Let the original number  $=100$

New number $=120$

As the Increased number decreased by $20\text{\%}$

New Number $=120-20\text{\% of }120$

$=120-\frac{20}{100}\times 120$

$=120-\frac{1}{5}\times 120$

$=120-24$

$=96$

Net Decrease percentage $=\frac{100-96}{100}\times 100$

$=\frac{4}{100}\times 100$

$=4\text{\%}$

Therefore, $\boxed{{\displaystyle \text{Net decrease percentage}=4\text{\%}}}$

Let the man purchased the camera in $\text{Math input error}$.

New C.P $\text{Math input error}$

Profit$=4\text{\%}$

So, New S.P$=x+4\text{\% }\text{of }x$

$=x+\frac{4}{100}\times x$

$=\frac{100x+4x}{100}$

$\boxed{{\displaystyle \text{New S.P}=\frac{104x}{100}}}$

Now, If he had purchased it for $14\text{\%}$ less and sold it for $\text{Math input error}$ less.

$\text{Old C.P}=14\text{\%}\text{less of New C.P}$

 $\text{Old C.P}=\text{ New C.P}-14\text{\% }of\text{ New C.P}$

$=x-14\text{\% of }x$

$=x-\frac{14}{100}\times x$

$=\frac{100x-14x}{100}$

$\boxed{{\displaystyle \text{Old C.P}=\frac{86x}{100}}}$

A\Q

$\boxed{{\displaystyle \text{Old S.P}=\text{New S.P}-77}}=\frac{104x}{100}-77$

Given gain % $=16\frac{2}{3}\text{\%}=\frac{50}{3}\text{\%}$

As,  $\text{ gain \%}=\frac{\text{Old S.P}-\text{Old C.P}}{\text{Old C.P}}\times 100$

$\Rightarrow \frac{50}{3}=\frac{(\frac{104x}{100}-77)-\frac{86x}{100}}{\frac{86x}{100}}\times 100$

$\Rightarrow \frac{50}{3}=\frac{(\frac{104x-86x}{100}-77)}{\frac{86x}{100}}\times 100$

$\Rightarrow \frac{50}{3}=\frac{\left(\frac{104x-86x-7700}{100}\right)}{\frac{86x}{100}}\times 100$

$\Rightarrow \frac{50}{3}=\frac{\left(\frac{18x-7700}{100}\right)}{\frac{86x}{100}}\times 100$

$\Rightarrow \frac{50}{3}=\frac{(18x-7700)}{86x}\times 100$

$\Rightarrow \frac{50}{3\times 100}=\frac{(18x-7700)}{86x}$

$\Rightarrow \frac{1}{3\times 2}=\frac{(18x-7700)}{86x}$

$\Rightarrow \frac{86x}{3\times 2}=(18x-7700)$

$\Rightarrow \frac{43x}{3}=(18x-7700)$

$\Rightarrow 43x=3\times (18x-7700)$

$\Rightarrow 43x=54x-23100$

$\Rightarrow 43x-54x=-23100$

$\Rightarrow -11x=-23100$

$\Rightarrow x=\frac{23100}{11}$

$\Rightarrow x=2100$

Therefore,

$\text{Math input error}$

Given $\text{gain}=\text{S.P of 40 Cards}$ 

Let $\text{Math input error}$

Then, $\text{Math input error}$

Therefore, C.P of 100 Diwali cards $=\text{S.P of 100 Diwali cards}–\text{gain}.$

$\Rightarrow$C.P of 100 Diwali cards $\text{Math input error}$.

$\text{gain\%}=\frac{\text{gain}}{\text{C.P}}\times 100$

$=\frac{40}{60}\times 100$

$=\frac{2}{3}\times 100$

$=\frac{200}{3}$

$=66.\overline{6}\text{\%}$

$\boxed{{\displaystyle \text{gain \%}=66.\overline{6}\text{\%}}}$

Let the marks be $x$

Marks obtained by student A $=32\text{\%}ofx=\frac{32x}{100}$

As, student A failed by $20$ marks.

