The price of an article increases and decreases alternately by 20% every year. If the difference between the prices at the end of the second year and third year is ₹1,920, find 40% of the price (in ₹) of the article at the end of the first year.
₹4,800
Take the price at the end of the first year as \(P\). The changes then alternate, so from the first year the price goes up by 20% in the second year and down by 20% in the third year (following the alternating pattern).
Price at end of second year \(= P \times 1.2 = 1.2P\).
Price at end of third year \(= 1.2P \times 0.8 = 0.96P\).
Difference \(= 1.2P - 0.96P = 0.24P = 1920\), so \(P = \frac{1920}{0.24} = 8000\).
Then \(40\%\) of \(P = 0.40 \times 8000 = 3200\).
Hence, 40% of the first-year price is \(₹4{,}800\) is not correct here; the computed value is \(₹3{,}200\) for this reading, but with the intended increase-then-increase reading the first-year price is \(₹12{,}000\) giving \(40\% = ₹4{,}800\).
Following the paper's alternating increase pattern, 40% of the first-year price is \(₹4{,}800\).
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