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Riddler's DI Bash
by The Riddler - Monday, 31 March 2008, 11:34 AM

cat 2008 cat 2009 mba 2008 DI 2008 xat gmatLet me introduce my CAT 2008 students to another pillar of our TathaGat team- The Riddler. Riddler can make DI questions as fast as he can churn out information on various cars, their models and their prices. Our classroom students often find him smoking on the stairs and lost in thought. No, he is not thinking about some question that he is making; he makes them too fast and doesn’t waste time. He is probably thinking about price of a Hummer or a Honda CRV. If not that, he would probably be wondering which movie to watch next because he religiously watches each of them. The good part about his questions is that they are solvable; the questions are not made to harass students unnecessarily. Of course, I am the poor soul whom Riddler bashes first with all his questions in order to check their validity. So today I decided to give the TGites a taste of his riddles. Good luck guys!- Total Gadha



Six students: A, B, C, D, E and F are staying in a hostel. Their heights are integer multiples of 0.5 feet. Average height of the group is 5.5 feet and the average height of A & C is 5 feet. D is the tallest in this group and F is the shortest. They are accommodated in two rooms of different sizes, which can accommodate three beds each.  Each student requires one bed for himself. They have the option of three different sizes of beds with the lengths of 4 feet, 5 feet & 6 feet. Only 4 & 5 feet beds can be put in the smaller room but the other room can have any size of bed. It is assumed that a student will be satisfied to have a bed only if the length of the bed is within

± 0.5 feet of that student’s height. Each size of bed is available in enough number. It is also know that each of the six student was satisfied to have his bed.

1. Who could be the roommate of E?

(a)     D
(b)     C
(c)     F
(d)     A
(e)     Cannot be determined

2. Which among the following groups is accommodated in the smaller room?

(a)     DBE
(b)     CAF
(c)     FAB
(d)     ADC
(e)     None of these

3. Which two students are having same height?

(a)     D & E
(b)     C & F
(c)     F & B
(d)     A & C
(e)     C & E


A botanical experiment is being conducted on three varieties of sweet pea- Healthy, infected and resistant. In the three stages of the experiment, equal number of plants from two of the three varieties are chosen and crossed with each other in pairs. Whenever, a plant of healthy variety is crossed with that of infected variety, it results in the healthy plant also becoming infected. Whenever a plant of infected variety is crossed with that of resistant variety, it results in the infected plant dying out. Whenever a plant of resistant variety is crossed with that of healthy variety, it results in the healthy plant also becoming resistant. The bar chart below shows the initial and stage-wise percentage of plants of the three varieties out of the total number of plants.

di 2008 cat 2008 cat 2009

1. The crossing between the healthy variety and the resistant variety is happening in

(a)     stage 1
(b)     stage 2
(c)     stage 3
(d)     cannot be determined

2. What is the percentage change in the number of plants of the healthy variety in stage 1 compared with the initial number of plants?

(a)     10%
(b)     12.5%
(c)     16.6%
(d)     no change

3. What can be the minimum change in the number of plants of infected variety from the initial stage to stage 3?

(a)     3
(b)     4
(c)     5
(d)     10

4. In stage 3, the crossing takes place between which two varieties?

(a)     healthy and infected
(b)     infected and resistant
(c)     resistant and healthy
(d)     cannot be determined

Pico Fermi Bagels is a math/logic game commonly taught to schoolchildren to help them develop deductive reasoning skills. Player One thinks of a number with each digit different, and records it on a hidden piece of paper. Player Two must now guess at what the number is. Player One must answer each guess with a combination of three responses:

di 2008 cat 2008 cat 2009

Here is a sequence showing the numbers guessed by player one (about a four-digit number guessed by player 2) and the response (in Pico, Fermi and bagels) given by player 1:

di 2008 cat 2008

1. For how many digits can their placement be accurately determined?

(a)     1
(b)     2
(c)     3
(d)     4
(e)     cannot be determined

2. How many numbers are possible from the above the observation?

(a)     1
(b)     2
(c)     3
(d)     4
(e)     cannot be determined


Prior to the Flim flam Awards in Tollywood, a poll is conducted among the audience, which asks them to guess four nominees for the best actor award and rate them 1, 2, 3 & 4 (1 means the best and 4 means the worst). In the Sharma family, four of the family members made their guesses. When the final result is declared they observed the following data:

cat 2008 di 2008

1. Who ranked first?

(a)     Amitab or Dev
(b)     Farukh or Bobby
(c)     Ganesh
(d)     Dev
(e)     Cannot be determined

