Train and 2 boys problem

Question:
2 boys are walking on railway bridge.
After they walked 2/3 of the bridge length, they observed that a train is coming from opposite to their way.
Each boy ran into a different directon with the speed of 10kmph.
Each boy escaped from train at the exact instant in which the train would have hit him.
What could be the speed of train?

Train and 2 boys problem

River cross problem

3 men and 3 ghosts are there on one side of river(say side1) and they have to reach another side of river(say side2).
And
1. There is only one boat.
2. The capacity of boat is two members at a time.
3. At any side of the river, if number of ghosts is greater than number of men, then the ghosts will eat the men.
How could they reach safely?

Initially at

Side1: 3 ghosts and 3 men
Side2: none

Step1: First send 2 ghosts to another side.
Side1 : 1 ghost and 3 men
Side2 : 2 ghost

Step2: Send back one ghost to side1.
Side1: 2 ghosts and 3 men.
Side2: 1 ghost.

Step3: Send 2 ghosts to Side2
Side1: 3 men
Side2: 3 ghosts

Step4: Send back 2 ghosts to Side1
Side1: 2 ghost and 3 men
Side2: 1 ghosts

Step5: Send 2 men to Side2
Side1: 1 ghost and 1 man
Side2: 2 ghosts and 2 men

Step6: Send back 1 ghost and a man to Side1
Side1: 2 ghost and 2 men
Side2: 1 ghost and a man

Step7: Send 2 men to Side2.
Side1: 2 ghosts
Side2: 1 ghosts and 3 men

Step8: Send back 1 ghost to Side1
Side1: 3 ghosts
Side2: 3 men

Step9: Send 2 ghosts to Side2
Side1: 1 ghost
Side2: 2 ghost and 3 men

Step10: Send back 1 ghost to Side1
Side1: 2 ghosts
Side2: 1 ghost and 3 men

Step11: Send 2 ghost to Side2
Side1: none
Side2: 3 ghosts and 3 men

River cross problem

Horse race problem

Questions:
You have 25 horses and you have to pick 3 fastest horses.
1. There are only 5 tracks.
2. In each track only one horse can run. So you can conduct each race with 5 horses only.
3. you don't have timer or stop watch.
How many minimum number of races can you conduct to pick 3 fastest horses?

Answer is 7.
Step1: Devide horses into 5 groups and Conduct race for each group. arrange them based on their Position as shown in figure.

Step2: Take fastest horse from each group conduct a race among them and assume that positions are as shown in below figure.

Step3: Arrange the groups as shown in figure based on their position.

Now H01 horse is the fastest horse.
Group4 and Group5 are not eligible for top 3.
H12 and H13 are not eligible for top3 because they are slower than H11(which is in top3).
H08 is not eligible for top3. because it is slower than H07(which is top3).

Step4: Now H02,H03,H06,H07 and H11 are in top2 and top3 position.
Conduct race among those and pick top2 horses.

So top 3 horses are H01 and top2 horses picked from step4.

Horse race problem

Measure 4 litres from 5 and 3

Q. You have a 3 litre jug and a 5 litre jug.
The jugs have no measurement lines.
How could you measure exactly 4 litre of water using only those two jugs ?

Initially
5 litre jug: 0 litres
3 litre jug: 0 litres

Step1: Fill the 5 litre jug and pour into 3 litre jug.
5 litre jug: 2 litres
3 litre jug: 3 litres

Step2: Empty the 3 litre jug.
5 litre jug: 2 litres
3 litre jug: 0 litres

Step3: Pour remaining 2 litres into 3 litres.
5 litre jug: 0 litres
3 litre jug: 2 litres

Step4: Fill 5 litre jug again.
5 litre jug: 5 litres
3 litre jug: 2 litres

Step5: Pour One litre into 3 litre jug because it has only one litre empty.
5 litre jug: 4 litres
3 litre jug: 3 litres
Finally you have 4 litres of water into the 5 litre of jug.

