Selina Solutions Concise Mathematics Class 6 Chapter 14: Fractions

Selina Solutions Concise Mathematics Class 6 Chapter 14 Fractions discusses the concepts related to Fractions, with illustrations for better learning. Subject experts have solved the questions clearly for every exercise. The solutions prepared in a simple language help to increase logical thinking among students. Selina Solutions is a key resource, which provides answers to all the queries of the students. Frequent practice of textbook problems is a must, for those who aspire to score more marks in the exams. Students can download the Selina Solutions Concise Mathematics Class 6 Chapter 14 Fractions PDF, from the below available links.

Chapter 14 of Selina Solutions explains basic topics like types of fractions and conversion of fractions. Various shortcut methods are also available to help students in grasping the concepts effortlessly, within a short duration.

Selina Solutions Concise Mathematics Class 6 Chapter 14: Fractions Download PDF

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Exercises of Selina Solutions Concise Mathematics Class 6 Chapter 14: Fractions

Exercise 14(A) Solutions

Exercise 14(B) Solutions

Exercise 14(C) Solutions

Exercise 14(D) Solutions

Exercise 14(E) Solutions

Access Selina Solutions Concise Mathematics Class 6 Chapter 14: Fractions

Exercise 14(A)

1. For each expression, given below, write a fraction:

(i) 2 out of 7 = ………

(ii) 5 out of 17 = …….

(iii) three-fifths = …….

Solution:

(i)The fraction for 2 out of 7 is written as 2 / 7

(ii) The fraction for 5 out of 17 is written as 5 / 17

(iii) The fraction for three-fifths is written as 3 / 5

2. Fill in the blanks:

(i) 5 / 8 is …………. fraction

(ii) 8 / 5 is ………. fraction

(iii) -15 / -15 is …………fraction

(iv) The value of 5 / 5 = ………..

(v) The value of 5 / -5 = ………….

Solution:

(i) 5 / 8 is proper fraction

(ii) 8 / 5 is improper fraction

(iii) -15 / -15 is improper fraction

(iv) The value of 5 / 5 = 1

(v) The value of 5 / -5 = -1

3. From the following fractions, separate:

(i) Proper fractions

(ii) Improper fractions:

2 / 9, 4 / 3, 7 / 15, 11 / 20, 20 / 11, 18 / 23 and 27 / 35

Solution:

(i) A fraction whose numerator is less than denominator is known as proper fractions

The proper fractions are 2 / 9, 7 / 15, 11 / 20, 18 / 23 and 27 / 35

(ii) A fraction whose numerator is greater than denominator is known as improper fractions

The improper fractions are 4 / 3 and 20 / 11

4. Change the following mixed fractions to improper fractions:

(i) Selina Solutions Concise Mathematics Class 6 Chapter 14 - 1

(ii)
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 2

(iii)
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 3

(iv) Selina Solutions Concise Mathematics Class 6 Chapter 14 - 4

Solution:

(i)
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 5

The conversion of mixed fraction to an improper fraction is shown below

Selina Solutions Concise Mathematics Class 6 Chapter 14 - 6= (2 × 5 + 1) / 5

= (10 + 1) / 5

= 11 / 5

(ii)
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 7

The conversion of mixed fraction to an improper fraction is shown below

Selina Solutions Concise Mathematics Class 6 Chapter 14 - 8= (3 × 4 + 1) / 4

= (12 + 1) / 4

= 13 / 4

(iii)
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 9

The conversion of mixed fraction to an improper fraction is shown below

Selina Solutions Concise Mathematics Class 6 Chapter 14 - 10= (7 × 8 + 1) / 8

= (56 + 1) / 8

= 57 / 8

(iv)
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 11

The conversion of mixed fraction to an improper fraction is shown below

Selina Solutions Concise Mathematics Class 6 Chapter 14 - 12= (2 × 11 + 1) / 11

= (22 + 1) / 11

= 23 / 11

5. Change the following improper fractions to mixed fractions:

(i) 100 / 17

(ii) 81 / 11

(iii) – 209 / 7

(iv) – 113 / 15

Solution:

(i) 100 / 17

The conversion of an improper fraction into mixed fractions is shown below

100 / 17 =
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 13

(ii) 81 / 11

The conversion of an improper fraction into mixed fraction is shown below

81 / 11 =
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 14

(iii) -209 / 7

The conversion of an improper fraction into mixed fraction is shown below

-209 / 7 =
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 15

(iv) -113 / 15

The conversion of an improper fraction into mixed fraction is shown below

-113 / 15 =
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 16

6. Change the following groups of fractions to like fractions:

(i) 1 / 3, 2 / 5, 3 / 4, 1 / 6

(ii) 5 / 6, 7 / 8, 11 / 12, 3 / 10

(iii) 2 / 7, 7 / 8, 5 / 14, 9 / 16

Solution:

(i) 1 / 3, 2 / 5, 3 / 4, 1 / 6

The conversion of fractions to like fractions is shown below

LCM of the denominator 3, 5, 4, 6 = 60

Selina Solutions Concise Mathematics Class 6 Chapter 14 - 17

= 2 × 3 × 1 × 5 × 2 × 1

= 60

1 / 3 = (1 × 20) / (3 × 20)

= 20 / 60

2 / 5 = (2 × 12) / (5 × 12)

= 24 / 60

3 / 4 = (3 × 15) / (4 × 15)

= 45 / 60

1 / 6 = (1 × 10) / (6 × 10)

= 10 / 60

Hence 1 / 3, 2 / 5, 3 / 4, 1 / 6 = 20 / 60, 24 / 60, 45 / 60, 10 / 60

(ii) 5 / 6, 7 / 8, 11 / 12, 3 / 10

The conversion of fractions to like fractions is shown below

LCM of the denominators 6, 8, 12, 10 = 120

Selina Solutions Concise Mathematics Class 6 Chapter 14 - 18

= 2 × 2 × 3 × 1 × 2 × 1 × 5

= 120

5 / 6 = (5 × 20) / (6 × 20)

