Wine Water
8L 32L
1 : 4
20 % 80% (original ratio)
30 % 70% (required ratio)
In ths case, the percentage of water being reduced when the mixture is being replaced with wine.
so the ratio of left quantity to the initial quantity is 7:8
Therefore ,
=> K = 5 Lit
2 : 3 4 : 5 5 : 7
= 72/180
= 80/180
= 75/180
=> 5 : 3
Therefore, the ratio is 5: 3
Let the capacity of the pot be 'P' litres.
Quantity of milk in the mixture before adding milk = 4/9 (P - 8)
After adding milk, quantity of milk in the mixture = 6/11 P.
6P/11 - 8 = 4/9(P - 8)
10P = 792 - 352 => P = 44.
The capacity of the pot is 44 liters.
From the given data,
let the initial quantity of the mixture = 5x
Then,
Then the initial quantity of the mixture = 5x = 5 x 16 = 80 lit.
It means (since 343 + 169 = 512)
Thus,
Thus the initial amount of wine was 120 liters.
Let the tin contain 5x litres of liquids
=> 5(4x - 36) = 2(x + 36)
=> 20x - 180 = 2x + 72
=> x = 14 litres
Hence, the initial quantity of mixture = 70l
Quantity of liquid B
=
= 50 litres.
Water in 60 gm mixture=60 x 75/100 = 45 gm. and Milk = 15 gm.
After adding 15 gm. of water in mixture, total water = 45 + 15 = 60 gm and weight of a mixture = 60 + 15 = 75 gm.
So % of water = 100 x 60/75 = 80%.
Given mixture = 48 lit
Milk in it = 48 x 5/8 = 30 lit
=> Water in it = 48 - 30 = 18 lit
Let 'L' lit of water is added to make the ratio as 3:5
=> 30/(18+L) = 3/5
=> 150 = 54 + 3L
=> L = 32 lit.
Let the initial amount of honey in the jar was K, then
or
Therefore, K = 1250
Hence initially the honey in the jar= 1.25 kg
Customer ratio of Milk and Water is given by
Milk :: Water
6.4 0
=> Milk : Water = 110 : 11 = 10 : 1
Therefore, the proportionate of Water to Milk for Customer is 1 : 10
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