Saturday, November 3, 2012

If 1.5 dogs eat 1.5 pounds of meat in 1.5 days, how much will 16 dogs eat in 4 days?

1.5 dogs eat 1.5 pounds of meat in 1.5 days this means that 1.5 dogs eat 1 pound of meat/day
or to get rid of the decimal, 3 dogs eat 2 pounds of meat/day so 16 dogs would eat 32/3 pounds of meat/day
because 2*16/3=32/3 so in 4 days they will have eaten 128/3 pounds of meat because 4*32/3=128/3

the answer is therefore 128/3 pounds of meat which can written as 42 2/3 .

we love comment so lets us know if any thing wrong or other way to approach the same :)

Sunday, September 30, 2012

How will you represent (3AC)14 in base 7 numeral system.

It can be done using this formula!
If a no X is represented in base 14 as
X = an a(n-1) a(n-2) .... a(0)
then in base 7 we can write it as
q=((2^1)a(1) + p/7)%7
r=((2^2)a(2) + q/7)%7
nth term=((2^n)a(n) + (n-1)th term/7)%7
(will go further because a no. in base 14 will require more digits in base 7).
The logic is simple, just based on properties of bases, and taking into account the fact that 7 is half of 14. Else it would have been a tedious task.
Eg. here it is given 3AC.
C =12;
so last digit is (2^0 * 12)%7 = 5
next digit is (2^1 * 10 + 12/7)%7 = (20+1)%7=21%7=0
next is 3;
next digit is (2^2 * 3 + 21/7)%7 = (12+3)%7=15%7=1
next is nothing(0);
next digit is (2^3 * 0 + 15/7)%7 = (0+2)%7=2%7=2
Hence, in base 7 number will be, 2105.

Now try how you can perform any base to any other base conversion ?

You are given n dice. Each die has m sides i.e has values from 1 to m. Given m,n,x calculate the probability that the sum of all the dice is greater than or equal to x.

Example  n=4 m=2 x=8 ans: 1/16=0.0625  n=2 m=6 x=3 ans:35/36=0.972

Monday, September 24, 2012

You are given two hourglasses. One measures 4 minutes and one measures 7 minutes. How would you measure exactly 9 minutes?

StepTime4 minute timer7 minute timer
10 minStartStart
24 minsFlip3 minutes left
37 mins1 minute leftFlip
48 minsStopFlip (1 minute left)
59 mins

Tuesday, August 21, 2012

You have 50 red marbles and 50 blue marbles. Your have two jars. If I am to choose a jar at random, what ratios of marbles would you put in each jar to increase the probability of me choosing a red marble?

Determine which jar has the heavier marbles in the least amount of weighings

You have 10 jars .Each jar has 100 marbles
9 of the 10 jars -> each marble weighs 1 gram each
last jar -> each marble weights 1.1 grams each
electronic weight scale
    - place what you want to weight on it
    - hit a button
    - get the total weight

find the defective jar ?

Answer:Take 1 marble from jar 1, 2 from jar 2, 3 from jar 3 and so u have 55 marbles in all
Weigh them all together...if total weight is 55.1....then jar 1 is the one with obese marbles...if 55.2..then jar 2...and so on...  again 55.3 can't be the combination of jar 1 and 2 as (.1 +.2 ) since defective jar is only one.

Wednesday, February 22, 2012

Three Doors, 1 Prize Puzzle , One of Toughest Puzzle on Probabality

You are on a gameshow and the host shows you three doors. Behind one door is a suitcase with $1 million in it, and behind the other two doors are sacks of coal. The host tells you to choose a door, and that the prize behind that door will be yours to keep.
You point to one of the three doors. The host says, "Before we open the door you pointed to, I am going to open one of the other doors." He points to one of the other doors, and it swings open, revealing a sack of coal behind it.
"Now I will give you a choice," the host tells you. "You can either stick with the door you originally chose, or you can choose to switch to the other unopened door."
Should you switch doors, stick with your original choice, or does it not matter?

Monday, February 20, 2012

Einstein's Riddle


There are no tricks, just pure logic, so good luck and don't give up.

1. In a street there are five houses, painted five different colors.

2. In each house lives a person of different nationality

3. These five homeowners each drink a different kind of beverage, smoke different brand of cigar and keep a different pet.  



   1. The British man lives in a red house.
   2. The Swedish man keeps dogs as pets.
   3. The Danish man drinks tea.
   4. The Green house is next to, and on the left of the White house.
   5. The owner of the Green house drinks coffee.
   6. The person who smokes Pall Mall rears birds.
   7. The owner of the Yellow house smokes Dunhill.
   8. The man living in the center house drinks milk.
   9. The Norwegian lives in the first house.
   10. The man who smokes Blends lives next to the one who keeps cats.
   11. The man who keeps horses lives next to the man who smokes Dunhill.
   12. The man who smokes Blue Master drinks beer.
   13. The German smokes Prince.
   14. The Norwegian lives next to the blue house.
   15. The Blends smoker lives next to the one who drinks water.


