SOLUTION: A solution of 75% pesticide is to be mixed with a solution of 35% pesticide to form 440 liters of a 43% solution. How many liters of the 75% solution must be used?

Click here to see ALL problems on Mixture Word Problems

Question 585605: A solution of 75% pesticide is to be mixed with a solution of 35% pesticide to form 440 liters of a 43% solution. How many liters of the 75% solution must be used?
Found 3 solutions by mananth, stanbon, josmiceli:
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website!
Acid percent ---------------- quantity
Pesticide type I 75.00% ---------------- x liters
Pesticide type II} 35.00% ------ 440 - x liters
Mixture 43.00% ---------------- 440

75.00%x+35.00%(440-x)=43.00% * 440
75x+35( 440-x)= 18920
75x+15400-35x=18920
75x-35x =18920-15400
40x= 3520
/ 40
x=88 liters 75.00% Pesticide type I
352 liters 35.00% Pesticide type II}

m.ananth@hotmail.ca

Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website!
A solution of 75% pesticide is to be mixed with a solution of 35% pesticide to form 440 liters of a 43% solution. How many liters of the 75% solution must be used?
----------------------
Equation:
active + active = active
0.75x + 0.35(440-x) = 0.43*440
-----
75x + 35*440 - 35x = 43*440
40x = 8*440
x = (1/5)*440
x = 88 liters (amt of 75% solution needed)
----
440-x = 352 liters (amt. of 35% solution needed)
================================================
Cheers,
Stan H.

Answer by josmiceli(19441)   (Show Source):
You can put this solution on YOUR website!
Let = liters of 75% pesticide needed
Let = liters of 35% pesticide needed
given:
(1)
(2)
-----------------------------
(2)
(2)
(2)
(2)
Multiply both sides of (1) by
and subtract (1) from (2)
(2)
(1)

and, since
(1)
(1)
(1)
(1)
88 liters of 75% pesticide are needed
352 liters of 35% pesticide are needed
check:
(2)
(2)
(2)
(2)
(2)
OK