**Balbharati Solutions Class 6 Mathematics Operations on Fractions**

Welcome to NCTB Solutions. Here with this post we are going to help 6th class students for the Solutions of Balbharati Class 6 Mathematics Book, Practice Set 9, 10, 11, 12 and 13, Operations on Fractions. Here students can easily find step by step solutions of all the problems for Operations on Fractions. Also our Expert Mathematic Teacher’s solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get Operations on Fractions solutions. Here all the solutions are based on Maharashtra State Board latest syllabus.

**Operations on Fractions all Question Solutions :**

**Practice Set – (9) :**

**Question no – (1) **

**Solution : **

**(i) 7 2/5**

= (7 × 5 + 2)/5

= (35 + 2)/5

= 37/5

**(ii) 5 1/6**

= (5 × 6 + 1)/6

= (30 + 1)/6

= 31/6

**(iii) 4 3/4**

= (4 × 4 + 3)/4

= (16 + 3)/4

= 19 /4

**(iv) 2 5/9**

= (2 × 9 + 5)/9

= (18 + 5)/9

= 23/9

**(v) 1 5/7**

= (1 × 7 + 5)/7

= (7 + 5)/7

= 12/7

**Question no – (2) **

**Solution : **

**(i) 30/7**

= (7 + 7 + 7 + 7 + 2)/7

= (4 × 7 + 2)/7

= 4 2/7

**(ii) 7/4**

= (4 + 3)/4

= (1 × 4 + 3)/4

= 1 3/4

**(iii) 15/12**

= (12 + 3)/12

= (1 × 12 + 3)/12

= 1 3/12

**(iv) 11/8**

= (8 + 3)/8

= (1 × 8 + 3)/8

= 1 3/8

**(v) 21/4**

= (4 + 4 + 4 + 4 + 4 + 1)/4

= (5 × 4 + 1)/4

= 5 1/4

**(vi) 20/7**

= (7 + 7 + 6)/7

= (2 × 7 + 6)/7

= 2 6/7

**Question no – (3) **

**Solution : **

**(i)** Given, total rice is 9 kg.

Total person is 5.

This, each person

= 9/5

= (5 + 4)/5

= 1 4/5 kg

Hence, for each person get 1 4/5 kg rice.

**(ii)** We have to make 5 shirts of the same size and we have given that 11 meters cloth.

To find how much cloth is needed for one shirt.

Now, total cloth = 11 meters

Make shirts = 5

Thus, for 1 shirt

= 11/5

= (2 × 5 + 1)/5

= 2 1/5 meters cloth.

Therefore, 2 1/5 meters cloth needed for one shirt.

