# Darn Word Problems!

1. Feb 26, 2010

### santhean

1. Suppose that a hospital charges you $220 per day if you sty in a private room and$166 / day if you stay in a room for 4 people. Assume that the amount you pay is the sum of a constant amount (for the services you receive) and an amount which varies inversely with the number of people the room will hold (to pay for your share of the room)
- a] Calculate the constant amount you pay for services and the proportionality constant in the variation function.
- b] What does the hospital charge for a room, exclusive of the services to the paitent?
- c] How much would you expect to pay / day if you stayed in
i. semi-private room holding 2 people?
ii. a ward holding 20 people?

2. I'm not sure what you put here for relevant equations? But this section deals with functions of more than one independent variable where in y=kx there is more than 1 x. Something like that.

3. The attempt at a solution
- a] total amount = k1 + amount x
amount x = k2 / number of people
then i plugged in for amount x and got --> total amount = k1 + k2/ number of people
so now a] is asking for the value of the two ks? But which numbers do i plug in? and can I combine k1 + k2 to make a k3?

- b] I think for this you use the equation - total amount = k2/ number of people but w/o solving a, i can't solve b

- c] this is plugging and chugging.. if I knew the equation from a]

Any help is appreciated.
Thanks so much in advance!

2. Feb 26, 2010

### Borg

santhean,

Welcome to Physics Forums. Your problem can be reworded to make it a little more understandable.

You have two equations. In the first equation, the total cost of $220 equals the services plus the cost of a room (Y) divided by a single person in the room. In the second equation, the total cost of$166 equals the services plus the cost of a room (Y) divided by four people in the room.

The cost of the services (X) is the same in both cases. For the first case, your equation would look like this:

\$220 = X + (Y/1)

What would the second equation look like?

Last edited: Feb 26, 2010
3. Feb 26, 2010

### Borg

santhean,

I reworded the above text but, the equation is the same. It's very early here and I guess that I'm not awake yet.