Crack The 10-Digits Password

In an attempt to further protect his secret chicken recipe, Colonel Sanders has locked it away in a safe with a 10-digit password.  Unable to resist his crispy fried chicken, you'd like to crack the code. 

Try your luck using the following information:

All digits from 0 to 9 are used exactly once.

No digit is the same as the position which it occupies in the sequence.

The sum of the 5th and 10th digits is a square number other than 9.

The sum of the 9th and 10th digits is a cube.

The 2nd digit is an even number.

Zero is in the 3rd position.

The sum of the 4th, 6th and 8th digits is a single digit number.

The sum of the 2nd and 5th digits is a triangle number.

Number 8 is 2 spaces away from the number 9 (one in between).

The number 3 is next to the 9 but not the zero.

Click here to know how to CRACK it! 

Cracking The 10-Digits Password

What was the challenge? 

Let's suppose the 10-digits password is ABCDEFGHIJ.

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Take a look at the given clues to crack down the password.

1. All digits from 0 to 9 are used exactly once.

2. No digit is the same as the position which it occupies in the sequence.

3. The sum of the 5th and 10th digits is a square number other than 9.
 
4. The sum of the 9th and 10th digits is a cube.
 
5. The 2nd digit is an even number.
 
6. Zero is in the 3rd position.
 
7. The sum of the 4th, 6th and 8th digits is a single digit number.
 
8. The sum of the 2nd and 5th digits is a triangle number.
 
9. Number 8 is 2 spaces away from the number 9 (one in between).
 
10. The number 3 is next to the 9 but not the zero. 

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STEPS : 

Respective Clues are in ().

1] Since, every digit is used only once (1), the sum of 2 digits can't exceed 
9 + 8 = 17 and minimum sum of 2 digits can be only 0 + 1 = 1.

2] So, the sum of 5th & 10th digits which is square (3) can be - 1, 4, 16.
The sum of 9th & 10th digits which is cube (4) can be - 1, 8.
The sum of 2nd & 5th digits which is triangle number (8) can be - 1, 3, 6, 10, 15.

3] As per (6), 0 is already at third position i.e. C = 0. Hence, 2nd, 5th, 9th or 10th digits can't be 0. Therefore, the sum of 5th & 10th or 9th & 10th or 2nd & 5th can't be 1. Revising list of possible values of those in step 2.

The sum of 5th & 10th digits which is square (3) can be -  4, 16.
The sum of 9th & 10th digits which is cube (4) can be -  8.
The sum of 2nd & 5th digits which is triangle number (8) can be -  3, 6, 10, 15.

4] Possible pairs of 9th & 10th digits - (1,7), (7,1), (2,6), (6,2), (3,5), (5,3)

If IJ = 71, then 5th digit E must be 4 - 1 = 3 & possible values of 2nd digit B are 0,3,7.
But as per (5), B must be even & C = 0 already.

IJ = 26 or 62 or 35 doesn't leave any valid value for 5th digit E.

If IJ = 53, then 5th digit E must be 1 & possible values of 2nd digit B are 2,5,9
That is B = 2 (5). But as per (2), no digit is the same as the position which it occupies in the sequence. So, the number 2 can't be at 2nd position where B is there.

Hence, IJ must be 17 i.e. I = 1, J = 7.

5] This leaves only 9 as valid value for 5th digit E so that (3) is true. 

So, E = 9 and hence B = 6 with sum of 2nd digit B and 5th digit E as 15.

6] So far, we have, B = 6, C = 0, E = 9, I = 1 and J = 7. 

As per (10), the number 3 is next to the 9 but not the zero. Hence, the 6th digit F must be 3. F = 3.

7] As per (9), the 8 is 2 spaces away from 9 i.e. here 5th digit E=9. With 3rd position already occupied by C = 0, the digit 8 must be at 7th position. 
So, G = 8. 

8] As per (7), D + F + H must be single digit. We have, F = 3 already, 
so possible values of D + H are 0, 1, 2, 3, 4, 5, 6.

