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: Here, we will assume \[x\] to be the number of years after 2001 such that the commodity X will cost 40 paise more than the commodity Y in the year \[2001 + x\]. We will find the increase in the price of the two commodities after \[x\] years, and the prices of the two commodities in the year \[2001 + x\]. Then, we will use the given information to form a linear equation in terms of \[x\]. We will solve this equation to find the value of \[x\], and use the value of \[x\] to find the required year.

Complete step-by-step answer:
Let \[x\] be the number of years after 2001 such that the commodity X will cost 40 paise more than the commodity Y in the year \[2001 + x\].
First, we will convert 40 paise and 15 paise to rupees using unitary method.
We know that 1 rupee is equal to 100 paise.
Therefore, we get
100 paise \[ = \] 1 rupee
Now, dividing both sides by 100, we get
1 paise \[ = \dfrac{1}{{100}}\] rupees
This is the value of 1 paise in rupees.
The value of 40 paise in rupees is the product of the value of 1 paise in rupees, and the number of paise.
Therefore, we will multiply the equation the get the value of 40 paise in rupees.
Multiplying both sides of the equation by 40, we get
\[\left( {1 \times 40} \right)\]paise \[ = \left( {\dfrac{1}{{100}} \times 40} \right)\] rupees
Thus, we get
40 paise \[ = \dfrac{{40}}{{100}}\]rupees
The value \[\dfrac{{40}}{{100}}\] is in fractions. We will divide 40 by 100 to convert it to decimals.
Therefore, we get
\[\dfrac{{40}}{{100}} = 0.4\]
Hence, 40 paise \[ = 0.4\]rupees.
The value of 15 paise in rupees is the product of the value of 1 paise in rupees, and the number of paise.
Therefore, we will multiply the equation the get the value of 15 paise in rupees.
Multiplying both sides of the equation by 15, we get
\[\left( {1 \times 15} \right)\]paise \[ = \left( {\dfrac{1}{{100}} \times 15} \right)\] rupees
Thus, we get
15 paise \[ = \dfrac{{15}}{{100}}\]rupees
The value \[\dfrac{{15}}{{100}}\] is in fractions. We will divide 15 by 100 to convert it to decimals.
Therefore, we get
\[\dfrac{{15}}{{100}} = 0.15\]
Hence, 15 paise \[ = 0.15\]rupees.
Now, we will find the increase in price of the commodities after \[x\] years.
The increase in price after \[x\] years is the product of the increase in price per year, and the number of years.
The price of commodity X increases by 40 paise, or Rs. \[0.4\] in 1 year.
Therefore, we get
Increase in price of commodity X after \[x\] years \[ = 0.4 \times x = 0.4x\]
The price of commodity Y increases by 15 paise, or Rs. \[0.15\] in 1 year.
Therefore, we get
Increase in price of commodity Y after \[x\] years \[ = 0.15 \times x = 0.15x\]
The price of commodity X in the year \[2001 + x\] is the sum of the price of commodity X in year 2001, and the increase in price of commodity X after \[x\] years.
Therefore, we get
Price of commodity X in the year \[2001 + x\] \[ = {\rm{Rs}}{\rm{.}}\left( {4.20 + 0.4x} \right)\]
The price of commodity Y in the year \[2001 + x\] is the sum of the price of commodity Y in year 2001, and the increase in price of commodity Y after \[x\] years.
Therefore, we get
Price of commodity Y in the year \[2001 + x\] \[ = {\rm{Rs}}{\rm{.}}\left( {6.30 + 0.15x} \right)\]
Now, we know that the commodity X will cost 40 paise more, or Rs. \[0.4\] more than the commodity Y in the year \[2001 + x\].
Therefore, we get the equation
\[{\rm{Rs}}{\rm{.}}\left( {4.20 + 0.4x} \right) = {\rm{Rs}}{\rm{.}}\left( {6.30 + 0.15x} \right) + {\rm{Rs}}{\rm{. }}0.4\]
This is a linear equation in one variable. We will solve this equation to find the value of the variable \[x\].
Rewriting the equation, we get
\[ \Rightarrow 4.2 + 0.4x = 6.3 + 0.15x + 0.4\]
Subtracting \[4.2\] and \[0.15x\] from both sides of the equation, we get
\[\begin{array}{l} \Rightarrow 4.2 + 0.4x - 0.15x - 4.2 = 6.3 + 0.15x + 0.4 - 0.15x - 4.2\\ \Rightarrow 0.25x = 6.3 + 0.4 - 4.2\end{array}\]
Adding and subtracting the terms in the expression, we get
\[ \Rightarrow 0.25x = 2.5\]
Dividing both sides of the equation by \[0.25\], we get
\[\begin{array}{l} \Rightarrow \dfrac{{0.25x}}{{0.25x}} = \dfrac{{2.5}}{{0.25}}\\ \Rightarrow x = 10\end{array}\]
Therefore, the commodity X will cost 40 paise more, or Rs. \[0.4\] more than the commodity Y 10 years after 2001.
Substituting \[x = 10\] in the expression \[2001 + x\], we get
\[\begin{array}{c} \Rightarrow 2001 + x = 2001 + 10\\ \Rightarrow 2001 + x = 2011\end{array}\]
\[\therefore \] The commodity X will cost 40 paise more, or Rs. \[0.4\] more than the commodity Y in the year 2011.
Thus, the correct option is option (b).

Note: We have formed a linear equation in one variable in terms of \[x\] in the solution. A linear equation in one variable is an equation of the form \[ax + b = 0\], where \[a\] and \[b\] are integers. A linear equation of the form \[ax + b = 0\] has only one solution.
We used a unitary method to solve the problem. Unitary method is a method where first, the per unit quantity is calculated, and then the number of units are multiplied. Here, we calculated the value of paise per rupee and multiplied it by 15 and 40 to get the value of 15 and 40 paise in rupees.