1

x = -6

x = 9

36

x = -12

x = 5

Let my present age be x years.
\(\therefore\) After 24 years, my age will be (x + 24) years.
4 years ago, my age was (x - 4) years.
According to the given condition, we have
(x + 24) = 3(x - 4)
\(\Rightarrow\) x + 24 = 3x - 12 \(\Rightarrow\) x - 3x = -12 - 24
\(\Rightarrow\) -2x = -36
\(\Rightarrow x={-36\over -2}=18\)
Thus, my present age is 18 years.

\({7y+4\over y+2}={-4\over 3}\)
By cross-multiplication, we have
3 x (7y + 4) = - 4 x (y + 2)
or 21y + 12 = - 4y - 8
Transposing 12 to RHS and (-4y) to LHS,
we have
21y + 4y = -8 - 12
or 25y = -20 or y = \({-20\over 25}={-4\over 5}\)
(Dividing both sides by 25)
\(\therefore y={-4\over 5}\)

Let the present age of Shaloo = 5x years
Let the present age of Preeti = 4x years
After 5 years, age of Shaloo = (5x + 5) years
Age of Preeti = (4x + 5) years
According to the condition, we have
\({(5x+5)\over (4x+5)}={6\over 5}\) By cross-multiplication, we have
5(5x + 5) = 6(4x + 5) or 25x + 25 = 24x + 30
Transposing 25 to RHS and 24x to LHS, we have
25x - 24x = 30 - 25 or x = 5
\(\therefore\) 5x = 5 x 5 = 25 and 4x = 4 x 5 = 20
Therefore, present age of Shaloo = 25 years
Present age of Preeti = 20 years.

Let the length of the rectangle = 1 cm
\(\therefore\) Since width of the rectangle (b) = \(3{3\over 4}\) and perimeter = 17 cm
We know that perimeter of a rectangle = 2(l + b) cm
Then, \(2(l+3{3\over 4})=17\) or \(2(l+{15\over 4})=17\)
or \(2l+(2\times {15\over 4})=17\) or \(2l+{15\over 2}=17\)
or \(2l=17-{15\over 2}\) (Transposing \({15\over 2}\) to RHS)
or \(2l={35-15\over 2}={19\over 2}\)
Dividing both side by 2, we have
\(i={19\over 2}\times {1\over 2}={19\over 4}=4{3\over 4}\) cm
\(\therefore\) Length of the rectangle = \(4{3\over 4}\)cm.

Twice of \({-5\over 3} =2\times (\frac{-5}{3})={-10\over 3}\)
Let the required number to be subtracted be x.
\(\therefore ({-10\over 3})-x={2\over 5}\)
Transposing \({-10\over 3}\) to RHS, we have
\(-x={2\over 5}+{10\over 3}\) or \(-x={6+10\over 15}={56\over 15}\)
\(\therefore x={-56\over 15}\)
Thus, the required number is \((\frac{-56}{15})\)

we have \(\frac { n }{ 2 } -\frac { 3n }{ 4 } +\frac { 5n }{ 6 } =21\)
\(\Rightarrow\) \(\frac { 6n-9n+10n }{ 12 } =21\) [\(\therefore\) LCM of 2,4 and 6 = 12]
\(\Rightarrow\) \(\frac { 16n-9n }{ 12 } =21\Rightarrow \frac { 7n }{ 12 } =21\)
\(\Rightarrow\) \(\frac { 7 }{ 12 } n\times \frac { 12 }{ 7 } =\frac { 21\times 12 }{ 7 } \)
\(\Rightarrow\) \(n=\frac { 21\times 12 }{ 7 } \)
\(\Rightarrow\)  n = 3 x 12 = 36, which is the required solution

Let Shobo's present age be x yr.
Shobe's mother's present age =6x
According to the question, 5 yr later
Shobo's age = \(\frac { 1 }{ 3 } \) Shobo's mother's present age
\(\Rightarrow\) x + 5 = \(\frac { 1 }{ 3 } \) x 6x \(\Rightarrow\) x + 5 = 2x \(\Rightarrow\) 5 = 2x - x 
\(\Rightarrow\) 5 =x \(\Rightarrow\) x = 5
Shobe's present age = x = 5 yr
and Shobo's mother's present age = 6x = 6 x 5 = 30 yr
Hence, Shobe's present age is 5 yr and her mother's age is 30 yr. 

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 present ages of Rahul and Haroon be 5x yr and 7x yr, respectively.
4 yr later, age of Rahul = (5x + 4)yr
and age of Haroon=(7x + 4)yr
According to the question,
(5x + 4) + (7x + 4) = 56
\(\Rightarrow\) 5x + 4 + 7x + 4 = 56 \(\Rightarrow\) 12x = 56 - 8
\(\Rightarrow\) 12x = 56 -8  [transposing 8 to RHS]
\(\Rightarrow\) 12x = 48 \(\Rightarrow\) x=\(\frac { 48 }{ 12 } \)=4 [dividing both sides by 12]
\(\therefore\)  Present age of Rahul = 5x = 5 x 4 = 20 yr
and present age of Haroon = 7x = 7 x 4 = 28 yr
Hence, their present ages are 20 yr and 28 yr.