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3 years ago

You have a 10% sugar solution and a 30% sugar solution, but you need a 25% sugar solution to fill your

Mathematics

1 answer:

Juli2301 [7.4K]3 years ago

4 0

<h3>Answers:</h3>

$x+y = \boxed{ \ 8 \ }$

$\boxed{ \ 0.10 \ }x+\boxed{ \ 0.30 \ }y = \boxed{ \ 2\ }$

$\boxed{ \ 2 }$ cups of the 10% sugar solution

$\boxed{ \ 6 }$ cups of the 30% sugar solution

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Work Shown:

x = number of cups of the 10% solution

y = number of cups of the 30% solution

The two mixes add together to get 8 cups since this is the total amount we want. Therefore, the first equation is simply x+y = 8

Solve for y to get y = 8-x. We'll use this equation later.

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The second equation is a bit more tricky.

0.10x = amount of pure sugar from just the 10% batch

0.30y = amount of pure sugar from just the 30% batch

0.10x+0.30y = total amount of pure sugar from both batches

25% of 8 = 0.25*8 = 2 = total amount of pure sugar required

0.10x+0.30y = 2 is the second equation

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Apply substitution to solve for x and y

0.10x+0.30y = 2 ..... start with the second equation

0.10x+0.30(8-x) = 2 ....... plug in y = 8-x

0.10x+2.4-0.30x = 2 ..... distribute

-0.20x + 2.4 = 2

-0.20x = 2-2.4 ....... subtract 2.4 from both sides

-0.20x = -0.4

x = -0.4/(-0.20) ..... divide both sides by -0.20

x = 2

Now that we know the value of x, we can find y.

y = 8-x

y = 8-2 ... plug in x = 2

y = 6

So we need 2 cups of the 10% solution and 6 cups of the 30% solution.

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As a check,

x+y = 2+6 = 8 works out

0.10*x+0.30*y = 0.10*2+0.30*6 = 0.2+1.8 = 2 works out as well

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A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flo

djverab [1.8K]

Answer:

a) $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) $31690.7 g/L$

Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is $\frac{dy}{dt}=R_{i}-R_{o}$ , where $R_{i}$ is the rate of salt entering and $R_{o}$ is the rate of salt going outside.

Then we have, $R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}$ , and

$R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}$

So we obtain. $\frac{dy}{dt}=4000-\frac{2y}{25}$ , then

$\frac{dy}{dt}+\frac{2y}{25}=4000$ , and using the integrating factor $e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}$ , therefore $(\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}$ , we get $\frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}$ , after integrating both sides $y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C$ , therefore $y(t)= 50000 +Ce^{\frac{-2t}{25}}$ , to find $C$ we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions $y(0)=10$ , so $10= 50000+Ce^{\frac{-0*2}{25}}$

$10=50000+C\\C=10-50000=-49990$

Finally we can write an expression for the amount of salt in the tank at any time t, it is $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) The tank will overflow due Rin>Rout, at a rate of $80 L/min-40L/min=40L/min$ , due we have 500 L to overflow $\frac{500L}{40L/min} =\frac{25}{2} min=t$ , so we can evualuate the expression of a) $y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7$ , is the salt concentration when the tank overflows