**Practice Set – (10) :**

**Question no – (1) **

**Solution : **

**(i) 6 1/3 + 2 1/3**

= 6 + 2 + 1/3 + 1/3

= 8 + 2/3

= 8 2/3

**(ii) 1 1/4 + 3 1/2**

= 1 + 3 + 1/4 + 1/2

= 4 + 1/4 + 2/4

= 4 + 3/4

= 4 3/4

**(iii) 5 1/5 + 2 1/7**

= 5 + 2 + 1/5 + 1/7

= 7 + 1 × 7/5 × 7 + 1 × 5/7 × 5

= 7 + 7/35 + 5/35

= 7 + 12/35

= 7 12/35

**(iv) 3 1/5 + 2 1/3**

= 3 + 2 + 1/5 + 1/3

= 5 + 1 × 3/5 × 3 + 1 × 5/3 × 5

= 5 + 3/15 + 5/15

= 5 + 8/15

= 5 8/15

**Question no – (2) **

**Solution : **

**(i) 3 1/3 – 1 1/4**

= 3-1 + 1/3 – 1/4

= 2 + 1 × 4/3 × 4 – 1 × 3/4 × 3

= 2 + 4/12 – 3/12

= 2 + 1/ 12

= 2 1/12

**(ii) 5 1/2 – 3 1/3**

= 5-3 + 1/2 – 1/3

= 2 + 1 × 3/2 × 3 – 1 × 2/3 × 2

= 2 + 3/6 – 2/6

= 2 + 1/6

= 2 1/6

**(iii) 7 1/8 – 6 1/10**

= 7-6 + 1/8 – 1/10

= 1 + 1 × 10/8 × 10 – 1 × 8 /10 × 8

= 1 + 10/80 – 8/80

= 1 + 2/80

= 1 + 1/40

= 1 1/40

**(iv) 7 1/2 – 3 1/5**

= 7-3 + 1/2 – 1/5

= 4 + 1 × 5/2 × 5 – 1 × 2/5 × 2

= 4 + 5/10 – 2 /10

= 4 + 3/10

= 4 3/10

**Question no – (3) **

**Solution : **

**(1)** Given, Suyash bought 2 1/2 kg of sugar and Ashish bought 3 1/2 kg.

Suyash = 2 1/2 kg bought sugar

Ashish = 3 1/2 kg bought sugar

Total = 2 1/2 + 3 1/2

= 2 + 3 + 1/2 + 1/2

= 5 + 2/2

= 5 + 1

= 6 kg bought sugar

Now, for 1 kg = ₹32

Thus, for 6 kg = 6 × 32 = ₹192

**(2) **Given, Aradhana grows potatoes in 2/5 part of her garden, greens in 1/3 part.

Let b part be grows brinjals in her garden.

Thus, 2/5 + 1/3 + b = 1

= 6/15 + 5/15 + b = 1 ; 11/15 + b = 15/15

= b = 15/15 – 11/15 = 4/15 part

Hence, 4/15 part be grows brinjals in her garden.

**Practice Set – (12) :**

**Question no – (1) **

**Solution : **

**(i) 7/5 × 1/4**

= 7/5 × 1/4

= 7/20

**(ii) 6/7 × 2/5**

= 6/7 × 2/5

= 12/35

**(iii) 5/9 × 4/9**

= 5/9 × 4/9

= 20/81

**(iv) 4/11 × 2/7**

= 4/11 × 2/7

= 8/77

**(v) 1/5 × 7/2**

= 1/5 × 7/2

= 7/10

**(vi) 9/7 × 7/8**

= 9/7 × 7/8

= 63/56

**(vii) 5/6 × 6/5**

= 5/6 × 6/5

= 30/30

= 1

**(viii) 6/17 × 3/2**

= 6/17 × 3/2

= 18/34

= 9/17

**Question no – (2) **

**Solution : **

Given, Ashokrao planted bananas on 2/7 of his field of 21 acres.

We must find out 2/7 of 21 acres

Thus, 21/1 * 2/7

= 21*2/1*7

= 42/7

= 6

Therefore, the area for the Banana plantation will be 6 acers

**Practice Set – (13) :**

**Question no – (1) **

**Solution : **

**(i) 7**

= It’s reciprocal is 1/7

**(ii) 11/3**

= It’s reciprocal is 3/11

**(iii) 5/13**

= It’s reciprocal is 13/5

**(iv) 2**

= It’s reciprocal is 1/2

**(v) 6/7**

= It’s reciprocal is 7/6

**Question no – (2) **

**Solution : **

**(i) 2/3 ÷ 1/4**

= 2/3 ÷ 1/4

= 2/3 × 4/1

= 8/3

**(ii) 5/9 ÷ 3/2**

= 5/9 ÷ 3/2

= 5/9 × 2/3

= 10/27

**(iii) 3/7 ÷ 5/11**

= 3/7 ÷ 5/11

= 3/7 × 11/5

= 33/35

**(iv) 11/12 ÷ 4/7**

= 11/12 ÷ 4/7

= 11/12 × 7/4

= 77/48

**Question no – (3) **

**Solution : **

Given, 420 students participating in the Swachh Bharat campaign.

They cleaned 42/75 part of the town, Sevagram.

Now, 42/75 ÷ 420

= 42/75 × 1/420

= 42 × 1/75 × 420

= 1/75 × 10

= 1/750

Hence, 1/750 part of Sevagram did each student clean.

**More Solutions : **

👉 Angles

👉 Integers

👉 HCF-LCM