You have a pack of 52 playing cards. You draw one card every day, and after drawing, you put the card back into the pack and shuffle it. What is the least number of days you'd need to draw cards in order for you to have a greater than 50% chance of drawing the Ace of Spades at least once? 36. This is a question about probability. Rather than calculating the probability of drawing the Ace of Spades at least once, it's easier to calculate the probability of never drawing it at all, which is (51/52) to the power of the number of days. To have a greater than 50% chance of drawing the Ace of Spades at least once, we want the probability of never drawing it to be less than 50%, or 0.5. So, we solve this equation for the number of days: (51/52)^x < 0.5. The solution to this equation is approximately 36.


A flock of birds is flying in a V formation. There are 25 birds on the left side of the V and 25 birds on the right side. If one bird from the left side moves to the right side, how many more birds will be on the right side than the left side? 2. When one bird moves from the left side to the right side, the left side loses one bird, and the right side gains one bird. So, the difference between the two sides increases by 2. Therefore, after the bird moves, there will be 2 (26 - 24 = 2) more birds on the right side than on the left side.


You borrowed a pen from your friend. This pen can write 500 words on average before running out of ink. After writing a 300-word letter to your friend, you want to know how much longer the pen can write. However, your friend told you that this pen writes twice as fast when the pen is warmer, which happened during the first 150 words. How many more words can you write with this pen? 50. If the pen writes twice as fast when it is warmer, it means it uses twice as much ink for the same number of words. So, the first 150 words you wrote with the warm pen actually consumed ink equivalent to 300 words. Adding the other 150 words written under normal conditions, you've used ink equivalent to 450 words. Therefore, you can write 500 - 450 = 50 more words.


Alice and Bob are playing a game with two dice. However, the dice are not fair, and the probability of rolling a 6 is 1/4, while the probability of rolling any other number is 1/8. Alice wins if the two dice show the same number; Bob wins if the numbers are different. What is the probability that Bob will win the game? 84.38%, 0.8438, 27/32. The probability of rolling a 6 is 1/4. The probability of rolling other 6 numbers is each 1/8. To calculate the probability of Bob winning, we need to first consider the outcomes where the numbers on the two dice are the same (Alice wins). The probability of rolling a 6 on one die is 1/4, and the probability of rolling any other number is 1/8. So, the probability of Alice winning is (1/4 * 1/4) + 6 * (1/8 * 1/8) = 5/32. Therefore, the probability that Bob will win the game is 1 - 5/32 = 27/32 (84.38%).


A vendor is pulling a cart full of watermelons weighing 100 kilograms. He needs to cross a desert to the other side to sell them. Due to the hot and dry climate of the desert, the water content of the watermelons decreases from 99% to 98%. What is the current weight of the watermelons? 50. The weight of the watermelons can be calculated as the weight of the water plus the weight of other solid flesh. Let M1 be the weight of the watermelons before, and M2 be the weight of the watermelons after. The weight of flesh before is (1 - 99%) * M1, and after is (1 - 98%) * M2. The weight of all other substances (Mt) remains the same before and after. Therefore, M2 = 0.5 * M1 = 50 kilograms. Hence, the current weight of the watermelons is 50 kilograms.


Suppose we have a game that assumes we have a fair coin. Two people who are playing can pick two patterns of coin flips. The players then flip a large coin a number of times and record each of the individual flips. Here, H stands for heads and T stands for tails. After the flips, they pay attention to the record of flips and decide which pattern came first in the sequence. If HTH came before HHT, player A won the game, otherwise, B won the game. What is the probability that player B won this game? 0.67, 66.67%, 2/3. Penney's game


You're participating in a race. You overtake the second person. What position are you in? 2. When you overtake the second person, you move into their position, which is second place.


How many times can you subtract 10 from 100? 1. After you subtract 10 from 100 once, you're no longer subtracting 10 from 100. You're subtracting 10 from 90, then 80, and so on.


