Answer:
0.98g+ 0.02m= 7.1224 gallons standard mixture
0.96g+ 0.04m= 6.25 gallons break oil
3.92 gallons of gasoline is required for 4 gallons the standard mixture
Step-by-step explanation:
If we have to use 6 gallons of gasoline this means that 98%= 6 gallons and 2% would be 6/98*2
Gasoline: Motor oil
6 : x
98 : 2
Using the product rule
x= 6*2/98= 0.1224 gallons of motor oil is required.
Let the gasoline be represented by g and motor oil by m then for the standard oil the equation would be
0.98g+ 0.02m= 7.1224 gallons standard mixture
For break in oil
Gasoline: Motor oil
6 : x
96 : 4
Using the product rule
x= 6*4/96= 0.25 gallons of motor oil is required.
Let the gasoline be represented by g and motor oil by m then for the break oil the equation would be
0.96g+ 0.04m= 6.25 gallons break oil
Part b. For 4 gallons of the "break in" mixture
96% of the 4 gallons of the "break in" mixture is gasoline = 3.84 gallons of gasoline
4% of 4 = 0.16 gallons of motor oil
Now the standard mixture 4 gallons would contain 98 % of gasoline and 2 % of motor oil
98% of 4 gallons= 3.92 gallons of gasoline
2% of 4 gallons= 0.08 gallons of motor oil.
and
are the two parallel bases