Ramu sells milk after adulterating X litres of pure milk with 0.75 lit...
Approach:
Let's assume the cost price of one litre of pure milk be Rs. x
To solve the problem, we need to follow the below steps:
- Calculate the selling price of milk sold after adulterating X litres of pure milk with 0.75 litres of tap water.
- Calculate the selling price of milk sold after adulterating X litres of pure milk with 0.75 litres of mineral water.
- Calculate the profit percentage in both the above cases.
- Formulate an equation using the given information.
- Solve the equation to get the value of x.
Calculation:
Step 1: Calculation of selling price of milk sold after adulterating X litres of pure milk with 0.75 litres of tap water.
- Let's assume that X litres of pure milk is adulterated with 0.75 litres of tap water.
- So, the total quantity of milk will be X + 0.75 litres.
- As Ramu always sells at cost price, the selling price of milk will be equal to the cost price of milk which is Rs. x per litre.
- Therefore, the selling price of X litres of milk after adulteration will be X * x.
- Similarly, the selling price of 0.75 litres of tap water will be 0.75 * 0 = 0.
- Hence, the total selling price of X litres of milk after adulteration with 0.75 litres of tap water will be X * x + 0.
Step 2: Calculation of selling price of milk sold after adulterating X litres of pure milk with 0.75 litres of mineral water.
- As given, the cost of mineral water is Rs. 12 per litre.
- So, the cost of 0.75 litres of mineral water will be 0.75 * 12 = Rs. 9.
- Therefore, the total cost of X litres of pure milk after adulteration with 0.75 litres of mineral water will be X * x + 9.
- As Ramu sells at cost price, the selling price of milk sold after adulterating X litres of pure milk with 0.75 litres of mineral water will also be X * x + 9.
Step 3: Calculation of profit percentage in both the above cases.
- Profit percentage in the first case will be (selling price - cost price) / cost price * 100
- Profit percentage in the second case will be (selling price - cost price) / cost price * 100 * 0.85 (as the profit reduces by 15%)
- Therefore, profit percentage in the first case will be (X * x + 0 - X * x) / X * x * 100 = 0%.
- And, profit percentage in the second case will be (X * x + 9 - X * x) / X * x * 100 * 0.85 = 0.15 * 100 = 15%.
Step 4: Formulation of equation using the given information.
- Profit percentage reduces by 15% when X litres of pure milk is adulterated with 0.75 litres of mineral water instead of tap water.
- Therefore, (profit percentage with mineral water - profit percentage with tap water) / profit percentage with tap water * 100 = 15.
- Substituting the values, we
Ramu sells milk after adulterating X litres of pure milk with 0.75 lit...
15%=9 rs
100%=60rs
Profit 60 with tap water
Profit 51 with mineral water
To make sure you are not studying endlessly, EduRev has designed CAT study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in CAT.