### #1 - Unlock The Distance Puzzle

Distances from you to certain cities are written below.
BERLIN 200 miles
PARIS 300 miles
ROME 400 milesAMSTERDAM 300 miles
CARDIFF ??? miles
How far should it be to Cardiff ?

100 miles. Each vowel is worth 300 and each consonant is worth -100. These are totalled in each city name to give the distance

### #2 - River Riddle

Four people need to cross a dark river at night.They have only one torch and the river is too risky to cross without the tourch. if all people cross simultanoesly then torch light wont be sufficient.Speed of each person of crossing the river is different.cross time for each person is 1 min, 2 mins, 7 mins and 10 mins. What is the shortest time needed for all four of them to cross the river ?

The initial solution most people will think of is to use the fastest person as an usher to guide everyone across. How long would that take? 10 + 1 + 7 + 1 + 2 = 21 mins. Is that it? No. That would make this question too simple even as a warm up question.

Let's brainstorm a little further. To reduce the amount of time, we should find a way for 10 and 7 to go together. If they cross together, then we need one of them to come back to get the others. That would not be ideal. How do we get around that? Maybe we can have 1 waiting on the other side to bring the torch back. Ahaa, we are getting closer. The fastest way to get 1 across and be back is to use 2 to usher 1 across. So let's put all this together.

1 and 2 go cross
2 comes back
7 and 10 go across
1 comes back
1 and 2 go across (done)

Total time = 2 + 2 + 10 + 1 + 2 = 17 mins

### #3 - Measuring Time Interview Problem

I possess two wires. Both of them have an inconstant thickness but both of them burns completely in sixty minutes. The problem is that I want to measure forty-five minutes while using these two wires.

How can I measure if cutting the wire in half is not possible due to non-homogeneous construction ?

45minutes

It is not as hard as it seems. I will burn one wire on both ends and the other wire at one end only. The first wire will burn completely in thirty minutes and at that very moment, I will burn the other end of the second wire and it will burn in fifteen minutes. Thus both of the wires will be burnt in 30 + 15 = 45 minutes.

### #4 - Cross Bridge Puzzle

Four people need to cross a rickety bridge at night. Unfortunately, they have only one torch and the bridge is too dangerous to cross without one. The bridge is only strong enough to support two people at a time. Not all people take the same time to cross the bridge. Times for each person: 1 min, 2 mins, 7 mins and 10 mins. What is the shortest time needed for all four of them to cross the bridge?

17 mins

The initial solution most people will think of is to use the fastest person as an usher to guide everyone across. How long would that take? 10 + 1 + 7 + 1 + 2 = 21 mins. Is that it? No. That would make this question too simple even as a warm up question.

Let’s brainstorm a little further. To reduce the amount of time, we should find a way for 10 and 7 to go together. If they cross together, then we need one of them to come back to get the others. That would not be ideal. How do we get around that? Maybe we can have 1 waiting on the other side to bring the torch back. Ahaa, we are getting closer. The fastest way to get 1 across and be back is to use 2 to usher 1 across. So let’s put all this together.

1 and 2 go cross
2 comes back
7 and 10 go across
1 comes back
1 and 2 go across (done)

Total time = 2 + 2 + 10 + 1 + 2 = 17 mins

### #5 - Labor Day Maths Puzzle

Manish drive to his office at 20km/hr. After reaching office, he realize that today is a holiday of 'Labor Day'.He went back at average speed of 30km/hr. Discounting the time spent in the stoppage what was his average speed of his journey ?

24km/hr (not 25km/hr which might be guessed by many)

D : distance traveled
T1 : time of going to office
T2 : time of returning back
Y : Average Speed

D = 20 * T1
T1 = D/20

D = 30 * T2
T2 = D/30

2D = Y(T2 + T1)
2D = Y(D/20 + D/30)
2D = Y(3D/60 + 2D/60)
2D = Y(5D/60)
Y = 120D/5D

=> Y = 24

### #6 - Maths Logical Problem

Accidentally, two trains are running in the opposite direction and enter a tunnel that is 200 miles long. A supersonic bird that has fled the lab and taken shelter in the tunnel starts flying from one train towards the other at a speed of 1000 mph. As soon as it reaches the second train, he starts flying back to avoid collision and meets the first train again at the other end. The bird keeps flying to and fro till the trains collide with each other.

