Percentage increase :
= $$\frac{51300 - 41800}{41800}$$   × 100
= $$\frac{9500}{41800}$$ × 100
= $$\frac{9500}{418}$$
= $$\frac{500}{22}$$
= $$\frac{250}{11}$$
= $$22\frac{8}{11}$$%

$$\left( {0.85\% {\text{ of }}405} \right) + \left( {2.25\% {\text{ of }}550} \right)$$
$$ = \left( {\frac{{85}}{{100}} \times \frac{1}{{100}} \times 405} \right) + $$     $$\left( {\frac{{225}}{{100}} \times \frac{1}{{100}} \times 550} \right)$$
$$\eqalign{ & = \frac{{225}}{{10000}}\left( {153 + 550} \right) \cr & = \left( {\frac{{225 \times 703}}{{10000}}} \right) \cr & = 15.8175 \cr} $$

Let number = x
According to question,
⇒ x - $$\frac{17x}{100}$$ = 498
⇒ 100x - 17x = 49800
⇒ 83x = 49800
⇒ x = $$\frac{49800}{83}$$ = 600

Let B + M + D = $$x$$
Then,
B = 25 of $$x$$ - 20
$$\eqalign{ & = \left( {\frac{{25x}}{{100}} - 20} \right) \cr & = \left( {\frac{x}{4} - 20} \right) \cr} $$
And D = 50
∴ $$\frac{x}{4}$$ - 20 + M + 50 = $$x$$
or M = $$\left( {\frac{{3x}}{{4}} - 30} \right)$$
So, marks in Maths cannot be determined.

The production of cycles rose to 48400 from 40000 in 2 years
⇒ Present production = 40000
⇒ After two years = 48000
⇒ Time = 2 years
⇒ Rate of increment = ?
According to question,
Production after 2 years = Present production $${\left( {1 + \frac{{\text{R}}}{{100}}} \right)^t}$$
$$\eqalign{ & \Rightarrow 48400 = 40000{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr & \Rightarrow \frac{{484}}{{400}} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr & \Rightarrow 1 + \frac{{\text{R}}}{{100}} = \frac{{22}}{{20}} \cr & \Rightarrow \frac{{\text{R}}}{{100}} = \frac{1}{{10}} \cr & \Rightarrow {\text{R}} = 10\% \cr} $$
⇔ Rate of increment = 10%

Let total number of students be x
Then,
$$\eqalign{ & \left( {100 - 76} \right)\% {\text{ of }}x = 204 \cr & \Rightarrow 24\% {\text{ of }}x = 204 \cr & \Rightarrow \frac{{24}}{{100}}x = 204 \cr & \Rightarrow x = \left( {\frac{{204 \times 100}}{{24}}} \right) \cr & \Rightarrow x = 850 \cr} $$

By mixture and allegation method.

4 : 1 = 5 units
5 units = 100
1 unit = 20
Pass candidate :
= 4 units
= 4 × 20
= 80

Candidates failed in English
= (100 - 60)%
= 40%
Candidate failed in Mamatics
= (100 - 70)%
= 30%
20% Candidate fail in Both.
Candidate only fail in English = 40 - 20 = 20%
Candidate only fail in Mamatics = 30 - 20 = 10%
Students passed in both subjects
= 100 - (20% Candidate fail in Both + Candidate only fail in English + Candidate only fail in Mamatics)
= 100 - (20 + 20 + 10)
= 50%
50% of students = 2500
Total students = $$\frac{{2500}}{{50}} \times 100$$
                      = 5000

Required percentage :
$$\eqalign{ & = \left( {\frac{5}{{2250}} \times 100} \right)\% \cr & = \frac{2}{9}\% \cr} $$

20% = $$\frac{1}{5}$$ = $$\frac{4}{5}$$
X  :   Y
4   :   5
Let X = 4a
    Y = 5a
Hence,
⇔ $$\frac{Y - X}{Y}$$
= $$\frac{5a - 4a}{5a}$$
= $$\frac{a}{5a}$$
= $$\frac{1}{5}$$
⇔ $$\frac{X}{X - Y}$$
= $$\frac{4a}{4a - 5a}$$
= $$\frac{4a}{- a}$$
= - 4
Hence required answer = $$\frac{1}{5}$$, - 4