SOLUTION: A chemist needs 10 liters of a 50% salt solution. All she has available is a 20% salt solution and a 70% salt solution. How much of each of the two solutions should she mix to ob
Algebra ->
Customizable Word Problem Solvers
-> Mixtures
-> SOLUTION: A chemist needs 10 liters of a 50% salt solution. All she has available is a 20% salt solution and a 70% salt solution. How much of each of the two solutions should she mix to ob
Log On
Question 21619: A chemist needs 10 liters of a 50% salt solution. All she has available is a 20% salt solution and a 70% salt solution. How much of each of the two solutions should she mix to obtain her desired solution? Answer by venugopalramana(3286) (Show Source):
You can put this solution on YOUR website! A chemist needs 10 liters of a 50% salt solution. All she has available is a 20% salt solution and a 70% salt solution. How much of each of the two solutions should she mix to obtain her desired solution?
LET X LITRES OF 20% SOLUTION BE REQUIRED...SINCE TOTAL IS 10 LITRES ,WE HAVE TO ADD 10-X LITRES OF 70% SOLUTION..SO TAKING A SALT BALANCE
SALT IN X LITRES OF 20 % SOLUTION......=X*20/100=0.2X
SALT IN 10-X LITRES OF 10 % SOLUTION.. =(10-X)*70/100=7-0.7X
SALT IN 10 LITRES OF FINAL 50 % SOLUTION =10*50/100=5
HENCE
0.2X+7-0.7X=5
0.7X-0.2X=7-5=2
0.5X=2
X=2/0.5=20/5=4 LITRES OF 20 % SWOLUTION AND 10-4=6 LITRES OF 70% ARE NEEDED TO BE MIXED TO GET 10 LITRES OF 50% SOLUTION