If red flower is 1/10 more than yellow flowers, then yellow flower is 1/__ less than red flowers? 11. Suppose there are 10 yellow flowers. If red flowers are 1/10 more than yellow flowers, then there would be 1/10 * 10 = 1 more red flower than yellow flowers. So, there are 10 + 1 = 11 red flowers. Now, we want to find out how much less the number of yellow flowers is compared to red flowers. There is 1 less yellow flower than red flowers. To express this as a fraction, we can write it as 1/11. So, the yellow flower is 1/11 less than red flowers.


We divide 50 into 4 natural numbers a,b,c,d such that 2a = b/2 and c+2=d-2. How many divisions we can make? 5. Let's first solve the given equations for a, b, c, and d in terms of one another.
From the first equation, 2a = b/2, we can write b = 4a.
From the second equation, c + 2 = d - 2, we can write d = c + 4.
Now, we know that a + b + c + d = 50. Substituting the expressions for b and d, we get:
a + 4a + c + (c + 4) = 50
5a + 2c = 46.
Now, we need to find the possible values of a and c that satisfy this equation and result in natural numbers for a, b, c, and d.
(a, b, c, d) = (0, 0, 23, 27) or (2, 8, 18, 22) or (4, 16, 13, 17) or (6, 24, 8, 12) or (8, 32, 3, 7).
Therefore, We have found 5 possible divisions of 50 into 4 natural numbers a, b, c, and d that satisfy the given conditions.


A train leaves a station and travels north at a speed of 75 mph. Three hours later, a second train leaves the same station and travels in the same direction at 90 mph. How long will it take for the second train to catch up to the first train? 15. When the second train departs, the first train is 225 miles ahead (75 mph * 3 hours). Each hour, the second train closes the gap by 15 miles (90 mph - 75 mph). Therefore, it will take 225 miles / 15 mph = 15 hours for the second train to catch up to the first train.


We roll a dice and count the number of times we roll. If we get an even number, we will reset the count to zero. We will stop if we get one. What is the expected number of final counts? 1.5. Let E be the expected number of final counts. E = 1P(x=1) + 2P(x=2) + 3P(x=3) + .. + nP(x=n) Where, P(x=1) = 1/6 + (1/2)*(1/6) + (5/6)*(1/2)*(1/6) + (5/6)*(5/6)*(1/2)*(1/6) + ... = 2/3
 Similarly, P(x=2) = (2/3)*(1/3) P(x=3) = (2/3)*(1/3)^2 P(x=n) = (2/3)*(1/3)^(n-1)

E = 1P(x=1) + 2P(x=2) + 3P(x=3) + .. + nP(x=n) = 1.5


100 people are boarding a plane in a line and each of them is assigned to one of the 100 seats on a plane. However, the first person in line forgot his boarding pass and as a result decided to sit down in a random seat. The second person will do the following: Sit in her seat if it still available. If her seat is not available, choose a random seat among the seats remaining and sit there. Each following person sits according to the same rules as the second person. What is the probability the 100th person will be able to sit in her assigned seat? 0.5, 50%, 1/2. The 100th person either gets her own seat, or the first person's (no other seat can be unoccupied when she gets there, since it would have been taken by its owner). The other one of those two seats has taken by someone else. They must have picked it at random, so it's equally likely to be either of the two. For anyone who has a chance of sitting in either the first or the last seat, both are available and equally likely, since picking either means everyone else (before the 100th) can take their assigned seat.


A group of police officers have breathalyzers displaying false drunkenness in 5% of the cases in which the driver is sober. However, the breathalyzers never fail to detect a truly drunk person. One in a thousand drivers is driving drunk. Suppose the police officers then stop a driver at random, and force the driver to take a breathalyzer test. It indicates that the driver is drunk. We assume you don't know anything else about him or her. How high is the probability he or she really is drunk? 1.96%, 0.0196. 1/50.95
0.019627=1.96%


A family has two children. They can be brothers/sisters/big brother and little sister/big sister and little brother with equal probability. If you see one of them is a boy, then what is the probability that the other one is a boy? 0.33, 33%, 1/3. Let's use the following notation for the children: B for a boy and G for a girl. There are four possible combinations of two children: BB, BG, GB, and GG. Since we know that one of them is a boy, we can eliminate the GG combination. So, we are left with three possibilities: BB, BG, and GB.
Now, we want to find the probability that the other child is a boy, given that one of them is a boy. This corresponds to the BB combination. Out of the three remaining possibilities, one of them is BB. Therefore, the probability that the other child is a boy is 1/3.


