C.I –S.I = P\({ \left( \frac { r }{ 100 } \right) }^{ 2 }\)= 5000 x \(\frac { 4 }{ 100 } \times \frac { 4 }{ 100 } \) = Rs 8
(ii) C.I –S.I = P\({ \left( \frac { r }{ 100 } \right) }^{ 2 }\left( 2+\frac { r }{ 100 } \right) \)
= 8000 x \(\frac { 5 }{ 100 } \times \frac { 5 }{ 100 } \) x \(\left( 3+\frac { 5 }{ 100 } \right) \)
= 20 x \(\frac { 61 }{ 20 } \) = Rs.61

Bacteria at the end of 3 hours
a = p\(\left( 1+\frac { a }{ 100 } \right) \left( 1-\frac { b }{ 100 } \right) \left( 1+\frac { c }{ 100 } \right) \)('-' because ‘decrease’)
= 10000\(\left( 1+\frac { 5 }{ 100 } \right) \left( 1-\frac { 8 }{ 100 } \right) \left( 1+\frac { 10 }{ 100 } \right) \)
= 10000 x \(\frac { 105 }{ 100 } \times \frac { 92 }{ 100 } \times \frac { 110 }{ 100 } \)
A = Rs.10626

Depreciated value = P\({ \left( 1-\frac { r }{ 100 } \right) }^{ n }\)
= 70000\({ \left( 1-\frac { 4 }{ 100 } \right) }^{ 2 }\)
= 70000 x \(\frac { 96 }{ 100 } \times \frac { 96 }{ 100 } \)
= Rs.64512

Let the population in 2016 be ‘P’.
Then, A = P\({ \left( 1+\frac { r }{ 100 } \right) }^{ n }\)
⇒ 238765 = p\({ \left( 1+\frac { 6 }{ 100 } \right) }^{ 2 }=P{ \left( \frac { 53 }{ 50 } \right) }^{ 2 }\)
⇒ P = 238765 x \(\frac { 50 }{ 53 } \times \frac { 50 }{ 53 } \)
∴ P = 212500
Let the population in 2020 be ‘A’
Then, A= P\({ \left( 1+\frac { r }{ 100 } \right) }^{ n }\)
∴ A = 238765\({ \left( 1+\frac { 6 }{ 100 } \right) }^{ 2 }\)
= 238765 x \(\frac { 53 }{ 50 } \times \frac { 53 }{ 50 } \)
= 95.506 x 53 x 53
A = 268276
∴ The population in the year 2016 is 212500 and that in the year 2020 is 268276.

(i) Amount, A = P\({ \left( 1+\frac { r }{ 100 } \right) }^{ n }\)
= 4000\({ \left( 1+\frac { 5 }{ 100 } \right) }^{ 2 }\)
= 4000 x \(\frac { 21 }{ 20 } \)x\(\frac { 21 }{ 20 } \)
A = Rs. 4410
∴C.I = A − P = 4410 – 4000 = Rs. 410
(ii) Amount, A = P\({ \left( 1+\frac { r }{ 100 } \right) }^{ 2n }\)= 5000\({ \left( 1+\frac { 4 }{ 200 } \right) }^{ 2\times \frac { 3 }{ 2 } }\) = 5000 x \(\frac { 51 }{ 50 } \times \frac { 51 }{ 50 } \times \frac { 51 }{ 50 } \)
= 51 × 10.2 × 10.2
= Rs. 5306.04
∴ C.I = A − P = Rs.5306.04 – Rs.5000
= Rs.306.04
(iii) A = P\(\left( 1+\frac { a }{ 100 } \right) \left( 1+\frac { b }{ 100 } \right) \)
= 3000\(\left( 1+\frac { 7 }{ 100 } \right) \left( 1+\frac { 8 }{ 100 } \right) \)
= 30000 x \(\frac { 107 }{ 100 } \times \frac { 108 }{ 100 } \)
= Rs.34668
∴ C.I = A − I = 34668 - 30000 = 4668.
(iv) A = P\({ \left( 1+\frac { r }{ 100 } \right) }^{ a }\left( 1+\frac { \frac { b }{ c } \times r }{ 100 } \right) =10000{ \left( 1+\frac { 8 }{ 100 } \right) }^{ 2 }\left( 1+\frac { \frac { 3 }{ 4 } \times 8 }{ 100 } \right) \)
= 10000\({ \left( \frac { 27 }{ 25 } \right) }^{ 2 }\left( \frac { 53 }{ 50 } \right) \)
= 10000 x \(\frac { 27 }{ 25 } \times \frac { 27 }{ 25 } \times \left( \frac { 53 }{ 50 } \right) \)
A = 12363.84
∴C.I = 12363.84 − 10,000
= Rs.2363.84

Let the maximum marks be x.
Now, 80% of x = 576
\(\frac { 80 }{ 100 } \)\(\times\) x = 576
⇒ x = 576 x \(\frac { 100 }{ 10 } \)
x = 720 marks
Therefore, the total marks in the examination = 720.

Increase in population = 25000 − 20000
= 5000
∴ Percentage increase in population =\(\frac { 5000 }{ 20000 } \)x 100
= 25%

Let his income be rs x.
Income after 10% increase is
\(100+100 \times \frac{10}{100}=Rs. 110\)
Now, income after 10% decrease is
\(110-110 \times \frac{10}{100}=110-11=Rs. 99\)
Net change in his income = 100 – 99 = 1
Percentage change \(=\frac{1}{100} \times 100 \%=1 \%\)
That is, income of the person is reduced by 1%.
Aliter
Let his income be Rs. 100
Income after 10% increase is
100 + 100 × \(\frac{10}{100}\) = Rs.110
Now, income after 10% decrease is,
110 – 110 × \(\frac{10}{100}\) = 110 – 11 = Rs. 99
∴ Net change in his income = 100 – 99 = 1
Percentage change = × = \(\frac{1 }{100}\) × 100% = 1% 
That is, income of the person is reduced by 1%.

Let the total votes polled be x.
Votes polled in favour in A = 58% of x =\(\frac { 58x }{ 100 } \)
Votes polled in favour of B = (100 − 58)% of x = 42% of x = \(\frac { 42x }{ 100 } \)
Given, Winning margin A− B = 192
That is, \(\frac { 58x }{ 100 } \)- \(\frac { 42x }{ 100 } \) = 192
⇒\(\frac { 16x }{ 100 } \)=192
x = 192 x \(\frac { 100 }{ 16 } \)
x = 1200 votes

Let the original price of Orid dhall be Rs x.
New price aft er price of 20% increase = x +\(\frac { 20 }{ 100 } x=\frac { 120x }{ 100 } \)
Given that, 96 = \(\frac { 120x }{ 100 } \)
∴ x = \(\frac { 96\times 100 }{ 200 } \)
∴ Original price of Orid dhall per kg, x = Rs. 80

Let the number be x.
x-\(\frac { 25 }{ 100 } x\) = 120
\(\frac { 100x-25x }{ 100 } \) = 120
\(\frac { 75x }{ 100 } \) = 120
x = \(\frac { 120\times 100 }{ 75 } \)
x = 160

x% of 600 = 450
\(\frac { x }{ 100 } \)× 600 = 450
\(x=\frac { 450 }{ 6 } \)
x = 75

Number of students who did not pass = 30% of boys + 15% of girls
=\(\frac { 30 }{ 100 } \) x 900 + \(\frac { 15 }{ 100 } \) x 600
= 270 + 90 = 360
∴ Percentage of students who did not pass =\(\frac { 360 }{ 1500 } \) x 100 = 24%