= 100 / 120

7 / 8 = (7 × 15) / (8 × 15)

= 105 / 120

11 / 12 = (11 × 10) / (12 × 10)

= 110 / 120

3 / 10 = (3 × 12) / (10 × 12)

= 36 / 120

Hence 5 / 6, 7 / 8, 11 / 12, 3 / 10 = 100 / 120, 105 / 120, 110 / 120, 36 / 120

(iii) 2 / 7, 7 / 8, 5 / 14, 9 / 16

LCM of the denominators 7, 8, 14, 16 = 112

Selina Solutions Concise Mathematics Class 6 Chapter 14 - 19

= 2 × 7 × 4 × 1 × 1 × 1 × 2

= 112

2 / 7 = (2 × 16) / (7 × 16)

= 32 / 112

7 / 8 = (7 × 14) / (8 × 14)

= 98 / 112

5 / 14 = (5 × 8) / (14 × 8)

= 40 / 112

9 / 16 = (9 × 7) / (16 × 7)

= 63 / 112

Hence 2 / 7, 7 / 8, 5 / 14, 9 / 16 = 32 / 112, 98 / 112, 40 / 112, 63 / 112

Exercise 14(B)

1. Reduce the given fractions to their lowest terms:

(i) 8 / 10

(ii) 50 / 75

(iii) 18 / 81

(iv) 40 / 120

(v) 105 / 70

Solution:

(i) 8 / 10

The fraction 8 / 10 can be simplified as below

8 / 10 = (8 ÷ 2) / (10 ÷ 2)

= 4 / 5

Hence 4 / 5 is the simplified form of 8 / 10

(ii) 50 / 75

The fraction 50 / 75 can be simplified as below

50 / 75 = (50 ÷ 25) / (75 ÷ 25)

= 2 / 3

Hence 2 / 3 is the simplified form of 50 / 75

(iii) 18 / 81

The fraction 18 / 81 can be simplified as below

18 / 81 = (18 ÷ 9) / (81 ÷ 9)

= 2 / 9

Hence 2 / 9 is the simplified form of 18 / 81

(iv) 40 / 120

The fraction 40 / 120 can be simplified as below

40 / 120 = (40 ÷ 40) / (120 ÷ 40)

= 1 / 3

Hence 1 / 3 is the simplified form of 40 / 120

(v) 105 / 70

The fraction 105 / 70 can be simplified as below

105 / 70 = (105 ÷ 35) / (70 ÷ 35)

= 3 / 2

Hence 3 / 2 is the simplified form of 105 / 70

2. State, whether true or false?

(i) 2 / 5 = 10 / 15

(ii) 35 / 42 = 5 / 6

(iii) 5 / 4 = 4 / 5

(iv) 7 / 9 = Selina Solutions Concise Mathematics Class 6 Chapter 14 - 20

(v) 9 / 7 = Selina Solutions Concise Mathematics Class 6 Chapter 14 - 21

Solution:

(i) 2 / 5 = 10 / 15

The given expression can be solved as below

2 / 5 = (10 ÷ 5) / (15 ÷ 5)

2 / 5 ≠ 2 / 3

Hence false

(ii) 35 / 42 = 5 / 6

The given expression can be solved as below

(35 ÷ 7) / (42 ÷ 7) = 5 / 6

5 / 6 = 5 / 6

Hence true

(iii) 5 / 4 = 4 / 5

The given expression can be solved as below

5 / 4 ≠ 4 / 5

Hence false

(iv) 7 / 9 = Selina Solutions Concise Mathematics Class 6 Chapter 14 - 22

The given expression can be solved as below

7 / 9 = (7 × 1 + 1) / 7

7 / 9 ≠ 8 / 7

Hence false

(v) 9 / 7 =
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 23

The given expression can be solved as below

9 / 7 = (7 × 1 + 1) / 7

9 / 7 ≠ 8 / 7

Hence false

3. Which fraction is greater?

(i) 3 / 5 or 2 / 3

(ii) 5 / 9 or 3 / 4

(iii) 11 / 14 or 26 / 35

Solution:

(i) 3 / 5 or 2 / 3

The given fractions can be simplified as follows

LCM of 5, 3 is 15

Hence 3 / 5 = (3 × 3) / (5 × 3)

= 9 / 15 and

2 / 3 = (2 × 5) / (3 × 5)

= 10 / 15

We know that

10 / 15 > 9 / 15 [Numerator is greater]

Thus, 2 / 3 > 3 / 5

Hence 2 / 3 is greater fraction

(ii) 5 / 9 or 3 / 4

The given expression can be simplified as follows

First convert the given expression into like fractions

So, 5 / 9 = (5 × 4) / (9 × 4)

= 20 / 36 and

3 / 4 = (3 × 9) / (4 × 9)

= 27 / 36

We know that

27 / 36 > 20 / 36 [Numerator is greater]

Thus 3 / 4 > 5 / 9

Hence 3 / 4 is greater fraction

(iii) 11 / 14 or 26 / 35

The given expression can be simplified as follows

First convert the given expression into like fractions

So, 11 / 14 = (11 × 5) / (14 × 5)

= 55 / 70 and

26 / 35 = (26 × 2) / (35 × 2)

= 52 / 70

We know that

55 / 70 > 52 / 70 [Numerator is greater]

Thus, 11 / 14 > 26 / 35

Hence 11 / 14 is greater fraction

4. Which fraction is smaller?

(i) 3 / 8 or 4 / 5

(ii) 8 / 15 or 4 / 7

(iii) 7 / 26 or 10 / 39

Solution:

(i) 3 / 8 or 4 / 5

The given expression can be simplified as follows

First convert the given expression into like fractions

So, 3 / 8 = (3 × 5) / (8 × 5)