Ankur's brother has two offsprings and one of the boy was born on Tuesday. What is the probability that Ankur's brother has two boys? Assume an equal chance of giving birth to either sex and an equal chance to giving birth on any day. 13/27, 0.4815, 48.15%. There can be four cases, i.e. either his brother has two boys, or one boy(elder) and one girl, one girl(elder) and one boy, or both of them are girls. And there are 7 equally likely possibilities for any one of the children of taking birth on Tuesday for all these four cases. Therefore, the total cases are 7*4- 1(as there is repetition of two boys two times). The possibility for having two boys will be 7*2- 1. Therefore, the answer would be 13/27.


You have 100 candidates, upon which there exists a total order. The candidates have appointments with you, one at a time, in a random order. From each interview, you can tell exactly how that candidate ranks amongst those you have already examined. At that point, you may either accept or reject the candidate. Any acceptance or rejection is permanent; no candidate, once rejected, may be reconsidered. Your objective is solely to accept the best candidate. What is the strategy for maximizing your probability of doing so, and what is that probability? 1/e, 0.3679, 36.79%. Secretary problem


In a 600 meters circular runway, Alice and Bob run from the same location clockwise. They meet with each other every 12 minitues. If they do not change the speed and run from the same location again while this time Alice runs counterclockwise, then they meet with each other every 4 minites. How long does Alice need to run one circle? 6 or 12. Let Alice's speed be A meters per minute and Bob's speed be B meters per minute. When they run in the same direction (clockwise), their relative speed is |A - B| meters per minute. When they run in opposite directions (Alice counterclockwise), their relative speed is A + B meters per minute.
When they run in the same direction, they meet every 12 minutes, so the distance they cover together in 12 minutes is 600 meters (one full circle). Therefore, we have:
12|A - B| = 600 (1)
When they run in opposite directions, they meet every 4 minutes, so the distance they cover together in 4 minutes is 600 meters (one full circle). Therefore, we have:
4(A + B) = 600 (2)Now we have two cases to consider:
Case 1: A > B In this case, we can rewrite equation (1) as: 12(A - B) = 600
Now, we can find A = 100. In this case, Alice's speed is 100 meters per minute, and she would need 6 minutes to run one circle (600 meters / 100 meters per minute).

Case 2: A < B In this case, we can rewrite equation (1) as: 12(B - A) = 600
We can solve for A = 50. In this case, Alice's speed is 50 meters per minute, and she would need 12 minutes to run one circle (600 meters / 50 meters per minute).
In conclusion, both 6 minutes and 12 minutes are correct answers, depending on whether Alice is faster or slower than Bob.


There are two circles. Circles A and B are externally tangent to each other, and there is no relative sliding at the point where the two circles touch. The radius of Circle A is 1/3 the radius of Circle B. Circle A rolls around Circle B one trip back to its starting point. How many times will Circle A revolve in total? 4. Let the radius of Circle A be r and the radius of Circle B be 3r. When Circle A rolls around Circle B, it traces a circular path with a radius equal to the sum of the radii of both circles, which is r + 3r = 4r.
The distance Circle A travels along this path is equal to the circumference of the circular path, which is 2 * pi * 4r = 8 * pi * r.
The number of times Circle A revolves is equal to the distance it travels divided by its own circumference. The circumference of Circle A is 2 * pi * r.
So, the number of times Circle A revolves is (8 * pi * r) / (2 * pi * r) = 8 / 2 = 4 times.


