Question

The price of commodity X increases by $40$paise every year, while the price of commodity Y increases by $15$ paise every year. If in $2001$, the price of commodity X was Rs. $4.20$and that of Y was Rs $6.30$, in which year commodity X will cost $40$paise more than the commodity Y?
(A) $2010$
(B) $2011$
(C) $2012$
(D) $2013$

Answer 1

Hint: Equating the nth term of the A.P. to find the value of n and so the year in which commodity X will cost more than Y by $40$ paise.

Formulas used:${A_n} = a + \left( {n - 1} \right)d$, where ‘a’ is the first term and ‘d’ is the common difference of A.P.

Complete Step by step solution:
Price of the commodity X in $2001$is Rs. $4.20$ or $420$paise and also it increases by $40$paise every year. Which implies
$420,460,500,.......$is an arithmetic progression.

Price of the commodity Y in $2001$ is Rs.$6.30$ or $630$paise and also it increases by $15$paise every year. Which implies
$630,645,660,.........$is an arithmetic progression.

Let after n years the price of the commodity X is more than that of the price of commodity Y by $40$ paise.
Then difference of nth term of the above two arithmetic progression will be equal to $40$
$ \Rightarrow {A_n} - A{'_n} = 40$
\[\left\{ {a + \left( {n - 1} \right)d} \right\} - \left\{ {a' + \left( {n - 1} \right)d'} \right\} = 40\] Where $a = 420,\,d = 40,\,a' = 630\,\,and\,\,d' = 15$
Substituting values in above equation we have,
\[\left\{ {420 + (n - 1)(40)} \right\} - \left\{ {630 + (n - 1)15} \right\} = 40\]
$ \Rightarrow \left( {420 + 40n - 40} \right) - \left( {630 + 15n - 15} \right) = 40$
$
   \Rightarrow (380 + 40n) - (615 + 15n) = 40 \\
   \Rightarrow 380 + 40n - 615 - 15n = 40 \\
   \Rightarrow 25n - 235 = 40 \\
 $
$
   \Rightarrow 25n = 40 + 235 \\
   \Rightarrow 25n = 275 \\
   \Rightarrow n = \dfrac{{275}}{{25}} \\
   \Rightarrow n = 11 \\
 $

Therefore, from above we see that after $11$ years price of commodity X is more than price of commodity Y by $40$ paise.
Hence, in $2001 + 11 = 2012$price of commodity X will be more than price of commodity Y by $40$ paise.

Note: For getting nth term of two different arithmetic progression we equate their nth term.