Question

A farmer bought a cow and a calf together for Rs.14000. After some time, he sold the cow at a profit of $20\%$ and the calf at a profit of $13\%$. On the whole, he got a profit of $17\%$. Find the cost price of the cow and the calf.
A. Rs.8000, Rs.6000
B. Rs.6000, Rs.6000
C. Rs.8000, Rs.8000
D. Rs.9000, Rs.5000

Answer 1

Hint: In this question, first of all, consider the cost price of the cow and the calf as variables. Then find out the profit amount and selling price of the cow and calf separately. Equate the total selling price of the cow and the calf to the sum of the total cost price and the total profit amount of the cow and the calf to get the required answer. So, use this concept to reach the solution to the given problem.

Complete step-by-step solution:
Let \[x\] be the cost price of the cow and \[y\] be the cost price of the calf.
Given that the cost price of cow and calf is Rs.14000.
So, we have \[x + y = 14000........................\left( 1 \right)\]
Also given that profit% on cow = $20\%$
So, amount of profit on cow \[ = \dfrac{{20x}}{{100}}\]
And the profit% on calf = $13\%$
So, amount of profit on calf \[ = \dfrac{{13y}}{{100}}\]
Given that the total profit% = $17\%$ of the total cost price i.e., Rs.14000
So, amount of profit on both cow and calf \[ = \dfrac{{17}}{{100}} \times 14000\]
We know that selling price = cost price + profit amount
So, selling price of cow = cost price of cow + profit amount on cow
                                       \[ = x + \dfrac{{20x}}{{100}}\]
And selling price of calf = cost price of calf + profit amount on calf
                                           \[ = y + \dfrac{{13y}}{{100}}\]
So, selling price of cow and calf = cost price of cow and calf + amount of profit on both cow and calf
\[ \Rightarrow \left( {x + \dfrac{{20x}}{{100}}} \right) + \left( {y + \dfrac{{13y}}{{100}}} \right) = 14000 + \left( {\dfrac{{17}}{{100}} \times 14000} \right) \\
   \Rightarrow x + 0.2x + y + 0.13y = 14000 + 2380 \\
   \Rightarrow 1.2x + 1.13y = 16380...................\left( 2 \right) \]
Multiplying equation (1) with 1.2 we get
\[ \Rightarrow 1.2x + 1.2y = 16800..................................\left( 3 \right)\]
On subtracting equation (3) from equation (2), we get
\[ \Rightarrow \left( {1.2x + 1.13y} \right) - \left( {1.2x + 1.2y} \right) = 16380 - 16800 \\
   \Rightarrow 1.13y - 1.2y = - 420 \\
   \Rightarrow - 0.07y = - 420 \\
  \therefore y = \dfrac{{ - 420}}{{ - 0.7}} = 6000 \]
On substituting \[y = 6000\] in equation (1). We get
\[ \Rightarrow x + 6000 = 14000 \\
  \therefore x = 14000 - 6000 = 8000 \]
Therefore, the cost price of the cow is Rs.8000 and the cost price of the calf is Rs.6000.
Thus, the correct option is A. Rs.8000, Rs.6000.

Note: \[x\% \] of \[y\] is given by \[\dfrac{x}{{100}} \times y\]. The sum of obtained cost price of cow and calf must be equal to Rs.14,000 or else our answer is wrong. Here we have used elimination method two find the obtained two equations in terms of \[x\& y\].
Here we could also use the substitution method by which we substitute the value of one variable in terms of the second variable and solve the equations.