### Brain Teasers

# Mice & Ropes

You have two identical ropes and three mice. Each mouse bites off the rope at a different biting speed. All three biting speeds are constant.

Mouse(A) bites off one rope in 60 minutes.

Mouse(B) bites off one rope in 90 minutes.

Mouse(C) bites off one rope in 180 minutes.

You do not have any time or distance measuring tools and the length of the rope is unknown. How can you measure 75 minutes?

Mouse(A) bites off one rope in 60 minutes.

Mouse(B) bites off one rope in 90 minutes.

Mouse(C) bites off one rope in 180 minutes.

You do not have any time or distance measuring tools and the length of the rope is unknown. How can you measure 75 minutes?

### Answer

Mark the two ropes as Rope(A) and Rope(B).1. Let Mouse(A) and Mouse(C) start biting off Rope(A) at the same time, starting each from one end (moving in opposite directions). At the same time, let Mouse(B) start biting off Rope(B).

Finding relative speeds between Mouse(A) and Mouse(C):

Distance = Speed * Time

X = S(A) * 60, where S(A) is biting speed for Mouse(A) and X is Rope(A) length

X = S(C) * 180, where S(C) is biting speed for Mouse(C) and X is Rope(A) length

Solving the two above equations gives:

S(A)/S(C) = 3:1, which means that for each 3 units Mouse(A) will move, Mouse (C) will move 1 unit.

Moving in opposite directions (assume Mouse(A) from left to right and Mouse(C) from right to left), the two mice will meet after Mouse(A) bites off three times what Mouse(C) bit off, which divides Rope(A) length in to 4 units.

Mouse(A) bitten off portion = 3/4 X (where X is Rope(A) length)

Mouse(C) bitten off portion = 1/4 X

Finding the time spent before Rope(A) is completely bitten off by the two mice:

If Mouse(A) needs 60 minutes to bite off X, How many minutes are needed to bite off 3/4 X?

Minutes needed = 3/4 * 60 = 45 minutes are needed to fully bite off Rope(A) by Mouse(A) and Mouse(C). At that point in time, Mouse(B) would have bitten off exactly half the length of Rope(B) (needing 90 minutes to bite off the full rope), and needs an additional 45 minutes to bite off the remaining rope half.

2. Let Mouse(B) continue biting off Rope(B) and make Mouse(C) start biting off the other end of Rope(B).

Finding relative speeds between Mouse(B) and Mouse(C):

Distance = Speed * Time

X/2 = S(B) * 90, where S(B) is biting speed for Mouse(B) and X/2 is the remaining half of Rope(B) length

X/2 = S(C) * 180, where S(C) is biting speed for Mouse(C) and X/2 is the remaining half of Rope(B) length

Solving the two above equations gives:

S(B)/S(C) = 2:1, which means that for each 2 units Mouse(B) will move, Mouse(C) will move 1 unit.

Moving in opposite directions (assume Mouse(B) from left to right and Mouse(C) from right to left), the two mice will meet after Mouse(B) bites off two times what Mouse(C) bit off, which divides the remaining half of Rope(B) length to 3 units.

Mouse(B) bitten off portion = 2/3 * X/2 = X/3

Mouse(C) bitten off portion = 1/3 * X/2 = X/6

Finding the time spent before the remaining half of Rope(B) is completely bitten off by the two mice:

If Mouse(B) needs 90 minutes to bite off X, how many minutes are needed to bite off X/3?

Minutes needed = 1/3 * 90 = 30 minutes are needed to fully bite off the remaining half of Rope(B) by Mouse(B) and Mouse(C).

At that point in time, 75 minutes would have passed (45 minutes consumed from biting off Rope(A) and another 30 minutes consumed from biting off Rope(B)).

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## Comments

Can you right that answer in english?

I'm sure I didn't think of all that woffle in getting the answer. In fact I don't think I did it that way at all, but it takes too long even for a shorter way.

Ummm, or you don't even need the third mouse. Start mouse A and mouse B on their ropes. As soon as mouse A finishes 60 mins has passed and you fold the remainder of Mouse B's rope in half (or at right angles so long as halfway is marked) and this should be 30 mins more worth of biting. Halving this is 15 mins so as soon as Mouse B is half way up this remainder ie. hits your fold voila 15 mins have passed therefore 60 + 15 = 75. In saying this, I do think your workings are quite incredible, well done.

A greater anwser even. Thanks.

I didn't solve it that way at all...I figure that you use one rope. Mouse A chews the entire rope in 60 minutes. Mouse B chews the entire rope in 90 minutes. Therefore, you allow both mice to chew at opposite ends of one rope. Since the speed is constant, both will chew exactly half of the rope, Mouse A in 30 minutes and Mouse B in 45 minutes. Total of 75 minutes.

forget my last answer..it's wrong

Another solution: After mice A and C have finished their rope, remove mouse B from the other rope and let mouse A finish the job. Since there is still 1/2 rope left, mouse A will finish it in 30 minutes, making a total of 75 minutes.

Well, there is even simpler solution, using only one mouse (A):

Let mouse A start eating first rope. Fold twice second rope and mark 1/4th of second rope. As soon as mouse A eat first rope (60min), let him eat second rope and stop when he reach first 1/4th mark. Since mouse A eats whole rope in 60min, 1/4th needs only 15min, and in total we have 75min

But since solution with only one mouse was probably not intended, maybe adding following condition would be better:

"Ropes are not uniform width, so for example it is possible that mouse A can eat 1st half of rope in 20min and 2nd half of rope in 40min"

Let mouse A start eating first rope. Fold twice second rope and mark 1/4th of second rope. As soon as mouse A eat first rope (60min), let him eat second rope and stop when he reach first 1/4th mark. Since mouse A eats whole rope in 60min, 1/4th needs only 15min, and in total we have 75min

But since solution with only one mouse was probably not intended, maybe adding following condition would be better:

"Ropes are not uniform width, so for example it is possible that mouse A can eat 1st half of rope in 20min and 2nd half of rope in 40min"

hahaha thats on bigggg answer

dssquared you are totally wrong you do know that right?

ok mouse b takes 90 min to chew a whole rope right?

so folding it in half would be 45min right?

and further more folding that would be22.5 min

so you would be 7.5 min late to watever you were going to

ok mouse b takes 90 min to chew a whole rope right?

so folding it in half would be 45min right?

and further more folding that would be22.5 min

so you would be 7.5 min late to watever you were going to

Simple solution, can even do it with just one rope. Just fold both in half, let mouse A do their half. That'd be 30 minutes. Then give the remaining half to mouse B, that'd be 45 minutes. 30+45 = 75 mins.

Sorry, meant just fold one in half.

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