Alligation or Mixture problems
The ratio of water and alcohol in two different containers is 2:3 and 4:5. In what ratio we are required to mix the mixtures of two containers in order to get the new mixture in which the ratio of alcohol and water be 7:5?
W 1 : A 1 W 2 : A 2 . . . . . W N : A N 2 : 3 4 : 5 5 : 7 W 1 W 1 + A 1 = 2 5 W 2 W 2 + A 2 = 4 9 W N W N + A N = 5 12 = 72/180 = 80/180 = 75/180 => 5 : 3 Therefore, the ratio is 5: 3
Let the quantity of the wine in the cask originally be x litres. Then, quantity of wine left in cask after 4 operations = [ x ❨ 1 - 8 ❩ 4 ] litres. x ∴ ❨ x(1 - (8/x))4 ❩ = 16 x 81 ⟹ ❨ 1 - 8 ❩ 4 = ❨ 2 ❩ 4 x 3 ⟹ ❨ x - 8 ❩ = 2 x 3 ⟹ 3x - 24 = 2x ⟹ x = 24.
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.
As given equal amounts of alloys are melted, let it be 1 kg. Required ratio of gold and silver = 5 13 + 5 8 8 13 + 3 8 = 105 103 . Hence, ratio of gold and silver in the resulting alloy = 105/103.
EXAMIANS