Quantity I: A container has 60 liters of a mixture of milk and water in the ratio 3:2. 20 liters of this mixture is replaced with pure water. If the mixture is sold for Rs 20 per liter and cost of pure milk is Rs 35 per liter then find the profit (in Rs)

Quantity II: A shopkeeper has two varieties(A & B) of tea worth ₹200/kg and ₹300/kg. He mixes variety A & B in the ratio 2:3 respectively. Find the price(in Rs) per kg of the resulting mixture.

In the given question, two quantities are given, one as ‘Quantity I’ and another as ‘Quantity II’. You have to determine relationship between two quantities and choose the appropriate option. (Compare only numeric values)

The correct answer is Quantity I > Quantity II.

To determine the relationship between the two quantities, let us calculate their numerical values step-by-step.

Analysis of Quantity I:

Initial volume of the mixture = 60 liters.
Ratio of milk to water = 3 : 2.

Initial quantity of milk = $\frac{3}{5}\times 60=36$ liters.
Initial quantity of water = $\frac{2}{5}\times 60=24$ liters.

Now, 20 liters of this mixture is removed.
Quantity of milk in 20 liters mixture = $\frac{3}{5}\times 20=12$ liters.
Quantity of water in 20 liters mixture = $\frac{2}{5}\times 20=8$ liters.

Remaining milk in the container = $36-12=24$ liters.
Remaining water in the container = $24-8=16$ liters.

Then, 20 liters of pure water is added to the container.
Total volume of the new mixture = 60 liters.
Total quantity of milk in the final mixture = 24 liters.
Total quantity of water in the final mixture = $16+20=36$ liters.

Cost of pure milk = Rs 35 per liter.
Since water is free of cost, total cost price (CP) of the mixture = cost of 24 liters of pure milk.
CP = $24\times 35=\text{Rs }840$.

Selling price (SP) of the final mixture at Rs 20 per liter:
SP = $60\times 20=\text{Rs }1200$.

Profit = $\text{SP}-\text{CP}=1200-840=\text{Rs }360$.
Thus, Quantity I = 360.

Analysis of Quantity II:

Variety A worth ₹200/kg and Variety B worth ₹300/kg are mixed in the ratio 2 : 3.

Price of the resulting mixture per kg = weighted average cost price
Price per kg = $\frac{(2\times 200)+(3\times 300)}{2+3}$

$\text{Price per kg}=\frac{400+900}{5}=\frac{1300}{5}=\text{Rs }260$.
Thus, Quantity II = 260.

Comparison:

Quantity I = 360
Quantity II = 260

Since $360>260$, we get Quantity I > Quantity II.