= 15 / 40 and

4 / 5 = (4 × 8) / (5 × 8)

= 32 / 40

We know that

15 / 40 < 32 / 40 [Numerator is smaller]

Thus, 3 / 8 < 4 / 5

Hence 3 / 8 is the smaller fraction

(ii) 8 / 15 or 4 / 7

The given expression can be simplified as follows

First convert the given expression into like fractions

So, 8 / 15 = (8 × 7) / (15 × 7)

= 56 / 105 and

4 / 7 = (4 × 15) / (7 × 15)

= 60 / 105

We know that

56 / 105 < 60 / 105 [Numerator is smaller]

Thus, 8 / 15 < 4 / 7

Hence 8 / 15 is the smaller fraction

(iii) 7 / 26 or 10 / 39

The given expression can be simplified as follows

First convert the given expression into like fractions

So, 7 / 26 = (7 × 3) / (26 × 3)

= 21 / 78 and

10 / 39 = (10 × 2) / (39 × 2)

= 20 / 78

We know that

20 / 78 < 21 / 78 [Numerator is smaller]

Thus, 10 / 39 < 7 / 26

Hence 10 / 39 is the smaller fraction

5. Arrange the given fractions in descending order of magnitude:

(i) 5 / 16, 13 / 24, 7 / 8

(ii) 4 / 5, 7 / 15, 11 / 20, 3 / 4

(iii) 5 / 7, 3 / 8, 9 / 11

Solution:

(i) 5 / 16, 13 / 24, 7 / 8

The given expression can be simplified as follows

Selina Solutions Concise Mathematics Class 6 Chapter 14 - 24

LCM of 16, 24, 8 = 2 × 2 × 2 × 2 × 3

= 48

Converting given expression into like fractions, we get

5 / 16 = (5 × 3) / (16 × 3)

= 15 / 48 and

13 / 24 = (13 × 2) / (24 × 2)

= 26 / 48 and

7 / 8 = (7 × 6) / (8 × 6)

= 42 / 48

Hence, fractions in descending order are 7 / 8, 13 / 24, 5 / 16

(ii) 4 / 5, 7 / 15, 11 / 20, 3 / 4

The given expression can be simplified as follows

Selina Solutions Concise Mathematics Class 6 Chapter 14 - 25

LCM of 5, 15, 20, 4 = 4 × 5 × 3

= 60

Converting the given expression into like fractions, we get

4 / 5 = (4 ×12) / (5 × 12)

=48 / 60 and

7 / 15 = (7 × 4) / (15 × 4)

= 28 / 60 and

11 / 20 = (11 × 3) / (20 × 3)

= 33 / 60 and

3 / 4 = (3 × 15) / (4 × 15)

= 45 / 60

Hence, fractions in descending order are 4 / 5, 3 / 4, 11 / 20, 7 / 15

(iii) 5 / 7, 3 / 8, 9 / 11

The given expression can be simplified as follows

Selina Solutions Concise Mathematics Class 6 Chapter 14 - 26

LCM of 5, 3, 9 = 3 × 3 × 5

= 45

Converting the given expression into like fractions, we get

5 / 7 = (5 × 9) / (7 × 9)

= 45 / 63 and

3 / 8 = (3 × 15) / (8 × 15)

= 45 / 120 and

9 / 11 = (9 × 5) / (11 × 5)

= 45 / 55

The fraction with the smallest denominator is the biggest fraction if the numerator is same

Hence, fractions in descending order are

45 / 55, 45 / 63, 45 / 120 i.e

9 / 11, 5 / 7, 3 / 8

6. Arrange the given fractions in ascending order of magnitude:

(i) 9 / 16, 7 / 12, 1 / 4

(ii) 5 / 6, 2 / 7, 8 / 9, 1 / 3

(iii) 2 / 3, 5 / 9, 5 / 6, 3 / 8

Solution:

(i) 9 / 16, 7 / 12, 1 / 4

The given fractions can be simplified as follows

Selina Solutions Concise Mathematics Class 6 Chapter 14 - 27

LCM of 16, 12, 4 = 48

Converting the given expression into like fractions, we get

9 / 16 = (9 × 3) / (16 × 3)

= 27 / 48 and

7 / 12 = (7 × 4) / (12 × 4)

= 28 / 48 and

1 / 4 = (1 × 12) / (4 × 12)

= 12 / 48

Hence, fractions in ascending order are

12 / 48, 27 / 48, 28 / 48 i.e

1 / 4, 9 / 16, 7 / 12

(ii) 5 / 6, 2 / 7, 8 / 9, 1 / 3

The given fractions can be simplified as follows

Selina Solutions Concise Mathematics Class 6 Chapter 14 - 28

LCM of 6, 7, 9, 3 = 3 × 3 × 2 × 7

= 126

Converting the given expression into like fractions, we get

5 / 6 = (5 × 21) / (6 × 21)

= 105 / 126 and

2 / 7 = (2 × 18) / (7 × 18)

= 36 / 126 and

8 / 9 = (8 × 14) / (9 × 14)

= 112 / 126 and

1 / 3 = (1 × 42) / (3 × 42)

= 42 / 126

Hence, fractions in ascending order are

36 / 126, 42 / 126, 105 / 126, 112 / 126 i.e

2 / 7, 1 / 3, 5 / 6, 8 / 9

(iii) 2 / 3, 5 / 9, 5 / 6, 3 / 8

The given fractions can be simplified as follows

Selina Solutions Concise Mathematics Class 6 Chapter 14 - 29

LCM of 3, 9, 6, 8 = 72

Converting the given expressions into like fractions, we get

2 / 3 = (2 × 24) / (3 × 24)

= 48 / 72 and

5 / 9 = (5 × 8) / (9 × 8)

= 40 / 72 and

5 / 6 = (5 × 12) / (6 × 12)

= 60 / 72 and

3 / 8 = (3 × 9) / (8 × 9)

