Question 1163883: The total mass of a mixture of two liquids is 2.4 kg and the total volume is 1000 cm³ if 1cm³ of one of the liquids weighs 2 grams and 1 cm³ of the other liquid weighs 3 grams what volume of each liquid is present and what mass of each liquid is present.
Found 2 solutions by htmentor, ikleyn:
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! The total mass is 2400 g, and the density of the mixture is
D = 2400 g/1000 cm^3 = 2.4 g/cc
The mass of liquid 1 is m1 = d1*V1 where d1 and V1 are the density and volume of #1
The mass of liquid 2 is m2 = d2*V2
Given: d1 = 2, d2 = 3
We know that: m1 + m2 = 2400 and V1 + V2 = 1000
Thus m1 = 2V1
2400 - m1 = 3(1000 - V1) -> 2400 - 2V1 = 3000 - 3V1 -> V1 = 600
Hence V2 = 400, m1 = 2*600 = 1200, m2 = 3*400 = 1200
Ans: V1 = 600 cm^3, V2 = 400 cm^3, m1 = m2 = 1200 g
Answer by ikleyn(52781)  (Show Source):
You can put this solution on YOUR website! .
Let x be the volume of the first liquid and y be the volume of the second liquid (in cm^3).
Then the total volume equation is x + y = 1000 cm^3, (1)
while the total mass equation is 2x + 3y = 2400 grams. (2)
From the first equation x = 1000 - y.
Substitute it into the second equation. You will get
2*(1000-y) + 3y = 2400
Simplify and find y.
2000 - 2y + 3y = 2400
y = 400.
Then from equation (1), x = 1000-400 = 600.
thus the volumes are 600 cm^3 and 400 cm^3 for the first and the second liquids, respectively.
The masses are 2*600 = 1200 grams and 3*400 = 1200 grams (the same masses for both liquids).
Solved, answered and explained.
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