Given, perimeter of swimming pool = 154 m
Let the breadth of the swimming pool be x m
According ro the question,
Length of swimming pool = 2 x (Breadth) + 2 = (2x + 2) m
\(\therefore\)  Perimeter of the swimming pool = 2 (Length + Breadth)
\(\Rightarrow\) 154 = 2 [ 2x + 2 + x]
\(\Rightarrow\) \(\frac { 154 }{ 2 } \) = 3x + 2
\(\Rightarrow\) 77 = 3x + 2 \(\Rightarrow\) 3x = 77 - 2
\(\Rightarrow\) 3x = 75 \(\Rightarrow\) x = \(\frac { 75 }{ 3 } \) \(\Rightarrow\)  x = 25
Breadth of the swimming pool = x = 25 m
and length of the swimming pool = 2x + 2 = 2 x  25 + 2
= 50 + 2 = 52m

Let the number of boys and girls in a class be 7x and 5x, respectively
According to the question,
Number of boys = Number of girls +8
\(\Rightarrow\) 7x = 5x + 8 \(\Rightarrow\) 7x - 5x = 8 [transposing 5x to LHS]
\(\Rightarrow\) 2x = 8 \(\Rightarrow\) x = \(\frac { 8 }{ 2 } \) = 4 [dividing both sides by 2]
\(\therefore\) Number of boys = 7x = 7 x 4 = 28
and number of girls = 5x = 5 x 4 = 20
\(\therefore\) Total classstrength = Number of boys + Number of girls
= 28 + 20 = 48

Let the number of currency notes of denominations
Rs 100, Rs. 50 and Rs. 10 be 2x, 3x and 5x, respectively.
The amount she has from Rs.100 notes
= 2x x 100 = 200x
The amount she has from Rs. 50 notes = 3x x 50 = 150x
The amount she has from Rs. 10 notes = 5x x 10 = 50x
According to the question,
Total amount = 400000
\(\Rightarrow\) 200x + 150x + 50x = 400000
\(\Rightarrow\) 400x = 400000
\(\Rightarrow\) \(x=\frac { 400000 }{ 400 } =1000\)
\(\therefore\) Number of Rs.100 notes = 2x = 2 x 1000 = 2000
N umber of Rs. 50 notes = 3x = 3 x 1000 = 3000
and number of Rs. 10 notes = 5x = 5 x 1000 = 5000
Hence, she has 2000, 3000 and 5000 notes of denominations Rs. 100, Rs.  50 and Rs. 10, respectively

We have, \({-2\over3}\times{3\over5}+{5\over2}-{3\over5}\times{1\over6}={-2\over3}\times{3\over5}-{3\over5}\times{1\over6}+{5\over2}\) [by associativity]
\(={3\over5}\times{-2\over3}+{3\over5}\times{-1\over6}+{5\over2}\)         [by commutativity]
\(={3\over5}\times({-2\over3}-{1\over6})+{5\over2}\)
[by distributivity, taking \({3\over5}\)  as common factor]
\(={3\over5}\times({-4-1\over6})+{5\over2} \ \ \ \ [\because LCM \ of \ 3 \ and \ 6=6] \)
\(\\ ={3\over5}\times({-5\over6})+{5\over2}={3\over5}\times {-5\over6}+{5\over2}={-1\over2}+{5\over2}={-1+5\over2}\)
\(\\ ={4\over2}=2\)

Number of above average students=\({1\over6}\) of the class student
Number of average students= \({1\over4}\)of the class student
\(\therefore\) Number of below average students  \(=1-[{1\over6}+{1\over4}]\) of the class student
\(=-[{2+3\over12}]=1-{5\over12}={7\over12}\)of the class student
Since, number of students in the class = 48
\(\therefore\)Number of below average students
\(={7\over12}\times48=28\)
So, Number of below average students= 28

We have \(\frac { 3x-5 }{ 17 } +\left[ \frac { 11-x }{ 76 } -\frac { 3 }{ 4 } \right] =\frac { 4+x }{ 2 } -13\)
\(\Rightarrow\) \(\frac { 3x }{ 17 } -\frac { 5 }{ 17 } +\frac { 11 }{ 76 } -\frac { x }{ 76 } -\frac { 3 }{ 4 } =2+\frac { x }{ 2 } -13\)
\(\Rightarrow\) \(\frac { 3x }{ 17 } -\frac { x }{ 76 } -\frac { x }{ 2 } =-11+\frac { 5 }{ 17 } -\frac { 11 }{ 76 } +\frac { 3 }{ 4 } \)
\(\Rightarrow\) \(\frac { 456x-34x-1292x }{ 2584 } \)
\(\frac { -56848+1520-748+3876 }{ 5168 } \)
\(\frac { -56848+1520-748+3876 }{ 5168 } \)
\(\Rightarrow\) \(\frac { 456x-1326x }{ 2584 } \frac { 52200 }{ 5168 } \Rightarrow \frac { -870x }{ 2584 } =\frac { -52200 }{ 5168 } \)
\(x=\frac { -52200 }{ 5168 } x\left( \frac { 2584 }{ 870 } \right) =\frac { 60 }{ 2 } =30\)