= 27 / 72

Hence, fractions in ascending order are

27 / 72, 40 / 72, 48 / 72, 60 / 72 i.e

3 / 8, 5 / 9, 2 / 3, 5 / 6

7. I bought one dozen bananas and ate five of them. What fraction of the total number of bananas was left?

Solution:

Given

Number of bananas bought = 1 dozen

We know there are 12 bananas in a dozen

Number of bananas eaten = 5

Number of bananas left = 12 – 5

= 7

Therefore, the required fraction is 7 / 12

8. Insert the symbol ‘=’ or ‘>’ or ‘<’ between each of the pairs of fractions, given below:

(i) 6 / 11 …. 5 / 9

(ii) 3 / 7 ….. 9 / 13

(iii) 56 / 64 …. 7 / 8

(iv) 5 / 12 …. 8 / 33

Solution:

(i) 6 / 11 …. 5 / 9

LCM of 11, 9 = 99

Converting the given expression into like fraction

We get

6 / 11 = (6 × 9) / (11 × 9)

= 54 / 99 and

5 / 9 = (5 × 11) / (9 × 11)

= 55 / 99

Therefore,

54 / 99 < 55 / 99 i.e

6 / 11 < 5 / 9

(ii) 3 / 7 ….. 9 / 13

LCM of 7, 13 = 91

Converting the given expression into like fraction

We get

3 / 7 = (3 × 13) / (7 × 13)

= 39 / 91 and

9 / 13 = (9 × 7) / (13 × 7)

= 63 / 91

Therefore,

39 / 91 < 63 / 91 i.e.

3 / 7 < 9 / 13

(iii) 56 / 64 …. 7 / 8

LCM of 64, 8 = 64

Converting the given expression into like fraction

We get

56 / 64 = (56 × 1) / (64 × 1)

= 56 / 64 and

7 / 8 = (7 × 8) / (8 × 8)

= 56 / 64

Therefore,

56 / 64 = 56 / 64 i.e.

56 / 64 = 7 / 8

(iv) 5 / 12 …. 8 / 33

LCM of 12, 33 = 132

Converting the given expression into like fractions

We get

5 / 12 = (5 × 11) / (12 × 11)

= 55 / 132 and

8 / 33 = (8 × 4) / (33 × 4)

= 32 / 132

55 / 132 > 32 / 132 i.e

5 / 12 > 8 / 33

9. Out of 50 identical articles, 36 are broken. Find the fraction of:

(i) The total number of articles and the articles broken.

(ii) The remaining articles and total number of articles.

Solution:

(i) Given

Total number of articles = 50

Number of articles broken = 36

Remaining articles = 50 – 36

= 14

The fraction of total number of articles and articles broken = 50 / 36

= 25 / 18

(ii) Given

Total number of articles = 50

Number of articles broken = 36

Remaining articles = 50 – 36

= 14

The fraction of remaining articles and total number of articles = 14 / 50

= 7 / 25

Exercise 14(C)

1. Add the following fractions:

(i) Selina Solutions Concise Mathematics Class 6 Chapter 14 - 30 and 3 / 8

(ii) 2 / 5, Selina Solutions Concise Mathematics Class 6 Chapter 14 - 31 and 7 / 10

(iii) Selina Solutions Concise Mathematics Class 6 Chapter 14 - 32, Selina Solutions Concise Mathematics Class 6 Chapter 14 - 33 and Selina Solutions Concise Mathematics Class 6 Chapter 14 - 34

(iv) Selina Solutions Concise Mathematics Class 6 Chapter 14 - 35, Selina Solutions Concise Mathematics Class 6 Chapter 14 - 36, and Selina Solutions Concise Mathematics Class 6 Chapter 14 - 37

(v) Selina Solutions Concise Mathematics Class 6 Chapter 14 - 38, 11 / 18 and Selina Solutions Concise Mathematics Class 6 Chapter 14 - 39

Solution:

(i)
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 40and 3 / 8

The given fractions can be added as follows

7 / 4 + 3 / 8 = (7 × 2) / (4 × 2) + 3 / 8

= 14 / 8 + 3 / 8

= 17 / 8

=
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 41

Hence,
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 42is the addition of given fractions

(ii) 2 / 5,
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 43and 7 / 10

The given fractions can be added as follows

2 / 5 + 33 / 15 + 7 / 10 = (2 × 6) / (5 × 6) + (33 × 2) / (15 × 2) + (7 × 3) / (10 × 3)

= 12 / 30 + 66 / 30 + 21 / 30

= 99 / 30

= 33 / 10

=
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 44

Hence,
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 45is the addition of given fractions

(iii)
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 46,
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 47and
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 48

The given fractions can be added as follows

15 / 8 + 3 / 2 + 7 / 4 = (15 × 1) / (8 × 1) + (3 × 4) / (2 × 4) + (7 × 2) / (4 × 2)

= 15 / 8 + 12 / 8 + 14 / 8

= 41 / 8

=
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 49

Hence,
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 50is the addition of the given fractions

(iv)
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 51,
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 52and
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 53

The given fractions can be added as follows

15 / 4 + 13 / 6 + 13 / 8 = (15 × 6) / (4 × 6) + (13 × 4) / (6 × 4) + (13 × 3) / (8 × 3)

= 90 / 24 + 52 / 24 + 39 / 24

= 181 / 24

=
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 54

Hence,
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 55is the addition of given fractions

(v)
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 56, 11 / 18 and
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 57

The given fractions can be added as follows

26 / 9 + 11 / 18 + 23 / 6 = (26 × 2) / (9 × 2) + 11 / 18 + (23 × 3) / (6 × 3)

= 52 / 18 + 11 / 18 + 69 / 18

= 132 / 18

= 22 / 3

=
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 58

Hence,
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 59is the addition of given fractions

