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(B) Let $n$ be the number of years after $2001$.The price of commodity $X$ after $n$ years is $420 + 40n$ paise.The price of commodity $Y$ after $n$ y

Vedclass · Accepted Answer

$2011$

Vedclass · Answer

(B) Let $n$ be the number of years after $2001$.The price of commodity $X$ after $n$ years is $420 + 40n$ paise.The price of commodity $Y$ after $n$ years is $630 + 15n$ paise.According to the problem,we want the price of $X$ to be $40 	ext{ paise}$ more than the price of $Y$:$420 + 40n = (630 + 15n) + 40$$420 + 40n = 670 + 15n$$40n - 15n = 670 - 420$$25n = 250$$n = 10$Thus,after $10$ years from $2001$,the condition will be met.The year is $2001 + 10 = 2011$.

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