Measure 4 litres from 5 and 3

Candles problem

Question:
You have 2 candles and matchbox.
Each candle take exactly 30 minutes to burn from one end to another end.
How could you burn 2 candles within 45 minutes if you don't have stopwatch?

Step1: Burn one candle upto 30 minutes.
Step2: Burn another candle both sides it takes 15 minutes to burn. So total time is 45 minutes.

Candles problem

Apples and oranges problem

Questions:
1. There are 3 boxes each one contains Apples,Oranges and "Apples and Oranges" respectively.
2. There are three labels "Apples", "Oranges" and "Apple and Oranges".
3. Each Box is labelled wrongly i.e, The box labelled with Apples doesn't contains apples.
How could you label correctly by opening only one box?

Open the Box labelled with "Apples and Oranges".
Case1:
1. if it has Apples then the box labelled with "Apples" must contain Oranges.
because if it has mixed fruits then the box labelled with "Oranges" should contain Oranges which is not wrong.
2. Other one labelled with "Oranges" has mixed fruits.

Case2:
1. if it has Oranges then the box labelled with "Oranges" must contain Apples.
because if it has mixed fruits then the box labelled with "Apples" should contain Apples which is not wrong label.
2. Other one labelled with "Apples" has mixed fruits.

Apples and oranges problem

Finding required number

Question: Find a number that satisfy below conditions?
1. If you divide with 10, you will be get 9 as remainder.
2. If you divide with 9, you will be get 8
3. If you divide with 8, you will be get 7
4. If you divide with 7, you will be get 6
5. If you divide with 6, you will be get 5
6. If you divide with 5, you will be get 4
7. If you divide with 4, you will be get 3
8. If you divide with 3, you will be get 2
9. If you divide with 2, you will be get 1

The required number is 2519
How to do this: First, find the LCM of 1,2,3,4,5,6,7,8,9,10 reduce 1 number to satisfy the given condition.

Finding required number

Finding poisoned can

Question:
There are 30 Cans which have milk.
1. Among those one can got fully poisoned.
2. So even if a man taste a 1/8th of single drop of that can's milk he will die exactly after 12 hours.
How many minimum number of rats you need to test and find the poisoned can?

answer is 4. 
If a man taste the 8th part of one drop he will die exactly after 12 hours. For the same thing rat will die less than 12 hours. because of it's low immunity power. 
Step1: Take first 15 cans and do as shown in figure.


After 12 hours:

1. If first Rat only dies then can-1 is poisoned. 
2. If second Rat only dies then can-2 is poisoned. 
3. If 3rd Rat only dies then can-3 is poisoned. 
4. If 4th Rat only dies then can-4 is poisoned. 
5. If 1st Rat and 2nd Rat will die then can-5 is poisoned and so on. 
at last if no rats are die then no can is poisoned. In this cases, proceed step 2. 
Step2: Do the same thing for the next 15 cans with  those rats and find the poisoned one. 

Finding poisoned can

finding odd one from 9 coins

Question :
1. You have 9 coins with same size.
2. Among those 1 Coin has less weight.
3. You have Weighting scale to find the Odd Coin.
In How many minimum number of times can you use Weighting scale to find the odd coin.?

Answer is 2.

Step1: Take 3 coins in one side of scale and 3 more coins in another side. and judge the scale
1. if both groups are equal then then take the remaining 3 coins to proceed Step2.
2. Otherwise take the group which has less weight to proceed Step2.

Step2: Now you have 3 coins. take one coin in one side and one more coin in another side and judge the scale.
1. if both are equal then then the 3rd coin is the odd one.
2. Other wise the coin which has less weight is the odd one.

finding odd one from 9 coins

100 switches problem

Question:
There are 100 switches numbered from 1 to 100. initially in off position.
1. First time I switched on each switch.
2. Second time I switched off switches which has even numbers.
3. Third time I put the switches in opposite position of their state which are numbered with 3 multiplier like 3, 6, 9 and so on.
4. fouth time I put the switches in opposite position of their state which are numbered with 4 multipliers like 4, 8, 12, 16 and so on.
5. I continued this procedure upto 100 times now How many switches are in on position and what are they?

