Correct option is A
Given:
· Initial milk = 50 L
· Supplies
5 L to each customer
· After each customer,
adds 5 L of water
· We need
milk % in 1 litre of solution bought by
5th customer
Concept Used:
We treat this as a
successive dilution problem. After each transaction:
· 5 L of the current solution (milk + water) is removed
· Then 5 L of
water is added
We repeatedly
remove a fraction of milk, since water doesn't affect milk amount directly.
Solution:
Step-by-step calculation:
Initial state:
Milk = 50 L
Water = 0 L
Total = 50 L
Milk fraction = 1
After 1st customer:
Removed 5 L milk (since it's pure milk)
Remaining milk = 45 L
Add 5 L water → Total = 50 L
Milk fraction = 45 / 50 = 0.9
After 2nd customer:
Removed 5 L × 0.9 = 4.5 L milk
Remaining milk = 45 – 4.5 = 40.5 L
Add 5 L water → Total = 50 L
Milk fraction = 40.5 / 50 = 0.81
After 3rd customer:
Removed 5 L × 0.81 = 4.05 L milk
Remaining milk = 40.5 – 4.05 = 36.45 L
Add 5 L water → Total = 50 L
Milk fraction = 36.45 / 50 = 0.729
After 4th customer:
Removed 5 L × 0.729 = 3.645 L milk
Remaining milk = 36.45 – 3.645 = 32.805 L
Add 5 L water → Total = 50 L
Milk fraction = 32.805 / 50 = 0.6561
5th customer:
Milk in 1 L = 0.6561 L
Percentage of milk = 0.6561 × 100 = 65.6%
Final Answer: (a) 65.6%