D + H can't be 0 (0+0), 1 (1+0), 2(1+1/2+0), 3(1+2/3+0), 4(2+2/3+1/4+0), 5(2+3/4+1/5+0) since digits are repeated with C=0, I=1, F=3 already.

Hence, D + H = 6. 

9] Now, D = 5 and H = 1 is not possible. Also, D/H can't be 0 or 6. Both D and H can't be 3 at the same time. D can't be 4 as D is at 4th position. 

Hence, D = 2, H = 4.

10] With B = 6, C = 0, D = 2, E = 9, F = 3, G = 8, H = 4, I = 1, J = 7, the only digit left for the first position A is 5. So A = 5.

CONCLUSION : 

A = 5, B = 6, C = 0, D = 2, E = 9, F = 3, G = 8, H = 4, I = 1, J = 7.

So, the 10-digits password ABCDEFGHIJ is 5602938417.

 

The Candle Light Study

Power went off when Vipul was studying. It was around 2:00 AM. He lighted two uniform candles of equal length but one thicker than the other. The thick candle is supposed to last four hours and the thin one one hours less. When he finally went to sleep, the thick candle was twice as long as the thin one.

For how long did Vipul study in candle light?

Well, he studied in candle light for....hours. Know here! 

The Candle Light Studying Hours

What was the question?

It's clear that, the thick candle lasts for 4 hours and thin one lasts only 3 hours.

If L is length of these candles (they are of equal length) then thick candle burns L/4 per hour and thin one burns L/3 per hour.

Let's assume Vipul studied for X hours.

In X hours, amount of thick candle burnt = XL/4

 

In X hours, amount of thin candle burnt = XL/3

 

After X hours, amount of thick candle remaining = L - XL/4

 

After X hours, amount of thin candle remaining = L - XL/3

 

Also, it is given that the thick candle was twice as long as the thin one when he finally went to sleep.

 

(L - XL/4) = 2(L - XL/3 )

 

L(1 - X/4) = 2L (1 - X/3)

(1 - X/4) = 2(1 - X/3)

 

(4 - X)/4 = 2(3 - X)/3

3(4 - X) = 8(3 - X)

12 - 3X = 24 - 8X

5X = 12

X=12/5

 

Converting 12/5 hours into minutes as X = 12/5 * 60 = 144 minutes.

 

Hence, Vipul must have studied for 2 hours and 24 minutes.

 

 

  

Bags of Marbles - Puzzle

There are three bags, each containing two marbles. Bag A contains two white marbles, Bag B contains two black marbles, and Bag C contains one white marble and one black marble. 

You pick a random bag and take out one marble, which is white. 

What is the probability that the remaining marble from the same bag is also white?

Here is the solution! 

Bags of Marbles Puzzle - Solution

What is the puzzle?

Well, the probability that the second marble is also white should be 1/2 not 2/3 as most of answers to this puzzle claims!

There is no way that you have picked up the bag B having 2 black marbles since the first marble you have drawn is white. Hence, you must have picked up Bag A or C.

Claim :  

You chose Bag A, first white marble. The other marble will be white

You chose Bag A, second white marble. The other marble will be white

You chose Bag C, the white marble. The other marble will be black

So 2 out of 3 possibilities are white.


Hence, 2/3 is the probability of the second ball is also white.


Correct Approach :

The probability of the particular bag being picked randomly is 1/3 initially. Once the bag picked and since the marble drawn from it is white, the bag must be either A (2 whites) or C (1 white, 1 black).

The probability that the bag you picked is A (or C) must be 1/2 and hence the probability that second marble is also white (or not white) has to be also 1/2.

How long was he walking?

Every day, Jack arrives at the train station from work at 5 pm. His wife leaves home in her car to meet him there at exactly 5 pm, and drives him home. 

One day, Jack gets to the station an hour early, and starts walking home, until his wife meets him on the road. They get home 30 minutes earlier than usual. How long was he walking? 

