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

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.

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

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

700cm

P = Rs. 3200, r = 2.5% n = 2years
\(\text { C.I }=P\left(1+\frac{r}{100}\right)^{n}-P\)
\( =3200\left(1+\frac{2.5}{100}\right)^{2}-3200 \)
\(=3200\left(\frac{102.5}{100}\right)^{2}-3200 \)
\(=32 \times \frac{1025}{100} \times \frac{1025}{100}-3200 \)
\(=32 \times \frac{41}{4} \times \frac{41}{4}-3200 \)
\(=2 \times 41 \times 41-3200 \)
= 3362 - 3200
= 162

Let x be the cost price of the article.
810 - x = x - 530
2x = 810 + 530
2x = 1340
x = 670
The cost price of the article is Rs. 670.

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 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 numberbe x.
Given 75% of x - 60% of x = 82.5
\( \frac{75}{100} x-\frac{60}{100} x =82.5 \)
\(\frac{x}{100}(75-60) =82.5 \)
\(\frac{x \times 15}{100} =82.5 \)
\(\therefore x=\frac{82.5 \times 100}{15}\)
x = 550
The required number is 550
\(20 \% \text { of } 550=\frac{20}{100} \times 550\)
= 110

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

\(\frac{Pnr}{100}\)
\(240 = \frac{3000\times2 \times r}{100}\\ = \frac{240}{60}\)
R = 4%

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

Let the marked price be Rs x.
Now, x + \(\frac { 18x }{ 100 } \) = 10502
\(\frac { 118x }{ 100 } \) = 10502
∴ Marked price, x = Rs. 8900.
GST at 18% = 8900 x \(\frac { 18 }{ 100 } \)
= Rs. 10502 – Rs. 8900
= Rs.1602

Total C.P of the car
= C.P + Overhead Expenses
= 24000 + 15000 +8500
= Rs. 263500
S.P = Rs. 258230
As S.P < C.P , there is a loss.
∴ Loss % = \(\left( \frac { loss }{ C.P } \times 100 \right) %\)%
= \(\left( \frac { 263500-258230 }{ 263500 } \times 100 \right) %\)%
= \(\left( \frac { 5270 }{ 263500 } \times 100 \right) %\)%
= 2%
∴ Loss percentage = 2%