If ₹ 3,750 amounts to ₹ 4,620 in 3 years at simple interest. Find:

(i) the rate of interest.

(ii) the amount of ₹ 7,500 in $5\dfrac{1}{2}$ years at the same rate of interest.

(i) Given:

P = ₹ 3,750

A = ₹ 4,620

T = 3 years

Let the rate of interest be $r$.

$\text{A = P + S.I.}\\[1em] \Rightarrow ₹ 4,620 = ₹ 3,750 + \text{S.I.}\\[1em] \Rightarrow \text{S.I.} = ₹ 4,620 - ₹ 3,750 \\[1em] \Rightarrow \text{S.I.} = ₹ 870$

And,

$\text{S.I.} = \Big(\dfrac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow ₹ 870 = \Big(\dfrac{₹ 3,750 \times r \times 3}{100}\Big)\\[1em] \Rightarrow ₹ 870 = \dfrac{11,250r}{100}\\[1em] \Rightarrow ₹ 870 = \dfrac{225r}{2}\\[1em] \Rightarrow r = \dfrac{2 \times 870}{225}\\[1em] \Rightarrow r = \dfrac{1740}{225}\\[1em] \Rightarrow r = \dfrac{116}{15}\\[1em] \Rightarrow r = 7\dfrac{11}{15}$

Hence, the rate of interest = $7\dfrac{11}{15}\%$.

(ii) Given:

P = ₹ 7,500

R = $\dfrac{116}{15}\%$

T = $5\dfrac{1}{2}$ years

= $\dfrac{11}{2}$ years

$\because \text{S.I.} = ₹ \Big(\dfrac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{S.I.} = ₹ \Big(\dfrac{7,500 \times 116 \times 11}{15 \times 2 \times 100}\Big)\\[1em] \Rightarrow \text{S.I.} = ₹ \dfrac{95,70,000}{3000}\\[1em] \Rightarrow \text{S.I.} = ₹ 3,190$

And,

$\text{A = P + S.I.}\\[1em] \Rightarrow\text{A} = ₹ 7,500 + ₹ 3,190\\[1em] \Rightarrow\text{A} = ₹ 10,690$

Hence, the amount = ₹ 10,690.