A 50-lb watermelon is 99%. What does it weigh after some
- Thread starter lwarner
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- Apr 12, 2005
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A very creative variation to the standard mixture problem.
Just find something to equate. You can use "Water" or "Not Water". Either way.
Water:
x is the new weight
(50-x) is the weight of the water evaporated
50 lbs*99%water - (50-x) lbs*100%water = x lbs*98%water
50*0.99 - (50-x) = x*0.98
50*0.99 - 50 + x = x*0.98
-50*0.01 = x*0.98 - x
-50*0.01 = -x*0.02
x = 50*0.01/0.02 = 25
25 pounds!! Wow. I didn't expect that.
Let's do it by equating NONwater and see if the result is the same.
99% water ==> 1% NONwater
50 lbs * 1%NONwater - (50-x) lbs*0%NONwater = x lbs * 2%NONwater
50*0.01 = x*0.02
x = 50*0.01/0.02 = 25
That was even easier!!
Note: Generally, the one with the zero (0) percent is the easier of the two.
Just find something to equate. You can use "Water" or "Not Water". Either way.
Water:
x is the new weight
(50-x) is the weight of the water evaporated
50 lbs*99%water - (50-x) lbs*100%water = x lbs*98%water
50*0.99 - (50-x) = x*0.98
50*0.99 - 50 + x = x*0.98
-50*0.01 = x*0.98 - x
-50*0.01 = -x*0.02
x = 50*0.01/0.02 = 25
25 pounds!! Wow. I didn't expect that.
Let's do it by equating NONwater and see if the result is the same.
99% water ==> 1% NONwater
50 lbs * 1%NONwater - (50-x) lbs*0%NONwater = x lbs * 2%NONwater
50*0.01 = x*0.02
x = 50*0.01/0.02 = 25
That was even easier!!
Note: Generally, the one with the zero (0) percent is the easier of the two.