Shortcut Trick
Initial Ratio (Milk : Water) = 7 : 3 (Total = 10 units)
Final Ratio = 1 : 1 → Equate Milk (since only water is added) → 7 : 7
Increase in Water units = 7 − 3 = 4 units
Given 10 units = 60 L ⇒ 1 unit = 6 L
Added Water = 4 units × 6 L = 24 L
∴ The correct answer is 24 L.
Alternate Method
Given: Total mixture = 60 L, Initial ratio (M:W) = 7 : 3, Final ratio (M:W) = 1 : 1.
⇒ Total parts in initial mixture = 7 + 3 = 10 units
⇒ Quantity of Milk = (7 ÷ 10) × 60 = 42 L
⇒ Quantity of Water = (3 ÷ 10) × 60 = 18 L
⇒ In the final 1 : 1 mixture, let the water added be x liters.
⇒ New Water content = 18 + x
⇒ Milk : Water = 42 : (18 + x) = 1 : 1
⇒ 42 = 18 + x
⇒ x = 42 − 18 = 24 L
∴ The correct answer is 24 L.
Additional Information
Constant Quantity Concept
When one component of a mixture is added or removed, the quantity of the other component remains constant. Always equate the parts of the constant component in the ratio calculations.
Ratio Balancing
To simplify calculations, multiply the final ratio by a factor that makes the non-changing substance equal in both ratios (e.g., if milk is 7 parts initially, make it 7 parts in the final ratio).
Percentage Method
Initial Milk % = 70%. In 1:1 ratio, Milk % = 50%. Using inverse variation: Total Volume1 × %1 = Total Volume2 × %2 ⇒ 60 × 70 = Vnew × 50 ⇒ Vnew = 84 L.