translate the word problem into fraction operations, calculate step by step, and include the correct unit.

Fraction of income spent = 310 + 14 + 110 + 225

Fraction of income saved = 1 − 73100 = 27100.

Hence, the man saved ₹ 1,944 in that month.

Class 6 Fractions

Multiple Choice Questions

Concept-check questions

13 solved items

Question 1View solution

In the given number line, identify the values of A and B.

Option 1: A = 23, B = 45.

Option 2: A = 23, B = 134.

Option 3: A = 3, B = 7.

Option 4: A = 2, B = 6.

Solution

Method: Apply the relevant fraction rule, check each option, and select the option that matches the calculation.

A is on the second of three equal divisions between 0 and 1, so A = 23.

B is on the third of four equal divisions after 1, so B = 134.

Hence, option 2 is correct.

Question 2View solution

In 35, 25, 34, 27 and 45, identify the like fractions.

Option 1: 25, 27.

Option 2: 35, 34.

Option 3: 35, 25.

Option 4: 35, 25, 45.

Solution

Method: Apply the relevant fraction rule, check each option, and select the option that matches the calculation.

Like fractions have the same denominator.

The fractions with denominator 5 are 35, 25 and 45.

Hence, option 4 is correct.

Question 3View solution

Out of 157, 158, 1511 and 1513, which is the largest fraction?

Option 1: 157.

Option 2: 158.

Option 3: 1511.

Option 4: 1513.

Solution

Method: Apply the relevant fraction rule, check each option, and select the option that matches the calculation.

For fractions with the same numerator, the fraction with the smallest denominator is the largest.

Hence, 157 is the largest and option 1 is correct.

Question 4View solution

Out of 715, 815, 1115 and 1315, which is the least fraction?

Option 1: 715.

Option 2: 815.

Option 3: 1115.

Option 4: 1315.

Solution

Method: Apply the relevant fraction rule, check each option, and select the option that matches the calculation.

For like fractions, the fraction with the smallest numerator is the least.

Hence, 715 is the least and option 1 is correct.

Question 5View solution

716 + 1316 − 516 is equal to:

Option 1: 1516.

Option 2: 2516.

Option 3: 0.

Option 4: 1.

Solution

Method: Apply the relevant fraction rule, check each option, and select the option that matches the calculation.

716 + 1316 − 516 = 1516.

Hence, option 1 is correct.

Question 6View solution

3123 − 1623 − 1523 is equal to:

Option 1: 123.

Option 2: 0.

Option 3: 13.

Option 4: −23.

Solution

Method: Apply the relevant fraction rule, check each option, and select the option that matches the calculation.

Convert the mixed numbers into improper fractions: 953 − 503 − 473 = −23.

Hence, option 4 is correct.

Question 7View solution

12 ÷ 12 = 1 and 0 ÷ 12 = 0. What is 12 ÷ 0?

Option 1: 1.

Option 2: 0.

Option 3: Not defined.

Solution

Method: Apply the relevant fraction rule, check each option, and select the option that matches the calculation.

Division by zero is not defined.

Hence, option 3 is correct.

Question 8View solution

14 × 14 ÷ 14 of 14 is equal to:

Option 1: 1.

Option 2: 16.

Option 3: 116.

Option 4: 14.

Solution

Method: Apply the relevant fraction rule, check each option, and select the option that matches the calculation.

Calculate “of” first: 14 of 14 = 116.

Then 14 × 14 ÷ 116 = 1.

Hence, option 1 is correct.

Question 9View solution

14 of 14 ÷ 14 × 14 is equal to:

Option 1: 1.

Option 2: 16.

Option 3: 116.

Option 4: 14.

Solution

Method: Apply the relevant fraction rule, check each option, and select the option that matches the calculation.

Calculate “of” first: 14 of 14 = 116.

Then 116 ÷ 14 × 14 = 116.

Hence, option 3 is correct.

Question 10View solution

If x3 = 15, then x5 is:

Option 1: 45.

Option 2: 9.

Option 3: 75.

Option 4: None of these.

Solution

Method: Apply the relevant fraction rule, check each option, and select the option that matches the calculation.

x3 = 15 gives x = 45.

Therefore, x5 = 455 = 9.

Hence, option 2 is correct.

Question 11View solution

Assertion (A): 7 ÷ 7 × 7 − 7 = 0.

Reason (R): 7 × 17 × 7 − 7 = 7 − 7 = 0.

Option 1: A is true, R is false.

Option 2: A is false, R is true.

Option 3: Both A and R are true.

Option 4: Both A and R are false.

Solution

Method: Apply the relevant fraction rule, check each option, and select the option that matches the calculation.

By BODMAS, 7 ÷ 7 × 7 − 7 = 1 × 7 − 7 = 0, so A is true.

The reason is also true and correctly explains the assertion.

Hence, option 3 is correct.

Question 12View solution

Assertion (A): If 12 of a number is 15, then 15 of the same number is 6.

Reason (R): The number is 30.

Option 1: A is true, R is false.

Option 2: A is false, R is true.

Option 3: Both A and R are true.

Option 4: Both A and R are false.

Solution

Method: Apply the relevant fraction rule, check each option, and select the option that matches the calculation.

If 12 of the number is 15, the number is 30.

Then 15 of 30 is 6, so both A and R are true.

Hence, option 3 is correct.

Question 13View solution

Statement 1: If x = 112 ÷ 312 × 213, then x = 1.

Statement 2: 112 ÷ 312 × 213 = 32 × 27 × 73 = 1.

Option 1: Both statements are true.

Option 2: Both statements are false.

Option 3: Statement 1 is true and Statement 2 is false.

Option 4: Statement 1 is false and Statement 2 is true.

Solution

Method: Apply the relevant fraction rule, check each option, and select the option that matches the calculation.

The calculation in Statement 2 is correct and gives x = 1.

Therefore, both statements are true.

Hence, option 1 is correct.

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Frequently Asked Questions

What is a proper fraction?

A proper fraction has a numerator smaller than its denominator, so its value is less than 1.

What is an improper fraction?

An improper fraction has a numerator greater than or equal to its denominator.

How do we convert a mixed fraction into an improper fraction?

Multiply the whole number by the denominator, add the numerator, and write the result over the same denominator.

How do we add unlike fractions?

Find the LCM of the denominators, convert the fractions into like fractions, and then add the numerators.

How do we divide fractions?

Multiply the first fraction by the reciprocal of the second fraction.

What does “of” mean in fraction problems?

In fraction calculations, “of” means multiplication.

Concept by Teacher Ritu, designed by Shaleen Shekhar.