Question

Beaker A contains 1 liter which is 30 percent oil and the rest is vinegar, thoroughly mixed up. Beaker B contains 2 liters which is 40 percent oil and the rest vinegar, completely mixed up. Half of the contents of B are poured into A, then completely mixed up. How much oil should now be added to A to produce a mixture which is 60 percent oil?

Answer 1

Answer:

1.25 liters of oil

Step-by-step explanation:

Volume in Beaker A = 1 L

Volume of Oil in Beaker A = 1*0.3 = 0.3 L

Volume of Vinegar in Beaker A = 1*0.7 = 0.7 L

Volume in Beaker B = 2 L

Volume of Oil in Beaker B = 2*0.4 = 0.8 L

Volume of Vinegar in Beaker B = 1*0.6 = 1.2 L

If half of the contents of B are poured into A and assuming a homogeneous mixture, the new volumes of oil (Voa) and vinegar (Vva) in beaker A are:

$V_{oa} = 0.3+\frac{0.8}{2} \\V_{oa} = 0.7 \\V_{va} = 0.7+\frac{1.2}{2} \\V_{va} = 1.3$

The amount of oil needed to be added to beaker A in order to produce a mixture which is 60 percent oil (Vomix) is given by:

$0.6*V_{total} = V_{oa} +V_{omix}\\0.6*(V_{va}+V_{oa} +V_{omix}) = V_{oa} +V_{omix}\\0.6*(1.3+0.7+V_{omix})=0.7+V_{omix}\\V_{omix}=\frac{0.5}{0.4} \\V_{omix}=1.25 \ L$

1.25 liters of oil are needed.