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# Algebra

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Ruth has a beaker containing a solution of 800 mL of acid and 200 mL of water. She thinks the solution is a little strong, so she drains 150 mL from the beaker, adds 150 mL of water, and stirs the solution. Ruth thinks the solution is still too strong, so again she drains 150 mL from the beaker, and adds 150 mL of water. How many mL of water are now in the beaker?

Mar 8, 2024

#1
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From the start, we know the porportion of the amount of water in the beaker, to the total amount of solution in the beaker.

The amount of acid drained away to the amount of water drained away is in porportion to their original amounts.

We know $$\frac{150}{200+800}$$of the total amount is drained away, so $$200*\frac{150}{200+800}=30$$.

So there is $$200-30=170$$mL left.

Adding 150mL water gives 320mL water.

We know that the total amount of solution didn't change, so there is $$1000-320=680$$mL acid.

Similarly, now $$320*\frac{150}{320+680}=48$$ mL drained away.

Subtracting this, and adding the 150 mL of water, we have 422mL of water now.

Mar 8, 2024

#1
+394
+2

From the start, we know the porportion of the amount of water in the beaker, to the total amount of solution in the beaker.

The amount of acid drained away to the amount of water drained away is in porportion to their original amounts.

We know $$\frac{150}{200+800}$$of the total amount is drained away, so $$200*\frac{150}{200+800}=30$$.

So there is $$200-30=170$$mL left.

Adding 150mL water gives 320mL water.

We know that the total amount of solution didn't change, so there is $$1000-320=680$$mL acid.

Similarly, now $$320*\frac{150}{320+680}=48$$ mL drained away.

Subtracting this, and adding the 150 mL of water, we have 422mL of water now.

hairyberry Mar 8, 2024