Volume of water in the bucket =24 \(\frac { 3 }{ 4 } \) litres =\(\frac { 99 }{ 4 } \)litres
Capacity of jug =\(\frac { 3 }{ 4 } \) litre
\(\therefore\) Number of jugs that can be filled to get the bucket emptied 
=\(\frac { 99 }{ 4 } \div \frac { 3 }{ 4 } =\frac { 99 }{ 4 } \times \frac { 4 }{ 3 } =\frac { 99\times 4 }{ 4\times 3 } =33\)
Hence, 33 jugs of\(\frac { 3 }{ 4 } \) litre can be filled to get the bucket emptied.

We have,
Cost of one note book = Rs 8\(\frac { 3 }{ 4 } \)=Rs \(\frac { 35 }{ 4 } \)
Total amount = Rs 131\(\frac { 1 }{ 4 } \)= Rs \(\frac { 525 }{ 4 } \)
\(\therefore\)  Number of notebooks =\(\frac { Total\ amount }{ Cost\ of\ one\ notebook } \)
=\(\frac { 525 }{ 4 } \div \frac { 35 }{ 4 } =\frac { 525 }{ 4 } \times \frac { 4 }{ 35 } =\frac { 525\times 4 }{ 4\times 35 } =15\)

We have,
Cost of 5\(\frac { 2 }{ 5 } \) kg of sugar = Rs 101\(\frac { 1 }{ 4 } \)
or, Cost of \(\frac { 27 }{ 5 } \)kg of sugar = Rs\(\frac { 405 }{ 4 } \)
or, Cost of 1 kg of sugar
= Rs\(\left( \frac { 405 }{ 4 } \div \frac { 27 }{ 5 } \right) \)= Rs \(\left( \frac { 405 }{ 4 } \times \frac { 5 }{ 27 } \right) \)
= Rs =\(\left( \frac { 405\times 5 }{ 4\times 27 } \right) \)= Rs \(\frac{ 75 }{ 4 }\) = Rs \(18\frac { 3 }{ 4 } \)
Hence, the cost of 1 kg of sugar is Rs 18 \(\frac { 3 }{ 4 } \)

∵ Car covers in 1 litre = 8.6 km
Car covers in 36.5 litres
= 8.6\(\times\)36.5
= 313.90
= 313.9 km

We have
\(5\times \frac { 3 }{ 20 } \times \frac { 2 }{ 15 } \)=\(\frac { 5 }{ 1 } \times \frac { 3 }{ 20 } \times \frac { 2 }{ 15 } =\frac { 5\times 3\times 2 }{ 1\times 20\times 15 } =\frac { 1\times 1\times 2 }{ 1\times 4\times 5 } \)
= \(\frac { 1 }{ 2\times 5 } =\frac { 1 }{ 10 } \)

In order to know who worked longer, we will compare fractions \(\frac { 7 }{ 12 }\)and​\(\frac { 3 }{ 4 } \)
We have,
(LCM of 12 and 4) = 12
Converting each fraction into an equivalent fraction with 12 as denominator, we have
\(\frac { 7 }{ 12 } =\frac { 7\times 1 }{ 12\times 1 } =\frac { 7 }{ 12 } \) and \(\frac { 3 }{ 4 } =\frac { 3\times 3 }{ 4\times 3 } =\frac { 9 }{ 12 } \)
∵        7 < 9
∵      \(\frac { 7 }{ 12 } <\frac { 9 }{ 12 }\) or \( \frac { 7 }{ 12 } <\frac { 3 }{ 4 } \)
Thus, Vaibhav finished colouring in longer time
Now, \(\frac { 3 }{ 4 } -\frac { 7 }{ 12 } =\frac { 9 }{ 12 } -\frac { 7 }{ 12 } =\frac { 9-7 }{ 12 } =\frac { 2 }{ 12 } =\frac { 1 }{ 6 } \)
Hence, Pramod finished colouring in\(\frac { 1 }{ 6 } \)hour more  time than Suval.

