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Involves Acid Solutions

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Hi Melissa,
 
We have a system of equations that needs to be solved. One acid solution has 4% acid, and the other has 11% acid. Say we have x cm3 of the first solution (4%) and y cm3 of the second solution (11%). When mixed, we want to have a total of 50 cm3, so
 
x + y = 50
 
The resultant solution must have 6.94% acid, and the solution is 50 cm3, so
 
0.04x + 0.11y = 50*0.0694
 
or just
 
0.04x + 0.11y = 3.47
 
Since x + y = 50, y = 50 - x. Substitute this into the above to obtain
 
0.04x + 0.11(50 - x) = 3.47
 
Multiplying out yields
 
0.04x + 5.5 - 0.11x = 3.47
 
Combine like terms, and move the 5.5 over to obtain
 
-.07x = -2.03
 
Dividing by -.07 yields
 
x = 29
 
Since x = 29, y = 50 - 29 = 21