Distances are unspecified. Speeds are unspecified, but constant.

Give a number which represents the answer in minutes.

He must be walking for....minutes. Click to know! 

Jack's Walking Duration in Journey!

What was the question?

It's important to think from wife's point of view in the case.

Ideally, had Jack somehow informed earlier his wife about his 1 hour early arrival then his wife's (and his as well) total 60 minutes in round trip would have been saved. 

30 minutes of her round trip are saved which means only 15 minutes of each leg of her trip must have been saved. That is she meets her husband only 15 minutes earlier on the day instead of 60 minutes earlier (if Jack had informed her earlier). 

Hence, Jack must be walking for 45 minutes.

Let's understand this with example.

Suppose wife needs exactly one hour to reach the station every day. She leaves home at 4 PM everyday and reach station at 5 PM & drive Jack home at 6 PM

On one day, Jack arrived at 4 PM and started walking whereas wife leaves home at the same time as usual. They reach home at 5:30 PM.

30 minutes of wife saved indicates that she took 45 minutes to meet husband (instead of 1 hour) at 4:45 PM (instead of 5PM, only 15 minutes earlier) and took him to home at 5:30 PM (instead of 6PM) in next 45 minutes (instead of 1 hour) thereby saving 15 + 15 = 30 minutes only.

Since, Jack started walking at 4 PM and meet her wife at 4:45 PM, he must be walking for 45 minutes.   

Sequential Order of Fruits

A person is asked to put in a basket one apple when ordered 'one', one guava when ordered 'two', one orange when ordered 'three' and is asked to take out from the basket one apple and one guava both when ordered 'four'.

The ordered sequence executed by the person is as follows : 

1 2 3 3 2 1 4 2 3 1 4 2 2 3 3 1 4 1 1 3 2 3 4 1. 

How many apples will be there in the basket at the end of the above order sequence ? 

(a) 4 (b) 3 (c) 2 (d) 12.

How many fruits will be there in the basket at the end of the above order sequence ? 

(a) 13 (b) 17 (c) 9 (d) 11

Click to know the answers! 

Fruits Collection in a Sequential Order

What was that order?

The ordered sequence executed by the person is as follows :

1 2 3 3 2 1 4 2 3 1 4 2 2 3 3 1 4 1 1 3 2 3 4.

The question asks the final count of apple in the basket.

One apple is placed in the basket on order of ONE and one apple is removed when ordered FOUR.

So,

the number of apples placed in basket = Number of 1's in order sequence.

and

the number of apples taken out of basket = Number of 4's in order sequence.

Then,

the number of apples remaining in the basket = Number of 1's in order sequence - Number of 4's in order sequence = 6 - 4 = 2.



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Q.1 How many apples will be there in the basket at the end of the above order sequence ? (a) 4 (b) 3 (c) 2 (d) 1 2.


ANSWER - (C) 2

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Each time 1, 2 or 3 ordered, one fruit is added to the basket but when ever 4 is ordered, two fruits are removed from it.

Total number of fruits = Number of 1's + Number of 2's + Number of 3's - 2 x (Number of 4's) = 6 + 6 + 7 - (2x4) = 19 - 8= 11.

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Q.2 How many fruits will be there in the basket at the end of the above order sequence ? (a) 13 (b) 17 (c) 9 (d) 11

ANSWER - (d) 11

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'The Cookie Crunchers' Among Bakers

Cassie, Jon, Luke, Maria, and Sahas baked a batch of 36 cookies, two-thirds of which were chocolate chip. The rest were plain. They each ate at least one right away. The cookies were so delicious, only one-and-a-half dozen are left, of which half are plain.

1. Cassie is allergic to chocolate.
2. Luke ate twice as many chocolate chip cookies as plain cookies.
3. Sahas and Maria each ate as many cookies as Luke and Cassie combined.
4. Sahas ate more chocolate chip cookies than Maria.

If all the cookies were eaten by the bakers themselves, how many cookies of each kind did each person eat?

Click here to know the final stats! 

