How many milligrams of pure water must be added to 6 mg of a 40% solution to obtain a 25% solution? Round to the nearest tenth.
2.8 mg
3.6 mg
3.1 mg
4.3 mg

Question

How many milligrams of pure water must be added to 6 mg of a 40% solution to obtain a 25% solution?  Round to the nearest tenth.  
    2.8 mg  
   3.6 mg  
   3.1 mg  
   4.3 mg

anonymous · Answer

3.6 mg

saifoo.khan · Answer

6x.4=2.4
.25p=2.4
p=2.4/.25=9.6
9.6-6=3.6 ms pure water necessary to be added..

amistre64 · Answer

eliminate some more :)


w     + 6      = t        ;*-.25
w(0) + 6(.4) = t(.25)



-w(.25)- 6(.25) = -t(.25) 
  w(0)   + 6(.4)  =   t(.25)
----------------------
-w(.25) +6(.15) = 0  ; solve for w

                   w = 90/25 = 18/5 = 3.6