Alok has three daughters. His friend Shyam wants to know the ages of his daughters. Alok gives him first hint.
1) The product of their ages is 72.
Shyam says this is not enough information Alok gives him a second hint.
2) The sum of their ages is equal to my house number.
Shyam goes out and looks at the house number and tells “I still do not have enough information to determine the ages”.
Alok admits that Shyam can not guess and gives him the third hint
3) The oldest girl likes strawberry ice cream.
Shyam is able to guess after the third hint. Can you guess what is the age of the oldest girl? 8. Below are all possibilities to get 72 from product of three different ages:
1 * 1 * 72 = 72
1 * 2 * 36 = 72
1 * 3 * 24 = 72
1 * 4 * 18 = 72
1 * 6 * 12 = 72
1 * 8 * 9 = 72
2 * 2 * 18 = 72
2 * 3 * 12 = 72
2 * 4 * 9 = 72
2 * 6 * 6 = 72
3 * 3 * 8 = 72
3 * 4 * 6 = 72

2) Sum of the ages is given
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

All sums are unique except 14. So the age sum must have been 14, otherwise, Shyam would have guessed the ages from hint 2 only.

So we have two possible combinations to get a sum of 14
2 + 6 + 6 = 14
3 + 3 + 8 = 14

3) Alok has the oldest girl (not two!!). So the ages must be 3, 3 and 8.


Two trains are on same track and they are coming toward each other. The speed of the first train is 50 km/h and the speed of the second train is 70 km/h. A bee starts flying between the trains when the distance between two trains is 100 km. The bee first flies from first train to second train. Once it reaches the second train, it immediately flies back to the first train … and so on until trains collide. Calculate the total distance travelled by the bee. Speed of bee is 80 km/h. 66.67, 200/3. Let the first train A move at u km/h.
Let the second train B move at v km/h.
Let the distance between two trains be d km
Let the speed of bee be b km/h

Therefore, the time taken by trains to collide = d/(u+v)
Now putting all the known values into the above equation, we get,
u = 50 km/hr
v = 70 km/hr
d = 100 km
b = 80 km/hr

Therefore, the total distance travelled by bee
= b*d/(u+v)
= 80 * 100/(50+70)
= 66.67 km (approx)


There are two empty bowls in a room. You have 50 white balls and 50 black balls. After you place the balls in the bowls, a random ball will be picked from a random bowl. Distribute the balls (all of them) into the bowls to maximize the chance of picking a white ball. 0.75, 75%, 3/4. First, let us assume that we divided the balls into jars equally so each jar will contain 50 balls.
So the probability of selecting a white ball will be = probability of selecting the first jar*probability of white ball in the first jar + probability of selecting the second jar*probability of white ball in the second jar
=(1/2)*(25/50)+(1/2)*(25/50)=0.5
Since we have to maximize the probability so we will increase the probability of white ball in the first jar and keep the second probability same mean equal to 1
so we add 49 white balls with 50 black balls in the first jar and only one white ball in the second jar
so the probability will be now= (1/2)*(49/99)+(1/2)*(1/1)=0.747
Therefore, probability of getting white ball becomes 1/2*1 + 1/2*49/99 which is approximately 3/4.


There are 1000 wine bottles. One of the bottles contains poisoned wine. A rat dies after one hour of drinking the poisoned wine. How many minimum rats are needed to figure out which bottle contains poison in hour. 10. We need to figure out in hour. We need 10 rats to figure out the poisoned bottle. The result is based on binary number system. We get 10 using ⌈ Log(2)1000 ⌉.

The idea is to number bottles from 1 to 1000 and write their corresponding binary numbers on the bottle. Each rat is assigned a position in the binary numbers written on bottles. Let us take an example. Rat 1 represents first bit in every bottle, rat 2 represents second bit and so on. If rat numbers 2, 4 and 6 die, then bottle number 42 (Binary 0000101010) is poisoned.


In a country, all families want a boy. They keep having babies till a boy is born. What is the expected ratio of boys and girls in the country? 1:1, 50:50. Assumptions: Probability of having a boy or girl is same. Also, the probability of next kid being a boy doesn’t depend on history.

The problem can be solved by counting expected number of girls before a baby boy is born.
Let NG be the expected no. of girls before a boy is born

Let p be the probability that a child is girl and (1-p)
be probability that a child is boy.
NG can be written as sum of following infinite series.
NG = 0*(1-p) + 1*p*(1-p) + 2*p*p*(1-p) + 3*p*p*p*(1-p) + 4*p*p*p*p*(1-p) +.....

