Answer:

Given that, Mohan bought a house at 2,50,000. He spent 50,000 for repairing it.

Cost Price of a house = 2, 50, 000 + 50, 000 = 3, 00, 000

Selling Price of a house = 3, 75, 000

\begin{gathered}\sf\implies \sf \: SellingPrice \: > \: Cost \: Price \\ \\ \end{gathered}

⟹SellingPrice>CostPrice

\begin{gathered}\sf\implies \sf \: Profit \: in \: this \: tranaction. \\ \\ \end{gathered}

⟹Profitinthistranaction.

Now,

\begin{gathered}\sf \: \sf \: Profit\% = \dfrac{Selling \: Price \: - \: Cost \: Price}{Cost \: Price} \times 100\% \\ \\ \end{gathered}

Profit%=

CostPrice

SellingPrice−CostPrice

×100%

\begin{gathered}\sf \: \sf \: Profit\% = \dfrac{375000 - 300000}{300000} \times 100\% \\ \\ \end{gathered}

Profit%=

300000

375000−300000

×100%

\begin{gathered}\sf \: \sf \: Profit\% = \dfrac{75000}{3000} \% \\ \\ \end{gathered}

Profit%=

3000

75000

%

\begin{gathered}\sf\implies \sf \: \bf \: Profit\% = 25 \: \% \\ \\ \\ \end{gathered}

⟹Profit%=25%

\large\underline{\sf{Solution-2}}

Solution−2

Given that,

Principal, P = Rs 4500

Rate of interest, r = 3 1/2 = 3.5 % per annum

Time period, n = 1 1/2 = 1.5 years

We know,

Simple interest (SI) received on a certain sum of money of P invested at the rate of r % per annum for n years is

\begin{gathered}\qquad\boxed{ \sf{ \:SI = \dfrac{P \times r \times n}{100} \: \: }} \\ \\ \end{gathered}

SI=

100

P×r×n

So, on substituting the values, we get

\begin{gathered}\sf \: SI = \dfrac{4500 \times 3.5 \times 1.5}{100} \\ \\ \end{gathered}

SI=

100

4500×3.5×1.5

\begin{gathered}\sf \: SI = 45 \times \dfrac{35}{10} \times \dfrac{15}{10} \\ \\ \end{gathered}

SI=45×

10

35

×

10

15

\begin{gathered}\sf \: SI = 9 \times \dfrac{35}{2} \times \dfrac{15}{10} \\ \\ \end{gathered}

SI=9×

2

35

×

10

15

\begin{gathered}\sf \: SI = 9 \times \dfrac{35}{2} \times \dfrac{3}{2} \\ \\ \end{gathered}

SI=9×

2

35

×

2

3

\begin{gathered}\sf \: SI = \dfrac{945}{4} \\ \\ \end{gathered}

SI=

4

945

\begin{gathered}\sf\implies \sf \: SI = 236.25 \\ \\ \end{gathered}

⟹SI=236.25

Thus,

\begin{gathered}\sf \: Amount = P + SI \\ \\ \end{gathered}

Amount=P+SI

\begin{gathered}\sf \: Amount = 4500 + 236.25 \\ \\ \end{gathered}

Amount=4500+236.25

\begin{gathered}\sf\implies \bf \: Amount = 4736.25 \\ \\ \end{gathered}

⟹Amount=4736.25