NCERT solutions for class 6 maths chapter7 Fractions

(a) Arya has divided the sandwich into 3 equal parts. So each person will get one part.

 

 

(b) Each boy receives 1 / 3 part

 

 

∴ The required fraction is 1 / 3

 

There are 24 hours in a day

We have 8 hours

Hence, the required fraction is 8 / 24

 

(i) The shaded portion is not half

Hence, this is not 1 / 2

(ii) Since the parts are not equal

The shaded portion is not 1 / 4

(iii) Since the parts are not equal

∴ The shaded portion is not 3 / 4

 

(i) Number of parts = 4

Shaded portion = 2

∴ Fraction = 2 / 4

(ii) Number of parts = 9

Shaded portion = 8

∴ Fraction = 8 / 9

(iii) Number of parts = 8

Shaded portion = 4

∴ Fraction = 4 / 8

(iv) Number of parts = 4

Shaded portion = 1

∴ Fraction = 1 / 4

(v) Number of parts = 7

Shaded portion = 3

∴ Fraction = 3 / 7

(vi) Number of parts = 12

Shaded portion = 3

∴ Fraction = 3 / 12

(vii) Number of parts = 10

Shaded portion = 10

∴ Fraction = 10 / 10

(viii) Number of parts = 9

Shaded portion = 4

∴ Fraction = 4 / 9

(ix) Number of parts = 8

Shaded portion = 4

∴ Fraction = 4 / 8

(x) Number of parts = 2

Shaded portion = 1

∴ Fraction = 1 / 2

 

Number of CDs Kristin bought from the market = 3

Number of CDs received as gifts = 5

Total number of CDs Kristin has = 3 + 5 = 8

∴ Fraction of CD she bought = 3 / 8

∴ The fraction of CDs received as gifts = 5 / 8

Natural numbers from 102 to 113 are

102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113

Total number of natural numbers given = 12

Number of prime numbers = 4 [103, 107, 109, 113]

∴ The required fraction = 4 / 12 = 1 / 3

 

Natural numbers from 2 to 12 are

2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12

Total number of natural numbers given = 11

Number of prime numbers = 5

∴ The required fraction = 5 / 11

 

Total number of dresses Kanchan has to dye = 30 dresses

Number of dresses she has finished = 20 dresses

∴ The required fraction = 20 / 30 = 2 / 3

 

Total number of circles in the figure = 8

Number of circles having Xs in them = 4

∴ The required fraction = 4 / 8 = 1 / 2

 

There are 60 minutes in 1 hour

∴ 1 hour = 60 minutes

Hence, the required fraction = 40 / 60

 

(a) 1 / 2, 1 / 4, 3 / 4, 4 / 4

Here, divide the number line from 0 to 1 into four equal parts

C = 2 / 4 = 1 / 2

B = 1 / 4

D = 3 / 4 and

E = 4 / 4 = 1

(b) 1 / 8, 2 / 8, 3 / 8, 7 / 8

Divide the number line from 0 to 1 into eight equal parts

B = 1 / 8

C = 2 / 8

D = 3 / 8

H = 7 / 8

(c) 2 / 5, 3 / 5, 8 / 5, 4 / 5

From the given number line, we have

C = 2 / 5

D = 3 / 5

E = 4 / 5

I = 8 / 5

 

(a) (7 × 4 + 3) / 4 = 31 / 4

∴ The improper form is 31 / 4

(b) (5 × 7 + 6) / 7 = 41 / 7

∴ The improper form is 41 / 7

(c) (2 × 6 + 5) / 6 = 17 / 6

∴ The improper form is 17 / 6

(d) (10 × 5 + 3) / 5 = 53 / 5

∴ The improper form is 53 / 5

(e) (9 × 7 + 3) / 7 = 66 / 7

∴ The improper form is 66 / 7

(f) (8 × 9 + 4) / 9 = 76 / 9

∴ The improper form is 76 / 9

(a) 20 / 3

∴ 20 / 3 =

(b) 11 / 5

∴ 11 / 5 =

(c) 17 / 7

∴ 17 / 7 =

(d) 28 / 5

∴ 28 / 5 =

(e) 19 / 6

∴ 19 / 6 =

(f) 35 / 9

∴ 35 / 9 =

 

(a)

   (i) The shaded portion is 1 / 2

   (ii) The shaded portion is 2 / 4 = (2 / 2) / (4 / 2) = 1 / 2

   (iii) The shaded portion is 3 / 6 = (3 / 3) / (6 / 3) = 1 / 2

   (iv) The shaded portion is 4 / 8 = (4 / 4) / (8 / 4) = 1 / 2

   Hence, all fractions are equivalent.