In order to know who solved less part of the exerci.se, we will compare \(\frac { 2 }{ 7 } \)and\(\frac { 4 }{ 5 } \)
We have,
LCM of denominators (i.e., 7 and 5), 7 \(\times\)5 = 35.
Converting each fraction into an equivalent fraction having 35 as its denominator, we have
\(\frac { 2 }{ 7 } =\frac { 2\times 5 }{ 7\times 5 } =\frac { 10 }{ 35 } \)and\(\frac { 4 }{ 5 } =\frac { 4\times 7 }{ 5\times 7 } =\frac { 28 }{ 35 } \)
∵10 < 28
∴ \(\frac { 10 }{ 35 } <\frac { 28 }{ 35 } \Rightarrow \frac { 2 }{ 7 } <\frac { 4 }{ 5 } \)
Hence, Mukul solved lesser part than Deeksha.

For any moving object,
Speed = \(\frac { distance }{ time } \)
In this question,
Distance covered = 300km
Time taken =3\(\frac { 1 }{ 2 } \) hours
=\(\frac { 7 }{ 2 } \) hours
∴ Speed =\(\frac { 300 }{ \frac { 7 }{ 2 } } \)=\(\frac { 300\times 2 }{ 7 } \)
=\(\frac { 600 }{ 7 } \) = 85.7 km/h

For rectangle,
Area = Length \(\times\) Breadth
Since, Length = 0.5 m
Breadth = 1.5 m
∴ Area = (1.5\(\times\)0.5)m²
= \(\left( \frac { 15 }{ 10 } \times \frac { 5 }{ 10 } \right) \)m²
= \(\frac { 75 }{ 100 } \)m²
=0.75 m²

7÷3.5=\(\frac { 7 }{ 3.5 } =\frac { 7\times 10 }{ 35 } \)
=\(\frac { 70 }{ 35 } \)=2
OR
7÷3.5=\(7\div \frac { 35 }{ 10 } \)
= \(7\times \frac { 10 }{ 35 } \)
= \(\frac { 70 }{ 35 } \)
=2

We have 
\(4\frac { 2 }{ 3 } -3\frac { 1 }{ 4 } +2\frac { 1 }{ 6 } \)=\(\frac { 14 }{ 3 } -\frac { 13 }{ 4 } +\frac { 13 }{ 6 } \)
= \(\frac { 14\times 4 }{ 3\times 4 } -\frac { 13\times 3 }{ 4\times 3 } +\frac { 13\times 2 }{ 6\times 2 } \)
∵ LCM of 3, 4 and 6 is 12, so we convert each fraction into an equivalent fraction with denominator 12
i,e  \(\frac { 56 }{ 12 } -\frac { 39 }{ 12 } +\frac { 26 }{ 12 } =\frac { 56-39+26 }{ 12 } \)
= \(\frac { 82-39 }{ 12 } =\frac { 43 }{ 12 } =3\frac { 7 }{ 12 } \)

(a) No. of students like to study English
= \(\frac { 1 }{ 5 } \times 40\)
= 8
(b) No. of students like to study Maths
= \(\frac { 2 }{ 5 } \times 40\)
= 2\(\times\) 8
= 16
(c) No. of students who like to study Science
=40 - (8 + 16)
= 40-24
= 16
In part (ii), we know
\(\frac { 2 }{ 5 } \times 40\) = 16
So,\(\frac { 2 }{ 5 } \)of the total number of students like to study Science

Total hours spend by Ritika for studies =11\(\frac { 1 }{ 3 } \) hours =\(\frac { 34 }{ 3 } \)hours
 Total hours devoted by Ritika for Hindi and Sanskrit 5\(\frac { 3 }{ 5 } \)= \(\frac { 28 }{ 5 } \)hours
Total hours devoted by Ritika for other subjects
= \(\frac { 34 }{ 3 } -\frac { 28 }{ 5 } \)
= \(\frac { 170-84 }{ 15 } =\frac { 86 }{ 15 } \)
=5\(\frac { 11 }{ 15 } \) hours