Let the C.P of one mobile phone be $$₹x$$.
S.P of first mobile phone = ₹18,750
S.P of second mobile phone = ₹14,250
Let the loss percent, incurred on the second mobile phone be $$p\ \%$$.
Profit percentage, incurred on the first mobile phone = $$\left(5\times p\%\right)=5p\ \%$$
Now, we will use the profit-and-loss percentage formula.
For the first mobile phone:
Profit percentage = $$\dfrac{\left(\text{S.P}\ -\ \text{C.P}\right)}{\text{C.P}}\times100$$
$$5p=\dfrac{\left(₹18,750-₹x\right)}{₹x}\times100$$ $$\longrightarrow\ i$$
For the second mobile phone:
Loss percentage = $$\dfrac{\left(\text{C.P}\ -\ \text{S.P}\right)}{\text{C.P}}\times100$$
$$p=\dfrac{\left(₹x-₹14,250\right)}{₹x}\times100$$ $$\longrightarrow\ ii$$
From equations $$i$$ and $$ii$$, we have to find the value of $$x$$.
Divide equation $$i$$ by equation $$ii$$,
$$\dfrac{5p}{p}=\dfrac{\dfrac{\left(₹18,750-₹x\right)}{₹x}\times100}{\dfrac{\left(₹x-₹14,250\right)}{₹x}\times100}$$
$$5=\dfrac{\left(18,750-x\right)}{\left(x-14,250\right)}$$
$$5\left(x-14,250\right)=\left(18,750-x\right)$$
$$5x-\left(5\times14,250\right)=18,750-x$$
$$5x+x=18,750+\left(5\times14,250\right)$$
$$6x=18,750+71,250=90,000$$
$$x=15,000$$
Hence, the C.P of one mobile phone is ₹15,000.
Total S.P = ₹18,750 + ₹14,250 = ₹33,000
Total C.P = $$₹2x$$ = $$₹\left(2\times15,000\right)=₹30,000$$
Overall Profit = Total S.P - Total C.P = ₹33,000 - ₹30,000 = ₹3,000
Overall Profit percentage = $$\dfrac{₹3,000}{₹30,000}\times100=10\%$$
Hence, the overall profit percentage of the transaction is 10%.
$$\therefore$$ The required answer is D.