2. Simplify:

(i) Selina Solutions Concise Mathematics Class 6 Chapter 14 - 60 – 13 / 16

(ii) Selina Solutions Concise Mathematics Class 6 Chapter 14 - 61Selina Solutions Concise Mathematics Class 6 Chapter 14 - 62

(iii) Selina Solutions Concise Mathematics Class 6 Chapter 14 - 63 + 3 / 14 – 13 / 21

(iv) Selina Solutions Concise Mathematics Class 6 Chapter 14 - 64 – 1 / 6 – Selina Solutions Concise Mathematics Class 6 Chapter 14 - 65

(v) 6 + 3 / 10 – Selina Solutions Concise Mathematics Class 6 Chapter 14 - 66

Solution:

(i)
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 67– 13 / 16

The given expression can be simplified as below

23 / 12 – 13 / 16 = (23 × 4) / (12 × 4) – (13 × 3) / (16 × 3)

= (92 – 39) / 48

= 53 / 48

=
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 68

Hence, simplified form of the given expression is
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 69

(ii)
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 70
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 71

The given expression can be simplified as below

11 / 4 – 11 / 6 = (11 × 6) / (4 × 6) – (11 × 4) / (6 × 4)

= (66 – 44) / 24

= 22 / 24

= 11 / 12

Hence, simplified form of the given expression is 11 / 12

(iii)
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 72+ 3 / 14 – 13 / 21

The given expression can be simplified as below

19 / 7 + 3 / 14 – 13 / 21 = (19 × 6) / (7 × 6) + (3 × 3) / (14 × 3) – (13 × 2) / (21 × 2)

= 114 / 42 + 9 / 42 – 26 / 42

= (114 + 9 – 26) / 42

= 97 / 42

=
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 73

Hence, simplified form of the given expression is
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 74

(iv)
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 75– 1 / 6 –
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 76

The given expression can be simplified as below

23 / 6 – 1 / 6 – 13 / 12 = (23 × 2) / (6 × 2) – (1 × 2) / (6 × 2) – 13 / 12

= 46 / 12 – 2 / 12 – 13 / 12

= (46 – 2 – 13) / 12

= 31 / 12

=
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 77

Hence, simplified form of the given expression is
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 78

(v) 6 + 3 / 10 –
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 79

The given expression can be simplified as below

6 / 1 + 3 / 10 – 23 / 15 = (6 × 30) / (1 × 30) + (3 × 3) / (10 × 3) – (23 × 2) / (15 × 2)

= 180 / 30 + 9 / 30 – 46 / 30

= (180 + 9 – 46) / 30

= 143 / 30

=
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 80

Hence, simplified form of the given expression is
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 81

Exercise 14(D)

1. Simplify:

(i) 3 / 7 × 2 / 5

(ii) 4 / 9 × 3 / 5

(iii) 5 / 12 × 8

(iv) 7 / 6 of 3 / 14

(v) Selina Solutions Concise Mathematics Class 6 Chapter 14 - 82 × Selina Solutions Concise Mathematics Class 6 Chapter 14 - 83

Solution:

(i) 3 / 7 × 2 / 5

The given expression can be simplified as below

3 / 7 × 2 / 5 = (3 × 2) / (7 × 5)

= 6 / 35

Hence, the simplified form of the given expression is 6 / 35

(ii) 4 / 9 × 3 / 5

The given expression can be simplified as below

4 / 9 × 3 / 5 = (4 × 3) / (9 × 5)

= (4 × 1) / (3 × 5)

= 4 / 15

Hence, the simplified form of the given expression is 4 / 15

(iii) 5 / 12 × 8

The given expression can be simplified as below

5 / 12 × 8 / 1 = (5 × 8) / (12 × 1)

= (5 × 2) / (3 × 1)

= 10 / 3

=
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 84

Hence, the simplified form of the given expression is
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 85

(iv) 7 / 6 of 3 / 14

The given expression can be simplified as below

7 / 6 × 3 / 14 = (1 × 1) / (2 × 2)

= 1 / 4

Hence, the simplified form of the given expression is 1 / 4

(v)
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 86×
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 87

The given expression can be simplified as below

Selina Solutions Concise Mathematics Class 6 Chapter 14 - 88×
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 89= (27 × 27) / (8 × 7)

= 729 / 56

=
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 90

Hence, the simplified form of the given expression is
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 91

2. Simplify:

(i) 2 / 3 ÷ Selina Solutions Concise Mathematics Class 6 Chapter 14 - 92

(ii) Selina Solutions Concise Mathematics Class 6 Chapter 14 - 93 ÷ 4 / 9

(iii) 1 ÷ 2 / 5

(iv) 4 / 9 ÷ 4 / 9

(v) Selina Solutions Concise Mathematics Class 6 Chapter 14 - 94 ÷ Selina Solutions Concise Mathematics Class 6 Chapter 14 - 95

Solution:

(i) 2 / 3 ÷
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 96

The given expression can be simplified as below

2 / 3 ÷ 6 / 5 = (2 × 5) / (3 × 6)

= 5 / 9

Hence, the simplified form of the given expression is 5 / 9

(ii)
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 97÷ 4 / 9

The given expression can be simplified as below

9 / 2 ÷ 4 / 9 = (9 × 9) / (2 × 4)

= 81 / 8

=
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 98

Hence, the simplified form of the given expression is
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 99

(iii) 1 ÷ 2 / 5

The given expression can be simplified as below

1 / 1 ÷ 2 / 5 = (1 × 5) / (1 × 2)

= 5 / 2

=
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 100

Hence, the simplified form of the given expression is
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 101

(iv) 4 / 9 ÷ 4 / 9

The given expression can be simplified as below

4 / 9 ÷ 4 / 9 = (4 × 9) / (9 × 4)

=1

Hence, the simplified form of the given expression is 1

(v)
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 102÷
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 103

