Question 1:

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 4 Operations on Fractions Class 6 Practice Set 10 Answers Solutions.

Question 2:

Add:
i. $6 \frac{1}{3}+2 \frac{1}{3}$
ii. $1 \frac{1}{4}+3 \frac{1}{2}$
iii. $5 \frac{1}{5}+2 \frac{1}{7}$
iv. $3 \frac{1}{5}+2 \frac{1}{3}$
Solution:
i. $6 \frac{1}{3}+2 \frac{1}{3}$

ii. $1 \frac{1}{4}+3 \frac{1}{2}$

iii. $5 \frac{1}{5}+2 \frac{1}{7}$

iv. $3 \frac{1}{5}+2 \frac{1}{3}$

Question 3:

Subtract:
i. $3 \frac{1}{3}-1 \frac{1}{4}$
ii. $5 \frac{1}{2}-3 \frac{1}{3}$
iii. $7 \frac{1}{8}-6 \frac{1}{10}$
iv. $7 \frac{1}{2}-3 \frac{1}{5}$
Solution:
i. $3 \frac{1}{3}-1 \frac{1}{4}$

ii. $5 \frac{1}{2}-3 \frac{1}{3}$

iii. $7 \frac{1}{8}-6 \frac{1}{10}$

iv. $7 \frac{1}{2}-3 \frac{1}{5}$

Question 4:

Solve:
i. Suyash bought $2\frac { 1 }{ 2 }$ kg of sugar and Ashish bought $3\frac { 1 }{ 2 }$ kg. How much sugar did they buy altogether? If sugar costs Rs 32 per kg, how much did they spend on the sugar they bought?

ii. Aradhana grows potatoes in $\frac { 2 }{ 5 }$ part of her garden, greens in $\frac { 1 }{ 3 }$ part and brinjals in the remaining part. On how much of her plot did she plant brinjals?

iii. Sandeep filled water in $\frac { 4 }{ 7 }$ of an empty tank. After that, Ramakant filled $\frac { 1 }{ 4 }$ part more of the same tank. Then Umesh used $\frac { 3 }{ 14 }$ part of the tank to water the garden. If the tank has a maximum capacity of 560 litres, how many litres of water will be left in the tank?
Solution:
i. Sugar bought by Suyash = $2\frac { 1 }{ 2 }$ kg
Sugar bought by Ashish = $3\frac { 1 }{ 2 }$ kg
∴ Total sugar bought by both

Cost of 1 kg of sugar = Rs 32
∴ Cost of 6 kg of sugar = 32 x 6
= Rs 192
∴ They bought 6 kg sugar altogether and the total money spent on sugar is Rs 192.

ii. Part of garden occupied by potatoes = $\frac { 2 }{ 5 }$
Part of garden occupied by greens = $\frac { 1 }{ 3 }$
Since brinjals are planted in the remaining part,
∴ (Part occupied by potatoes) + (part occupied by greens) + (part occupied by brinjals) = 1 entire garden.
∴ Part of garden occupied by brinjals = 1 – (part of garden occupied by potatoes + part of garden occupied by greens)

∴ Aradhana planted brinjals on $\frac { 4 }{ 15 }$ part of her plot.

iii. Part of tank filled by Sandeep = $\frac { 4 }{ 7 }$
Part of tank filled by Ramakant = $\frac { 1 }{ 4 }$

Since maximum capacity of tank is 560 litres
∴ Quantity of water left in tank = $\frac { 17 }{ 28 }\times560$ = 340 litres
∴ The quantity of water left in the tank is 340 litres.

Question 5:

How to do this subtraction: $4 \frac{1}{4}-2 \frac{1}{2}$ ? Is it same as $\left[4-2+\frac{1}{4}-\frac{1}{2}\right]$ ? (Textbook pg. no. 23)
Solution:
$4 \frac{1}{4}-2 \frac{1}{2}$

$\left[4-2+\frac{1}{4}-\frac{1}{2}\right]$

The subtraction $4 \frac{1}{4}-2 \frac{1}{2}$ is the same as $\left[4-2+\frac{1}{4}-\frac{1}{2}\right]$.