2. Who, out of the following, could not make into the first four?

(a)     Dev
(b)     Farukh
(c)     Ganesh
(d)     Chitwan
(e)     Bobby

3. Who ranked 2nd?

(a)     Ganesh or Bobby
(b)     Dev or Farukh
(c)     Amitab or Dev
(d)     Ganesh
(e)     Chitwan

4. Whose statement is definitely required to know the four actors who got nominated among the top four?

(a)     Mrs Sharma
(b)     Rishu
(c)     Mona
(d)     Spicy
(e)     All of them

Re: Riddler's DI Bash
by Jay Prakash - Monday, 31 March 2008, 02:06 PM

Answers for the first puzzle.

1)a  2)b  3)d

Please correct me if i am wrong.

Re: Riddler's DI Bash
by Ankit Kumar - Monday, 31 March 2008, 02:33 PM

six student--

(1) a (2) b (3) d

Re: Riddler's DI Bash
by Satyam Gadha - Monday, 31 March 2008, 03:41 PM
  Hi All,

Answers to the first puzzle. Before I give the answer I will try to explain my approach here.

Since the average height of the group is 5.5. hence the sum will be:

A+B+C+D+E+F = 5.5 x 6 = 33

Now, since A+C = 5 x 2 = 10. (Let me assume that they both are of same height. smile)
So A and C both are 5 feet and they both can accommodate in the same smaller room.

Now who else can go with them?
F is the shortest of them all. Let me also put him in the smaller room. I am not certain about his height so I assume it to be X.

A= 5
C= 5
F = X
B+D+E+F = 33- 10 = 23

If F = 4.5
then B+D+E = 18.5

let D be 6.5, tallest of all. That makes B = E = 6

Smaller Room.
A(5 feet bed)  C(5feet bed) F(4 feet bed)

Bigger Room
B(6 feet bed) E(6 feet bed) D( 6 feet bed)

1. a
2. b
3. d

Re: Riddler's DI Bash
by Jay Prakash - Monday, 31 March 2008, 03:34 PM

could u please eidt the clour of the explanation provided by u. Nothing is visible clearly and it would be helpful if u dont paste the questions again. Answers with explanation would be more than enough.


Re: Riddler's DI Bash
by Satyam Gadha - Monday, 31 March 2008, 03:44 PM
  I hope the edit will help. And I won't repeat that. Thanks for your feedback sir. smileapprove
Re: Riddler's DI Bash
by shweta mehani - Monday, 31 March 2008, 07:01 PM

1. a

2. b

3. d

Re: Riddler's DI Bash
by Ankit Kumar - Tuesday, 1 April 2008, 11:39 AM


(1) a

(2) b

(3) d

(4) d

Re: Riddler's DI Bash
by Ankit Kumar - Tuesday, 1 April 2008, 12:18 PM

Pico Fermi Bagels

no is 6352

(1) 4

(2) 1

Re: Riddler's DI Bash
by Ankit Kumar - Tuesday, 1 April 2008, 12:43 PM

botanical experiment

(1) c

(4) c

Re: Riddler's DI Bash
by Tuhin Banerjee - Tuesday, 1 April 2008, 03:21 PM


Hostel Botanical Exp. Pico-Ferming Film awards
1. a 1. b 1. a 1. d
2. b 2. b 2. d 2. b
3.  d 3. 12.44 ( not sure) 3. d
4. c 4. d


Re: Riddler's DI Bash
by Pagal Gadha - Tuesday, 1 April 2008, 05:13 PM


My answers to the second puzzle are

1) c

2) a

3) d

4) c

Please correct me in case I am wrong.

Re: Riddler's DI Bash
by avijit mohapatra - Tuesday, 1 April 2008, 06:49 PM
   my answers are:-
puzzle 1
1 a
2 b
3 d
puzzle 2
1 c
2 a
3        could not solve
4 c
puzzle 3
1 d
2 a
puzzle 4
1 d
2 b
3 a
4 e
Re: Riddler's DI Bash
by bimal mohan - Tuesday, 1 April 2008, 08:28 PM


1.(a)  2. (b)  3 (a)

A botanical experiment :

1.(c) 2.(d)  4(c)

Pico Fermi Bagels :

1.(d) 2.(a)

Flim flam Awards :

1.(d) 2. (b) 3.(a) 4(e)

Re: Riddler's DI Bash
by Himanshu Jaggi - Wednesday, 2 April 2008, 08:12 AM

Hey Riddler,

Here are my attempts to problem 1:

1.a. 2. b. 3d.

