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At the recent BrainBashers Festival, the 100 metres heats were closely monitored.
Each contestant had to run in two races so that the average place could be determined.
Only one runner finished in the same place in both races.
Alan was never last. Charles always beat Darren. Brian had at least one first place. Alan finished third in at least one of the races. Both Darren and Charles had a second place.
To explain  since Charles always beat Darren and Darren had a second place, one race must have been Charles first and Darren second. Brian therefore won the other race with Charles second. Since only one runner finished in the same place in both races, this must have been Alan in third.
Puzzle 351
Practical Peter was asked to cut a 99 foot rope into three smaller, equal length ropes.
However, as usual, Pete couldn't find his measuring tape so he guessed!
When he finally did find his tape (it was under his hat), he discovered that:
A) the second piece of rope was twice as long as the first piece, minus 35 feet (i.e. 2 x first,  35).
B) the third piece of rope was half the length of the first, plus 15 feet (i.e. 0.5 x first, + 15)
Answer:
First = 34 feet.
Second = 33 feet.
Third = 32 feet.
This question can be solved easily using algebra, if we call the length of the first rope A, we have:
Rope 1 = A
Rope 2 = 2 x A  35
Rope 3 = 1 ÷ 2 x A + 15
The three ropes add to 99 feet, so:
99 = Rope 1 + Rope 2 + Rope 3
99 = A + (2 x A  35) + 1 ÷ 2 x A + 15
99 = 3.5 x A  20
Adding 20 to both sides we have:
119 = 3.5 x A
So:
A = 119 ÷ 3.5
A = Rope 1 = 34 feet
Rope 2 = 2 x 34  35 = 68  35 = 33 feet
Rope 3 = 1 ÷ 2 x A + 15 = 1 ÷ 2 x 34 + 15 = 17 + 15 = 32 feet
Just to check:
Rope 1 + Rope 2 + Rope 3 = 34 + 33 + 32 = 99.
As required.
Puzzle 352
10 prisoners are locked up in individual cells, unable to see, speak or communicate in any way with each other. There is a exercise room with a single light, that is initially off and the prisoners cannot see the light from their own cell.
Every day, the warden picks a prisoner at random, and that prisoner goes to the exercise room.
While there, the prisoner can choose to switch the light on or off, and are not allowed to leave a message.
At any point, any prisoner can claim that all 10 prisoners have been to the exercise room. If they are wrong then all 10 prisoners will be locked up forever! However, if they are correct all of the prisoners are set free.
Before the random picking begins, the prisoners are allowed to discuss a plan. What is their best plan to determine when all 10 prisoners have visited the exercise room?
[Ref: ZAWX]
Direct Link: www.brainbashers.com?ZAWX
Hint: Remember that once the picking starts, there is no method of communication.
Answer:
One person is chosen as a Counter and will only ever turn the light off, and they will count the number of times they do this.
Each of the other prisoners will only turn the light on once. They will only touch the light if it's already off  and only if they've not yet turned it on.
Ever time the Counter goes to the exercise room, if the light is on again, they'll know a new person has entered the room, and they'll turn the light off.
Once the Counter has turned the light off 9 times, they will know that the other 9 prisoners have entered the room, and therefore can ask for everyone to be released.
For 10 prisoners, on average, this would take around 91 days.
For 20 prisoners, on average, this would take around 381 days.
For 50 prisoners, on average, this would take over 6 years.
For 100 prisoners, on average, this would take over 27 years.
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