SOLUTION: How many liters of each of a 15% acid solution and a 75% acid solution must be used to to produce 90 liters of a 25% acid solution? ( round to two decimal places if necessary.)
Algebra.Com
Question 693921: How many liters of each of a 15% acid solution and a 75% acid solution must be used to to produce 90 liters of a 25% acid solution? ( round to two decimal places if necessary.)
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
How many liters of each of a 15% acid solution and a 75% acid solution must be used to to produce 90 liters of a 25% acid solution? ( round to two decimal places if necessary.)
----------------------
Equation:
acid + acid = acid
0.15x + 0.75(90-x_ = 0.25*90
-----------------------------
15x + 75*90 - 75x = 25*90
-80x = -50*90
x = (5/8)90
x = 56.25 liters (amt. 15% solution needed)
-----
90-x + 33.75 liters (amt. of 75% solution needed)
=============================================
Cheers,
Stan H.
==============
RELATED QUESTIONS
How many liters each of a
25% acid solution and a
75% acid solution must be used to... (answered by josgarithmetic)
How many liters each of a 15% acid solution and a 60% acid solution must be used to... (answered by josgarithmetic)
How many liters of each of a 10% acid solution and a 25% acid solution must be used to... (answered by stanbon)
how many liters each of a 15% acid solution and a 25% acid solution must be used to... (answered by josgarithmetic,amalm06,Alan3354)
How many liters each of a 25% acid solution and a 65% acid solution must be used to... (answered by josgarithmetic,ikleyn,greenestamps)
How many liters each of a
25%
acid solution and a
85%
acid solution must be used to (answered by josgarithmetic,rothauserc)
How many liters each of a 15% acid solution and a 65% acid solution must be used to... (answered by ikleyn)
How many liters each of a 25% acid solution and a 70% acid solution must be used to... (answered by ikleyn)
How many liters each of a 40% acid solution and a 65% acid solution must be used to... (answered by checkley79)