5 lb Q30 page 613

[kpkanupriyakhmi]

Etienne began to eat 20 cookies at exactly the same time Jacques began making more cookies, one at a time, at a
constant rate of 16 cookies per hour. If Etienne ate 20 cookies per hour, after how many hours were there no
cookies?
I didn't understand how to do such a question?

[tommywallach]

Hey Kp,

It's a rates question. It's written TERRIBLY though (where did you find it?). Here's what it meant.

There are 20 cookies. Etienne begins eating them at a rate of 20 cookies per hour. At the exact same time that Etienne begins eating, Jacques begins making cookies at the rate of 16 cookies per hour, and adds them to Etienne's pile.

Given all this, the cookies are disappearing at a rate of 4 per hour (because 20 get eaten but 16 are replaced, leaving 4 gone). So if there were 20 to start with, it will take 5 hours to get rid of them.

You could also do it like this:

Rate * Time = Work

Etienne 20 x y
Jacques 16 x y - 20

Then just solve to the two equations for x:

20x = y
16x = y - 20

Personally, I find this more confusing than just the logic I showed you above, but it works because we know the times are the same (they start at the same moment, and will stop at the same moment), and we know Jacques has to make 20 fewer cookies than Etienne will eventually eat (because Etienne will eat the 20 she started with AS WELL AS whichever cookies Jacques makes).

Hope that helps!

-t

[n00bpron00bpron00b]

5lbs :D

[kpkanupriyakhmi]

How could you make out this:- if there were 20 to start with, it will take 5 hours to get rid of them,i.e. how could you make out he will have to finish 20 cookies since when one is eating another is cooking?

[tommywallach]

Yeah. It's a bad 5 lber. I'll be talking to someone! :)

Consider it hour by hour.

Hour 0: 20 cookies
Hour 1: 16 cookies made, so 36; 20 eaten --> 16
Hour2: 16 cookies made, so 32: 20 eaten --> 12
etc.

See how we lose 4 every hour?

-t

[kpkanupriyakhmi]

TY.

[tommywallach]

Glad to help!

-t