Let cost of a single ticket be Rs. $$x$$
The, cost of monthly return ticket :
$$\eqalign{ & = 125\% \,\,{\text{of}}\,\,{\text{Rs}}.x \cr & = {\text{Rs}}{\text{. }}\left( {\frac{{125}}{{100}}x} \right) \cr & = {\text{Rs}}{\text{. }}\frac{{5x}}{4} \cr} $$
Cost of monthly return ticket with extension :
$$\eqalign{ & = 105\% \,\,{\text{of}}\,\,{\text{Rs}}.\frac{{5x}}{4} \cr & = {\text{Rs}}{\text{. }}\left( {\frac{{105}}{{100}} \times \frac{{5x}}{4}} \right) \cr & = {\text{Rs}}{\text{. }}\frac{{21x}}{{16}} \cr} $$
$$\eqalign{ & \refore \frac{{21x}}{{16}} = 84 \cr & \Rightarrow x = \left( {\frac{{84 \times 16}}{{21}}} \right) \cr & \Rightarrow x = 64 \cr} $$

Total marks obtained by student
= 55% of 800
= $$\frac{55}{100}$$ × 800
= 440
∴ Marks scored in English
= 15% of 440
= $$\frac{15}{100}$$ × 440
= 66

Let original price be Rs. $$x$$
New final price :
= 110% of (110% of Rs. 100)
= Rs. $$\left( {\frac{{110}}{{100}} \times \frac{{110}}{{100}} \times 100} \right)$$
= Rs. 121
∴ Total increase :
= (121 - 100)%
= 21%

Height of tree 2 years ago :
$$\eqalign{ & = \frac{{540}}{{{{\left( {1 + \frac{{20}}{{100}}} \right)}^2}}}cm \cr & = \left( {540 \times \frac{5}{6} \times \frac{5}{6}} \right)cm \cr & = 375\,\,cm \cr} $$

Let original number = x
According to question,
x × $$\frac{90}{110}$$ × $$\frac{110}{100}$$ = (x - 50)
x × $$\frac{99}{100}$$ = x - 50
99x = 100x - 5000
x = 5000
Hence original number = 5000

Quantity of alcohol in 5 litres solution = 20% of 5 litres = 1 litre
Quantity of alcohol removed = 20% of 2 litres = 400 ml
Quantity of alcohol in new solution :
= (1000 - 400)ml
= 600 ml
∴ Strength of alcohol in new solution
$$\eqalign{ & = \left( {\frac{{600}}{{5000}} \times 100} \right)\% \cr & = 12\% \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{}}\,{\text{sum}}\,{\text{paid}}\,{\text{to}}\,{\text{y}}\,{\text{per}}\,{\text{week}}\,{\text{be}}\,Rs.\,z \cr & {\text{Then}},\,z + 120\% \,of\,z = 550 \cr & \Rightarrow z + \frac{{120}}{{100}}z = 550 \cr & \Rightarrow \frac{{11}}{5}z = 550 \cr & \Rightarrow z = {\frac{{550 \times 5}}{{11}}} = 250 \cr} $$

Let Vinay's salary be Rs. $$x$$
Then,
75% of 5% of $$x$$ = 1687.50
⇒ $$\frac{75}{100}$$ × $$\frac{5}{100}$$ × $$x$$ = 1687.50
⇒ $$x$$ = $$\left( {\frac{{1687.50 \times 100 \times 100}}{{75 \times 5}}} \right)$$
⇒ $$x$$ = 45000

Given (550% of 250) ÷ 275 = (?)
⇒ (?) = $$\frac{550 × 250}{100}$$   ÷ 275
⇒ (?) = (55 × 25) ÷ 275
⇒ (?) = $$\frac{55 × 25}{275}$$
⇒ (?) = 5

Failed students in Mamatics = 34%
Failed students in English = 42%
Student failed in both subject math and english = 20%
Student only fail in Math = 34 - 20 = 14%
Student only fail in English = 42 - 20 = 22%
Percentage of passed students in both subjects :
= 100 - [ student fail in maths + student fail in english + student fail in both subject]
= [100 - (14 + 20 + 22)]
= 44%