Scorecard of 'The Cookie Crunchers'

What was the puzzle?

Cassie, Jon, Luke, Maria, and Sahas baked a batch of 36 cookies, two-thirds of which were chocolate chip. That means there were 24 chocolate cookies and 12 plain cookies.

The cookies were so delicious, only one-and-a-half dozen are left, of which half are plain. So, 18 cookies are left out of which 9 were plain and other 9 were chocolate.

That is out of 24 chocolate cookies, 24 - 9 = 15 are eaten and out of 12 plain cookies, 12 - 9 = 3 are eaten. 

Now, taking a look at the hints given.

1. Cassie is allergic to chocolate.
2. Luke ate twice as many chocolate chip cookies as plain cookies.
3. Sahas and Maria each ate as many cookies as Luke and Cassie combined.
4. Sahas ate more chocolate chip cookies than Maria. 


They each ate at least one right away. Since Cassie is allergic to chocolates,  must had chosen plain cookie. Luke too must have eaten at least 1 plain cookie otherwise total of cookies eaten by him along with chocolate cookies would be 0.

Sahas and Maria ate same number of cookies but Sahas ate more chocolate cookies than Maria. That means Maria must have eaten more plain cookies than Sahas to balance her number of cookies with the number of cookies eaten by Sahas.

Since there are only 3 plain cookies eaten with 1 of them eaten by Cassie, other 2 must be eaten by Sahas and Luke each while Sahas of Jon had no plain cookies.

With that Luke must had eaten 2 chocolate and 1 plain cookies.

Luke and Cassie together ate to 2 + 1 + 1 = 4 cookies. Hence, Sahas and Maria too ate 4 cookies. Sahas ate 4 chocolate cookies and Maria ate 1 plain and 3 chocolate cookies.

Total of 0 + 2 + 4 + 3 = 9 chocolate cookies eaten by Cassie + Luke + Sahas + Maria together. Hence, rest of chocolate cookies i.e. 15 - 9 = 6 eaten by Jon. 

The final stats are - 

Cassie :  Chocolate - 0          Plain - 1

Luke   :  Chocolate - 2          Plain - 1

Sahas :  Chocolate - 4          Plain - 0

Maria  :  Chocolate - 3          Plain - 1

Jon     :  Chocolate - 6          Plain - 0  

The Bird Watching Program : Puzzle

Abel, Mabel, and Caleb went bird watching. Each of them saw one bird that none of the others did. Each pair saw one bird that the third did not. And one bird was seen by all three. Of the birds Abel saw, two were yellow. Of the birds Mabel saw, three were yellow. Of the birds Caleb saw, four were yellow. 

How many yellow birds were seen in all? How many non-yellow birds were seen in all?

Click here for SOLUTION! 

The Bird Watching Program Puzzle : Solution

What was the puzzle?

Let's find the total count of birds including yellow and non-yellow.

1. Each of them saw one bird that none of the others did. Individuals - Abel, Mabel, and Caleb. So, Birds = 3.

2. Each pair saw one bird that the third did not. Pairs - Abel-Mabel, Abel-Caleb, and Mabel-Caleb. So, Birds = 3.

3. And one bird was seen by all three. Abel-Mabel-Caleb. Birds = 1.

Therefore, total of 3 + 3 + 1 = 7 birds were seen.

It's clear that, everyone saw 4 birds viz. one individually, two when in pairs and 1 with all along. 

Since Caleb saw 4 yellow birds, Caleb must not had saw any of non-yellow birds.

The 4 yellow birds that Caleb saw must be as below.

1. The one Caleb saw individually. 

2. The one that Caleb saw in pair with Abel-Caleb.

3. The one that Caleb saw in pair with Mabel-Caleb. 

4. The one that Caleb saw with all.

Now, the two yellow birds that Abel saw must be the one that is pointed by (2) above and other pointed by (4) above.

That's make sure that the pair Abel-Mabel saw a non-yellow bird. 