What is the total distance that the supersonic bird has traveled till the trains collided?

Let us consider the length of the tunnel first; which is 200 miles. Now, the trains are running on the same speed which means that they will collide at the center of the tunnel and will take an hour to reach the center. Now the bird is travelling at a speed of 1000 mph and it is flying for an hour (since the trains will take an hour to collide). Thus the bird will travel 1000 miles in the process.

### #7 - Infosys IQ Interview Question

You are frying fishes on a pan which can accommodate only 2 fishes at one time. It takes 5 minutes to fry one side of the fish.

What is the shortest time in which you can fry 3 fishes in the same pan ?

15 minutes.

Put 2 fishes in the pan. Fry for 5 minutes and then take out one and turn the other one. Also put in the third fish. Now fry for 5 minutes. The fish that you turned must be ready now so take it off while turning the other one. Also put in the fish which was fried from one side only in the pan. Fry them for 5 minutes and all the three fishes are fried now.

### #8 - Measure Time Interview Puzzle

There are two candles. Both will only burn exactly for an hour. How will you use these two candles to measure forty-five minutes?

Burn one candle from both the ends and simultaneously burn the other candle from just one end. In half an hour, the first candle would have been burnt fully and the second one would have been burnt half. Now light the other end of the second candle as well. In this way, the second candle will take only half the time (30/2 = 15) to burn fully.

Thus, you will have measured forty five minutes.

### #9 - Drive Time Maths Puzzle

Alpha and Beta have same amount of petrol in their respective cars. Alpha can drive for 4 hours whereas Beta can drive for 5 hours with the amount of petrol left.

Both of them start driving at the same time. After a few hours, they find out that the amount of petrol remaining in one of the cars is four times the other car.

Can you calculate for how long did they drive the cars?

3 3/4 hours

Assume that the initial amount of petrol in each of the petrol tanks be P.
Now, let us assume that they drove for H hours.

In that case, the amount of petrol used by one car in H hours will be = P*H /4
The amount of petrol used by the other car will be = P*H/5

Therefore, the petrol left in one car = P - P*H / 4
Petrol left in the other car = P - P*H / 5

Now, according to the question,
P - P*H / 5 = 4 (P - P*H / 4)
=> H = 15 / 4 = 3 3/4 hours.

### #10 - Maths Time Puzzle

A large water tank has two inlet pipes (a large one and a small one) and one outlet pipe. It takes 3 hours to fill the tank with the large inlet pipe. On the other hand, it takes 6 hours to fill the tank with the small inlet pipe. The outlet pipe allows the full tank to be emptied in 9 hours.

What fraction of the tank (initially empty) will be filled in 0.64 hours if all three pipes are in operation? Give your answer to two decimal places (e.g., 0.25, 0.5, or 0.75).

In one hour,

Soln 1:
the large inlet pipe fills 1 / 3 of the tank;
the small inlet pipe fills 1 / 6 of the tank;
the outlet pipe empties 1 / 9 of the tank; and therefore
all three pipes together fill [ (1 / 3) + (1 / 6) − (1 / 9) ] of the tank.

Fraction of the tank that will be filled in 0.64 hours =
0.64 [ (1 / 3) + (1 / 6) − (1 / 9) ] = 0.25.

Soln 2:
Let V be the total volume of the tank. From equation (1),

Flow Rate (large inlet pipe) = V / 3
Flow Rate (small inlet pipe) = V / 6
Flow Rate (outlet pipe) = V / 9.

Substituting in equation (2),

Rate of Accumulation in tank = (V / 3) + (V / 6) − (V / 9).

Using the above result in equation (1),

Time required to fill the complete tank = V / [ (V / 3) + (V / 6) − (V / 9) ].

Note that V cancels out on simplifying the above expression.

Fraction of the tank that will be filled in 0.64 hours =

0.64 [ (1 / 3) + (1 / 6) − (1 / 9) ] = 0.25.