A = Rs. 10,050
n = 5 years
r = 10%
P = ?
For calculating principal with the given data, we proceed as follows.
We know that,
\(I=\cfrac { Pnr }{ 100 } \) 
A = P + I
\(A=P\left( 1+\cfrac { nr }{ 100 } \right) \) 
Therefore, \(10,050=P\left( 1+\cfrac { 10\times 5 }{ 100 } \right) \) 
= \(P(1+\cfrac { 50 }{ 100 } )\) 
= \(P(\cfrac { 150 }{ 100 } )\) 
=  \(P(\cfrac { 3 }{ 2 } )\) 
Therefore, \(P=10,050\times \cfrac { 2 }{ 3 } =3350\) 
Hence, Principal = Rs.3,350.

Principal (P) = Rs. 52,000
Time (n) = 4 years
Interest = Amount – Principal
= 79,040 − 52,000 = 27,040
We find the Simple Interest \((I)=\cfrac { Pnr }{ 100 } \) 
Therefore,\(27040=\cfrac { 52000\times r\times 4 }{ 100 } \) 
Rate of interest he borrowed \((r)=\cfrac { 27040\times 100 }{ 52000\times 4 } =13\%\)

Given the rate of interest (r) = 10% per annum
Time (n) = 2 years
We know that Simple Interest (I) = \(\cfrac { Pnr }{ 100 } \) 
\(750=\cfrac { P\times 2\times 10 }{ 100 } \) 
Therefore, Principal \((P)=\cfrac { 750\times 100 }{ 2\times 10 } =3750\) 
Therefore, Kumaravel has borrowed a sum of Rs.3,750.

The fraction in the Fig. 2.3, that is shaded = \(\cfrac { 2 }{ 4 } =\cfrac { 1 }{ 2 } \)
That is half of the Fig. 2.3 is shaded blue.
So, the percentage of shaded portion = \(\cfrac { 1 }{ 2 } \times 100\%=50\%\)
Thus, 50% of the Fig. s shaded

\(0.05=\cfrac { 5 }{ 100 } \times 100\%=5\%\)