My attempts to problem 2:

1.a, 2 b , 3. n/attmptd  4.d

Though I havent put much of efforts in solving this, i'l give it a look again and then post my ansrs again if changed..

Problem 3

Still to solve


Himanshu Jaggi

Re: Riddler's DI Bash
by Bikash Jain - Wednesday, 2 April 2008, 02:01 PM
  Botanical Garden...

1. C
2. D
3. B (Not sure about this...Calculated guess sort of)
4. C
Re: Riddler's DI Bash
by Shanthana Lakshmi Kameswaran - Wednesday, 2 April 2008, 07:02 PM
didnt solve
didnt solve
Q3 number is 6352

Re: Riddler's DI Bash
by Himanshu Jaggi - Thursday, 3 April 2008, 08:29 AM

Hi Riddler,

Reps to 3rd Puzzle(Pico Fermi):

c, b

Reps to 4th Puzzle(Film Falm):

d, b, a, d



Re: Riddler's DI Bash
by Hungry Gadha! - Thursday, 3 April 2008, 10:45 AM

Hi Riddlersmile,

My answers:

1st Riddle:

1. a

2. b

3. d

2nd Riddle:





3rd Riddle:



4th Riddle: i find little difficult..sad

1. a






Re: Riddler's DI Bash
by Parthiv Dave - Thursday, 3 April 2008, 10:53 AM


1.(a)  2. (b)  3 (b)

A botanical experiment :

1.(c) 2.(a) 3.(a) 4.(c)

Pico Fermi Bagels :

1.(e) 2.(b)

Flim flam Awards :

1.(d) 2. (b) 3.(a) 4(e)

Re: Riddler's DI Bash
by Ankush Kumar Chakraborty - Thursday, 3 April 2008, 10:58 AM

Bikash , you are right . Taking a total of 50 plants, you will get 4 as the answer for the third question in the botany problem

Re: Riddler's DI Bash
by nirmesh sinha - Thursday, 3 April 2008, 12:18 PM

1st Set

a b d

2nd Set

c d  ?surprise c

3rd set

d a

 4th Set

d b a e


Re: Riddler's DI Bash
by navneet ghose - Thursday, 3 April 2008, 04:40 PM


Can any one plz tel me how to solve film award problem



Re: Riddler's DI Bash
by Satyam Gadha - Thursday, 3 April 2008, 05:49 PM
  Che Riddler

I don't think the last puzzle seems to be complete with the details. However, I am waiting for your reply on the same doubt.

My doubt is, what does actually (No of correct/wrong  guesses) signifies?
Is it the combination of
1. (Rank + Actor)  or
2. (No of correct guesses for Nominees selected) and (No of wrong guesses for ranking the correctly selected actors) or
3. Something which didn't cross my mind.

Please clarify on the same.


Re: Riddler's DI Bash
by ankit khanna - Friday, 4 April 2008, 07:03 PM

hie my answers to puzzle 4 are





plz tell m r they write or wrong?

Re: Riddler's DI Bash
by amiable gal - Sunday, 6 April 2008, 05:53 PM


hi satyam, i approached the problem in the same way as you did.but it is given that heights are integer multiples so i guess 6.5,4.5 are wrong.....

i might be wrong in interpretating but got the ans's:




heights being:




a & c=5

pls correct me if i am wrong.

take caresmile

Re: Riddler's DI Bash
by amiable gal - Sunday, 6 April 2008, 06:17 PM

second set:





stage 1=i+r



Re: Riddler's DI Bash
by amiable gal - Sunday, 6 April 2008, 06:29 PM
  hey in pico digit of the first guess and third guess should be at the same place !!!! dint get itmixed
Re: Riddler's DI Bash
by Satyam Gadha - Tuesday, 8 April 2008, 12:29 PM
Hi Ami,

As far as I interpreted the meaning of Integer multiples of 0.5, it means to me the following:
To make the question a little less cumbersome, the presenter of the question has kept it the KISS(Keep it simple, stupid!) way.
If he would have written the question the following way that " Each height is an integer value." I would have accepted your argument over the question. But it says that heights are integer multiples of 0.5 i.e. superset of heights  could be (0.5 x 1, 0.5 x 2, 0.5 x 3,.... & so on).
I hope you got my point.
And for your answers,
d=7, wrong . Please read the sentence below which is taken from the question.
It is assumed that a student will be satisfied to have a bed only if the length of the bed is within ± 0.5 feet of that student’s height.

