You need to use the idea that the concentration of a solute = amount of solute / volume of solution.
In symbols:
#c=n/v" "color(red)((1))#
The total amount of solute before and after the dilution remains the same.
Rearranging #color(red)((1))# gives:
#n=cxxv" "color(red)((2))#
Now we can set up 2 simultaneous equations from the information given. We have 2 equations and 2 unknowns so we can solve it.
From #color(red)((2))# we can write:
#(60xxV_(60))+(15xxV_(15))=650xx38#
#:.60V_(60)+V_(15)=24,700" "color(red)((3))#
Where #V_(60)# and #V_(15)# refer to the volumes of the #60%# and #15%# solutions respectively.
I am using the #%# as the concentration.
Since we are told the total volume we can write:
#V_(60)+V_(15)=650" "color(red)((4))#
So #color(red)((3))# and #color(red)((4))# are our simultaneous equations.
Rearranging #color(red)((4))rArr#
#V_(60)=650-V_(15)#
We can now substitute this value for #V_(60)# back into #color(red)((3))rArr#
#60(650-V_(15))+15V_(15)=24,700#
#:.39,000-60V_(15)+15V_(15)=24,700#
#:.60V_(15)-15V_(15)=39,000-24,7000#
#:.45V_(15)=14,300#
#:.V_(15)=(14,300)/45=color(red)(317.8"ml")#
Substituting this value back into #color(red)((4))rArr#
#V_(60)+317.8=650#
#:.V_(60)=650-317.8=color(red)(332.2"ml")#