Feel free to comment your answers and share to your friends.

Thursday, February 9, 2012

Manager and Engineer Puzzle

The FBI has surrounded the headquarters of the Norne corporation. There are n people in the building. Each person is either an engineer or a manager. All computer files have been deleted, and all documents have been shredded by the managers. The problem confronting the FBI interrogation team is to separate the people into these two classes, so that all the managers can be locked up and all the engineers can be freed. Each of the n people knows the status of all the others. The interrogation consists entirely of asking person i if person j is an engineer or a manager. The engineers always tell the truth. What makes it hard is that the managers may not tell the truth. In fact, the managers are evil geniuses who are conspiring to confuse the interrogators.
  1. Under the assumption that more than half of the people are engineers, can you find a strategy for the FBI to find one engineer with at most n-1 questions?
  2. Is this possible in any number of questions if half the people are managers?
  3. Once an engineer is found, he/she can classify everybody else. Is there a way to classify everybody in fewer questions?

How many sequences of n coin tosses have no consecutive heads?

Thursday, January 12, 2012

You have a circle with radius r. What is the probability that a point lies more towards the circumference than towards the centre.

Probability of point lying close to circumference would be mean point is in the circle with area from r/2 from center till outside in the figure below it would be in blue part of the circle.

Probability = Area covered by circle from point r/2 towards circumference/ Area of the circle
In the above given circle formula would be = Area of blue circle/ Area of circle
= pi r pow 2/ 4 / pi r 2 = 1/4

minimize the time taken by all 4 to cross the bridge.

There is a bridge and to travel on it there should be a torch light available.
Only two persons can cross the bride at a time. There are 4 persons on the other side of the bridge. They require 1,2,5,10 minutes respectively to cross the bridge. How much time should it take for all 4 to go to the other side of the bridge.

Remember there is only one torch light. So if two persons cross the bridge with torch light, one has to come back with torch light for others to use torch light. If person with 1 min and 10 minutes cross the bridge together, they take 10 minutes together to cross the bridge(maximum of the time taken by two people).

Your criteria is to minimize the time taken by all 4 to cross the bridge.

How many regions (open or closed) can n straight lines give at the maximum?

Let f(n) denote the number of regions for n lines no three of which coincide in one point. This latter condition will guarantee the maximum. Obviously, f(0)=1, f(1)=2, f(2)=4, f(3)=7, etc.

Now, given n-1 lines, there are already f(n-1) regions. With the introduction of the n-th line, imagine the points of intersection of the pre-existing n-1 lines fall on one side of the n-th line. So, on one side of the n-th line -- where the points of intersection lie -- you still have f(n-1) regions, whereas on the other side of the n-th line, you now have n regions. So, for these n lines before us, the total number f(n) of regions must be equal to f(n-1)+n.

So, f(n)=f(n-1)+n. With the initial condition(s) given earlier, this gives us f(n)=1+n(n+1)/2.

find the optimal start point on the track such that you never run out of fuel and complete circuit.

You have a circular track containing fuel pits at irregular intervals. The total amount of fuel available from all the pits together is just sufficient to travel round the track and finish where you started. Given the the circuit perimeter, list of each fuel pit location and the amount of fuel they contain, find the optimal start point on the track such that you never run out of fuel and complete circuit.

Wednesday, January 4, 2012

You have a dice which has numbers from 1..6. Another dice is provided to you empty; you need to place numbers on this dice so that whenever both dices are used there is an equal probablity of getting a sum between 1.. 12

otally there are 6*6 possible combinations from two dices. Since we need the sum from 1..12 with equal probability there should be three combinations among these 36 which sum up to each number in 1 to 12 to achieve a equal probability for all sums.

Other Dice Numbers Presently
_ _ _ _ _ _

etting 1 we need 0 on the other dice
Getting 12 we need 6 on the other dice

State of Numbers on other dice now
0 6 _ _ _ _

We need to get SUM=1 in three different combinations and only possibility is 1 and 0 on both dices so put two more 0's on second dice and we have three ways to get 1.

State of numbers on other dice now 0 6 0 0 _ _

Now all the sum's from 7 to 12 can be achieved with one of the numbers be 6 and using the other numbers from first dice. Eg:: 7 can be 1 and 6 from dice 1 and dice2 and if we put two more sixes in dice2 then again 1 and other 6's on dice2 give us 3 ways for seven
Similarly for 8 we can 2 from dice1 and the three sixes one by one we will have 3 ways for 8 and so on for all numbers from 7 to 12. Again we can use the three 0's on dice2 in the similar way we have used three sixes above for getting three ways for sum 1 to 6.

State of numbers on other dice now 0 6 0 0 6 6