Operations & Word Problems

Depth · one concept from first principles · step 2 of 4

Track grams of salt through two pourings

Beaker A holds 300300 g of salt water that is a%a\% salt, and beaker B holds 100100 g of salt water that is b%b\% salt. You pour 100100 g of the salt water from beaker A into beaker B and stir well. Then you pour 5050 g of the salt water from beaker B into beaker A. Write the concentration of the salt water in beaker A in terms of aa and bb.

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Your answer
How to solve
Strategy Identify Subproblems — Percentages can't simply be added, but grams of salt can. So follow the grams of salt and the grams of water through each pour, and only turn them back into a percentage at the end.
1STEP 1

Salt in the first pour

100 g of A's water at a% carries a g of salt. B already has b g of salt in its 100 g.

a/100×100=a, b/100×100=b
2STEP 2

Beaker B after mixing

B now has 200 g of water containing a+b g of salt.

(a+b)/200×100=(a+b)/2 (%)
3STEP 3

Salt in the pour back

50 g of B's mixture carries a quarter of B's salt, because 50 g is a quarter of 200 g.

(a+b)/2×50/100=(a+b)/4
4STEP 4

Beaker A's totals

A kept 200 g at a%, which holds 2a g of salt, and receives 50 g more.

salt=2a+(a+b)/4=(9a+b)/4, water=200+50=250
5STEP 5

Back to a percentage

Divide salt by total weight and multiply by 100.

(9a+b)/4÷250×100=(9a+b)/10
Answer
(9a+b)/10%
If a=b, nothing should change, and indeed (9a+a)/10=a. The answer is a weighted average of a and b with most weight on a, which fits A being mostly its own water.
Takeaway

Percents don't add, but grams of salt do: follow the salt.

  • Salt in the first pour
  • Beaker B after mixing
  • Salt in the pour back
  • Beaker A's totals
  • Back to a percentage
Where next?
Another one like thissuggested

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▶ Practice — 12 problems

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