(b)

   (i) The shaded portion is 4 / 12 = (4 / 4) / (12 / 4) = 1 / 3

   (ii)The shaded portion is 3 / 9 = (3 / 3) / (9 / 3) = 1 / 3

   (iii) The shaded portion is 2 / 6 = (2 / 2) / (6 / 2) = 1 / 3

   (iv) The shaded portion is 1 / 3

   (v) The shaded portion is 6 / 15 = (6 / 3) / (15 / 3) = 2 / 5

   All the fractions in their simplest form are not equal

   Hence, they are not equivalent fractions.

 

(a) 48 / 60 = (12 × 4) / (12 × 5)

= 4 / 5

(b) 150 / 60 = (30 × 5) / (30 × 2)

= 5 / 2

(c) 84 / 98 = (14 × 6) / (14 × 7)

= 6 / 7

(d) 12 / 52 = (3 × 4) / (13 × 4)

= 3 / 13

(e) 7 / 28 = 7 / (7 × 4)

= 1 / 4

 

Total number of pencils Ramesh had = 20

Number of pencils used by Ramesh = 10

∴ Fraction = 10 / 20 = 1 / 2

Total number of pencils Sheelu had = 50

Number of pencils used by Sheelu = 25

∴ Fraction = 25 / 50 = 1 / 2

Total number of pencils Jamaal had = 80

Number of pencils used by Jamaal = 40

∴ Fraction = 40 / 80 = 1 / 2

Therefore, each has used up an equal fraction of pencils, i.e. 1 / 2

 

(a) 1 / 2

(b) 4 / 6 = (4 / 2) / (6 / 2)

= 2 / 3

(c) 3 / 9 = (3 / 3) / (9 / 3)

= 1 / 3

(d) 2 / 8 = (2 / 2) / (8 / 2)

= 1 / 4

(e) 3 / 4

(i) 6 / 18 = (6 / 6) / (18 / 6)

= 1 / 3

(ii) 4 / 8 = (4 / 4) / (8 / 4)

= 1 / 2

(iii) 12 / 16 = (12 / 4) / (16 / 4)

= 3 / 4

(iv) 8 / 12 = (8 / 4) / (12 / 4)

= 2 / 3

(v) 4 / 16 = (4 / 4) / (16 / 4)

= 1 / 4

The following are the equivalent fractions

(a) and (ii) = 1 / 2

(b) and (iv) = 2 / 3

(c) and (i) = 1 / 3

(d) and (v) = 1 / 4

(e) and (iii) = 3 / 4

 

(a) Given

2 / 7 = 8 / ☐

2 × ☐ = 7 × 8

☐ = (7 × 8) / 2

= 28

(b) Given

5 / 8 = 10 / ☐

☐ = (8 × 10) / 5

= 16

(c) Given

3 / 5 = ☐ / 20

☐ = (3 × 20) / 5

= 12

(d) Given

45 / 60 = 15 / ☐

☐ = (15 × 60) / 45

= 20

(e) Given

18 / 24 = ☐ / 4

☐ = (18 × 4) / 24

= 3

 

(a) We require denominator 20

Let M be the numerator of the fractions

∴ M / 20 = 3 / 5

5 × M = 20 × 3

M = (20 × 3) / 5

= 12

Therefore, the required fraction is 12 / 20

(b) We require numerator 9

Let N be the denominator of the fractions

∴ 9 / N = 3 / 5

3 × N = 9 × 5

N = (9 × 5) / 3

= 15

Therefore the required fraction is 9 / 15

(c) We require denominator 30

Let D be the numerator of the fraction

∴ D / 30 = 3 / 5

5 × D = 3 × 30

D = (3 × 30) / 5

= 18

Therefore the required fraction is 18 / 30

(d) We require numerator 27

Let N be the denominator of the fraction

∴ 27 / N = 3 / 5

3 × N = 5 × 27

N = (5 × 27) / 3

= 45

Therefore the required fraction is 27 / 45

 

 