Putting p = 1/2 and (1-p) = 1/2 in above formula.
NG    = 0*(1/2) + 1*(1/2)2 + 2*(1/2)3 + 3*(1/2)4  +  4*(1/2)5 + ...
1/2*NG = 0*(1/2)2 + 1*(1/2)3 + 2*(1/2)4 + 3*(1/2)5  +  4*(1/2)6 + ...
NG - NG/2 =  1*(1/2)2 + 1*(1/2)3 + 1*(1/2)4 + 1*(1/2)5  +  1*(1/2)6 + ...

Using sum formula of infinite geometrical progression with
ratio less than 1
NG/2 = (1/4)/(1-1/2) = 1/2
NG = 1
So Expected Number of number of girls = 1

Since the expected number of girls is 1 and there is always a baby boy, the expected ratio of boys and girls is 1:1.


A car has 4 tyres and 1 spare tyre. Each tyre can travel a maximum distance of 20000 miles before wearing off. What is the maximum distance the car can travel before you are forced to buy a new tyre? You are allowed to change tyres (using the spare tyre) an unlimited number of times. 25000. Divide the lifetime of the spare tyre into 4 equal part i.e., 5000 and swap it at each completion of 5000 miles distance.
Let four tyres be named as A, B, C and D and spare tyre be S.

5000 KMs: Replace A with S. Remaining distances (A, B, C, D, S) : 15000, 15000, 15000, 15000, 20000.
10000 KMs: Put A back to its original position and replace B with S. Remaining distances (A, B, C, D, S) : 15000, 10000, 10000, 10000, 15000.
15000 KMs: Put B back to its original position and replace C with S. Remaining distances (A, B, C, D,  ) : 10000, 10000, 5000, 5000, 10000.
20000 KMs: Put C back to its original position and replace D with S. Remaining distances (A, B, C, D, S) : 5000, 5000, 5000, 0, 5000.
25000 KMs: Every tyre is now worn out completely.

All tyres are used to their full strength.


You have 15 Rs with you. You go to a shop and shopkeeper tells you price as 1 Rs per chocolate. He also tells you that you can get a chocolate in return of 3 wrappers. How many maximum chocolates you can eat? 22. Buy and eat 15 chocolates
Return 15 wrappers and get 5 more chocolates.
Return 3 wrappers, get 1 chocolate and eat it (keep 2 wrappers)
Now we have 3 wrappers. Return 3 and get 1 more chocolate.

So total chocolates = 15 + 5 + 1 + 1


A Lady (L) bought an item of Rs 100 from the Shopkeeper (C). She paid him through a 500 Rs Note. Realizing that he did not have change, the shopkeeper C got change for that note from another shopkeeper (S) and paid Rs 400 to the Lady.
After a few days, S realized that the note is fake, And this railed at C and took 500 Rs back from him.
So in this whole process how much money did C lose in the end? 500. The total loss for shopkeeper = 500 ( given back to the person who had provided the change )

Consider a transaction box, the lady came with a counterfeit 500 Rs note which can be considered of 0 value.
Now the lady took the item (cost of the item 100 Rs ) and 400 Rs (the change given by shopkeeper(C) to the lady)
from the transaction box, total of 500 Rs. Now the equivalent amount should be lost by someone, thus shopkeeper(C) lost 500 Rs. Another shopkeeper(S) gave 500 Rs and took back the same amount hence no loss for him.


A man is allocated a task. He doubles the task done everyday. If the man completely does the task in 20 days, how many days did it take for the man to complete 25% of the task? 18. 100% of task = 20 days
As he doubles the task everyday. So, 50% of task = 19 days
25% of task = 18 days.


Four people need to cross a rickety bridge at night. Unfortunately, they have one torch and the bridge is too dangerous to cross without a torch. The bridge can support only two people at a time. All the people don’t take the same time to cross the bridge. Time 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. To reduce the amount of time, we should find a way for 10 and 7 to go together as they are the slowest among all these. 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. This brings us closer to the solution. So let’s put all this together.

