Math, asked by golu2516, 11 months ago

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". ​

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Answers

Answered by knjroopa
3

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

Answered by amitnrw
2

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

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