(a) Given numerator = 9

∴ 9 / D = 36 / 48

D × 36 = 9 × 48

D = (9 × 48) / 36

D = 12

Hence, the equivalent fraction is 9 / 12

(b) Given, denominator = 4

∴ N / 4 = 36 / 48

N × 48 = 4 × 36

N = (4 × 36) / 48

= 3

Hence, the equivalent fraction is 3 / 4

 

(a) Given 5 / 9 and 30 / 54

We have 5× 54 = 270

9 × 30 = 270

5 × 54 = 9 × 30

Hence, 5 / 9 and 30 / 54 are equivalent fractions

(b) Given 3 / 10 and 12 / 50

We have 3 × 50 = 150

10 × 12 = 120

3 × 50 ≠ 10 × 12

Hence, 3 / 10 and 12 / 50 are not equivalent fractions

(c) Given 7 / 13 and 5 / 11

We have 7 × 11 = 77

5 × 13 = 65

7 × 11 ≠ 5 × 13

Hence, 7 / 13 and 5 / 11 are not equivalent fractions

 

(i) 250 / 400

= (5 × 50) / (8 × 50)

= 5 / 8

25 / 40 and 30 / 48 are two more fractions

(ii) 180 / 200

= (9 × 20) / (10 × 20)

= 9 / 10

18 / 20 and 27 / 30 are two more fractions

(iii) 660 / 990

= (2 × 330) / (3 × 330)

= 2 / 3

20 / 30 and 200 / 300 are two more fractions

(iv) 180 / 360

= (1 × 180) / (2 × 180)

= 1 / 2

20 / 40 and 30 / 60 are two more fractions

(v) 220 / 550

= (2 × 110) / (5 × 110)

= 2 / 5

20 / 50 and 40 / 100 are two more fractions

∴ The equivalent fractions are

(i) 250 / 100 = (d) 5 / 8

(ii) 180 / 200 = (e) 9 / 10

(iii) 660 / 990 = (a) 2 / 3

(iv) 180 / 360 = (c) 1 / 2

(v) 220 / 550 = (b) 2 / 5

 

(a) Here, the numerators are the same. So, the fraction having lesser denominator is greater

∴ 1 / 6 < 1 / 3

(b) 3 / 4 = (3 × 3) / (4 × 3)

= 9 / 12

2 / 6 = (2 × 2) / (6 × 2)

= 4 / 12

Between 4 / 12, 9 / 12

Both fractions have the same denominators. So, the fraction having greater numerator will be greater

∴ 9 / 12 > 4 / 12

3 / 4 > 2 / 6

(c) Here, the numerators are the same. So, the fraction having lesser denominator is greater

∴ 2 / 3 > 2 / 4

(d) We get 6 / 6 = 1 and 3 / 3 = 1

So, 6 / 6 = 3 / 3

(e) Here, the numerators are the same. So, the fraction having lesser denominator is greater

∴ 5 / 6 < 5 / 5

 

(a) 5 / 9, 4 / 5

Convert these fractions into like fractions

5 / 9 = (5 / 9) × (5 / 5)

= 25 / 45

4 / 5 = (4 / 5) × (9 / 9)

= 36 / 45

∴ 25 / 45 ≠ 36 / 45

Hence, 5 / 9 is not equal to 4 / 5

(b) 9 / 16, 5 / 9

Convert into like fractions

9 / 16 = (9 / 16) × (9 / 9)

= 81 / 144

5 / 9 = (5 / 9) × (16 / 16)

= 80 / 144

∴ 81 / 144 ≠ 80 / 144

Hence, 9 / 16 is not equal to 5 / 9

(c) 4 / 5, 16 / 20

16 / 20 = (4 × 4) / (5 × 4)

= 4 / 5

∴ 4 / 5 = 16 / 20

Hence, 4 / 5 is equal to 16 / 20

(d) 1 / 15, 4 / 30

4 / 30 = (2 × 2) / (15 × 2)

= 2 / 15

∴ 1 / 15 ≠ 4 / 30

Hence, 1 / 15 is not equal to 4 / 30

 

(a) Here, the numerators are the same. So, the fraction having lesser denominator is greater

∴ 1 / 2 > 1 / 5

(b) 2 / 4 = 1 / 2 and 3 / 6 = 1 / 2

∴ 2 / 4 = 3 / 6

(c) 3 / 5 = (3 × 3) / (5 × 3)

= 9 / 15

2 / 3 = (2 × 5) / 3 × 5)

= 10 / 15

Here, between 9 / 15 and 10 / 15 both have the same denominators. Hence, the fraction having greater numerator will be greater.

