Help with mixture problem?

[匿名]

2020-08-26 5:23 pm

A 10% alcohol solution and a 40% alcohol solution are mixed to form 80ml of a 35% solution. How much of each mixture was needed?

Please show how to solve and give answer.

回答 (2)

[匿名]

2020-08-26 5:42 pm

✔ 最佳答案

Method 1: Equation with one unkown

Let v ml be the volume of 10% alcohol solution needed.
Then, the volume of 40% of alcohol solution needed = (80 - v) ml

10%v + 40%(80 - v) = 35%(80)
10v + 3200 - 40v = 2800
30v = 400
v = 13 and 1/3
80 - v = 66 and 2/3

The answers:
Volume of 10% alcohol solution needed = (13 and 1/3) ml
Volume of 40% alcohol solution needed = (66 and 2/3) ml

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Method 2: Equations with two unknowns.

Let x ml be the volume of 10% alcohol solution needed, and y ml be the volume of 40% alcohol solution needed.
x + y = 80 …… [1]
10%x + 40%y = 35%(80) …… [2]

From [2]:
10x + 40y = 2800
x + 4y = 280 …… [3]

[3] - [1]:
3y = 200
y = 66 and 2/3

Plug y = 66 and 2/3 into [1]:
x + (66 and 2/3) = 80
x = 13 and 1/3

The answers:
Volume of 10% alcohol solution needed = (13 and 1/3) ml
Volume of 40% alcohol solution needed = (66 and 2/3) ml

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Wayne DeguMan

2020-08-26 5:35 pm

0.1a + 0.4b = 0.35(a + b)

a + b = 80

so, 0.1a + 0.4(80 - a) = 0.35(80)

i.e. 32 - 0.3a = 28

Hence, 0.3a = 4

=> a = 40/3 so, b = 200/3

so, 13 1/3 ml of 10% solution is added to 66 2/3 ml of 40% solution to get an 80 ml solution at 35%

:)>

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