Multiply 24 and 42. We get 1008.
Both, in 2.4 and 4.2, there is 1 digit to the right of the decimal point.
So, count 1 + 1 = 2 digits starting from the rightmost digit (i.e., 8) in 1008 and move towards left.
We get 10.08.

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 \(\frac { 3 }{ 28 } \)

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 \(\frac { 3 }{ 4 } \)

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

\(\frac { 14 }{ 25 } \times \frac { 35 }{ 51 } \times \frac { 34 }{ 49 } \)=\(\frac { 14\times 35\times 34 }{ 25\times 51\times 49 } =\frac { 2\times 1\times 2 }{ 5\times 3\times 1 } =\frac { 4 }{ 15 } \)

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

2.73÷1.3=\(\frac { 2.73 }{ 1.3 } =\frac { 2.73 }{ 1.30 } \)
= \(\frac { 275 }{ 130 } \)
= \(\frac { 21 }{ 10 } \)
=2.1

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

4.26 is greater

1.6, 20, 4.0

83.33 paise

1\(\frac{1}{2}\) of \(\frac{16}{5}\)

Given, width of picture = 7\(\frac{3}{5}\)cm
\(=\frac{7\times5+3}{5}=​​\frac{35+3}{5}=\frac{38}{5}\)cm
and width of the frame = 7\(\frac{3}{10}\)cm=\(\frac{7\times 10+3}{10}\)
=\(\frac{70+3}{10}=\frac{73}{10}\)cm
ஃ To fit the picture in frame, picture should be trimmed
= Width of the picture - Width of the frame
\(=\frac{38}{5}-\frac{73}{10}=\frac{38\times 2-73\times 1}{10}\)   [∵ LCM of 5 and 10=10]
=\(\frac{76-73}{10}=\frac{3}{10}\)cm
Hence,\(\frac{3}{10}\)ern picture should be trimmed.

Given, length of a rectangular sheet of paper,
l=12\(\frac{1}{2}\)cm=\(\frac{12\times 2+1}{2}=\frac{25}{2}\)cm
Breadth of a rectangular sheet of paper,
b=\(10\frac{2}{3} and =\frac{10\times 3+2}{3}=\frac{32}{3}\)cm
Now, perimeter of a rectangular sheet of paper
\(=2(l+b)=2(\frac{25}{2}+\frac{32}{2})\)
=\(2(\frac{25\times 3+ 32\times 2}{6})\)   [∵ LCM of 2 and 3=6]
=2\((\frac{75+64}{6})=2(\frac{139}{6})=\frac{139}{6}=46\frac{1}{3}cm\)
Hence, the perimeter of a rectangular sheet is 46\(\frac{1}{3}\)cm.