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Meena.IF thumbnail
17th Anniversary Thumbnail Dazzler Thumbnail
Posted: 17 years ago
#31
Solution to:
There are several ways to solve this problem.
Note: In advance, you do not know whether the coin with different weight is heavier or lighter!
Number the coins from 1 up to 12. Perform the following three weighings:

Left side Right side
Weighing 1: 1 2 3 10 4 5 6 11
Weighing 2: 1 2 3 11 7 8 9 10
Weighing 3: 1 4 7 10 3 6 9 12


Call the outcome of a weighing "L" if the left side is most heavy, call the outcome "R" if the right side is most heavy, and call the outcome "B" if the left and right sides have the same weight. Then the following outcomes are possible:

Weighing 1 Weighing 2 Weighing 3 Different coin
L L L 1 heavier
L L R 3 heavier
L L B 2 heavier
L R L 10 heavier
L R B 11 lighter
L B L 6 lighter
L B R 4 lighter
L B B 5 lighter
R L R 10 lighter
R L B 11 heavier
R R L 3 lighter
R R R 1 lighter
R R B 2 lighter
R B L 4 heavier
R B R 6 heavier
R B B 5 heavier
B L L 9 lighter
B L R 7 lighter
B L B 8 lighter
B R L 7 heavier
B R R 9 heavier
B R B 8 heavier
B B L 12 lighter
B B R 12 heavier

Note: outcomes LRR, RLL, and BBB are not possible.
jasunap thumbnail
19th Anniversary Thumbnail Rocker Thumbnail Engager Level 1 Thumbnail
Posted: 17 years ago
#32
gosh arjun, i read the answer....but i have not understood it yet!!! mind boggling!!
suram thumbnail
18th Anniversary Thumbnail Voyager Thumbnail
Posted: 17 years ago
#33


sssss appa thalai suthudhey!!!! 😃


Meena.IF thumbnail
17th Anniversary Thumbnail Dazzler Thumbnail
Posted: 17 years ago
#34
I will try to explain more clearly.
Take twelve coins and number it from 1 to 12.
Do three weighing as shown in the solution.
Left side Right side
Weighing 1: 1 2 3 10 4 5 6 11
Weighing 2: 1 2 3 11 7 8 9 10
Weighing 3: 1 4 7 10 3 6 9 12

Let us take for all the three weighing Left side is heavy. The only common piece in left is 1. So it should be heavier. There is no common piece in right side.

Let us take for all the three weihings Right side is heavier. The only common piece in left side 1 is lighter. There is no common piece in right side.

This is how it goes.
Meena.IF thumbnail
17th Anniversary Thumbnail Dazzler Thumbnail
Posted: 17 years ago
#35
Mr. Wagle goes to work by a bus. One day he falls asleep when the bus still has twice as far to go as it has already gone.

Halfway through the trip he wakes up as the bus bounces over some bad potholes. When he finally falls asleep again, the bus still has half the distance to go that it has already travelled. Fortunately, Mr. Wagle wakes up at the end of his trip.

What portion of the total trip did Mr. Wagle sleep?
jasunap thumbnail
19th Anniversary Thumbnail Rocker Thumbnail Engager Level 1 Thumbnail
Posted: 17 years ago
#36
that is maths again....and fraction as well, i give up!!!
Meena.IF thumbnail
17th Anniversary Thumbnail Dazzler Thumbnail
Posted: 17 years ago
#37
Let someone try that. Here is ne for those who want non-maths. A simple one. May be tricky.
This is a Riddle:
How can you distribute three apples to two fathers and their two sons, giving a whole apple to each.
ginny007 thumbnail
17th Anniversary Thumbnail Explorer Thumbnail
Posted: 17 years ago
#38
hey tht wd be a grandfather,father and a son..so tht means 3 ppl and 3 apples...
Meena.IF thumbnail
17th Anniversary Thumbnail Dazzler Thumbnail
Posted: 17 years ago
#39

Originally posted by: ginny007

hey tht wd be a grandfather,father and a son..so tht means 3 ppl and 3 apples...

Ya.. correct.. Y cant u try the one in the previous page?

Meena.IF thumbnail
17th Anniversary Thumbnail Dazzler Thumbnail
Posted: 17 years ago
#40
Another one.
There are three lights (bulbs) in a room whose switches are in another room.
You are allowed to enter both the room once and switch on or off the lights.
How will you find out which is the switch for which bulb.
Remember that you can enter each room only once.
Edited by arjun_kk - 17 years ago

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