艾易艾斯(aeas)数学逻辑讲解-4

There are two acid solutions S1 and S2, such that S1 contains 50% acid and S2 contains 70% acid. What quantities of S1 and S2 should be mixed to get 10 liters of a solution that contains 54% acid.

有S1 和 S2 两种酸溶液，S1中酸的浓度为50%，S2中酸的浓度为70%. 需要多少S1 和 S2 溶液混合才能得到10升浓度为54%的溶液？

Answer:

8 liters of S1 and 2 liters of S2

Step 1

Let the quantities of solutions mixed together be x and y. The first equation, according to the question, is x + y = 10 as their quantities sum up to a total of 10 litres.

Step 2

The quantity of acid in one litre of the first solution will be equal to 50/100 *1 = 0.5liters

Step 3

Similarly, the quantity of acid in one litre of the second solution will be equal to 70/100*1=0.7liters

Step 4

The quantity of acid in one litre of the final mixture is 54/100*1=0.54liters

Step 5

Since, the first and second solutions are mixed in their respective quantities to 10 litres of the final equation, we can say that x × 0.5 + y × 0.7 = 10 × 0.54

or, 0.5x + 0.7y = 5.4

Step 6

We now have two linear equations in two variables. Let us solve them:

x + y = 10

or, x = 10 – y    ——(1)

0.5x + 0.7y = 5.4 ——(2)

Substitute the value of x in equation (2)

0.5(10 – y) + 0.7y = 5.4

or, 5 – 0.5y + 0.7y = 5.4

or, -0.2y = -0.4

or, y = 2

Substitute the value of y in equation (1)

x = 10 – 2 = 8

Step 7

Hence, 8 litres of solution S1 and 2 litres of solution S2 are mixed together to form the given solution.