A New Word Every Day!

During a six-day period from Monday through Saturday, Eliza Pseudonym and her friends Anna, Barbra, Carla, Delilah, and Fiona have subscribed to an internet mailing list that features a new word every day. 

No two women subscribed on the same day. On each day during the six-day period, a different word has been featured (abulia, betise, caryatid, dehisce, euhemerism, and floruit, in some order). 

From the clues below, determine the day on which each woman subscribed, and the day on which each word was featured.

1. Exactly one of the women has a name beginning with the same letter of the alphabet as the word featured on the day that she subscribed to the mailing list.

2. The word "caryatid" was featured precisely two days prior to Fiona joining the mailing list.

3. Carla joined the mailing list on Friday.

4. Anna signed up for the mailing list precisely one day after the word "euhemerism" was highlighted.

5. Wednesday's word did not end with the letter "e".

6. Barbra subscribed precisely three days after the word "dehisce" was featured.

Here is every word of woman of the day! 

Assigned Word to the Woman of the Day

What was the challenge?

As we know, Eliza Pseudonym and her friends Anna, Barbra, Carla, Delilah, and Fiona have subscribed to an internet mailing list that features a new word every day. No two women subscribed on the same day. On each day during the six-day period, a different word has been featured (abulia, betise, caryatid, dehisce, euhemerism, and floruit, in some order).

And given clues are - 

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1. Exactly one of the women has a name beginning with the same letter of the alphabet as the word featured on the day that she subscribed to the mailing list.

2. The word "caryatid" was featured precisely two days prior to Fiona joining the mailing list.

3. Carla joined the mailing list on Friday.

4. Anna signed up for the mailing list precisely one day after the word "euhemerism" was highlighted.

5. Wednesday's word did not end with the letter "e".

6. Barbra subscribed precisely three days after the word "dehisce" was featured.

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Let's make a table like below and fill it one by one as per clues.

 STEPS :  

1] As per (3), Carla joined the mailing list on Friday. And as per (6), Barbra's subscription day must be Thursday, Friday or Saturday with word 'dehisce' on Monday or Tuesday or Wednesday. 

But as per (5), ''dehisce'' can't be on Wednesday and Carla has already joined on Friday, hence Barbra must have joined on Thurday with word "dehisce" featured on Monday.

2] Only possible location for "caryatid" and Fiona for (2) to be true are Thursday and Saturday respectively. That's because Monday is already 'occupied' by "dehisce" so Fiona can't be on Wednesday. And as we can see, Thursday and Friday already 'occupied' by Barbra and Carla.

3] With that, for (4) to be true, only possible location for "euhemerism" is Tuesday and for Anna is Wednesday.

4] As per (1), Delilah and Eliza can't be on Monday & Tuesday at a time as that will violate (1). Therefore, Delilah must be on Tuesday and Eliza on Monday.

 
5] As per (5), "betise can't appear on Wednesday. 

Suppose it appears on Friday. 

CASE 1 : The word "abulia" with Anna on Wednesday & "floruit" with Fiona on Saturday. This violates (1), as there would be 2 women have names beginning with the same letter of the alphabet as the word featured on the day that she subscribed to the mailing list.

CASE 2 :  The word "floruit" with Anna on Wednesday & "abulia" with Fiona on Saturday. Again, this is against (1), as there is no woman has a name beginning with the same letter of the alphabet as the word featured on the day that she subscribed to the mailing list.

Therefore, "betise" must be appearing on Saturday, "floruit" on Friday and "abulia" on Wednesday with Anna.

CONCLUSION :

The final table looks like -

Story of Four High School Friends

Four high school friends, one named Cathy, were about to go to college. Their last names were Williams, Burbank, Collins, and Gunderson. Each enrolled in a different college, one of them being a state college. 

From the clues below determine each person's full name, and the college he or she attended.

1. No student's first name begins with the first letter as her or his last name, and no students first name's last letter is the same as his last name's last letter.

2. Neither Hank or Williams went to the community college.

3. Alan, Collins, and the student who went to the university all lived on the same street. The other student lived two blocks away.

4. Gladys and Hank lived next door to each other.

5. The private college accepted Hank's application, but he decided he could not afford to go there.

Here is ANALYSIS of the story! 

Analysing The Story of Four High School Friends

What was the story?

GIVEN DATA : 

First Name : Hank, Gladys, Cathy, Alan 
Last Name : Collins, Burbank, Gunderson, Williams 
College       :  State College, University, Community College, Private College 

HINTS : 

1. No student's first name begins with the first letter as her or his last name, and no students first name's last letter is the same as his last name's last letter.