Initially all are off position.

Step1: put all switches in on. So all are in on position.

Step2: put all even number off switch in off position.

Step3: put the switches in opposite position for multipliers of 3 like 3, 6, 9, 12 ...

Step4: put the switches in opposite position for multipliers of 4 like 4, 8, 12, 16 ...

Step5: put the switches in opposite position for multipliers of 5 like 5, 10, 15, 20....

Now Observe that switches 1,2,3,4 and 5 are will never be changed their positions for remaining future attempts.

Switch 1 and 4 are in on Position.
The factors of 4 are: 1,2 and 4.
The number off factors of 4 is 3.
So we will attempt the 4th switch 3 times(odd number).
So it is opposite to the initial position.
that is why it is in on position.

Switch 2,3,and 5 are in off position.
2,3 and 5 has even number of factors.(for 2 => 1 & 2; for 3 => 1 & 3; for 5 => 1 & 5)
So the number of attempts on these switches are even.
that is why They are in initial position.

All non-perfect squares numbers have even number of factors that is they are in initial position.
All perfect squares numbers have odd number of factors. So switches 1,4,16,25,36,49,64,81 and 100 are in on position.

100 switches problem

bulbs and switches problem

Question:
There are 3 bulbs in one room and 3 switches in another room.
Each bulb is connected to one of the switches.
you are in the room which has switches and all switches are initially in off position.
you can not see the light of another room.
How could you figure out which switch connected to which bulb by entering only one time into the room which has bulbs ?

Step1: Switch on the first switch and wait few minutes and switch off that. 
Step2: Switch on second switch and enter into the room. 
Step3: Touch all bulbs with hand. 
How to Figure:
1. Now the bulb which is giving light is connected to Second bulb.
2. The bulb which has hot temperature is connected to First switch.
3. The Other bulb is connected to 3rd switch.

bulbs and switches problem

7 persons and diamonds

Two persons have diamonds and when they distributed each other they have one remained. If they distributed with three persons equally then also one diamond will be remained.If they distributed with four persons equally then also one diamond will be remained.
If they distributed with five persons equally then also one diamond will be remained.If they distributed with six persons equally then also one diamond will be remained.
If they distributed with seven persons then all members got equal number of diamonds.
Totally how many diamonds are there?

answer is 301.

7 persons and diamonds

bridge cross problem

Queston :
There are 4 persons need to cross the bridge.
Unfortunately, they have only one flash light and it has enough light for only 17 minutes.
One can cross the bridge in 1 minute, another in 2 minutes, the third in 5 minutes, and fourth in 10 minutes to cross the bridge.
Only two members can cross the bridge at a time. How could they cross the bridge within 17 minutes?

Initially:
Side1: 1 min, 2min, 5min, 10min. 
Side2: none 
Totaltime: 0min 

Step1: First send 1 minute and 2 minute guys to another side(takes 2 minutes).    
Side1: 5min, 10min 
Side2: 1min, 2min 
Totaltime: 2min 

Step2: Send back 1min guy to Side1. 
Side1: 1min, 5min, 10min 
Side2: 2min 
Totaltime: 3min

Step3: send 10 minute and 5 minute guys to side2(takes 10 minutes). 
Side1: 1min 
Side2: 10min, 5min, 2min 
Totaltime: 13min 

Step4: Send back 2min guy to Side1. 
Side1: 1min,2min 
Side2: 10min, 5min 
Totaltime: 15min 

Step5: Send 2min guy and 1 min guy to Side2(takes 2min).        
Side1: none        
Side2: 10min, 5min, 1min, 2min     
Totaltime: 17min 

bridge cross problem