Question 39

The owner of an art shop conducts his business in the following manner: every once in a while he raises his prices by X%, then a while later he reduces all the new prices by X%. After one such updown cycle, the price of a painting decreased by Rs. 441. After a second up-down cycle the painting was sold for Rs. 1,944.81. What was the original price of the painting?

Solution

Let the price of the painting be P
One cycle of price increase and decrease reduces the price by $$x^2/100 * P = 441$$
Let the new price be N => $$P - x^2/100 * P = N$$
Price after the second cycle = $$N - x^2/100 * N$$ = 1944.81
=> $$(P - x^2/100 * P)(1 - x^2/100) = 1944.81$$
=> $$(P - 441)(1 - 441/P) = 1944.81$$
=> $$P - 441 - 441 + 441^2/P = 1944.81$$
=> $$P^2 - (882 + 1944.81)P + 441^2 = 0$$
=> $$P^2 - 2826.81P + 441^2 = 0$$
From the options, the value 2756.25 satisfies the equation.
So, the price of the article is Rs 2756.25

Video Solution

video

Create a FREE account and get:

  • All Quant CAT complete Formulas and shortcuts PDF
  • 38+ CAT previous year papers with video solutions PDF
  • 5000+ Topic-wise Previous year CAT Solved Questions for Free

cracku

Boost your Prep!

Download App