Rate is the same, ratio of lice to minutes is the same, so the number of mice is the same. The numbers don’t work out evenly, so the mice need to be willing to share and eat some of each others half-eaten lice. Yum.
100…lol if 6 mice eat 6 lice in 6 minuites then 1 mice eats a lice in 1 minuite…provided that each mice have same abilities witch i think is true in this occasion
So either
6 mice eat 16.667 lice in 100 minutes (strategically planned on the mice’s part where 4 mice get 2/3 of a lice and 2 mice get 2 -1/3 pieces of a lice to complete the last 4 in the last 4 minutes)
Or if we say the mice must eat lice in whole numbers only…
6 mice eat 96 lice in 96 minutes, so it takes 1 more mice to get the other 4 in under 100 minutes.
We also assumed the mice had similar throat capacity and stomach volume, and they didn’t get full after only 6 minutes.
if 6 mice eat 6 lice in 6 minutes, and presumably each mouse eats one louse, starting simultaneously, then it takes a mouse 6 minutes to eat a louse. In predator prey ecology, this is called “handling time.”
So if a mouse takes 6 minutes to eat one louse, then a mouse could eat 16.667 mice in 100 minutes (100/6 = 16.667). At that rate, 6 mice could eat 100 lice (100/16.667 = 6) in 100 minutes.
6 mice eat 6 lice in 6 minutes, so 1 mouse eats 1 louse in 6 minutes. Divide 6 minutes into 100 minutes gives 16 2/3 mice, but since you can’t have 2/3 mouse, it’s 17 mice.
The way is written, I assume that each of 6 mice eat 6 lice in 6 minutes, which lead us to assume that only one mouse is anough to eat 100 lice in 100 minutes, since from assumption above, each mice each a lice in a rate of 1 each minute.
If you assume that it takes all 6 mice to eat 6 lice in six minutes, then it takes 6 mice to eat 100 lice in 100 minutes
There are two ways to approach this.
1) 1 mouse eats 1 louse in 6 minutes, so in 100 minutes, 1 mouse could eat 16 and 2/3 lice. Thus, 6 mice could eat all 100 lice in 100 minutes.
2) 6 mice eat 1 louse in 1 minute, so 6 mice could eat 100 lice in 100 minutes.
Theoretically, 1 mouse.
But is it physically possible for 1 mouse to eat 100 lice?
I would go for 100 mice, enjoying 1 lice each.
Layman’s solution:
6 mice eat 6 lice in 6 minutes
ie the 6 mice eat one louse per minute (louse is the singular of lice)
and hence the 6 mice eat 100 lice in 100 minutes
6 mice.
6
6 mice.
6
Rate is the same, ratio of lice to minutes is the same, so the number of mice is the same. The numbers don’t work out evenly, so the mice need to be willing to share and eat some of each others half-eaten lice. Yum.
mice with lice inversely proportional…
mice with minutes directly proportional…
answer= = 6 * 100/6 * 6/100 = 6 mice…
a bit tricky !!
If 6 mice can eat 6 lice in 6 minutes, (1 lice per minute) then it still only takes 6 mice to eat 100 lice in 100 minutes.
Assuming of course that they don’t get full, but they’re mice. They’ve got fast metabolisms.
the other way around…
mice & lice directly proportional…
mice & time inversely proportional…
result stays the same, ie.; = 6 mice…
At one louse/minute for the 6 mice… so the answer is still 6 mice!
600 mice
100…lol if 6 mice eat 6 lice in 6 minuites then 1 mice eats a lice in 1 minuite…provided that each mice have same abilities witch i think is true in this occasion
1 eat 1/6 lice in 1 minute
1 eat 100/6 lice in 100 minutes
6 eat 100 lice in 100 minutes
so the same 6 mice, cool man
A couple of ways to interpret, but
1 mice eats 1 lice in 6 minutes.
1 mice eats 16.667 lice in 100 minutes
So either
6 mice eat 16.667 lice in 100 minutes (strategically planned on the mice’s part where 4 mice get 2/3 of a lice and 2 mice get 2 -1/3 pieces of a lice to complete the last 4 in the last 4 minutes)
Or if we say the mice must eat lice in whole numbers only…
6 mice eat 96 lice in 96 minutes, so it takes 1 more mice to get the other 4 in under 100 minutes.
We also assumed the mice had similar throat capacity and stomach volume, and they didn’t get full after only 6 minutes.
Are they pack rats or what?
if 6 mice eat 6 lice in 6 minutes, and presumably each mouse eats one louse, starting simultaneously, then it takes a mouse 6 minutes to eat a louse. In predator prey ecology, this is called “handling time.”
So if a mouse takes 6 minutes to eat one louse, then a mouse could eat 16.667 mice in 100 minutes (100/6 = 16.667). At that rate, 6 mice could eat 100 lice (100/16.667 = 6) in 100 minutes.
6
6 mice eat 6 lice in 6 minutes, so 1 mouse eats 1 louse in 6 minutes. Divide 6 minutes into 100 minutes gives 16 2/3 mice, but since you can’t have 2/3 mouse, it’s 17 mice.
The way is written, I assume that each of 6 mice eat 6 lice in 6 minutes, which lead us to assume that only one mouse is anough to eat 100 lice in 100 minutes, since from assumption above, each mice each a lice in a rate of 1 each minute.
If you assume that it takes all 6 mice to eat 6 lice in six minutes, then it takes 6 mice to eat 100 lice in 100 minutes
how can we send a puzzle on SmartKit?
Hi Tsopi- email me at info (at) smart-kit.com, thanks
There are two ways to approach this.
1) 1 mouse eats 1 louse in 6 minutes, so in 100 minutes, 1 mouse could eat 16 and 2/3 lice. Thus, 6 mice could eat all 100 lice in 100 minutes.
2) 6 mice eat 1 louse in 1 minute, so 6 mice could eat 100 lice in 100 minutes.
six mice could eat 100 lice in 100 minutes. same as the other problem just giving the mice more time to eat.
Fat mice!
100