The price of a commodity (in rupees per kilogram) is 100+ 0.1n, on the nth day…
2024
The price of a commodity (in rupees per kilogram) is 100+ 0.1n, on the nth day of 2007 (n = 1, 2,---- 100) and then remains constant. On the other hand, the price of another commodity (in rupees per kilogram) is 89+ 0.15n, on the nth day of 2007 (n = 1,2,..........,365). On which date in 2007. Will the prices of these two commodities be equal?
- A.
April 11
- B.
May 20
- C.
April 10
- D.
May 21
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Correct answer: B
Solution:
First commodity: price on the nth day is 100 + 0.1n for n = 1 to 100, and then it remains constant at the value on the 100th day.
Compute the price at n = 100: 100 + 0.1×100 = 110. So after day 100 the first commodity's price is 110.
Second commodity: price on the nth day is 89 + 0.15n (defined for n = 1..365). To find when prices are equal, solve 89 + 0.15n = 110.
Solve for n: 0.15n = 110 − 89 = 21, so n = 21 / 0.15 = 140.
Convert the 140th day of 2007 to a calendar date. Cumulative days by month in 2007 (not a leap year): Jan 31, Feb 59, Mar 90, Apr 120. Day 121 is May 1, so day 140 is 140 − 120 = 20 → May 20.
Therefore the prices of the two commodities are equal on May 20, 2007.