Let x be the maximum mark of the examination.
The students gets 31% of total Marks  \(=\frac{31}{100} x\)
\( \text { Given } \frac{31 x}{100}+12 =35 \% \text { of } x \)
\(\frac{31 x}{100}+12 =\frac{35 x}{100} \)
\(\frac{35 x}{100}-\frac{31 x}{100} =12 \)
\(\frac{x}{100}(35-31) =12 \)
\(\frac{4 x}{100} =12 \)
\(x =\frac{12 \times 100}{4}=300 \)
The maximum mark of the examination = 300

Let x be the initial number of mangoes.
Nurnber of Rotten mangoes = 10%
Number of good mangoes
\(=90 \% \text { of } x=\frac{90}{100} x\)
Percentage of sold mangoes \(=33 \frac{1}{3} \%\)
Percentage of unsold mangoes \(=100-33 \frac{1}{3} \%\)
\(=66 \frac{2}{3} \%\)
\( \frac{66 \frac{2}{3}}{100} \times \frac{90}{100} x =240 \)
\(\frac{200}{3 \times 100} \times \frac{90}{100} x =240 \)
\(\frac{2}{3} \times \frac{9}{10} \times x =240 \)
\(=\frac{240 \times 3 \times 10}{2 \times 9} =400 \)
The vendor bought 400 mangoes initially.

Let the numerator be x and the denominator be y.
The original fraction = x/y
Given the numerator of a fraction is increased by 50%
\( \therefore \mathrm{Nr} =x+50 \% \text { of } x \)
\(=x+\frac{50}{100} x=x+\frac{1}{2} x \\ \mathrm{Nr} =\frac{3}{2} x \)
Also given the denominator is decreased by 20 %
\( \text { Dr } =y-20 \% \text { of } y \)
\(=y-\frac{20}{100} y=y-\frac{1}{5} y=\frac{4}{5} y \)
By given data
\( \frac{\frac{3}{2} x}{\frac{4}{5} y}=\frac{3}{5} \Rightarrow \frac{3}{2} x \times \frac{5}{4 y}=\frac{3}{5} \)
\(\frac{x}{y}=\frac{8}{25} \)
The original fraction is 8/25

Let x and y be the increased and decreased percentage
The change in that number \(=\left(x+y+\frac{x y}{100}\right) \%\)
\( =\left(25+(-20)+\frac{25(-20)}{100}\right) \% \)
\(=\left(25-20-\frac{500}{100}\right) \)
\(=(25-20-5) \% \)
= 0%
There is no percentage changes in that number.

Let the number be x.
Given: x - 2 % of x = 80
\( x-\frac{20}{100} \times x=80 \)
\(x\left(1-\frac{20}{100}\right) =80 \)
\(x\left(\frac{80}{100}\right) =80 \)
\(\therefore x =\frac{80 \times 100}{80}\)
x = 100
The required number is 100.

Let the number be x.
Given x + 18% of x = 236
\( x+\frac{18}{100} x =236 \)
\(x\left(1+\frac{18}{100}\right) =236 \)
\(x\left(\frac{100+18}{100}\right) =236 \)
\(x\left(\frac{118}{100}\right) =236 \)
\(\therefore x =\frac{236 \times 100}{118} \)
= 200
The required number is 200.

Decrease value = Original price - Sold Price
= Rs. 3,00,00 - 2,00,00
= Rs. 1,00,000
\(\therefore \text { Percentage of decrease }=\frac{100000}{300000} \times 100\)
\(=33 \frac{1}{3} \%\)

(i) 500
ii) Cricket -45,Volleyball-42, Badminton -33

Let the number be x.
\( \frac{32}{100} \times x =48 \)
\(x =\frac{48 \times 100}{32}=150\)
The number is 150

(i) \(\text { one half }=\frac{1}{2} \times 100 \%=50 \%\)
(ii) \(\frac{7.5}{10} \times 100 \%=75 \%\)
(iii) The word pure means 100%
(iv) \(\frac{48}{50} \times 100 \%=96 \%\)
(v) \(\frac{2}{3} \times 100 \%=66 \frac{2}{3} \%\)