∴ 3 / 5 < 2 / 3

(d) Here, 2 / 8 = 1 / 4

As, 3 / 4 and 1 / 4 have the same denominators. Hence, the fraction having greater numerator will be greater

∴ 3 / 4 > 2 / 8

(e) Here, the denominators are the same. So, the fraction having greater numerator will be greater

∴ 3 / 5 < 6 / 5

(f) Here, the denominators are the same. So, the fraction having greater numerator will be greater

∴ 7 / 9 > 3 / 9

(g) We know 2 / 8 = 1 / 4

Hence, 1 / 4 = 2 / 8

(h) 6 / 10 = (3 × 2) / (5 × 2)

= 3 / 5

Between 3 / 5 and 4 / 5

Both have the same denominators. So, the fraction having greater numerator will be greater

∴ 6 / 10 < 4 / 5

(i) 3 / 4 = (3 × 2) / (4 × 2)

= 6 / 8

Between 6 / 8 and 7 / 8

Both have same denominators. So, the fraction having greater numerator will be greater

∴ 3 / 4 < 7 / 8

(j) 6 / 10 = (3 × 2) / (5 × 2)

= 3 / 5

∴ 6 / 10 = 3 / 5

(k) 5 / 7 = (5 × 3) / (7 × 3)

= 15 / 21

∴ 5 / 7 = 15 / 21

 

 

Rafiq exercised = 3 / 6 of an hour

Rohit exercised = 3 / 4 of a hour

3 / 6, 3 / 4

Convert these into like fractions

3 / 6 = (3 × 2) / (6 × 2)

= 6 / 12

3 / 4 = (3 × 3) / (4 × 3)

= 9 / 12

Clearly, 9 / 12 > 6 / 12

∴ 3 / 4 > 3 / 6

Therefore, Rohit exercised for a longer time than Rafiq.

 

(a) 2 / 12 = (1 × 2) / (6 × 2)

= 1 / 6

(b) 3 / 15 = (1 × 3) / (5 × 3)

= 1 / 5

(c) 8 / 50 = (4 × 2) / (25 × 2)

= 4 / 25

(d) 16 / 100 = (4 × 4) / (25 × 4)

= 4 / 25

(e) 10 / 60 = (1 × 10) / (6 × 10)

= 1 / 6

(f) 15 / 75 = (1 × 15) / (5 × 15)

= 1 / 5

(g) 12 / 60 = (1 × 12) / (5 × 12)

= 1 / 5

(h) 16 / 96

= (1 × 16) / (6 × 16)

= 1 / 6

(i) 12 / 75 = (4 × 3) / (25 × 3)

= 4 / 25

(j) 12 / 72 = (1 × 12) / 6 × 12)

= 1 / 6

(k) 3 / 18 = (1 × 3) / (6 × 3)

= 1 / 6

(l) 4 / 25

Totally there are 3 groups of equivalent fractions.

1 / 6 = (a), (e), (h), (j), (k)

1 / 5 = (b), (f), (g)

4 / 25 = (c), (d), (i), (l)

 

Total number of students in Class A = 25

Students passed in first class in Class A = 20

Hence, fraction = 20 / 25

= 4 / 5

Total number of students in Class B = 30

Students passed in first class in Class B = 24

Hence, fraction = 24 / 30

= 4 / 5

∴ An equal fraction of students passed in first class in both the classes

Total number of pages a book has = 100 pages

Lalita read = 2 / 5 × 100 = 40 pages

Ila read = 25 pages

∴ Ila read less than Lalita.

