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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.

Bridge

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Four friends A, B, C, D
Need to cross a bridge..
A maximum of 2 people can cross at a time..
It is night and they have just 1 lamp.
People that cross the bridge must carry the lamp to see the way..
A pair must walk together at the speed of slower person…

Speeds of
A: 1 minute to cross bridge
B: 2 minutes to cross bridge
C: 5 minutes to cross bridge
D: 10 minutes to cross bridge

Now question is… “what is the total minimum time required by all 4 friends to cross the bridge….?

See Solution : Bridge

Person A : 1 minute
Person B : 2 minutes
Person C : 5 minutes
Person D : 10 minutes

Person C and D are the slowest guys, if they don’t walk together that it self will make it 15 minutes, In this case (the best way to save time is) :
Initially A,B,C,D  ========
First A& B will cross the bridge and A will come back  => time taken 2 + 1 => 3
A,C,D ======== B
Now C & D should cross the bridge together and B will come back => time taken 10 + 2 => 12
Now A & B will cross the bridge together => time taken 2
total time 3 + 12 + 2 => 17

Supersonic bee

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Two trains enter a tunnel 200 miles long, traveling at 100 mph at the same time from opposite directions. As soon as they enter the tunnel a supersonic bee flying at 1000 mph starts from one train and heads toward the other one. As soon as it reaches the other one it turns around and heads back toward the first, going back and forth between the trains until the trains collide in a fiery explosion in the middle of the tunnel. How far did the bee travel?

See Solution : Supersonic bee
This puzzle is a little tricky one. One’s thinking about solving this problem goes like this “ok, so i just need to sum up the distances that the bee travels…” but then you quickly realize that its a difficult (not impossible) summation.
The tunnel is 200 miles long. The trains meet in the middle traveling at 100 mph, so it takes them an hour to reach the middle. The bee is traveling 1000 mph for an hour (since its flying the whole time the trains are racing toward one another) – so basically the bee goes 1000 miles.

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.

Daughter’s ages

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Two MIT math grads bump into each other at Fairway on the upper west side. They haven’t seen each other in over 20 years.

The first grad says to the second: “how have you been?”
Second: “great! i got married and i have three daughters now”
First: “really? how old are they?”
Second: “well, the product of their ages is 72, and the sum of their ages is the same as the number on that building over there..”
First: “right, ok.. oh wait.. hmm, i still don’t know”
Second: “oh sorry, the oldest one just started to play the piano”
First: “wonderful! my oldest is the same age!”

How old are the daughters ?

See Solution : Daughter's ages

We know that there are 3 daughters whose ages multiply to 72. Let’s look at the possibilities…

Ages:          Sum of ages:
1 1 72            74
1 2 36            39
1 3 24            28
1 4 18            23
1 6 12            19
1 8 9             18
2 2 18            22
2 3 12            17
2 4 9             15
2 6 6             14
3 3 8             14
3 4 6             13
After looking at the building number the second man still can’t figure out what their ages are, so that means that the sum of the ages (or building number) must be 14, since that is the only sum that has more than one possibility. Finally the man discovers that there is an oldest daughter. That rules out the “2 6 6” possibility since the two oldest would be twins. Therefore, the daughters ages must be “3 3 8”.

Red and blue marbles

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You have two jars, 50 red marbles and 50 blue marbles. You need to place all the marbles into the jars such that when you blindly pick one marble out of one jar, you maximize the chances that it will be red. When picking, you’ll first randomly pick a jar, and then randomly pick a marble out of that jar. You can arrange the marbles however you like, but each marble must be in a jar.

See Solution : Red and blue marbles

Lets say, we put all the red marbles in jar A and all blue marbles in jar B. Then the probability of getting a red marble is :

jar A : (1/2)*1 = 1/2 (selecting the jar A = 1/2, red marble from jar A = 50/50)
jar B : (1/2)*0 = 0 (selecting the jar B = 1/2, red marble from jar B = o/50)
So probability of getting red marble is 1/2 . Now as we need to maximize the P(getting a red marble), we have to increase the prob of getting a red marble in jar B. If we select jar A, then getting a red marble is guaranteed, but it will also be guaranteed if there is only one red marble in that jar, then also the probability of getting a red marble from jar A is 1/1=1. So now we can place remaining 49 red marbles in jar B, so it increases the prob of getting red marbles in jar B.

So the maximum probability will be :
jar A : (1/2)*1 = 1/2 (selecting the jar A = 1/2, red marble from jar A = 1/1)
jar B : (1/2)*(49/99) = 0 (selecting the jar B = 1/2, red marble from jar B = 49/99)
Total probability = 74/99 (~3/4)

Ants on a triangle

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There are three ants on a triangle, one at each corner. At a given moment in time, they all set off for a different corner at random. What is the probability that they don’t collide ?

See Solution : Ants on a triangle

Solution 1:
Let the three ants are a, b, c.

There are two cases when they will not collide, the one is when they all move clockwise and the other is when they all move anticlockwise.
They will collide if any two ants move towards each other, at the same time the third ant can move in clockwise or in anticlockwise. so for each pair there are 2 such cases. And there are 3 pairs possible (a,b), (b,c) and (c,a). So total 3*2 = 6 cases when they will collide.

So probability that they will not collide is 2/(2+6) i.e. 1/4

Solution 2 :

Consider the triangle ABC. We assume that the ants move towards different corners along the edges of the triangle.

Total no. of movements: 8

A->B, B->C, C->A; A->B, B->A, C->A; A->B, B->A, C->B; A->B, B->C, C->B; A->C, B->C, C->A; A->C, B->A, C->A; A->C, B->A, C->B; A->C, B->C, C->B

Non-colliding movements: 2

A->B, B->C, C->A; A->C, B->A, C->B

(i.e. the all ants move either in the clockwise or anti-clockwise direction at the same time)

So probability of not colliding = 2/8 = 1/4

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

Zeroes in 100 factorial

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How many trailing zeroes are there in 100! (100 factorial) ?

See Solution : Zeroes in 100!
For every factor of 10, there will be one trailing zero similarly for every factor of 5 there will be one trailing zero (as 5*2 =10, and there are enough number of 2’s). But we have to take care of repetitions.

10, 20,…., 90 = 9 zeros

100 = 2 zeros

5, 15, 25……95 = 10 zeros

and 1 extra 5 in each of 25, 50 and 75 = 3 zeros

so total 9+2+10+3 = 24 zeros.

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