5th grade word problem of the week

Question

A generator needs a fixed amount of fuel to keep it running 2 hours a day. After running for 6 days, the generator fuel tank was 2/3 full. After 11 more days, 3 1/2 liters of fuel were left in the tank. If the tank was originally full, how much fuel can the tank hold?

Answer 1

Let F0 = liters of fuel in full tank (L)

      F = liters of fuel remaining in tank (L)                        Note: F0 - F = liters of fuel used by generator

      R = fuel volumetric rate needs to run generator (L/hr)

      t = time generator runs (hr)

 

Note: 2 hrs/day generator run time

 

  F0 - F = R*t   

 

1) F0 - F = R*(6 days)*(2 hr/day)

     

    F = (2/3)*F0       (substitute into equation)

 

    F0 -  (2/3)*F0 = 12*R

 

 (1/3)* F0 = 12*R 

 

  F0 = 12*3*R = 36*R

 

2) F0 - 3 1/2 = R*(6 +11)*2 = 34*R      (substitute F0 = 36*R and solve for R)

 

    36*R  - 3 1/2  = 34*R 

 

     36*R  =  34*R + 3 1/2

 

     2*R = 3 1/2 = 7/2

 

    R = 7/4 L/hr

 

plug R value into equation 1) and solve for F0:

 

F0 = 36*R = 36*(7/4) = 9*7 = 63 L