\(\frac{2}{5} \times \frac{-3}{7}-\frac{1}{14}-\frac{3}{7} \times \frac{3}{5}=\frac{2}{5} \times \frac{-3}{7}-\frac{3}{7} \times \frac{3}{5}-\frac{1}{14}\)
\(=\frac{2}{5} \times \frac{-3}{7}+\left(\frac{-3}{7}\right) \times \frac{3}{5}-\frac{1}{14}\)
\(=\frac{-3}{7}\left(\frac{2}{5}+\frac{3}{5}\right)-\frac{1}{14}\)
\(=\frac{-3}{7} \times 1-\frac{1}{14}=\frac{-6-1}{14}=\frac{-1}{2}\)

Let the present ages of Anu and Raj be 4x years and 5x years respectively.
After eight years. Anu’s age = (4x + 8) years;
After eight years, Raj’s age = (5x + 8) years.
Therefore, the ratio of their ages after eight years = \(\frac{4 x+8}{5 x+8}\)
This is given to be 5 : 6
Therefore, \(\frac{4 x+8}{5 x+8}=\frac{5}{6}\)
Cross-multiplication gives 6 (4x + 8) = 5 (5x + 8)
or 24x + 48 = 25x + 40
or 24x + 48 – 40 = 25x
or 24x + 8 = 25x
or 8 = 25x – 24x
or 8 = x
Therefore, Anu’s present age = 4x = 4 × 8 = 32 years
Raj’s present age = 5x = 5 × 8 = 40 years

4z + 3 = 6 + 2z
Transposing 3 to RHS, we have
4z = 6 - 3 + 2z or 4z = 3 + 2z
Transposing 2z to LHS, we have
4z - 2z = 3 or 2z = 3
Dividing both sides by 2, we have z = \(\frac { 3 }{ 2 } \)

We have,
\({-4\over5}\times1=1\times {-4\over5}=-{4\over5}\)
This is in the form of a \(\times\) 1= 1 \(\times\) a = a, where a =-\(4\over5\)
So, 1 is multiplicative identity 

49 \({4\over5}\)  hec =  \({249\over5}\)  hec
Each share = \({1\over3}\times{249\over5}hec={83\over5}hec=16{3\over5}hec\)

Let Sahil’s present age be x years.

It is given that this sum is 66 years
Therefore, 4x + 10 = 66
This equation determines Sahil’s present age which is x years. To solve the equation,
we transpose 10 to RHS,
4x = 66 – 10
or 4x = 56
or \(x=\frac{56}{4}=14\)
Thus, Sahil’s present age is 14 years and his mother’s age is 42 years. (You may easily check that 5 years from now the sum of their ages will be 66 years.)

We find the mean of the given rational numbers.
\(\left(\frac{1}{4}+\frac{1}{2}\right) \div 2=\left(\frac{1+2}{4}\right) \div 2=\frac{3}{4} \times \frac{1}{2}=\frac{3}{8}\)
\(\frac{3}{8}\) lies between \(\frac{1}{4} \text { and } \frac{1}{2}\) 
This can be seen on the number line also.

(a) Let volume of the water in second tank be x L³
Volume of the water in first tank = 2x L³
Now, as per the given condition, we have
2x - 25 = x + 25
\(\Rightarrow\)   2x - x = 25 + 25
\(\Rightarrow\) x = 50 L
\(\therefore\) Volume of water in second tank = 50 L³ tank
volume of the water in first tank is 2 x 50 =100L³
(b) Check On putting x = 50 in the obtained Eq. (i), we get 
2 x 50 - 25 = 50 + 25 
\(\Rightarrow\) 100 - 25 = 75 \(\Rightarrow\) 75 =75
\(\therefore\) LHS = RHS
So, the solution of the given equation is correct
(c) In a scale, the balance from both the sides is  needed. In such a manner, a linear equation is also solved, where balance from the LHS and RHS is needed 

Seema gets allowance = \(1\over2\)of Huma's share 
Hubna gets allowance =\(1{2\over3}\) of Seema's share
\(={5\over3}of \ seema's \ share\)
\(\\={5\over3}of {1\over2}of \ Huma's \ sh are \)
\(\\ [\because Seema's \ share ={1\over2}of \ Huma's share]\)
\(\\ ={5\over3}\times{1\over2}={5\over6}of \ Huma's \ share\)
But Huma, Hubna and Seema get total share
= Rs 2016
\(\therefore\) Huma's share + Hubna's share + Seema's share
= Rs 2016
1 of Huma's share + \(5\over6\)of Huma's share +\(1\over2\) of Huma's share =Rs 2016
\(So,({1+{5\over6}+{1\over2}}) of \ huma's \ share=Rs2016\)
\(\\ \Rightarrow ({6+5+3\over6})of \ Huma's s\ share \)
\(\\ \Rightarrow {14\over6}of \ Huma's \ share = RS2016\)
\(\\ \therefore Huma's \ share =Rs2016\div {14\over6}\)
\(\\ =Rs2016\times{6\over14}=144\times6=864=432 and \ Hubna's \ share ={5\over6}of 864=5\times144=720\)