SOLUTION: I have a word problem here and I having trouble figuring out the let statements and equations to solve the problem. The word problem is: A boat has two motors, one large and one

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Question 1185630: I have a word problem here and I having trouble figuring out the let statements and equations to solve the problem. The word problem is:
A boat has two motors, one large and one small. If only the large motor is running. a tank of fuel lasts 2 h. If only the small motor is running, the tank of fuel lasts 4 h.
How long will the fuel last if both motors are running?

Found 2 solutions by Theo, greenestamps:
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
what i get is the following.

rate * time = quantity.

the rate is the rate that the motor empties the tank per hour.
the time is how long it takes to empty the tank.

the large motor empties the tank in 2 hours.
rate * 2 = 1 empty tank.
therefore, it is emptying 1/2 of the tank in one hour.

the small motor empties the tank in 4 hours.
rate * 4 = 1 empty tank.
therefore, it is emptying 1/4 of the tank in one hour.

when they work together, their rates are additive.
you get (1/2 + 1/4) * time = 1 empty tank.
combine fractions to get 3/4 * time = 1 empty tank.
solve for time to get time = 4/3 hours.

you figure it has to be less than 2 hours because the little motor can empty the tank in 2 hours all by itself.
with help from the big tank, the total time has to be less than 2 hours.

in 4/3 hours, the little motor empties 1/4 * 4/3 = 2/6 = 1/3 of the tank.
in the same 4/3 hours, the large motor empties 1/2 * 4/3 = 4/6 = 2/3 of the tank.

their combined work empties 1/3 + 2/3 = 1 = all of the tank in 4/3 hours.

solution makes sense.

i would go with 4/3 hours to empty the tank when both motors are running.

if you let x equal the rate of the large motor and y equal the rate of the small motor, then your equations become:

x * 2 = 1 and y * 4 = 1.
solve for the variables to get:
x = 1/2 and y = 1/4

those are the rates that the large and small motors are emptying the tank.

hen they work together, the formula becomes:

(x + y) * T = 1

since x = 1/2 and y = 1/4, you get:
(1/2 + 1/4) * T = 1
solve for T to get T = 3/4 hours.

let me know if you need more information as to how this was done.

the rates of x and y are how much of the tank has been emptied in one hour.
T is the time it takes to empty the tanks.





Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The question as posted asks the student to use an appropriate "let" statement and equation(s) to solve the problem. The number asked for is the number of hours the tank of fuel will last if both motors are running, so that is the logical choice for the variable.

Let x = # of hours the tank of fuel lasts if both motors are running.

This is a variation of a standard "working together" problem; the standard algebraic solution method is to work with the fractions of the job the two workers do individually and together. For this problem, the "job" is emptying the fuel tank.

1/4 = fraction of the job the small motor does in 1 hour
1/2 = fraction of the job the large motor does in 1 hour
1/x = fraction of the job the two motors together do in 1 hour

Logically, then, the equation for solving the problem says the fraction of the job done in 1 hour is the sum of the fractions of the job done by each motor in 1 hour:

1%2F4%2B1%2F2=1%2Fx

Start solving the equation by multiplying everything by the least common denominator, 4x:

x%2B2x=4
3x=4
x=4%2F3

ANSWER: The two motors running together take 4/3 hours, or 1 1/3 hours, or 1 hour 20 minutes, to empty the fuel tank