 

(a) The first circle shows 3 shaded parts out of 8 equal parts. Hence, the fraction is 3 / 8

The second circle shows 6 shaded parts out of 8 equal parts. Hence, the fraction is 6 / 8

The third circle shows 4 shaded parts out of 8 equal parts. Hence, the fraction is 4 / 8

The fourth circle shows 1 shaded part out of 8 equal parts. Hence, the fraction is 1 / 8

The arranged fractions are:

1 / 8 < 3 / 8 < 4 / 8 < 6 / 8

(b) The first square shows 8 shaded parts out of 9 equal parts. Hence, the fraction is 8 / 9

The second square shows 4 shaded parts out of 9 equal parts. Hence, the fraction is 4 / 9

The third square shows 3 shaded parts out of 9 equal parts. Hence, the fraction is 3 / 9

The fourth square shows 6 shaded parts out of 9 equal parts. Hence, the fraction is 6 / 9

The arranged fractions are:

3 / 9 < 4 / 9 < 6 / 9 < 8 / 9

(c) Each unit length should be divided into 6 equal parts to represent the fractions 2 / 6, 4 / 6, 8 / 6 and

6 / 6 on number line. These fractions can be represented as follows:

5 / 6 > 2 / 6

3 / 6 > 0

1 / 6 < 6 / 6

8 / 6 > 5 / 6

 

(a) Here both fractions have the same denominators. So, the fraction with the greater numerator is the highest factor

∴ 3 / 6 < 5 / 6

(b) Multiply by 4

1 / 7 = (1 × 4) / (7 × 4)

= 4 / 28

Multiply by 7

1 / 4 = (1 × 7) / (4 × 7)

= 7 / 28

Here 4 < 7

∴ 1 / 7 < 1 / 4

(c) Here both fractions have the same denominators. So, the fraction with the greater numerator is the highest factor

∴ 4 / 5 < 5 / 5

(d) Here both numerators are the same. So, the fraction having less denominator will be the highest factor

∴ 3 / 7 < 3 / 5

 

(a) 5 / 8 < 6 / 8

Here, the denominators are the same. So, the fraction having greater numerator is the highest factor

(ii) 5 / 8 > 2 / 8

Here, the denominators are the same. So, the fraction having greater numerator is the highest factor

(iii) 6 / 13 > 6 / 18

Here, the numerators are the same. So, the fraction having lesser denominator will be the highest factor

(iv) 5 / 25 > 3 / 25

Here, the denominators are the same. So, the fraction having greater numerator is the highest factor

(v) 9 / 50 < 9 / 45

Here, the numerators are the same. So, the fraction having lesser denominator will be the highest factor

 

Wall space painted by Shubham in a room = 2 / 3

Wall space painted by Madhavi in a room = 1 / 3

Total space painted by both = (2 / 3 + 1 / 3)

= (2 + 1) / 3

= 3 / 3

= 1

∴ Shubham and Madhavi together painted 1 complete wall in a room.

 

(a) Given 7 / 10 – ▯ = 3 / 10

▯ = 7 / 10 – 3 / 10

▯ = (7 – 3) / 10

▯ = 4 / 10

▯ = 2 / 5

(b) Given ▯ – 3 / 21 = 5 / 21

▯ = 5 / 21 + 3 / 21

▯ = (5 + 3) / 21

▯ = 8 / 21

(c) Given ▯ – 3 / 6 = 3 / 6

▯ = 3 / 6 + 3 / 6

▯ = (3 + 3) / 6

▯ = 6 / 6

▯ = 1

(d) Given ▯ + 5 / 27 = 12 / 27

▯ = 12 / 27 – 5 /27

▯ = (12 – 5) / 27

▯ = 7 /27

 

Fraction of oranges given to Javed = 5 / 7

Fraction of oranges left in the basket = 1 – 5 / 7

= 7 / 7 – 5 / 7

= (7 – 5) / 7

= 2 / 7

 

(a) 1 / 18 + 1 / 18

= (1 + 1) / 18

= 2 / 18

= 1 / 9

(b) 8 / 15 + 3 / 15

= (8 + 3) / 15

= 11 / 15

(c) 7 / 7 – 5 / 7

= (7 – 5) / 7

= 2 / 7

(d) 1 / 22 + 21 / 22

= (1 + 21) / 22

= 22 / 22

= 1

(e) 12 /15 – 7 / 15

= (12 – 7) / 15

= 5 / 15

= 1 / 3

(f) 5 / 8 + 3 / 8

= (5 + 3) / 8

= 8 / 8

= 1

(g) 1 – 2 / 3

= 3 / 3 – 2 / 3

= (3 – 2) / 3

= 1 / 3

(h) 1 / 4 + 0

= 1/ 4

(i) 3 – 12 / 5

= 15 / 5 – 12/ 5

= (15 – 12) / 5

= 3 / 5

 

