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Intec Interview Puzzle
Posted on :05-03-2016
Q1. A train is moving from station A to station B at 15 mph. Another train is moving in the opposite direction from station B to A at a speed of 20 mph. A vulture is flying from Station A to station B at 25 mph. When the vulture reaches at the train moving from Station B to A, it starts flying back and flies at the same speed till it reaches the train moving from station A to B and starts flying in the reverse direction again. The vulture keeps flying to and fro till both the trains collide with each other.
What is the total distance that is traveled by the vulture?
To solve this question, we will have to use the concept of relative speed.
Let us assume that the distance between the station A and B is D Miles
The relative speed at which the train are approaching each other is 20 + 15 = 35 mph
When the train collides with each other, the sum of the distance covered will be the distance between the stations i.e. D.
Now Speed = Distance/Time
Which means that the trains will collide in d/35 hours after their start
Another constant things we have here is the speed of the vulture i.e. 25 mph.
Thus by the time the trains collide, the vulture would have covered
25*(D/35) miles = 5D/7 Miles
Q2. A new company has launched a range of juicy candies. There are two flavors of which these juice candies will be available for retail - Peach and Grape. The toffees are packed in three different boxes with one containing peach candies, one containing grape candies and one containing a mixture of both. It is informed that the packaging person has accidentally mislabeled all the three boxes.
If you are asked to label them correctly, how many candies will you have to pick and from which jar in order to label every box correctly?
There are many ways in which you can think a solution for the problem. Your approach should be to minimize the efforts. What if you took out one candy from the box labeled Peach and Grape?
What you have to consider is the fact that every box is labelled incorrectly. Thus if you pick a candy from the box labelled with Peach and Grape, it is evident that the box will not have a mixture of both (as it is labelled wrong). Thus you will get to know which candies are present in that box.
Suppose the candy you take out turns out to be peach flavored, then the box labeled as Grapes is the one which contains a mixture of both peach and grapes. This is because it is labelled incorrectly and since we have already figured out the box with Peach candies, we can be confident of the fact. And then the jar labelled as peach will be the one containing grape flavored candies.
If the candy you take out from the mixed labeled box is Grape flavored, you can use the same logic to find out the rest.
Consider picking a candy from the other two boxes instead of what is told in the solution and you will find out that you will have to take out a candy from each of the boxes before being sure.
Q3. You are a king of an empire. You have a servant working in your palace. He works all the seven days and you only pay him in the form of gold bar. You must pay the worker for his work every day at the end of the day.
If you are only able to make two breaks in the gold bar, how will you pay the servant if the servant works for the equal time every day and thus equal amount must be paid at the end of the day?
If you are thinking of how to cut in such a way that you are able to make seven equal pieces, you are thinking in the wrong direction. You should give more stress on how to make the transaction with the worker.
You have to break the gold bar two times so you have the following sizes of the bar:
1/7 ---- Bar A
2/7 ---- Bar B
4/7 ---- Bar C
When the day 1 ends:
Give Bar A (You remain with Bar B and Bar C; Worker possess - Bar A)
When the day 2 ends:
Give Bar B and take the Bar A back (You remain with Bar A and Bar C; Worker possess - Bar B)
When the day 3 ends:
Give Bar A (You remain with Bar C; Worker possess - Bar A and Bar B)
When the day 4 ends:
Give Bar C and take back Bar A and Bar B (You remain with Bar A and Bar B; Worker possess - Bar C)
When the day 5 ends:
Give Bar A (You remain with Bar B; Worker possess - Bar A and Bar C)
When the day 6 ends:
Give Bar B and take back Bar A (You remain with Bar A; Worker possess - Bar B and Bar C)
When the day 7 ends:
Give Bar A (You remain with no bar; Worker possess - Bar A, Bar B and Bar C)
Thus by the end of seven days, you have taken care of everything.
Q4. Can you determine how many times do the minute and hour hands of a clock overlap in a day?
24 times is the most common answer however that is completely wrong. The right answer will be 22 times.
Let us prove it by some simple mathematics.
Assume that it takes T hours for the minute hand to complete T laps. The hour hand will complete T/12 Laps in the same time.
Consider the situation when the hour hand and the minute hand will overlap for the first time; the minute hand would have completed one lap extra than the hour hands.
Thus, T =T/12 + 1
Considering the above expression we know that the first overlap will take place after t = 12/11 hours i.e. 1:05 am.
Similarly, the second overlap will take place when the minute hand would have completed two more laps than the hour hand.
Considering there are X laps,
T = T/12 + X
Everyone knows that a day comprises of 24 hours. Putting that in the equation we get
24 = 24/12 + X
Solve it and you will get
X = 22
Hence both of the hands will overlap 22 times in 24 hours.
Let us give you with those exact timings as well.
The hands will overlap at 12:00, 1:05, 2:10, 3:15, 4:20, 5:25, 6:30, 7:35, 8:40, 9:45, and 10:50. Do consider the fact that there will be no 11:55. It becomes 12:00.
Q5. In an old town, there lived a pot maker who wanted to sell the last stock of his craft to a merchant. When the merchant asked him how many pots he had, he replied that he was unable to count after 100. But he gave him certain clues to figure out for himself:
If the number of pots are divided by two, there will be one left.
If they are divided by three, there will be one left.
If they are divided by four, there will be one left.
If they are divided by five, there will be one left.
If they are divided by six, there will be one left.
If they are divided by seven, there will be one left.
If they are divided by eight, there will be one left.
If they are divided by nine, there will be one left.
If they are divided by ten, there will be one left.
But if they are divided by eleven, there will be no pot left.
Can you find the number of pots, the pot maker possess?
You may solve this question with different methods but we are only giving away one.
What you have to do in that method is look for a number x to which all the numbers from 2 to 10 divide evenly. Multiplying the numbers together (2*3*4*5*6*7*8*9*10) will give you a big number but we have to find a smaller possibility. What we can do is find the prime factors, a subset of which we can use to form a number from 2 to 10.
2*2*2*3*3*5*7 will certainly give us what we need. The product is 2520 and it is the lowest possible number which can be evenly divided by 2-10.
Now if we add 1 to the number, the result will be 2521 which will satisfy all the conditions from 2 to 10. But will not hold true in case of 11.
If you calculate, you will see that you can multiply 2520 by any integer and add 1 to it to satisfy all the conditions except the divisibility by 11. What we need is an integer Y that can be multiplied by 2520 and if we add 1 to the product, it becomes divisible by 11.
2520/11 leaves a remainder 1. Now let us keep increasing the number of 2520. Suppose if we divide two 2520 by 11, the remainder then will be 1 + 1 = 2. Ten 2520 when divided by 11 will give 10 as remainder and if we add 1 to them, it will be completely divisible. Also it will satisfy all the conditions from 2-10.
Thus our answer is 25201.