SOLUTION: how many mL of each of 45% solution and a 15% solution should be mixed in order to obtain 25 mL of a 30% solution?

Question 324683: how many mL of each of 45% solution and a 15% solution should be mixed in order to obtain 25 mL of a 30% solution?
Found 2 solutions by stanbon, modi.kk91:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
how many mL of each of 45% solution and a 15% solution should be mixed in order to obtain 25 mL of a 30% solution?
----------
Equation:
active + active = active
0.45x + 0.15(25-x) = 0.30(25)
---
Multiply thru by 100 to get:
45x + 15*25 - 15x = 30*25
30x = 15*25
x = 12.5 mL (amt. of 45% solution needed in the mix)
---
25-x = 12.5 mL (amt of 15% solution needed in the mix)
================
Cheers,
Stan H.
Answer by modi.kk91(20) (Show Source): You can put this solution on YOUR website!
solved by allegation rule

15% 45%

30%

45-30=15 and 30-15=15
ratio=1:1=====1x+1x=2x

ans= 1x=12.5
1x=12.5
2x=25
x=12.5
RELATED QUESTIONS
How many mL of each 45% solution and a 15% solution should be mixed in order to obtain... (answered by richwmiller)
How many ml each of a 15% acetic acid solution and a 30% acetic acid solution must be... (answered by stanbon)
a 20% HCl solution and a 70% HCl solution are to be mixed to obtain 900 ml of a 30 % HCl... (answered by Alan3354)
How many ML of a 25% saline solution should be added to 30 mL of a 5% saline solution to... (answered by josgarithmetic)
In a chemistry lab the concentration of one solution is 10% HCU and that of a second... (answered by robertb)
In a chemistry labratory the concentration of one solution is 10% HCl and that of a... (answered by SadieKhan)
How many mL of 10% HCL solution should be combined with 60% HCL solution in order to get... (answered by nerdybill)
How many mL's of a 10% alcohol solution and how many mL's of a 20% solution must be mixed (answered by ankor@dixie-net.com,greenestamps)
How many ml of a 20% solution should be added to 50 ml of a 40% solution to obtain a 25%... (answered by Alan3354,Boreal)