Concentration 4419

We added 300 g of water to 400 g of salt solution, reducing the solution's concentration by 5%. How much water did the original solution contain, and what was its concentration?

Correct answer:

m = 1406.44 g
p₁ = 28.4406 %

Step-by-step explanation:

m = 400 + w p_{1} = 400 / m \cdot 100 p_{2} = 400 / (m + 300) \cdot 100 p_{2} = p_{1} - 5 400 / (m + 300) \cdot 100 = 400 / m \cdot 100 - 5 400 \cdot m \cdot 100 = 400 \cdot 100 \cdot (m + 300) - 5 \cdot (m + 300) \cdot m 400 \cdot m \cdot 100 = 400 \cdot 100 \cdot (m + 300) - 5 \cdot (m + 300) \cdot m 5 m^{2} + 1500 m - 12000000 = 0 5 . . . prime number 1500 = 2^{2} \cdot 3 \cdot 5^{3} 12000000 = 2^{8} \cdot 3 \cdot 5^{6} GCD (5, 1500, 12000000) = 5 m^{2} + 300 m - 2400000 = 0 a = 1; b = 300; c = - 2400000 D = b^{2} - 4 a c = 30 0^{2} - 4 \cdot 1 \cdot (- 2400000) = 9690000 D > 0 m_{1, 2} = \frac{- b \pm D}{2 a} = \frac{- 3 0 0 \pm 9 6 9 0 0 0 0}{2} = \frac{- 3 0 0 \pm 1 0 0 9 6 9}{2} m_{1, 2} = - 150 \pm 1556.438242 m_{1} = 1406.438241627 m_{2} = - 1706.438241627 m = m_{1} = 1406.4382 = 1406.44 g

Our quadratic equation calculator calculates it.

p_{1} = \frac{4 0 0}{m} \cdot 100 = \frac{4 0 0}{1 4 0 6 . 4 3 8 2} \cdot 100 ≐ 28.4406 % Verifying Solution: p_{2} = \frac{4 0 0}{m + 3 0 0} \cdot 100 = \frac{4 0 0}{1 4 0 6 . 4 3 8 2 + 3 0 0} \cdot 100 ≐ 23.4406 % Δ = p_{1} - p_{2} = 28.4406 - 23.4406 = 5 %

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Showing 1 comment:

Emcl

I think this answer is wrong, since original solution has 400g, how can it has 1406g water. we can set original concentration is x,then 400x/(300+400)=x-5%, then x is 11.67%.

2 years ago Like

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