Math Algebra Problem: Empty Fuel Tank in 4.8 Hours

  • Thread starter Thread starter Max0007
  • Start date Start date
  • Tags Tags
    Algebra
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
13 replies · 2K views
Max0007
Messages
66
Reaction score
1

Homework Statement


Two pumps of different sizes can empty an entire fuel tank in 4.8 hours. Used alone, the larger pump would empty the tank 4 hours less than would the smaller pump. If using only the smallest pump How long will it take to empty the tank?

Homework Equations


X = small pump
X-4hours = bigger pump

Y = 4.8 hours

This is what I understand so far. I am not sure what to do next.

The Attempt at a Solution


4.8 = x - 4

would that mean small pump needs 8.8 hours? am I doing this right?
 
Last edited:
Physics news on Phys.org
What is Y and where did 4.8 come from?

You should have something like ## 1/X+1/Y=1/4##
And
##Y=X-4##
 
Max0007 said:

Homework Statement


Two pumps of different sizes can empty an entire fuel tank in 4 hours. Used alone, the larger pump would empty the tank 4 hours less than would the smaller pump. If using only the smallest pump How long will it take to empty the tank?

Homework Equations


X = small pump
X-4hours = bigger pump

Y = 4.8 hours
RUber asked about Y and where the 4.8 hours came from, a reasonable question, since the number 4.8 doesn't appear anywhere in the problem description above.

A variable should represent an unknown quantity whose value you are trying to find.
I can't think of any good reason why you should give this value the name Y.
Max0007 said:
This is what I understand so far. I am not sure what to do next.

The Attempt at a Solution


4.8 = x - 4

would that mean small pump needs 8.8 hours? am I doing this right?
 
This is like a rate-time-distance problem. The rate that the smaller pump empties the tank is 1/X tanks per hour. The rate that the larger pump empties the tank is 1/(X-4) tanks per hour. How many tanks (of the same size) can the smaller pump empty in 4 hours? How many tanks (of the same size) can the larger pump empty in 4 hours? Together they empty 1 tank in 4 hours. Can you write an equation describing this?

Chet
 
Mark44 said:
RUber asked about Y and where the 4.8 hours came from, a reasonable question, since the number 4.8 doesn't appear anywhere in the problem description above.

A variable should represent an unknown quantity whose value you are trying to find.
I can't think of any good reason why you should give this value the name Y.
My apology, I accidentally deleted the .8 while typing. I edited the OP. it takes 4.8 hours with both pump together.
 
Chestermiller said:
This is like a rate-time-distance problem. The rate that the smaller pump empties the tank is 1/X tanks per hour. The rate that the larger pump empties the tank is 1/(X-4) tanks per hour. How many tanks (of the same size) can the smaller pump empty in 4 hours? How many tanks (of the same size) can the larger pump empty in 4 hours? Together they empty 1 tank in 4 hours. Can you write an equation describing this?

Chet
Both pump together can empty it in 4.8 hours. I edited the OP sorry for the confusion.
 
  • Like
Likes   Reactions: SammyS
OK. Substitute 4.8 hours for 4 hours. Are you able to answer my questions now?

Chet
 
Chestermiller said:
OK. Substitute 4.8 hours for 4 hours. Are you able to answer my questions now?

Chet
I am sorry but I do not understand why it is 1/X, why is the 1 needed?
 
Max0007 said:
I am sorry but I do not understand why it is 1/X, why is the 1 needed?
If you drive your car 1 mile in X minutes, then your rate of travel is 1/X miles per minute (rate = distance/time). If you empty 1 tank in X hours, then your rate of emptying tanks is 1/X tanks per hour.

Chet
 
  • Like
Likes   Reactions: Max0007
Chestermiller said:
If you drive your car 1 mile in X minutes, then your rate of travel is 1/X miles per minute (rate = distance/time). If you empty 1 tank in X hours, then your rate of emptying tanks is 1/X tanks per hour.

Chet
Oh so it should simply be 1/x + 1/x-4 = 1/Y
1/x + 1/x-4 = 1/4.8

Now I really have no idea how to factorize this and get the X value.

The only thing I can thing off is doing something like this:

1(4.8)(x-4) + 1(x*4.8) / (x)(x-4)(4.8) = 1(x)(x-4) / (x)(x-4)(4.8)

4.8x - 19.2 + 4.8x ((x)(x-4)(4.8))= x^2 - 4x / (x)(x-4)(4.8)

and I am stuck here.
 
Max0007 said:
Oh so it should simply be 1/x + 1/x-4 = 1/Y
1/x + 1/x-4 = 1/4.8

Now I really have no idea how to factorize this and get the X value.

The only thing I can thing off is doing something like this:

1(4.8)(x-4) + 1(x*4.8) / (x)(x-4)(4.8) = 1(x)(x-4) / (x)(x-4)(4.8)

4.8x - 19.2 + 4.8x ((x)(x-4)(4.8))= x^2 - 4x / (x)(x-4)(4.8)

and I am stuck here.
You seem to have the right idea, but you have to do the algebra correctly.

Chet
 
  • Like
Likes   Reactions: Max0007
Chestermiller said:
You seem to have the right idea, but you have to do the algebra correctly.

Chet
Thanks a lot for the help and the time :)

@everyone who helped and took the time to read and support me with this, thank you every much :).
 
Chestermiller said:
You seem to have the right idea, but you have to do the algebra correctly.

Chet
Okay so I am getting x1= 1.6 and x2 = 12, the answer should be 12 right? since it can't really be under 4.