The given expression can be simplified as below

Selina Solutions Concise Mathematics Class 6 Chapter 14 - 104÷
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 105= (7 × 4) / (3 × 7)

= 4 / 3

=
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 106

Hence, the simplified form of the given expression is
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 107

3. Simplify:

(i) 1 / 4 of Selina Solutions Concise Mathematics Class 6 Chapter 14 - 108 ÷ 3 / 5

(ii) Selina Solutions Concise Mathematics Class 6 Chapter 14 - 109 × 1 / 2 ÷ Selina Solutions Concise Mathematics Class 6 Chapter 14 - 110

(iii) Selina Solutions Concise Mathematics Class 6 Chapter 14 - 111 × 0 × Selina Solutions Concise Mathematics Class 6 Chapter 14 - 112

(iv) 3 / 4 × Selina Solutions Concise Mathematics Class 6 Chapter 14 - 113 ÷ 3 / 7 of Selina Solutions Concise Mathematics Class 6 Chapter 14 - 114

(v) Selina Solutions Concise Mathematics Class 6 Chapter 14 - 115 ÷ 2 / 7 of Selina Solutions Concise Mathematics Class 6 Chapter 14 - 116 × 2 / 3

Solution:

(i) 1 / 4 of
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 117÷ 3 / 5

The given expression can be simplified as follows:

1 / 4 × 16 / 7 ÷ 3 / 5 = 4 / 7 ÷ 3 / 5

= (4 × 5) / (7 × 3)

= 20 / 21

Hence, 20 / 21 is the simplified form of the given expression

(ii) Selina Solutions Concise Mathematics Class 6 Chapter 14 - 118 × 1 / 2 ÷
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 119

The given expression can be simplified as follows

Selina Solutions Concise Mathematics Class 6 Chapter 14 - 120× 1 / 2 ÷
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 121= 5 / 4 × 1 / 2 × 3 / 4

= 5 / 8 × 3 / 4

= 15 / 32

Hence, 15 / 32 is the simplified form of the given expression

(iii)
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 122× 0 × Selina Solutions Concise Mathematics Class 6 Chapter 14 - 123

The given expression can be simplified as follows

Selina Solutions Concise Mathematics Class 6 Chapter 14 - 124× 0 × Selina Solutions Concise Mathematics Class 6 Chapter 14 - 125 = (43 × 0 × 43) / (7 × 0 × 8)

= 0

Hence, 0 is the simplified form of the given expression

(iv) 3 / 4 ×
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 126÷ 3 / 7 of
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 127

The given expression can be simplified as follows

3 / 4 ×
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 128÷ 3 / 7 of
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 129= 3 / 4 × 4 / 3 ÷ 9 / 8

∵ 3 / 7 of
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 130= 3 / 7 × 21 / 8 = 9 / 8

= 3 / 4 × 4 / 3 × 8 / 9

= 8 / 9

Hence, 8 / 9 is the simplified form of the given expression

(v)
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 131÷ 2 / 7 of
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 132× 2 / 3

The given expression can be simplified as follows

[2 / 7 of
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 133= 2 / 7 × 4 / 3 = 8 / 21]

We get

= 9 / 4 ÷ 8 / 21 × 2 / 3

= 9 / 4 × 21 / 8 × 2 / 3

= 63 / 16

=
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 134

Hence,
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 135is the simplified form of the given expression

4. Simplify:

(i) 5 – (8 / 11 – Selina Solutions Concise Mathematics Class 6 Chapter 14 - 136)

(ii) 1 / 2 ÷ (7 / 8 – 3 / 5)

(iii) Selina Solutions Concise Mathematics Class 6 Chapter 14 - 137 ÷ (Selina Solutions Concise Mathematics Class 6 Chapter 14 - 138 + Selina Solutions Concise Mathematics Class 6 Chapter 14 - 139)

(iv) (Selina Solutions Concise Mathematics Class 6 Chapter 14 - 140
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 141) ÷ 1 / 2

(v) 4 / 7 ÷ (1 / 3 ×
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 142)

Solution:

(i) 5 – (8 / 11Selina Solutions Concise Mathematics Class 6 Chapter 14 - 143)

The given expression can be simplified as below

5 – (8 / 11 –
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 144) = 5 – (8 / 11 – 36 / 11)

= 5 – (8 – 36) / 11

= 5 – (-28 / 11)

On further calculation, we get

= 5 / 1 + 28 / 11

= 83 / 11

=
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 145

Hence,
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 146is the simplified form of the given expression

(ii) 1 / 2 ÷ (7 / 8 – 3 / 5)

The given expression can be simplified as below

1 / 2 ÷ (7 / 8 – 3 / 5) = 1 / 2 ÷ (5 × 7 – 8 × 3) / 40

= 1 / 2 ÷ (35 – 24) / 40

= 1 / 2 ÷ (11 / 40)

= 1 / 2 × 40 / 11

We get

= 20 / 11

=
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 147

Hence,
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 148is the simplified form of the given expression

(iii)
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 149 ÷ (Selina Solutions Concise Mathematics Class 6 Chapter 14 - 150 + Selina Solutions Concise Mathematics Class 6 Chapter 14 - 151)

The given expression can be simplified as below

Selina Solutions Concise Mathematics Class 6 Chapter 14 - 152 ÷ (Selina Solutions Concise Mathematics Class 6 Chapter 14 - 153 + Selina Solutions Concise Mathematics Class 6 Chapter 14 - 154) = 7 / 3 ÷ (11 / 2 + 15 / 4)

= 7 / 3 ÷ (2 × 11 + 1 × 15) / 4

On further calculation, we get

= 7 / 3 ÷ (22 + 15) / 4

= 7 / 3 ÷ (37 / 4)

= 7 / 3 × 4 / 37

= 28 / 111

Hence, 28 / 111 is the simplified form of the given expression

(iv) (Selina Solutions Concise Mathematics Class 6 Chapter 14 - 155
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 156) ÷ 1 / 2

