Quarrel Forum

etherkye · #1 (**permalink**) 10 Mar 2005, 02:58

3 brothers go into a shop, each needed a key cut and a shoe repared. if there are 2 assistants, and it takes 5 mins to cut a key, and 15 mins to repare a shoe, what is the minimyum time they can fix all shoes and cut all keys. please state what each assistant does as well as totoal time.

hoih · #2 (**permalink**) 10 Mar 2005, 03:30

if there are 2 assistants there's 1 "leader" as well, so 3*5+3*15 = 1 hour

etherkye · #3 (**permalink**) 10 Mar 2005, 03:31

you have some bizzare ideas. WRONG

hoih · #4 (**permalink**) 10 Mar 2005, 03:32

no dude I'm right, if you say assistants, there must be a boss who's working too!

etherkye · #5 (**permalink**) 10 Mar 2005, 03:33

your answer of 1 hour is wrong. the boss is on lunch.

hoih · #6 (**permalink**) 10 Mar 2005, 03:47

no he's not! lol ok, I'll calc when he's @ lunch
2*5 = 10 10
1*5 = 5 1*15= 15 the first is 10 minutes busy and the second 20
the first will do the shoe of 15 and is busy 25 minutes
second do the shoe of 15 and is busy for 35 minutes
so uuh 35 minutes

lord_vdc · #7 (**permalink**) 10 Mar 2005, 03:49

well there are 3 shoes to repare
so 2 shoes takes 15 minutes and 1 shoe takes 15 minutes
that makes already 30 minutes
but there are 3 key's and that takes 10 minutes and 5 minutes so
15 + 5 +15 makes 35 minutes

etherkye · #8 (**permalink**) 10 Mar 2005, 05:32

err, your first bit confused me. WRONG.

ourevildictator · #9 (**permalink**) 11 Mar 2005, 06:17

Lets try 20 min
each do 1 shoe and 1 key

IanCWU · #10 (**permalink**) 11 Mar 2005, 07:57

20 min doesn't work because there's only 2 assistants.

My answer: 30 minutes - 1 assistant does two shoes, other assistant does all three keys and last shoe.