(a) Total number of parts each rectangle has = 5

No. of shaded parts in the first rectangle = 1, i.e., 1 / 5

No. of shaded parts in the second rectangle = 2, i.e., 2 / 5

No. of shaded parts in the third rectangle = 3, i.e., 3 / 5

Clearly, the fraction represented by the third rectangle = Sum of the fractions represented by the first and second rectangles

Hence, 1 / 5 + 2 / 5 = 3 / 5

(b) Total number of parts each circle has = 5

We may observe that the first, second and third circles represent 5, 3 and 2 shaded parts out of 5 equal parts, respectively. Clearly, the fraction represented by the third circle is the difference between the fractions represented by the first and second circles.

Hence, 5 / 5 – 3 / 5 = 2 / 5

(c) Here, we may observe that the first, second and third rectangles represent 2, 3 and 5 shaded parts out of 6 equal parts, respectively. Clearly, the fraction represented by the third rectangle is the sum of fractions represented by the first and second rectangles.

Hence, 2 / 6 + 3 / 6 = 5 / 6

 

(a) ▯ – 5 / 8 = 1 / 4

▯ = 1 / 4 + 5 / 8

▯ = [(1 × 2 + 5)] / 8

▯ = 7 / 8

(b) ▯ – 1 / 5 = 1 / 2

▯ = 1 / 2 + 1 / 5

▯ = [(1 × 5) + (1 × 2)] / 10

▯ = (5 + 2) / 10

▯ = 7 / 10

(c) 1 / 2 – ▯ = 1 / 6

▯ = 1 / 2 – 1 / 6

▯ = [(1 × 3) – (1 × 1)] / 6

▯ = (3 – 1) / 6

▯ = 2 / 6

▯ 1 / 3

 

Time taken by Jaidev to walk across the school ground =
= 11 / 5 minutes

Time taken by Rahul to walk across the school ground = 7 / 4 minutes

Convert these fractions into like fractions.

11 / 5 = 11 / 5 × 4 / 4

= (11 × 4) / (5 × 4)

= 44 / 20

7 / 4 = 7 / 4 × 5 / 5

= (7 × 5) / (4 × 5)

= 35 / 20

Clearly, 44 / 20 > 35 / 20

11 / 5 > 7 / 4

∴ Rahul takes less time than Jaidev to walk across the school ground.

Difference = 11 / 5 – 7 / 4

= 44 / 20 – 35 / 20

= 9 / 20

Hence, Rahul walks across the school ground by 9 / 20 minutes.

 

Fraction of Asha’s bookshelf = 5 / 6

Fraction of Samuel’s bookshelf = 2 / 5

Convert these fractions into like fractions.

5 / 6 = 5 / 6 × 5 / 5

= (5 × 5) / (6 × 5)

= 25 / 30

2 / 5 = 2 / 5 × 6 / 6

= (2 × 6) / (5 × 6)

= 12 / 30

25 / 30 > 12 / 30

5 / 6 > 2 / 5

∴ Asha’s bookshelf is more full than Samuel’s bookshelf.

Difference = 5 / 6 – 2 / 5

= 25 / 30 – 12 / 30

= 13 / 30

 

Distance of the school from house = 9 / 10 km

Distance she travelled by bus = 1 / 2 km

Distance walked by Nandini = Total distance of the school – Distance she travelled by bus

= 9 / 10 – 1 / 2

= [(9 × 1) – (1 × 5)] / 10

= (9 – 5) / 10

= 4 / 10

= 2 / 5 km

∴ Distance walked by Nandini is 2 / 5 km

 

Total length of wire = 7 / 8 metre

Length of one piece of wire = 1 / 4 metre

Length of the other piece of wire = Length of the original wire and this one piece of wire

= 7 / 8 – 1 / 4

= [(7 × 1) – (1 × 2)] / 8

= (7 – 2) / 8

= 5 / 8

∴ Length of the other piece of wire = 5 / 8 metre

 

