If a shopkeeper first raises the price of jewellery by x% then he decreases…
2024
If a shopkeeper first raises the price of jewellery by x% then he decreases the new price by x%. After one such up down cycle, the price of a jewellery decreased by Rs. 1200. After a second up down cycle the jewellery was sold for Rs. 27648. What was the original price of the jewellery ?
- A.
Rs. 24000
- B.
Rs. 36000
- C.
Rs. 44000
- D.
Rs. 30000
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Correct answer: D
Answer: Rs. 30000
Let the original price be a. If the price is increased and then decreased by the same percentage x%, the net multiplicative factor for one up-down cycle is
(1 + x/100)(1 - x/100) = 1 - x^2/10000.
After the first cycle the price becomes a(1 - x^2/10000). The decrease after the first cycle is a - a(1 - x^2/10000) = a·x^2/10000 = 1200.
Thus x^2/10000 = 1200/a, so the factor after one cycle equals 1 - 1200/a.
After two cycles the price is a(1 - 1200/a)^2, and we are given a(1 - 1200/a)^2 = 27648.
Expand and rearrange: a(1 - 2400/a + 1440000/a^2) = 27648 ⇒ a^2 - 30048a + 1440000 = 0.
Factor the quadratic: (a - 30000)(a - 48) = 0, so a = 30000 or a = 48.
Reject a = 48 because then 1 - 1200/48 = 1 - 25 = -24, which would make the price negative after one cycle (not meaningful).
Therefore the original price is Rs. 30000. Check: 1 - 1200/30000 = 0.96, and 30000 × 0.96^2 = 30000 × 0.9216 = 27648, which matches the given final price.