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

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

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 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

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

Distance towards East denoted (+ve) sign and Distance towards West denoted (-ve) sign.
Since first Ridhaan goes toward East 25 km
So distance covered by him = + 25 km
Again, he goes towards West.
So, distance covered by him on same road
= -35 km.
So,position of Ridhaan from A= + 25 - 35
= -10 km.
(b) Value: Addition of Integers.
(c) Value: A rolling stone can gather no moss.

Given, Ravi can walk 3\(\frac{1}{3}\)km in 1h. 
i.e.  3\(\frac{1}{3}\)km/h =\(\frac{(3\times 3)+1}{3}\)
=\(\frac{9+1}{3}=\frac{10}{3}\)km/h
∵ Distance between Ravi's and his office = 10 km
∵ Time=\(\frac{Distance\ between\ Ravi' s\ and\ his\ office}{Ravi' s\ speed\ in\ 1h}\)
=\(\frac{10 km}{\frac{10}{3}km/h}=\frac{10}{1}\times \frac{3}{10}\) [∵division is reverse of multiplication]
=\(\frac{30}{10}\)=3h
Hence, Ravi reaches office in 3 h

Given, 2\(\frac{1}{3}\) of 4\(\frac{7}{9}\)of \(\frac{1}{2}\)
∵ \(2 \frac{1}{3}=\frac{(2\times 3)+1}{3}=\frac{6+1}{3}=\frac{7}{3}\)
\(4\frac{7}{9}=\frac{(4\times 9)+7}{9}=\frac{36+7}{9}=\frac{43}{9}\)
∴ \(\frac{7}{3}\)of \(\frac{43}{9}\) of \(\frac{1}{2}\)
Now, \(\frac{7}{3}\) of \(\frac{43}{9}\times= \frac{7}{3}\times \frac{43}{9}=\frac{7\times 43}{3\times 9}=\frac{301}{27}\)
and again,\(\frac{301}{27}\) of \(\frac{1}{2}=\frac{301}{27}\times \frac{1}{2}=\frac{301\times 1}{27\times2}=\frac{301}{54}=5\frac{31}{54}\)
Hence, \(\frac{7}{3}\)of \(\frac{43}{9}\)of \(\frac{1}{2}\)=5\(\frac{31}{54}\)

Rs 1208.7

12.004 is smaller

We have, 2.03 or 2.30
First, we compare the digits on the left of the decimal point starting from the leftmost digit. Here, both the digits to the left at the decimal point are same. i.e. 2. So, we compare the digits on the right of the decimal point starting from the tenth place.
Here, 0 < 3
So, 2.03 < 2.30
Hence, 2.30 is greater.

We have, \(6 \div 5 \frac{1}{3}=6 \div (​​\frac{5\times 3+1}{3})\)
=\((6\div \frac{15+1}{3})=6 \div \frac{16}{3}\)
=6\(\times\frac{3}{16}\)           [∵ reciprocal of \(\frac{16}{3}\)is \(\frac{3}{16}\)]
=\(\frac{6\times 3}{16}=\frac{18}{16}=\frac{9}{8}=1\frac{1}{8}\)  [dividing numerator and denominator by 2]

The complete table is shown as below:

The complete table is shown as below:

From the given figure, it is clear that the directions East and West are opposite to each other.
According to the question,

If moving towards East is a positive integer, then moving towards West is represented by a negative integer.
Now, distance moved towards West = - 30 km 
and distance moved towards East = 20 km
\(\therefore\) Rita's final position from A = 20 + (- 30) = 20 - 30
= -10 km (i.e. West)

Given, Jack's score ofIst round = 25
Score of IInd round = - 5
Score of IIIrd round = - 10
Score of IVth round = 15
and score of Vth round = 10
Now, total score = Sum of score of all rounds
= 25 + (- 5) + (-10) + 15 + 10 = 25 - 5 -10 + 15 + 10
= (25 + 15 + 10) + (- 5 -10)
=(40+10)+(-15) =50-15=35
Hence, the total score of Jack was 35.

Number line with all numbers is shown below:

From the above number line, the temperature of the indicated places are as follow:

Along first row =\(\frac{4}{11}\)+\(\frac{9}{11}\)+\(\frac{2}{11}\)=\(\frac{4+9+2}{11}=\frac{15}{11}\)
Along second row= \(\frac{3}{11}\)+\(\frac{5}{11}\)+\(\frac{7}{11}\)=\(\frac{3+5+7}{11}\)=\(\frac{15}{11}\)
Along thi.rd row =\(\frac{8}{11}\)+\(\frac{1}{11}\)+\(\frac{6}{11}\)=\(\frac{8+1+6}{11}=\frac{15}{11}\)
Along first column =\(\frac{4}{11}\)+\(\frac{3}{11}\)+\(\frac{8}{11}\)=\(\frac{4+3+8}{11}=\frac{15}{11}\)
Along second column = \(\frac{9}{11}\)+\(\frac{5}{11}\)+\(\frac{1}{11}\)=\(\frac{9+5+1}{11}=\frac{15}{11}\)
Along third column =\(\frac{2}{11}\)+\(\frac{7}{11}\)+\(\frac{6}{11}\)=\(\frac{2+7+6}{11}=\frac{15}{11}\)
Along one diagonal = \(\frac{4}{11}\)+\(\frac{5}{11}\)+\(\frac{6}{11}\)=\(\frac{4+5+6}{11}=\frac{15}{11}\)
and along other di.agonal =\(\frac{2}{11}\)+\(\frac{5}{11}\)+\(\frac{8}{11}\)=\(\frac{2+5+8}{11}=\frac{15}{11}\)
Since, sum of the numbers in each row, in each column and along the diagonals are same.
Hence, it is a magic square.

We have,

Now, completing the given number line is as follows:

Hence, the position of required letters is shown in the following table: