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Heavier ball

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You have 8 balls all look identical (in shape, color etc.). All of them have same weight except one defective ball which is heavier than others. You can use a two sided balance system (not the electronic one). Find the minimum no. of measures required to separate the defective ball and the way you separate it.

If you have solved it then try it for N no. of identical balls with one defective ball.

See Solution : Heavier Ball

The minimum no. of measures required are 2. Now can you solve it… ? Give it a try…

Here is the solution which tells you that how can you do it in 2 measurements.

Divide 8 balls into groups of 3, 3 and 2.

First weigh :
Weigh the two groups of 3 ball, now there are two possibilities
a) They are balanced (all 6 balls are of equal weight)
b) One side is heavier then the other.

case (a) :
The group of 2 has the defective ball. Weigh them with one on each side of balancing machine. The side which has more weigh has the heavier/defective ball.

case (b) :
We got the three balls of the side which is heavier. Now take any two balls and weigh them, then again there are two cases.
(i) They both are of equal weight.
(ii) One is heavier than the other.

for case (b)(i) :
The third ball is heavier/defective than all other balls

for case (b)(ii) :
The heavier ball is defective.
——————————————–

Solution for N no. of balls.

If there is 1 ball, No measurement is required.
If there are 2-3 balls, we need 1 measurement.[(3^0 +1) to 3^1] If there are 4-9 balls, we need 2 measurement. [(3^1 +1) to 3^2]

Now if there are N balls
and lets say it lies between (3^(x-1) +1) to 3^x
then x = [logN/log3] = Greatest integer of logN to the base 3.
Then minimum no. of measures is equal to x = [logN/log3]

Wires on fire

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A guy has two wires of varying thickness, which each burns in 60 minutes. He actually wants to measure 45 mins. How can he measure 45 mins using only these two wires. (he can’t cut the one wire in half because the wires are non-homogeneous and he can’t be sure how long it will burn)

See Solution : Wires on fire

He will burn the first wire at both the ends and the second wire at one end. After half an hour, the first one burns completely and at this point of time, he will burn the other end of the second wire so now it will take 15 mins more to completely burn.. so total time is 30+15 i.e. 45mins.

Jelly beans

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You have three jars that are all mislabeled. One contains peanut butter jelly beans, another grape jelly jelly beans and the third has a mix of both (not necessarily half-half mix). How many jelly beans would you have to pull out and out of which jars, to find out how to fix the labels on the jars?

Labels on jars are as follows
Jar 1 : Peanut butter
Jar 2 : Grape
Jar 3 : P.b. / Grape

See Solution : Jelly beans

Only one jelly bean from the p.b./grape jar will do the trick.

The trick here is to realize that every jar is mislabeled. Therefore you know that the peanut butter jelly bean jar is not the peanut butter jelly bean jar and the same goes for the rest.
You also need to realize that it is the jar labeled p.b./grape, labelled as the mix jar, that is your best hope. If you choose a jelly bean out of there, then you will know whether that jar is peanut butter or grape jelly jelly beans. It can’t be the mix jar because i already said that every jar is mislabeled.
Once you know that jar 3 is either peanut butter, or grape jelly, then you know the other jars also. If it is peanut butter, then jar 2 must be mixed because it can’t be grape (as its labeled) and it can’t be peanut butter (that’s jar 3). Hence jar 1 is grape.
If jar 3 is grape, then you know jar 1 must be the mix because it can’t be p.b. (as its labeled) and it can’t be grape (that’s jar 3). Hence jar 2 is peanut butter.

Palindrome dates

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This year on October 2, 2001, the date in MMDDYYYY format will be a palindrome (same forwards as backwards).
10/02/2001
When was the last date that this occurred on?See Solution : Palindrome dates

One year can have only one palindrome as the year fixes the month and date too, so the year has to be less than 2001 since we already have the palindrome for 10/02. It can’t be any year in 1900 because that would result in a day of 91, same for 1800 down to 1400. it could be a year in 1300 because that would be the 31st day. So whats the latest year in 1300 that would make a month? When i first solved it, 12th month came to my mind as we have to find the latest date, so i thought it would be 1321. But we have to keep in mind that we want the maximum year in 1300 century with a valid date, so lets think about 1390 that will give the date as 09/31, is this a valid date… ? No, because September has on 30 days, so last will be the 31st August. Which means the correct date would be 08/31/1380.

Balls in a bag

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You have 20 blue balls and 14 red balls in a bag. You put your hand in and remove 2 at a time. If they’re of the same color, you add a blue ball to the bag. If they’re of different colors, you add a red ball to the bag. (assume you have a big supply of blue & red balls for this purpose. note: when you take the two balls out, you don’t put them back in, so the number of balls in the bag keeps decreasing). What will be the color of the last ball left in the bag?

Once you tackle that, what if there are 20 blue balls and 13 red balls to start with?

See Solution : Balls in a bag

There are 3 possible cases of removing the two balls…

a) If we take off 1 RED and 1 BLUE, in fact we will take off 1 BLUE
b)If we take off 2 RED, in fact we will take off 2 RED (and add 1 BLUE)
c) If we take off 2 BLUE, in fact we will take off 1 BLUE
So In case of (a) or (c), we are only removing one blue ball, but we always take off red balls two by two.

1) 20 Blue, 14 Red balls

If there are 14 (even) number of red balls, we can not have one single red ball left in the bag, so the last ball will be blue.

2) 20 Blue, 13 Red balls

Now as the no. of red balls is odd, there will be one single red ball in the bag with other blue balls, and whenever we remove 1 red and 1 blue ball, we end up taking off only the blue ball. So the red ball will be the last ball in the bag.

Calendar cube

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A man has two cubes on his desk. Every day he arranges both cubes so that the front faces show the current day of the month. what numbers are on the faces of the cubes to allow this?

See Solution : Calendar cube
We have two cubes, that means 12 faces or 12 numbers one on each face. The all possible dates are 1 to 31, that includes 11 and 22. So 1 and 2 should be there on both cubes. It means we need 12 digits (0-9 and 1, 2). Now if we see the distribution of numbers on each faces, we have 1 and 2 on both cubes. For 30 we need 0 and 3 on different cubes. So lets say, first cube has 1, 2, 3, 4, 5, 6 and other one has 0, 1, 2, 7, 8, 9. It looks fine, but we we notice, the man uses both the cubes for each day, so how do we show 07, 08, 09 ???

So that means we need to have 0 on both the cubes, but that makes it 13, but there are only 12 faces. Thats the trick in this question, we can place cubes upside down too, now which is the number we can use both the ways, yes its 6, it can be used as 9 and then we can have all the possible dates.

so first cube : 0, 1, 2, 3, 4, 5

second cube : 0, 1, 2, 6, 7, 8

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