The given expression can be simplified as below

(Selina Solutions Concise Mathematics Class 6 Chapter 14 - 157
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 158) ÷ 1 / 2 = (31 / 8 – 18 / 5) ÷ 1 / 2

By taking LCM, we get

= [(31 × 5 – 18 × 8) / (8 × 5)] ÷ 1 / 2

= (155 – 144) / 40 ÷ 1 / 2

= (11 / 40) ÷ 1 / 2

By calculating further, we get

= 11 / 40 × 2 / 1

= 11 / 20

Hence, 11 / 20 is the simplified form of the given expression

(v) 4 / 7 ÷ (1 / 3 ×
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 159)

The given expression can be simplified as below

4 / 7 ÷ (1 / 3 ×
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 160) = 4 / 7 ÷ (1 / 3 × 14 / 5)

= 4 / 7 ÷ (14 / 15)

On further calculation, we get

= 4 / 7 × 15 / 14

= 2 / 7 × 15 / 7

= 30 / 49

Hence, 30 / 49 is the simplified form of the given expression

5. Simplify

(i) (1 / 2 + 1 / 3) ÷ (1 / 4 – 1 / 6)

(ii) (24 / 35 ÷ 6 / 7 + 5 / 9) × 3 / 4

(iii) 3 / 4 of Selina Solutions Concise Mathematics Class 6 Chapter 14 - 161 – 2 / 3 of Selina Solutions Concise Mathematics Class 6 Chapter 14 - 162

(iv) 7 / 30 of (1 / 3 + 7 / 15) ÷ (5 / 6 – 3 / 5)

(v) Selina Solutions Concise Mathematics Class 6 Chapter 14 - 163Selina Solutions Concise Mathematics Class 6 Chapter 14 - 164 × Selina Solutions Concise Mathematics Class 6 Chapter 14 - 165 + Selina Solutions Concise Mathematics Class 6 Chapter 14 - 166

Solution:

(i) (1 / 2 + 1 / 3) ÷ (1 / 4 – 1 / 6)

The given expression can be simplified as follows

(1 / 2 + 1 / 3) ÷ (1 / 4 – 1 / 6) = [(3 + 2) / 6] ÷ [(3 – 2) / 12]

On further calculation, we get

= (5 / 6) ÷ (1 / 12)

= 5 / 6 × 12 / 1

= 5 × 2

= 10

Hence, the simplified form of the given expression is 10

(ii) (24 / 35 ÷ 6 / 7 + 5 / 9) × 3 / 4

The given expression can be simplified as follows

(24 / 35 ÷ 6 / 7 + 5 / 9) × 3 / 4 = (24 / 35 × 7 / 6 + 5 / 9) × 3 / 4

= (4 / 5 + 5 / 9) × 3 / 4

By taking LCM, we get

= [(36 + 25) / 45] × 3 / 4

= (61 / 45) × 3 / 4

We get

= 61 / 60

=
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 167

Hence, the simplified form of the given expression is
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 168

(iii) 3 / 4 of
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 169– 2 / 3 of
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 170

The given expression can be simplified as below

3 / 4 of
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 171– 2 / 3 of
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 172= 3 / 4 of 49 / 8 – 2 / 3 of 9 / 4

= 3 / 4 × 49 / 8 – 2 / 3 of 9 / 4

= 147 / 32 – 2 / 3 × 9 / 4

We get

= 147 / 32 – 3 / 2

On taking LCM, we get

= [(147 – 48)] / 32

= 99 / 32

=
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 173

Hence, the simplified form of the given expression is
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 174

(iv) 7 / 30 of (1 / 3 + 7 / 15) ÷ (5 / 6 – 3 / 5)

The given expression can be simplified as below

7 / 30 of (1 / 3 + 7 / 15) ÷ (5 / 6 – 3 / 5) = 7 / 30 of [(5 + 7) / 15] ÷ [(25 – 18) / 30]

We get

= 7 / 30 of (4 / 5) ÷ (7 / 30)

= 7 / 30 × 4 / 5 × 30 / 7

= 4 / 5

Hence, the simplified form of the given expression is 4 / 5

(v)
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 175 Selina Solutions Concise Mathematics Class 6 Chapter 14 - 176 × Selina Solutions Concise Mathematics Class 6 Chapter 14 - 177 + Selina Solutions Concise Mathematics Class 6 Chapter 14 - 178

The given expression can be simplified as below

Selina Solutions Concise Mathematics Class 6 Chapter 14 - 179
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 180×
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 181+
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 182= 5 / 2 – 7 / 2 × 7 / 4 + 5 / 2

= 5 / 2 – 49 / 8 + 5 / 2

= 5 / 2 + 5 / 2 – 49 / 8

By taking LCM, we get

= (20 + 20 – 49) / 8

= (40 – 49) / 8

= – 9 / 8

= –
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 183

Hence, –
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 184is the simplified form of the given expression

Exercise 14(e)

1. From a rope of Selina Solutions Concise Mathematics Class 6 Chapter 14 - 185 m long, Selina Solutions Concise Mathematics Class 6 Chapter 14 - 186 m is cut off. Find the length of the remaining rope

Solution:

Given

Length of the rope =
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 187m

Length of cut off rope =
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 188m

Remaining rope =
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 189
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 190

We get

= 21 / 2 – 37 / 8

By taking LCM, we get

= (84 – 37) / 8

= 47 / 8 m

=
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 191

Hence, the length of the remaining rope is
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 192m

2. A piece of cloth is 5 metre long. After washing, it shrinks by 1 / 25 of its length. What is the length of the cloth after washing?

