At a village trade fair a man buys a horse and a camel together for Rs 51,250. He sold the horse at a profit of 25 % and the camel at a loss of 20%. Ifhe sold both the animals at the the same price, then the cost price of the cheaper animal was Rs ______________.

Options

A. 6600
B. 7500
C. 25000
D. 20000

Answer

To solve the problem, let's break it down step by step.

### Step 1: Define Variables
Let the cost price of the horse be \( H \) and the cost price of the camel be \( C \). According to the problem, we have:
\[ H + C = 51,250 \]

### Step 2: Calculate Selling Price of the Horse
The horse is sold at a profit of 25%. Therefore, the selling price (SP) of the horse can be calculated as:
\[ SP_{horse} = H + 0.25H = 1.25H \]

### Step 3: Calculate Selling Price of the Camel
The camel is sold at a loss of 20%. Therefore, the selling price of the camel can be calculated as:
\[ SP_{camel} = C - 0.20C = 0.80C \]

### Step 4: Set Selling Prices Equal
According to the problem, both animals are sold at the same price:
\[ SP_{horse} = SP_{camel} \]
Substituting the expressions from Steps 2 and 3:
\[ 1.25H = 0.80C \]

### Step 5: Express C in Terms of H
From the equation \( 1.25H = 0.80C \), we can express \( C \) in terms of \( H \):
\[ C = \frac{1.25H}{0.80} = \frac{1.25}{0.80}H = 1.5625H \]

### Step 6: Substitute C in the Total Cost Equation
Now substitute \( C \) in the total cost equation \( H + C = 51,250 \):
\[ H + 1.5625H = 51,250 \]
\[ 2.5625H = 51,250 \]

### Step 7: Solve for H
Now, solve for \( H \):
\[ H = \frac{51,250}{2.5625} \]
Calculating this gives:
\[ H = 20,000 \]

### Step 8: Calculate C
Now substitute \( H \) back to find \( C \):
\[ C = 51,250 - H = 51,250 - 20,000 = 31,250 \]

### Step 9: Identify the Cheaper Animal
Since \( H = 20,000 \) and \( C = 31,250 \), the cost price of the cheaper animal (the horse) is:
\[ \text{Cost price of the cheaper animal} = 20,000 \]

### Final Answer
The cost price of the cheaper animal was Rs 20,000.

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