Question

Question :
A milkman has two type of milk in the 1 st container the % of milk is 80% and in the 2 nd container the percentage of milk is 60% .If the mixes 28 liters of the milk of first container to the 32 liter of milk of the 2 nd container,then the % of milk in the mixture is:

Answer 1

Answer:

Since, 29 liter of milk solution having 80 % of milk is mixed with the 32 liters of milk having 60 % of milk,

Thus, the total quantity of mixture = 29 + 32 = 61 liters,

Also, The milk in resultant solution = The quantity of milk in 80 % of milk solution + The quantity of milk in 60 % of milk solution

= 80 % of 29 + 60% of 32

=\frac{80\times 29}{100}+\frac{60\times 32}{100}=

100

80×29

+

100

60×32

=\frac{2320}{100}+\frac{1920}{100}=

100

2320

+

100

1920

=\frac{4240}{100}=42.40\text{ liters}=

100

4240

=42.40 liters

Thus, the percentage of milk in resultant milk solution = \frac{42.40}{61}\times 100

61

42.40

×100

=\frac{4240}{61}=69.5081967\approx 69.51\%=

61

4240

=69.5081967≈69.51%