1. ## Using Two Pumps

A 5 horsepower (hp) pump can empty a pool in 5 hours. A smaller, 2hp pump empties the same pool in 8 hours. The pumps are used together to begin emptying this pool. After 2 hours, the 2hp pump breaks down. How long will it take the larger pump to empty the pool?

SIDE QUESTIONS:

(1) Is this question similar to math questions about working alone and working together?

(2) How do you set up an equation for such a question?

I got lost when the 2hp pump broke down after 2 hours.

2. Originally Posted by symmetry
A 5 horsepower (hp) pump can empty a pool in 5 hours. A smaller, 2hp pump empties the same pool in 8 hours. The pumps are used together to begin emptying this pool. After 2 hours, the 2hp pump breaks down. How long will it take the larger pump to empty the pool?
The first pump pumps 1/5 of the pool per hour,

The second pump pumps 1/8 of the pool per hour.

Together they pump 1/5+1/8 = 13/40 of the pool per hour.

In two hours working togethe they have emptied 2*(13/40)=13/20 of the
pool, leaving 7/20 to be emptied by the first pump alone at 1/5 of the pool
per hour. So it will take (7/20)/(1/5) hours to finish emptying the pool,
which is 7/4 hours = 1 hour 45 min.

SIDE QUESTIONS:

(1) Is this question similar to math questions about working alone and working together?
yes

(2) How do you set up an equation for such a question?
I do it as shown above.

RonL

3. Originally Posted by symmetry
A 5 horsepower (hp) pump can empty a pool in 5 hours. A smaller, 2hp pump empties the same pool in 8 hours. The pumps are used together to begin emptying this pool. After 2 hours, the 2hp pump breaks down. How long will it take the larger pump to empty the pool?

SIDE QUESTIONS:

(1) Is this question similar to math questions about working alone and working together?

(2) How do you set up an equation for such a question?

I got lost when the 2hp pump broke down after 2 hours.
Hello,

to (1): yes

to (2):

let p be the content of the pool.

the 5hp-pump empties in 1 hour $\displaystyle \frac{p}{5}$ of the content.

I got lost when the 2hp pump broke down after 2 hours.[/QUOTE]

Hello,

to (1): yes

to (2):

let p be the content of the pool.

the 5hp-pump empties in 1 hour $\displaystyle \frac{p}{5}$ of the content.

the 2hp-pump empties in 1 hour $\displaystyle \frac{p}{8}$ of the content.

Together and alone after 2 hours:

$\displaystyle \left(\frac{p}{5}+\frac{p}{8} \right)\cdot 2+\frac{p}{5} \cdot x=p$

Divide the equation by p, expand the brackets:
$\displaystyle \frac{13}{20}+\frac{1}{5} \cdot x=1$. Solve for x. I've got:
x = 7/4 hours = 1 h 45 min.

EB