Solution:

Given

Length of piece of cloth = 5 metre

After washing, it shrinks by = 1 / 25 of its length

Hence, the shrinked cloth can be calculated as below

Shrinked cloth = 1 / 25 of 5 m

= 1 / 5 m

Hence, length of cloth after washing can be calculated as below

Length of cloth after washing = 5 – 1 / 5

By taking LCM, we get

= (25 – 1) / 5

= 24 / 5

=
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 193m

Hence the length of cloth after washing is
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 194m

3. I bought wheat worth Rs Selina Solutions Concise Mathematics Class 6 Chapter 14 - 195, rice worth Rs Selina Solutions Concise Mathematics Class 6 Chapter 14 - 196 and vegetables worth Rs Selina Solutions Concise Mathematics Class 6 Chapter 14 - 197. If I gave a hundred-rupee note to the shopkeeper; how much did he return to me

Solution:

Given

Wheat = Rs
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 198

Rice = Rs
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 199

Vegetables = Rs
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 200

Hence, total amount used to purchase the goods can be calculated as below

Total amount of goods = Rs (
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 201+
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 202+
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 203)

We get

= 25 / 2 + 103 / 4 + 41 / 4

= Rs 194 / 4

Hence, money returned by shopkeeper can be calculated as below

Money returned = Rs (100 – 194 / 4)

By taking LCM, we get

= Rs (400 – 194) / 4

= Rs 103 / 2

= Rs
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 204

Hence, money returned by the shopkeeper is Rs
Selina Solutions Concise Mathematics Class 6 Chapter 14 - 205

4. Out of 500 oranges in a box, 3 / 25 are rotten and 1 / 5 are kept for some guests. How many oranges are left in the box?

Solution:

Given

Number of oranges in a box = 500

Rotten oranges out of 500 = 3 / 25

Oranges for guests out of 500 = 1 / 5

Rotten oranges = 3 / 25 of 500

We get,

= 3 / 25 × 500

= 60

Oranges for guests = 1 / 5 of 500

We get,

= 1 / 5 × 500

= 100

Oranges left in box = 500 – 60 – 100

= 340

Hence, 340 oranges are left in the box

5. An ornament piece is made of gold and copper. Its total weight is 96 g. If 1 / 12 of the ornament is copper, find the weight of gold in it.

Solution:

Given

Weight of an ornament = 96 g

Weight of copper = 1 / 12 of 96 g

Weight of copper = 1 / 12 × 96

We get,

= 8 g

Weight of gold = 96 – 8

= 88 g

Hence, the weight of gold in ornament is 88 g

6. 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:

Given

Half of work is done on Monday and one-third on Tuesday by a girl

Let total work done by a girl is 1

Work done on Monday = 1 / 2

Work done on Tuesday = 1 / 3

Hence, remaining work done on Wednesday to complete the work is calculated as below

Remaining work done = 1 – [(1 / 2 + 1 / 3)]

Taking LCM, we get

= 1 – [(3 + 2) / 6]

= 1 – 5 / 6

= (6 – 5) / 6

= 1 / 6

Hence, work done by a girl on Wednesday to complete is 1 / 6

7. A man spends 3 / 8 of his money and 8 still has Rs 720 left with him. How much money did he have at first?

Solution:

Given

Man spends 3 / 8 of his money

Let us assume a man has Rs 1

Amount spent = 3 / 8 of 1

We get,

= Rs 3 / 8

Amount left = 1 – 3 / 8

We get,

= (8 – 3) / 8

= Rs 5 / 8

Since 5 / 8 of his total money = Rs 720

∴ Total money = Rs (720 × 8) / 5

= Rs 5760 / 5

= Rs 1152

Hence, total money a man has is Rs 1152

8. In a school, 4 / 5 of the students are boys, and the number of girls is 100. Find the number of boys

Solution:

Given

Total number of girls = 100

Number of boys = 4 / 5

Let us assume the total number of boys and girls be x

Total number of boys = 4 / 5 of x

We get,

= 4x / 5

According to question, total strength of school can be calculated as below

x – (4x/5) = 100

(5x – 4x)/5 = 100

x = 500

Number of boys = Total strength – Girls

= 500 – 400

= 400

Hence number of boys are 400

9. After finishing 3 / 4 of my journey, I find that 12 km of my journey is covered. How much distance is still left to be covered?

Solution:

Let x km be the total journey

Given that total distance covered = 3 / 4 of the journey is 12 km

According to the question, the distance covered can be calculated as below

3 / 4 of x = 12 km

x = 4 / 3 × 12

x = 4 × 4

x = 16 km

Remaining distance = 16 – 12

= 4 km

Hence, 4 km of distance is left to be covered

10. 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:

Let the total journey = x km

Given total distance covered = 15 km

Journey left = 1 / 4 of x

Hence, according to the question, the total distance of journey can be calculated as below

1 / 4 of x = x – 15

x – x / 4 = 15

By calculating further, we get

(4x – x) / 4 = 15

3x / 4 = 15

3x = 60

x = 60 / 3

x = 20

Hence, the full length of the journey is 20 km

11. In a particular month, a man earns Rs 7, 200. Out of this income, he spends 3 / 10 on food, 1 / 4 on house rent, 1 / 10 on insurance and 2 / 25 on holidays. How much did he save in that month?

Solution:

Given

Money earned by a man in a particular month = Rs 7200

Amount spend on food, house rent, insurance and holidays by him are 3 / 10, 1 / 4, 1 / 10 and 2 / 25 respectively

Amount spend on food = 3 / 10 of 7200

= 3 / 10 × 7200

= 3 × 720

= Rs 2160

Amount spend on house rent = 1 / 4 of 7200

= 1 / 4 × 7200

= Rs 1800

Amount spend on insurance = 1 / 10 of 7200

= 1 / 10 × 7200

= Rs 720

Amount spend on holidays = 2 / 25 of 7200

= 2 / 25 × 7200

= Rs 576

Total amount spend = Rs (2160 + 1800 + 720 + 576)

= Rs 5256

Amount saved by man = 7200 – 5256

= Rs 1944

Hence, amount saved by a man in a month is Rs 1944

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