A chemist needs to mix a 75% saltwater solution with a 50% saltwater solution to obtain 10 gallons of a 60% mixture. How much of each soltion is required?
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We are mixing some stronger liquid (75%) with some
weaker liquid (50%) to get a medium strength
liquid (60%).
Make this chart:
Amount of Percentage expressed Amt. of
Solution liquid as a decimal saltwater
--------------------------------------------------------------
Stronger liquid
Weaker liquid
Medium liquid
Fill in the percentages:
Amount of Percentage expressed Amt. of
Solution liquid as a decimal saltwater
--------------------------------------------------------------
Stronger liquid .75
Weaker liquid .50
Medium liquid .60
There are to be 10 gallons of medium strength liquid.
Let x = the number of gallons of stronger liquid.
Then 10-x = the number of gallons of weaker liquid.
Fill those amounts in
Amount of Percentage expressed Amt. of
Solution liquid as a decimal saltwater
--------------------------------------------------------------
Stronger liquid x .75
Weaker liquid 10-x .50
Medium liquid 10 .60
Fill in the amount of saltwater by multiplying the amount of
liquid by the percentage:
Amount of Percentage expressed Amt. of
Solution liquid as a decimal saltwater
--------------------------------------------------------------
Stronger liquid x .75 .75
Weaker liquid 10-x .50 .50(10-x)
Medium liquid 10 .60 .60(10)
Get the equation from
Amount of saltwater in stronger + Amount of saltwater in weaker = Amount of saltwater in final medium strength
.75x + .50(10-x) = .60(10)
Solve that and get x=4 gallons of stronger strength liquid
Then 10-x = 10-4 = 6 gallons of weaker strength liquid.
Note as a partial check: Since the final medium strength
liquid (60%) is more like the weaker strength liquid (50%)
than it is like the stronger strength liquid (75%), we
would expect there to be more of the weaker than the stronger.
And that is what we ended up with, since 6 gallons is more
than 4 gallons.
Edwin
