Please enter the following details
Free Daily Targets & Mocks
Study Material & Formulas PDFs
1 on 1 Mentorship
Question 76
The price of an article was increased by p % and then the new price was decreased by p%. If the last price was ₹ 100, the original price was
Solution
Let original price = Rs. $$x$$
Price after $$p\%$$ increase = $$x+\frac{p}{100}\times x=Rs.$$ $$(x+\frac{px}{100})$$
Price after $$p\%$$ decrease = $$(x+\frac{px}{100})-(\frac{p}{100})(x+\frac{px}{100})$$
According to ques, => $$x+\frac{px}{100}-\frac{px}{100}-\frac{p^2x}{10000}=100$$
=> $$x(1-\frac{p^2}{10000})=100$$
=> $$x=Rs.$$ $$\frac{100}{1-\frac{p^2}{100^2}}$$
=> Ans - (A)
