Question

At a certain small town, p gallons of gasoline are needed per month for each car in town. At this rate, if there are r cars in town, how long, in months, will q gallons last?
(a) pq/r
(b) qr/p
(c) r/pq
(d) q/pr
(e) pqr

Answer 1

Simplest way is by using an example for yourself

Say, p = 100
r = 50
q = 5000

Using this in the statement, you get. 100 gallons are needed per month for each car. If there are 500 cars in town, how long will 5000 gallons last?

Calculating this:

\[5000\div(100\times50)\] = \[5000\div5000\] = 1

Change this into your letters, you get

\[q \div \left( p \times r \right)\]

If you think about it, this means that you have a limit of gasoline. You multiply the amount of cars with the amount of gasoline they need each month, and then divide those two values.

Answer d is correct.

Answer 2

Little mistype in my sentence: Using this in the statement, you get. 100 gallons are needed per month for each car. If there are ***50*** cars in town, how long will 5000 gallons last?

Answer 3

i think it is qr/p

Answer 4

pgal/(1month/1car)=qgal/(xmonth/rcars)
(1car x pgal)/1month = (rcars x q gals)/ x months
solve for x

Answer 5

you had the set up right but i think you misinterpret the end q/(p x r) = qr/p

Answer 6

Your last statement isn't true though. If you had x/(y*z) or, to give another example, 10/(5*2) = 1

then the statement x*z/y isn't correct : 10*2/5 = 4

Answer 7

i think you are required to plug in the 500 cars somewhere and properly otherwise the method won't work. I have to go right now so i can't do it now