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, \(\boxed{\text{Net decrease percentage}=4 \text{%}}\)

Let the man purchased the camera in \(₹ x\).

New C.P \(= ₹ x\)

Profit\(= 4\text{%}\)

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

\(=x + \frac{4}{100} \times x\)

\(=\frac{100x + 4x}{100} \)

\(\boxed{\text{New S.P}=\frac{104x }{100} }\)

Now, If he had purchased it for \(14\text{%}\) less and sold it for \(₹ 77\) less.

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

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

\(= x-14\text{% of }   x\)

\(=x-\frac{14}{100} \times x\)

\(=\frac{100x-14x}{100} \)

\(\boxed{\text{Old C.P}=\frac{86x}{100}}\)

A\Q

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

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

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

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

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

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

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

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

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

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

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

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

\(\Rightarrow43x =3\times \left(18x -7700 \right)\)

\(\Rightarrow43x =54x -23100 \)

\(\Rightarrow43x -54x = -23100\)

\(\Rightarrow-11x = -23100\)

\(\Rightarrow x= \frac{23100}{11}\)

\(\Rightarrow x=2100\)

Therefore,

\(\boxed{\text{The man  purchased the camera for }₹2100}\)

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

Let \(\text{S.P of 100 Diwali cards =} ₹ 100\)

Then, \(\text{gain}=₹ 40\)

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

\(\Rightarrow \)C.P of 100 Diwali cards \( =100-40= ₹60\).

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

\(= \frac{40}{60}\times 100\)

\(= \frac{2}{3}\times 100\)

\(= \frac{200}{3}\)

\(=66.\bar{6} \text{%}\)

\(\boxed{\text{gain %} =66.\bar{6} \text{%}}\)

Let the marks be \(x\)

Marks obtained by student A \(= 32 \text{%} of x = \frac{32x}{100}\)

As, student A failed by \(20\) marks.

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

Marks obtained by student B \(= 42 \text{%} of x = \frac{42x}{100}\)

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, \(\boxed{\text{Maximum marks} = 500}\)

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

By substituting value of \(x\), we get

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

\(= 32\times 5+20\)

\(= 32\times 5+20\)

\(= 160+20\)

\(= 180\)

\(\boxed{\text{Pass marks}=180}\)

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

\(= \frac{180}{500}\times 100\)

\(=\frac{180}{5}\)

\(=36\)

Therefore, \(\boxed{\text{Pass percentage} =36 \text{%}}\)

Let \(\text{Initial Price of sugar }=₹ 100\)

Then, \(\text{Increased Price} = 100+25 \text{%} of 100=125\)

In order to not increase the expenditure housewife need to reduce the consumption of sugar of \(= 125-100=₹25\)

\(\text{Percentage of sugar consumption reduce}\)

\(=\frac{25}{125}\times 100\)

\(=\frac{1}{5}\times 100\)

\(=\frac{1}{1}\times 20\)

\(=20 \text{%}\)

Therefore, Percentage of sugar consumption reduce\(\boxed{=20 \text{%}}\)

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

\(=24\text{%}-20\text{%}\)

\(=31\text{%}\)

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

(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 \(=\boxed{69\text{%}}\)

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

It is given that \(345\) students passed in both subjects.

Therefore, 

\(69\text{%}= 345\)

\(\Rightarrow 1\text{%}= \frac{345}{69}\)

\(\Rightarrow 1\text{%}= 5\)

\(\Rightarrow 100\text{%}= 5 \times 100 =500\)

\(\boxed{\text{Total number of students}=500}\)

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 \(6\text{%}\).

\(\text{C.P of }  100 \text{ eggs}= ₹ 80\)

By unitary method,

\(\text{C.P of }  1 \text{ egg}= ₹ \frac{80}{100}\)

So, \(\text{C.P of }  50 \text{ eggs}= ₹ \frac{80}{100}\times 50= ₹ 40\)

\(\text{S.P of }  50 \text{ eggs}=40+10\text{%} of 40=₹44\)

And \(\text{C.P of }  45 \text{ eggs}= ₹ \frac{80}{100}\times 45= ₹ 36\)

\(\text{S.P of }  45 \text{ eggs}=36+6\text{%} of 36=₹38.16\)

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

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

\(=\frac{2.16}{80}\times 100\)

\(=\frac{216}{8000}\times 100\)

\(=\frac{216}{80}\)

\(=\frac{27}{10}\)

\(=2.7\text{%}\)

\(\boxed{\text{Total profit % in the transaction}=2.7\text{%}}\)