2. Neither Hank or Williams went to the community college.

3. Alan, Collins, and the student who went to the university all lived on the same street. The other student lived two blocks away.

4. Gladys and Hank lived next door to each other.

5. The private college accepted Hank's application, but he decided he could not afford to go there.   

STEPS :  

1] Let's make a table like below and fill it as per hints.

  
2] As per Hint (3), we have, Alan, Collins and student going to the university as 3 different students.

3] As per (1), Collins can't be Cathy & as per (4), Cathy can't be at no.3 as in that case no blocks will be left for (4) to be true. Hence Cathy must be at no.4.

4] As per (2), Hank and Williams are two different students. And as per (4), Gladys and Hank must be at 2 & 3 (and anyhow these are only blocks left for them) but order yet to be known. So, Williams is certainly not at 3 or 2. And as per (1), Gladys can't be Gunderson or Collins or Williams & Hank can't be Burbank.  Therefore, Hank must be Collins at 2 and Gladys must be Burbank at 3.

5] Now, as per (1), Alan can't be Gunderson hence must be Williams and Cathy must be Gunderson. 

6] As per (2), neither Hank nor Williams went to community college, hence Cathy Gunderson must be. And as per (5), Hank didn't choose private college hence must have chosen state college while Alan Williams must be in private college.

7] Therefore, the final table looks like as below.

 

The Fearsome Logical Challenge

You and your two friends Pip and Blossom are captured by an evil gang of logicians. In order to gain your freedom, the gang’s chief, Kurt, sets you this fearsome challenge.

The three of you are put in adjacent cells. In each cell is a quantity of apples. Each of you can count the number of apples in your own cell, but not in anyone else’s. You are told that each cell has at least one apple, and at most nine apples, and no two cells have the same number of apples.

The rules of the challenge are as follows: 

The three of you will ask Kurt a single question each, which he will answer truthfully ‘Yes’ or ‘No’. Every one hears the questions and the answers. He will free you only if one of you tells him the total number of apples in all the cells.

    Pip: Is the total an even number?

    Kurt: No.

    Blossom: Is the total a prime number?

    Kurt: No

You have five apples in your cell. What question will you ask?

THIS should be the question that you need to ask!

Logical Response to The Fearsome Challenge

What was the challenge?

Remember, all you have to do is that ask one crucial question to logicians and not necessarily deduce the total count of apples.

Since, each cell has 1 to 9 apples and no two cells have same number of apples, the lowest count of apple is 1 + 2 + 3 = 6 and the highest count would be 7 + 8 + 9 = 24.

That is the total number of apples could be between 6 to 24.

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Now, Pip and Blossom already have gathered some information about the total.

1. The total is not an even number - Hence, only numbers  7,9,11,13,14,15,17,19,21,23 can represent the total count.

2. The Total is not a prime number - Out of the number above, only 9, 15, 21 are non-prime number.

Hence, the total count must be among 9,15 or 21.

Now, your task is easier. All you need to ask the Kert below question -

"Is total is 15?"
 
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CASE 1 : Total is really 15 -

Then Kert would reply with YES to your question and all of you know the total now.

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CASE 2 : Total is 9 -

The Kert's answer to your question would be NO.

If the total is 9 and you have 5 apples then rest of 4 apples must be distributed among Pip and blossom as (1,3) or (3,1) but can't be (2,2) since no 2 cells can have same number of apples.

Now, the friend having 1 apple (or 3 apples) can think that the total can't be 21 as in that case other 2 must have total of 20 (or 18) apples. But the maximum that other two can have is 9 + 8 = 17 apples.

So any of them can deduce that the total is 9.

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CASE 3 : Total is 21 -

Since you have 5 apple other 2 must be having total of 16 apples. One of them must be having 7 apples and other having 9 apples.

The friend having 9 apples can easily deduce the count as 21 since 9 as a total count is impossible in the case as for that the other must have 0 apples.

And the friend with 7 apples know that other can't have 1 + 1 or 2 + 0 (as per given data) apples in order to have total count of 9. Hence, he too can deduce that the total must be 21.

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To conclude, depending on the what Kert answers to your question and the count of apples that each of other 2 friends have one of them (or you too if count is 15) can deduce the total number of apples easily. And eventually, logicians have to set you free as promised.

The Ping Pong Puzzle

Three friends (A, B and C) are playing ping pong. They play the usual way: the winner stays on, and the loser waits his/her turn again. At the end of the day, they summarize the number of games that each of them played:

A played 10
B played 15
C played 17.

Who lost the second game? 

This person played & lost the second game! 

Participant of the Second Game!

How games were played?

A played 10, B played 15 and C played 17 games. So total number of presences are 10 + 15 + 17 = 42. Every 2 presences form a game. Hence, the number of games played are 42/2 = 21.