Steps:
1 and 2 cross the bridge and move to the other side.
Now 2 comes back with the torch from the other side.
7 and 10 crosses the bridge and 2 remain to this side only.
Now 1 comes back with the torch from the other side.
At last, 1 and 2 cross the bridge and we are done.
Total time taken = 2 + 2 + 10 + 1 + 2 = 17 mins


A man fell in a 50m deep well. He climbs 4 meters up and slips 3 meters down in one day. How many days would it take for him to come out of the well? 47. On the first day, the man climbs 4 meters up and slips 3 meters down, therefore he only climbs 1 meter up totally.
On the second day, again he climbs 1 meter up, so the total distance climbed is 2 meters till the second day. Therefore, the man climbs 1 meter every day.
Now as per the above pattern, on the 46th day, he must have climbed 46 meters. So on the 47th day, he climbs full (46 + 4) 50 meters and after that, he will not slip as he is already out of the well so the answer is 47 days.


In thirty minutes on a deserted road, there is a 95% chance of seeing a vehicle. What are the probabilities of seeing a vehicle in ten minutes? 63.15%, 0.6315. To make the question more understandable, we can state that the probability of seeing any other vehicle in the next 30 minutes is 95%, or, more precisely and practically, the probability (%) of not seeing any other vehicle is 5%. So, we need to find 3 consecutive times without seeing another vehicle for 10 minutes for it to be 30 minutes before we see another one.
Let the probability of not seeing a vehicle in 10 minutes be P(not10) = P
Therefore, the probability of not seeing a vehicle in 30 minutes,
P(not30) = P(not10)*P(not10)*P(not10)= P*P*P = P^3
Given that, there is a 95% chance of seeing a vehicle (i.e., 0.95) in 30 minutes:
We can find the probability of seeing any other vehicle in the next 30 minutes is,
1 – P(not30) = 0.95
= 1 – P^3 = 0.95
=  P^3 = 1 – 0.95 = 0.05
P = 3√0.05 = 0.3684
Therefore, the probability of seeing a vehicle in the next 30 minutes is  P = 0.3684
Therefore, the probability of seeing a vehicle in 10 minutes, = 1- P = 1 – 0.3684 = 0.6315


Ram used to arrive at the railway station every day at 6 pm from work. As soon as Ram arrived at the station, his wife Sita too arrives at the station by car to pick him up. Both drive back home. One day Ram arrived at the station one hour early and thought of taking a walk back to home. He started walking when after covering a specific distance, he meets with Sita. Both drive back home and this time they reached 30 minutes earlier than the usual time. How long has Ram been walking?
It is known that Sita drove every day at a uniform speed. 45. Since they reached home 30 minutes earlier, Sita must have saved 15 minutes in each trip i.e., up and down trip, as the speed of the car was uniform.
This means that had Sita drove for 15 more minutes, she would have met Ram at the station.
As we know Ram arrived at the station at 5 that day, and they must meet at 6, it means that Sita meets Ram at 5:45.
Therefore, Ram must have walked 45 minutes before meeting Sita.


On an Island, there is an airport that has an unlimited number of identical air-planes. Each air-plane has a fuel capacity to allow it to fly exactly 1/2 way around the world, along a great circle. The planes have the ability to refuel in flight without loss of speed or spillage of fuel. Though the fuel is unlimited, the island is the only source of fuel. You can ignore the time and fuel consumption of refuelling.
What is the minimum number of air-planes required to get one air plane all the way around the world assuming that all of the air planes must return safely to the airport? 3. The minimum is three.
Start three planes from the airport (yours and two others). After going 1/8 around, one plane refuels both yours and the other plane with 1/4 of a tank each, leaving it with 1/4 of a tank which is precisely enough to take it home.
Two planes continue with a full tank again.
Reaching 1/4 around the other plane transfers 1/4 tank of fuel to yours, leaving it with 1/2 tank, to take it precisely home again.
Your plane is now full at 1/4 and can go to 3/4 with that fuel.
The first returning plane refuels and start off the other way meeting your plane at its 3/4 position, where your plane is empty and the meeting one is half empty. Sharing their fuel they have both 1/4 tank left which can take them to the 7/8 position.
The remaining plane has arrived there in the meanwhile with 3/4 tank giving 1/4 to the other now empty planes.
All three planes have now 1/4 and can return home.