Since Mabel saw 3 yellow birds, 2 of them are as pointed by (3) and (4) above, the third one must be the yellow bird that she saw alone. So far, we discovered there must be 5 yellow birds (4 seen by Caleb and one seen by Mabel individually) at least.

So, far we know Abel saw a yellow bird as in pair Abel-Caleb, one more yellow bird with all and a non-yellow bird as in pair Abel-Mabel. Hence, the fourth bird that Abel saw must be a non-yellow bird since as per data Abel saw only 2 yellow birds. 

Confusing Ride on the Ferris Wheel

There are 10 two-seater cars attached to a Ferris wheel. The Ferris wheel turns so that one car rotates through the exit platform every minute. 

The wheel began operation at 10 in the morning and shut down 30 minutes later. 

What's the maximum number of people that could have taken a ride on the wheel in that time period?

Here is count of people enjoying the ride!

Peoples Enjoying Ferris Wheel Ride

What was the puzzle?

Let's simplify the situation by naming 10 cars as A, B, C, D, E, F, G, H, I, J.

Suppose the Car A is at the exit platform at 10:00 AM. Obviously it can be 'loaded' with 2 peoples say A1-A2.

At 10:01 AM, the Car B will be at the exit platform which can be 'loaded' with 2 more peoples say B1-B2.

So continuing this way, at 10:09 AM the car J will be loaded with peoples J1-J2.

At 10:10 AM, the Car A will be again at the exit platform and now A1-A2 can get off the board to allow 2 more new peoples A3-A4 to get on the board.

At 10:11 AM, the Car B will be at the exit platform where B3-B4 will replace B1-B2. 

Continuing this way, at 10:19 AM the J1-J2 in the Car J will be replaced by J3-J4.

So far, 10 x 4 = 40 different peoples would have enjoyed the ride. 

It's clear that every car takes 10 minutes to be at the exit platform after once it goes through it. That's how Car A is at the exit platform at 10:10 AM, 10:20 AM and it can be at 10:30 AM as well when the wheel is supposed to be shut down.

Now, since wheel needs to be shut down 10:30 AM, emptying all the cars must be started except Car A for the reason stated above.

Therefore, at 10:20, peoples A5-A6 replace A3-A4. Thereafter, every car should be emptied. So, far 40 + 2 = 42 different people have boarded on the cars of the wheel.

So, at 10:21 AM, the Car B is emptied, at 10:22 AM, the Car C should be emptied and so on.

At 10:29 AM, J3-J4 of Car J get out of the car and finally at 10:30 AM, A5-A6 get out of the Car A. 

Now, the ferris wheel can be shut down with no one stuck at any of cars.

To conclude, 42 different peoples can enjoy the ride in given time frame.

Aliens Meeting on the Earth

100 aliens attended an intergalactic meeting on earth.

73 had two heads,
28 had three eyes,
21 had four arms,
12 had two heads and three eyes,
9 had three eyes and four arms,
8 had two heads and four arms,
3 had all three unusual features.

How many aliens had none of these unusual features?

These are Aliens not having unusual features! 

Normal Aliens in the Meeting on the Earth

Read given data first!

Let's simplify the process with Venn diagram like below.

1. We know, 3 had all three unusual features.

2. 12 had two heads and three eyes, 9 had three eyes and four arms, 8 had two heads and four arms. Here, 3 having all unusual features too need to be counted in each of Aliens' counts above. 

So, Aliens having 2 unusual features of 2 heads & 3 eyes = 12 - 3 = 9.

Aliens having 2 unusual features of 3 eyes & 4 arms = 9 - 3 = 6.

Aliens having 2 unusual features of 2 heads & 4 arms = 8 - 3 = 5.

3. We know, 73 had two heads, 28 had three eyes, 21 had four arms.

Therefore, the number of Aliens having only 2 heads as an only unusual feature is 73 - (9 + 3 + 5) = 56.  

The number of Aliens having 3 eyes as an only unusual feature is 
28 - (9 + 3 + 6) = 10.