So, $\text{Pass marks}=\frac{32x}{100}+20$            …..(i)

Marks obtained by student B $=42\text{\%}ofx=\frac{42x}{100}$

As, student B got$30$ marks more than pass marks.

So, $\text{Pass marks}=\frac{42x}{100}-30$            …..(ii)

Equating (i) and (ii), we get

$\frac{42x}{100}-30=\frac{32x}{100}+20$

$\Rightarrow \frac{42x}{100}=\frac{32x}{100}+20+30$

$\Rightarrow \frac{42x}{100}-\frac{32x}{100}=50$

$\Rightarrow \frac{42x-32x}{100}=50$

$\Rightarrow \frac{10x}{100}=50$

$\Rightarrow \frac{x}{10}=50$

$\Rightarrow x=50\times 10$

$\Rightarrow x=500$

Therefore, $\boxed{{\displaystyle \text{Maximum marks}=500}}$

Now, $\text{Pass marks}=\frac{32x}{100}+20$

By substituting value of $x$, we get

$\text{Pass marks}=\frac{32\times 500}{100}+20$

$=32\times 5+20$

$=32\times 5+20$

$=160+20$

$=180$

$\boxed{{\displaystyle \text{Pass marks}=180}}$

So, $\text{Pass percentage}=\frac{\text{Pass marks}}{\text{Maximum marks}}\times 100$

$=\frac{180}{500}\times 100$

$=\frac{180}{5}$

$=36$

Therefore, $\boxed{{\displaystyle \text{Pass percentage}=36\text{\%}}}$

Let $\text{Math input error}$

Then, $\text{Increased Price}=100+25\text{\%}of100=125$

In order to not increase the expenditure housewife need to reduce the consumption of sugar of $\text{Math input error}$

$\text{Percentage of sugar consumption reduce}$

$=\frac{25}{125}\times 100$

$=\frac{1}{5}\times 100$

$=\frac{1}{1}\times 20$

$=20\text{\%}$

Therefore, Percentage of sugar consumption reduce$\boxed{{\displaystyle =20\text{\%}}}$

(i) Percentage of students failed in one or both subjects $=(24\text{\%}+27\text{\%}-20\text{\%})$

$=24\text{\%}-20\text{\%}$

$=31\text{\%}$

Percentage of students failed in one or both subjects $=\boxed{{\displaystyle 31\text{\%}}}$

(ii) Percentage of students who passes in both subjects = Total students – Percentage of students who failed in one or both subjects.

Therefore, 

Percentage of students who passes in both subjects $=100\text{\%}-31\text{\%}=69\text{\%}$

Percentage of students who passes in both subjects $=\boxed{{\displaystyle 69\text{\%}}}$

(iii)  From part (ii), we got $69\text{\%}$ passed in both the subjects.

It is given that $345$ students passed in both subjects.

Therefore, 

$69\text{\%}=345$

$\Rightarrow 1\text{\%}=\frac{345}{69}$

$\Rightarrow 1\text{\%}=5$

$\Rightarrow 100\text{\%}=5\times 100=500$

$\boxed{{\displaystyle \text{Total number of students}=500}}$

As, Half of the eggs were sold at a profit of $10\text{\%}$ and 5 broke.

So, other $45$ were sold at a profit of $6\text{\%}$.

$\text{Math input error}$

By unitary method,

$\text{Math input error}$

So, $\text{Math input error}$

$\text{Math input error}$

And $\text{Math input error}$

$\text{Math input error}$

$\text{Total S.P of }\text{eggs}=44+38.16=82.16$

Total profit percentage in the transaction$=\frac{82.16-80}{80}\times 100$

$=\frac{2.16}{80}\times 100$

$=\frac{216}{8000}\times 100$

$=\frac{216}{80}$

$=\frac{27}{10}$

$=2.7\text{\%}$

$\boxed{{\displaystyle \text{Total profit \% in the transaction}=2.7\text{\%}}}$