1) Find the difference between simple interst and compoundinterest on

1.	1) Find the difference between simple interst and compoundinterest on Rs. 6250 in 2 years at 4 p.c.p.a.
Answer» $\bf \: \underline{Given} :$ ↴ $\sf \implies Principal = Rs. \: 6250$ $\sf \implies Time = 2 \: years$ $\sf \implies Rate = 4 \: \%$ $\bf \: \underline{To \: find} :$ ↴ $\sf \implies Compound \: interest - Simple \: interest = \: ?$ $\bf \: \underline{\underline{Solution}}$ $\sf \: First \: of \: all, let's \: find \: the \: value \: of \: (S.I) \: Simple \: interest.$ $\sf \implies \boxed{ \sf \: \pink{S.I = \dfrac{Principal \times Rate \times Time }{100} }}$ $\sf \implies\sf \:S.I = \dfrac{6250 \times 4 \times 2 }{100}$ $\sf \implies\sf \:S.I = \dfrac{625 \times 4 \times 2 }{10}$ $\sf \implies\sf \:S.I = \dfrac{625 \times 4 }{5}$ $\sf \implies\sf \:S.I = 125 \times 4$ $\red{\sf \implies\sf \: S.I = Rs. \: 500}$ $\sf \: Now, we \: will \: find \: the \: value \: of \: (C.I) \: Compound \: interest.$ $\sf \: For \: that, we \: have \: \: to \: find \: the \: amount.$ $\sf \implies \boxed{ \sf \: \pink{Amount = P \bigg( 1 + \dfrac{R }{100} \bigg)^{n} }}$ $\sf \implies \sf 6250 \bigg( 1 + \dfrac{4 }{100} \bigg)^{2}$ $\sf \implies \sf 6250 \bigg( 1 + \dfrac{1}{25} \bigg)^{2}$ $\sf \implies \sf 6250 \bigg(\dfrac{25 + 1}{25} \bigg)^{2}$ $\sf \implies \sf 6250 \bigg(\dfrac{26}{25} \bigg)^{2}$ $\sf \implies \sf 6250 \times \dfrac{676}{625}$ $\sf \implies \sf 10 \times 676$ $\red{\sf \implies \sf Amount = Rs. \: 6760}$ $\sf \: Now, Compound \: interest = Amount - Principal$ $\sf \implies \sf Rs. \: 6760 \: - \: Rs. \: 6250$ $\red{\sf \implies \sf Compound \: interest = Rs. \: 510}$ $\sf \: Now, difference \: between \: C.I \: and \: S.I \: is :$ $\sf \implies \sf Compound \: interest \: - \: Simple \: interest$ $\sf \implies \sf Rs. \: 510 \: - \: Rs. \: 500$ $\red{\sf \implies \sf Rs. \: 10}$ $\underline{ \boxed{ \pink{ \sf \: Hence, the \: difference \: between \: C.I \: and \: S.I = Rs. \: 10}}}$