A farmer sold a cow and an ox for ₹ 800 and got a profit of 20% on the cow and 25% on the ox. If he sells the cow and the ox for ₹ 820 he gets a profit of 25% on the cow and 20% on the ox. The individual cost price of the cow and the ox is

### Concept & Simultaneous Equations When given two different combinations of profit percentages and their corresponding total selling prices, we can establish a system of linear equations to solve for the individual cost prices. ### Step-by-Step Solution * Let the Cost Price of the cow be $C$ and the ox be $O$. * From the first condition (Selling Price = ₹ 800): $1.20C + 1.25O = 800$ (Equation 1) * From the second condition (Selling Price = ₹ 820): $1.25C + 1.20O = 820$ (Equation 2) * Multiply both equations by 100 to remove decimals: $120C + 125O = 80000$ $125C + 120O = 82000$ * Add both equations together: $245C + 245O = 162000 \Rightarrow 245(C + O) = 162000 \Rightarrow C + O = \frac{162000}{245} \approx 661.22$ * Subtract the first equation from the second equation: $5C - 5O = 2000 \Rightarrow C - O = \frac{2000}{5} = 400$ * Now we have a simple system: $C + O = 661.22$ $C - O = 400$ * Add these two simplified equations: $2C = 1061.22 \Rightarrow C = 530.61$ * Substitute $C$ back to find $O$: $530.61 - O = 400 \Rightarrow O = 130.61$ ### Exam Strategy & Shortcut Whenever you see symmetrical coefficients (e.g., $aX + bY$ and $bX + aY$), adding and subtracting the equations is the fastest way to reduce them to simple $X + Y$ and $X - Y$ formats, bypassing messy direct substitution. ### Common Pitfall Attempting to solve the system by substitution straight from the decimal equations, which leads to highly complex fraction calculations and increased risk of arithmetic errors. ### Final Answer Therefore, the correct answer is **₹ 530.60, ₹ 130.60 (approx)**.