The number of Aliens having 4 arms as an only unusual feature is 
21 - (5 + 3 + 6) = 7.

4. So, we have 56 + 9 + 10 + 3 + 5 + 6 + 7 = 96 Aliens having at least one unusual feature. 

Therefore, the number of Aliens not having any of unusual features is 
100 - 96 = 4.

Crack And Win $500,000 - Puzzle

You are in a game show with four other contestants. The objective is to crack the combination of the safe using the clues, and the first person to do so will win $500,000.

The safe combination looks like this:

??-??-??-??

A digit can be used more than once in the code, and there are no leading zeroes.

Here are the clues:

1. The third set of two numbers is the same as the first set reversed.

2. There are no 2's, 3's, 4's or 5's in any of the combination.

3. If you multiply 4 by the third set of numbers, you will get the fourth set of numbers.

4. If you add the first number in the first set with the first number in the second set you will get 8.

5. The second set of numbers is not greater than 20.

6. The second number in the second set multiplied by the second number in the fourth set equals one higher than the numbers in the first set.

And get moving, I think another contestant has almost figured it out!

Click here for the SOLUTION! 

Crack And Win $500,000 Puzzle - Solution

What was the puzzle?

We know, the safe has a lock having 4 sets of 2 digits as -

?? - ?? - ?? - ??

Since, leading zeros are not allowed any of the set can't be started with 0 like 01, 07, 09 etc. 

Take a look at the clues given - 

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1. The third set of two numbers is the same as the first set reversed.

2. There are no 2's, 3's, 4's or 5's in any of the combination.

3. If you multiply 4 by the third set of numbers, you will get the fourth set of numbers.

4. If you add the first number in the first set with the first number in the second set you will get 8.

5. The second set of numbers is not greater than 20.

6. The second number in the second set multiplied by the second number in the fourth set equals one higher than the numbers in the first set.


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STEPS :

1] As per clue (2), digits 2, 3, 4 or 5 aren't allowed in any set. That means only digits 0, 1, 6, 7, 8, 9 are allowed for sure.

2] For (3) to be true, possible combinations of 3rd and 4th sets are - 

10 x 4 = 40
11 x 4 = 44
16 x 4 = 64
17 x 4 = 68
18 x 4 = 72
19 x 4 = 76

The third set can't be 10 for 2 reasons. - 1} As per clue (1), the first set will be 01 and leading 0's are not allowed. 2} The 4th set will have digit 4 which is not allowed.

The third set can't start with 2X, 3X, 4X or 5X as those digits aren't allowed. Moreover, it can't be started with 6X, 7X, 8X as in that case the value of the 4th set will exceed it's maximum possible value of 96 (if digit 2 was allowed) or 76.

Out of all above combinations, only 17 x 4 = 68 and 19 x 4 = 76 are valid combination as rest of combinations have digits that aren't allowed.

So, one thing is sure that the first digit of the third set is 1. And hence the first digit of first set also must be 1 as per clue (1).

3] As per clue (5), the second set can't exceed 20. It can't start with 0. Hence, possible values ranges from 10 to 19. That means, the first digit of the second set is also 1.

4] Now as per clue (4), if you add the first number in the first set with the first number in the second set you will get 8. That means the first number of first set is 8 - 1 = 7. 

As of now, the code looks like : 71 - 1? - 1? - ??

5] If 7 is the first digit of first set then as per clue (1), 7 itself is second digit of the third set.

Now, the code looks as : 71 - 1? - 17 - ??

6] As per clue (3) and rightly deduced as a possible combinations for 3rd & 4th set in STEP 2, the 4th set must be 17 x 4 = 68.

With that, code turns into : 71 - 1? - 17 - 68.

7] Finally, as per clue (6), the second number in the second set multiplied by the second number in the fourth set equals one higher than the numbers in the first set. This multiplication i.e. ? x 7 must be equal to 71 + 1 = 72. 
Hence, ? = 9.

The final code looks like : 71 - 19 - 17 - 68.