(a) 2 / 3 + 4 / 3

= (2 + 4) / 3

= 6 / 3

= 2

1 / 3 + 2 / 3

= (1 + 2) / 3

= 3 / 3

= 1

2 / 3 – 1 / 3

= (2 – 1) / 3

= 1 / 3

4 / 3 – 2 / 3

= (4 – 2) / 3

= 2 / 3

1 / 3 + 2 / 3

= (1 + 2) / 3

= 3 / 3

= 1

Hence, the complete given box is

(b) 1 / 2 + 1 / 3

= [(1 × 3) + (1 × 2)] / 6

= (3 + 2) / 6

= 5 / 6

1 / 3 + 1 / 4

= [(1 × 4) + (1 × 3)] / 12

= (4 + 3) / 12

= 7 / 12

1 / 2 – 1 / 3

= [(1 × 3) – (1 × 2)] / 6

= (3 – 2) / 6

= 1 / 6

1 / 3 – 1 / 4

= [(1 × 4) – (1 ×3)] / 12

= (4 – 3) / 12

= 1 / 12

1 / 6 + 1 / 12

= [(1 × 2) + 1] / 12

= (2 + 1) / 12

= 3 / 12

= 1 / 4

Hence, the complete given box is

 

 

Fraction of cake Naina got =
= 3 / 2

Fraction of cake Najma got =
= 4 / 3

The total amount of cake given to both of them = 3 / 2 + 4 / 3

= [(3 × 3) + (4 × 2)] / 6

= (9 + 8) / 6

= 17 / 6

=

 

Ribbon length bought by Sarita = 2 / 5 metre

Ribbon length bought by Lalita = 3 / 4 metre

The total length of the ribbon bought by both of them = 2 / 5 + 3 / 4

Taking LCM 20,

= [(2 × 4) + (3 × 5)] / 20

= (8 + 15) / 20

= 23 / 20 metre

∴ The total length of the ribbon bought by both Sarita and Lalita is 23 / 20 metre

 

(a) 2 / 3 + 1/ 7

Taking LCM

[(2 × 7) + (1 × 3)] / 21

= (14 + 3) / 21

= 17 / 21

(b) 3 / 10 + 7 / 15

Taking LCM 30,

= [(3 × 3) + (7 × 2)] / 30

= (9 + 14) / 30

= 23 / 30

(c) 4 / 9 + 2/ 7

Taking LCM 63,

= [(4 × 7) + (2 × 9)] / 63

= (28 + 18) / 63

= 46 / 63

(d) 5 / 7 + 1 / 3

Taking LCM 21,

= [(5 × 3) + (1 × 7)] / 21

= (15 + 7) / 21

= 22 / 21

(e) 2 / 5 + 1 / 6

Taking LCM 30,

= [(2 × 6) + (1 × 5)] / 30

= (12 + 5) / 30

= 17 / 30

(f) 4 / 5 + 2 / 3

Taking LCM 15,

= [(4 × 3) + (2 × 5)] / 15

= (12 + 10) / 15

= 22 / 15

(g) 3 / 4 – 1 / 3

Taking LCM 12,

= [(3 × 3) – (1 × 4)] / 12

= (9 – 4) / 12

= 5 / 12

(h) 5 / 6 – 1 / 3

Taking LCM 6,

= [(5 × 1) – (1 × 2)] / 6

= (5 – 2) / 6

= 3 / 6

= 1 / 2

(i) 2 / 3 + 3 / 4 + 1 / 2

Taking LCM 12,

= [(2 × 4) + (3 × 3) + (1 × 6)] / 12

= (8 + 9 + 6) / 12

= 23 / 12

(j) 1 / 2 + 1 / 3 + 1 / 6

Taking LCM 6,

= [(1 × 3) + (1 × 2) + (1 × 1)] / 6

= (3 + 2 + 1) / 6

= 6 / 6

= 1

(k)

= [(3 × 1) + 1] / 3 + [(3 × 3) + 2] / 3

= (3 + 1) / 3 + (9 + 2) / 3

= 4/ 3 + 11 / 3

= (4 + 11) / 3

= 15 / 3

= 5

(l)

= [(3 × 4) + 2] / 3 + [(3 × 4) + 1] / 4

= 14 / 3 + 13 / 4

= [(14 × 4) + (13 × 3)] / 12

= (56 + 39) / 12

= 95 / 12

(m) 16 / 5 – 7 / 5

= (16 – 7) / 5

= 9 / 5

(n) 4 /3 – 1 / 2

Taking LCM 6,

= [(4 × 2) – (1 × 3)] / 6

= (8 – 3) /6

= 5 / 6