Let's take into consideration the minimum number of games that a player can play. For that, he need to loose every game that he has played. That is, if he has played first game then he must have out in second but replaced looser of second in third game. In short, he must have played odd numbered of games like 1,3,5,7,9,11,13,15,17,19,21.That's 11 games in total.

And if he had made debut in second game then he must had played even numbered games like 2,4,6,8,10,12,14,16,18,20. That's 10 games in total.

Since, in the case only A has played 10 games, he must have made debut in second game where he lost that game to make comeback in 4th game thereby replacing looser of third game.

The Wedding Anniversary Puzzle

Recently I attended the twelfth wedding anniversary celebrations of my good friends Mohini and Jayant. Beaming with pride Jayant looked at his wife and commented, ‘At the time we were married Mohini was 3/4 of my age, but now she is only 5/6 th.
We began to wonder how old the couple must have been each at the time of their marriage!

Can you figure it out?

Know their ages! 

The Wedding Anniversary Puzzle's Solution

What was the puzzle?

Let 'x' be the age of the Jayant & 'y' be the age of Mohini 12 years ago.

So 12 years ago,

y = (3/4) x          .........(1)

And now 12 years later, the proportion is - 

y + 12 = (5/6) (x+12)  

Putting (1) in above,

(3/4) x + 12 = (5/6) x + 10

(1/12) x = 2

x = 24

Again putting this value in (1),

y = (3/4) 24 = 18

So, the age of Jayant was 24 and that of Mohini was 18 at the time of marriage. And now after 12 years, they are 36 and 30 years old respectively.  

"Your Surname Tells My Shirt's Color!"

Mr. Yellow, Mr. Black and Mr. Brown, three best friends since kindergarten meet in a function after 5 years. The three of them are wearing either a Yellow, Black or Brown shirt.
After giving each other a friendly hug, Mr. Black says, "Hey, did you notice that we are wearing a different colored shirt than our names!"
The man wearing a Brown shirt said, "Wow I certainly did not notice that but you are right Mr. Black."
Based on this conversation, can you find our who was wearing which colored shirt? 

 
Know color of each person's shirt here! 
  

Surnames And Shuffling of Shirts

Read this conversation first!

One thing is sure that Mr.Black is not wearing the BLACK shirt. He is also not wearing BROWN as well as we found the other person wearing the BROWN colored shirt is agreed with the statement made by Mr.BLACK in a conversation. So only color left is the YELLOW & Mr. Black must be wearing that.

Now, the colors left are BROWN and BLACK and persons left are Mr. Yellow and Mr.Brown. Since, each of them wearing different color than his surname, Mr. Brown must be wearing the BLACK shirt. And hence, Mr. Yellow is wearing the BROWN shirt! 

To conclude, Mr.Black is wearing YELLOW shirt, Mr. Brown is wearing BLACK shirt & Mr. Yellow must be wearing BROWN shirt.

 

Clues From Talk About Boats

At the local model boat club, four friends were talking about their boats. There were a total of eight boats, two in each color, red, green, blue and yellow. 

1. Each friend owned two boats. 

2. No friend had two boats of the same color.

3. Alan didn't have a yellow boat.

4. Brian didn't have a red boat, but did have a green one.

5. One of the friends had a yellow boat and a blue boat and another friend had a green boat and a blue boat. 

6. Charles had a yellow boat. 

7. Darren had a blue boat, but didn't have a green one.

Can you work out which friend had which colored boats? 

Know here step by step process of finding owners! 

Source 

Owners Of Boats

What's the task given? 

First let's rewrite all the clues here.

1. Each friend owned two boats. 

2. No friend had two boats of the same color.

3. Alan didn't have a yellow boat.

4. Brian didn't have a red boat, but did have a green one.

5. One of the friends had a yellow boat and a blue boat and another friend had a green boat and a blue boat. 

6. Charles had a yellow boat. 

7. Darren had a blue boat, but didn't have a green one.

Let's make 2 set of 4 colors of boats. Below is the table with owners in row & color of boats in columns.  

Below are clues which giving clear idea of owner of particular colored boat.

3. Alan didn't have a yellow boat.

4. Brian didn't have a red boat, but did have a green one.

6. Charles had a yellow boat.

7. Darren had a blue boat, but didn't have a green one.

We will fill this table one by one as per clues given.

Now, consider the first part of this clue.

5. One of the friends had a yellow boat and a blue boat.

Now Alan won't have this combination as he doesn't own yellow boat at all. The Brian already had green boat, so he too can't have this combination as in that case he would own 3 boats.

Let's assume Charles had this combination then other would have boats as below.