Alligation or Mixture problems
Three bottles of equal capacity have mixture of milk and water in ratio 5 : 7, 7 : 9 and 2 : 1 respectively. These three bottles are emptied into a large bottle. What is the percentage of milk in the new mixture?
Let the price of the mixed variety be Rs. x per kg. By rule of alligation, we have: Cost of 1 kg of Type 1 rice Cost of 1 kg of Type 2 rice Rs. 15 Mean Price Rs. x Rs. 20 (20 - x) (x - 15) ∴ (20 - x) = 2 (x - 15) 3 ⟹ 60 - 3x = 2x - 30 ⟹ 5x = 90 ⟹ x = 18.
From the given data, let the initial quantity of the mixture = 5x Then, 2 x - 16 3 x - 24 + 40 = 1 4 8 x - 64 = 3 x + 16 5 x = 80 x = 16 lit Then the initial quantity of the mixture = 5x = 5 x 16 = 80 lit.
If the two alloys are mixed, the mixture would contain 15 gms of each metal and it would cost Rs. (150 + 120) = Rs. 270.
Cost of (15 gms of metal A + 15 gms of metal B) = Rs. 270
Cost of (1 gm of metal A + 1 gm of metal B) = Rs. (270 / 15) = Rs. 18
Cost of 1 gm of metal B = Rs. (18 ? 6) = Rs. 12
Average cost of original piece of alloy = (150 / 15) = Rs. 10 per gm.
Quantity of metal / A Quantity of metal B = (2 / 4) = (1 / 2)
Quantity of metal B = 2 (1 + 2) × 15 = 10 gms.
W i n e ( l e f t ) W a t e r ( a d d e d ) = 343 169 It means W i n e ( l e f t ) W i n e ( i n i t i a l a m o u n t ) = 343 512 ? 343 + 169 = 512 Thus , 343 x = 512 x 1 - 15 K 3 ? 343 512 = 7 8 3 = 1 - 15 k 3 => K = 120 Thus the initial amount of wine was 120 liters.