Question

Rohit got profit of 23/2 by selling his old car. however he realized that had he sold it for ₹8100 more his profit would be 38.5 percent. at what price did he buy the car​

Answer 1

$\huge{\underline{\purple{\rm{Solution:}}}}$

$\small{\underline{\red{\rm{Let:}}}}$

• $\rm{The\:cost\: price\:car\:be\:x}$

$\small{\underline{\red{\rm{To\: Find:}}}}$

• $\rm{\implies The\: cost\: price\:of\:car}$

$\small{\underline{\red{\rm{Given\:in\:1st\: Case:}}}}$

• $\rm{The\: profit\:on\: selling\:car=11.5\%}$

$\rm\purple{\implies SP=\dfrac{100+G\%}{100}*CP}$

$\rm{\implies SP=\dfrac{100+11.5}{100}*x}$

$\rm{\implies SP=\dfrac{111.5}{100}*x}$

$\rm{\implies SP=\dfrac{111.5x}{100}}$

• $\rm{The\:car\:was\:sold\:Rs.8100\:more}$

$\rm\green{\implies SP=\dfrac{111.5x}{100}+8100}$

$\small{\underline{\red{\rm{Given\:in\:2nd\: Case:}}}}$

• $\rm{The\: profit\:on\: selling\:car=38.5\%}$

$\rm{\implies SP=\dfrac{100+38.5}{100}*x}$

$\rm{\implies SP=\dfrac{138.5}{100}*x}$

$\rm\green{\implies SP=\dfrac{138.5x}{100}}$

• $\rm{Now, \:according\:to\: above\: information}$

$\rm{\implies \dfrac{138.5x}{100}=\dfrac{111.5x}{100}+8100}$

$\rm{\implies \dfrac{138.5x}{100}-\dfrac{111.5x}{100}=8100}$

$\rm{\implies \dfrac{138.5x-111.5x}{100}=8100}$

$\rm{\implies \dfrac{138.5x-111.5x}{100}=8100}$

$\rm{\implies \dfrac{27x}{100}=8100}$

$\rm{\implies \cancel{27}x=\cancel{8100}*100}$

$\rm{\implies x=300*100}$

$\rm\purple{\implies x=Rs.30000}$

Hence,

$\rm\orange{\implies The\: cost\: price\:of\:car=Rs.30000}$