Two liquid containing (X and Y) mixture mixed in the ratio 2:3.X contains 40% sugar in water while Y contains alcohol in water. After adding 20 gram sugar in the final mixture ratio of sugar, alcohol and water becomes 8:5:16.Given that all ratio is calculated on weight. Find the weight alcohol in mixture "Y".
Answers
Answer:
25 grams
Step-by-step explanation:
Given Two liquid containing (X and Y) mixture mixed in the ratio 2:3.X contains 40% sugar in water while Y contains alcohol in water. After adding 20 gram sugar in the final mixture ratio of sugar, alcohol and water becomes 8:5:16.Given that all ratio is calculated on weight. Find the weight alcohol in mixture "Y"
Let the weight of mixture in x = 100 x grams
So weight of sugar in X = 40 x grams
weight of water in X = 60 x grams
Now weight of Y = 150 x grams
So let weight of alcohol in = p grams
So weight of water in = (150 x - p) grams
Now from the question we get
40 x + 20 : p : 150 x - p + 60 x = 8 : 5 : 16
p/210 x - p = 5/16
16 p = 1050 x - 5 p
21 p = 1050 x
p = 50 x
Now 40 x + 20 /p = 8/5
400 x = 200 x + 100
200 x = 100
x = 1/2
Now we have Y = p = 50 x 1/2 = 25 grams
So weight of alcohol in mixture Y = 25 grams
X contains 0.4X sugar
X contains 0.6X water
Y contains ay alcohol
y contains (1-a)y water
X : Y = 2:3
X /Y = 2/3
Y = 1.5X
putting these
aY = 1.5aX alcohol
(1-a)Y =(1-a)(1.5X) water
total water = 0.6X + (1-a)(1.5X)
= 2.1X - 1.5aX
total sugar = 0.4X + 20
Alcohol =1.5aX
sugar, alcohol and water becomes 8:5:16
water = 2 sugar
2.1X - 1.5aX = 0.8X + 40
=> 1.3X -1.5aX = 40
0.4X+20 / 1.5ax = 8/5
=> 1.5aX = (2X + 100)/8
1.3X - (2X + 100)/8 = 40
=> 10.4X -2X -100 = 320
=> 8.4X = 420
=>X = 50
Y = 1.5X = 75
1.3X -1.5aX = 40
65 - 75a = 40
=> 75a = 25
a = 1/3
Y contains ay alcohol = 75/3 = 25 gm
option a