Place value of 5 in 103.2523 - Place value of 2 in 115.552
= 0.0500 - 0.002
= 0.050 - 0.002 
= \(\frac { 50 }{ 1000 } -\frac { 2 }{ 1000 } \)
= \(\frac { 50-2 }{ 1000 } =\frac { 48 }{ 1000 } \)
= 0.048

Weight of 1 box = 3\(\frac { 3 }{ 4 } \)kg =\(\frac { 15 }{ 4 } \)kg
\(\therefore\) Weight of 40 boxes =\(\left( \frac { 15 }{ 4 } \times 40 \right) \)kg = \(\left( \frac { 15 }{ 4 } \times \frac { 40 }{ 1 } \right) \) kg
=\(\frac { 15\times 40 }{ 4\times 1 } \)kg
= 150 kg
Hence, weight of the carton is 150 kg.

\(4\frac { 5 }{ 6 } -2\frac { 3 }{ 8 } +3\frac { 7 }{ 12 } \)
=\(\frac { 6\times 4+5 }{ 6 } -\frac { 2\times 8+3 }{ 8 } =\frac { 3\times 12+7 }{ 12 } \)

= \(\frac { 29 }{ 6 } -\frac { 19 }{ 8 } +\frac { 43 }{ 12 } \)
= \(\frac { 29\times 4 }{ 6\times 4 } -\frac { 19\times 3 }{ 8\times 3 } +\frac { 43\times 2 }{ 12\times 2 } \)
 [∵ LCM of 6,8,12 is 2\(\times\)3\(\times\)2\(\times\)2=24]
= \(\frac { 116 }{ 24 } -\frac { 57 }{ 24 } +\frac { 86 }{ 24 } =\frac { 116-57+86 }{ 24 } =\frac { 202-57 }{ 24 } \)
= \(\frac { 145 }{ 24 } =6\frac { 1 }{ 24 } \)

LCM of 16 and 12 = (4\(\times\)4\(\times\)3) =

∴ \(\frac { 15 }{ 16 } -\frac { 11 }{ 12 } =\frac { 15\times 3 }{ 16\times 3 } =\frac { 11\times 4 }{ 12\times 4 } \)
[Converting each fraction to an equivalent fraction with denominator 48]
= \(\frac { 45 }{ 48 } -\frac { 44 }{ 48 } =\frac { 45-44 }{ 48 } =\frac { 1 }{ 48 } \)

1 \(\times\) 1 = 1
as reciprocal of 1 is 1

\(\frac { 4 }{ 9 } \div \frac { 2 }{ 3 } \)=\(\frac { 4 }{ 9 } \times \frac { 3 }{ 2 } \)
= \(\frac { 2 }{ 3 } \)

\(2\frac { 3 }{ 5 } \times 3\)=\(\frac { 13 }{ 5 } \times 3\)

= \(\frac { 39 }{ 5 } =7\frac { 4 }{ 5 } \)

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,
Product of two numbers = 20\(\frac { 5 }{ 7 } \)=\(\frac { 145 }{ 7 } \)
One of the number =6\(\frac { 2 }{ 3 } \)=\(\frac { 20 }{ 3 } \)
The other number = (Product of the numbers ÷ One of the numbers)
= \(\frac { 145 }{ 7 } \div \frac { 20 }{ 3 } \)
=\(\frac { 145 }{ 7 } \times \frac { 3 }{ 20 } =\frac { 145\times 3 }{ 7\times 20 } =\frac { 29\times 3 }{ 7\times 4 } =\frac { 87 }{ 28 } =3\frac { 3 }{ 28 } \)
Hence, the other number is 3 \(\frac { 3 }{ 28 } \)

Total weight of the fruits purchased by Sameera is \(\left( 3\frac { 1 }{ 2 } +4\frac { 3 }{ 4 } \right) \)kg
Now, \(3\frac { 1 }{ 2 } +4\frac { 3 }{ 4 } =\frac { 7 }{ 2 } +\frac { 19 }{ 4 } \)
= \(\frac { 7\times 2 }{ 2\times 2 } =\frac { 19\times 1 }{ 4\times 1 } \)
= \(\frac { 14 }{ 4 } +\frac { 19 }{ 4 } =\frac { 14+19 }{ 4 } =\frac { 33 }{ 4 } =8\frac { 1 }{ 2 } 2\)
Hence, total weight 8\(\frac { 1 }{ 4 } \) is kg

3.25÷0.5=\(\frac { 3.25 }{ 0.5 } =\frac { 3.25 }{ 0.50 } \)
=\(\frac { 325 }{ 50 } \)=6.5
OR
3.25÷0.5= \(\frac { 325 }{ 100 } \times \frac { 10 }{ 5 } \)
= \(\frac { 65 }{ 100 } \times \frac { 10 }{ 1 } \)
= \(\frac { 65 }{ 10 } \)
=6.5

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