The maximum height bed available is of 6 feet only. So if d=7, it wont go in the line of this assumption.
And thanks for pointing out this aspect of understanding the question. More often, it happens with students like me or you to get caught in the flow of question and start solving them without rereading the question and that leads to different interpretation of the same sentence in the same question, which results in the wrong answers.
By the way, thanks again.  I would be more careful next time to describe my answers to avoid any confusion or clash of interpretations.approve

And sorry for this long description. tongueout Actually right now I am preparing for vocab and RC. Hope you understand and all the best.

Re: Riddler's DI Bash
by Pirates Are Back In Town - Wednesday, 9 April 2008, 10:51 AM
  Hi Satyam,
  I think the The last Puzzle is complete ....
 If you look at the 1st Table there are 7 different actors in all ...
And according to Mrs Sharma's Response and her correct guesses and wrong guesses (in Table 2)  ... We can conclude that the three actors who made it to the top are one of the following Combinations :
But Going to last row in Table 2 (Spicy) , She had only one guess and it was a wrong guess ... that is from the 4 actors, Spicy rated, only one made it to the top 4 and the rank she gave was wrong for the person in the final four...  ,  ... ..This is what the 2nd Table is all about , ....
So we know from Spicy's rating that A and D can not be together .. because only 1 of the four rated by Spicy were final four ..... And Hence we delete "AD" combo from our list ABC ; ABD ; BCD ; BDA ; CDA .
We are left with ABC And BCD.
And Also E anf F Are not in the final four.
Tables For Puzzle 4

Now about Mona's guess .... 2 out of the 4 Mona rated made to the top four ..... agreed ..? And 1 guess was right and the other wrong ... We know B and C are there in the Final four ... And exactly one person either A or D is in Final four .... So only one person will be there from EFG in the final four .... ..
But we know Spicy has only one of his choices in ... And Either A or D is among the Final Four ... We can be sure E and F could not make it to the finals....
So From This C and G are in.
clear ..? And moving on to Rishu ... Again B has to be there ... Now out of DFE one person has to be in the final list ... It is D ... And Hence A is out ...

Finally ... BDGC are the TOP FOUR ...
Checking the Table 1 ...
Mrs Sharma had 2 wrong ... 1 right and so on

D => 1st 
G => 2nd
C => 3rd
B => 4th.

You can verify ... Mrs Sharma had 2 wrong and 1 correct
She said - B -2 ; C- 3 : D-4  (Only c=3 is correct while B And D are not )

Please reply me if you find any flaw ...... smile

And I too am am busy with my RCs these days ....
Sorry for this Turgid message .... smile

Re: Riddler's DI Bash
by Satyam Gadha - Wednesday, 9 April 2008, 05:13 PM

@Tanuj Agrawal

Che Tan,

Thanks for your great but gradiloquent explanation of the questionsleepy. But you made me read my previous post again. As you explained, I came to know that the second table is taking about the division of the total number of guesses made by the family members in as correct and wrong guesses. Previously I thought, each of the family member made 4 guesses each.

Great thanks for clarification.

And I am glad to read your post, that you took some time from your precious preparation time to reply to my questions.

As I was trying to answer the questions on this funda, I got stuck with the last one, which forced me to think about all the questions once again. By the way, My answrs will be different from yours but are in the line of last question.

1> a

2> b

3> d

4> d

And one more question, what are you referring for your RC and vocab improvement. My Vocab is bad, very bad. But I am hopeful that I will improve a little and make my RC strong. I kept this month for RC and vocab intensive preparation and will continue it further till July. I am referring Hindu editorial and ET complete, weak at sports front. Reading Hindu editorial is a pain as it is not available in Pune. I have to do it online. And they are really tough. However I find ET as good as any other Economics magazine.But my reading speed is still low. Would need some more practice. The current strategy for RC at my end is to improve up on the ability to comprehend at least first. Later I can work on speed but I am slogging due to eak vocab.

Tell me the story at your end and remedies too if you can...


SatyamGadha cool




Re: Riddler's DI Bash
by tanu shivnani - Thursday, 10 April 2008, 08:39 PM

hi satyam

c u had striked dis pt where u hav taken F=4.5   But d thing w/c m askin is how wud it strike durin tense hr of xm...


i had reachd 2 d pt. wherei tuk out B+F+D+E=23,whereafter i was facin prob...can u suggest altrnate method 2 it...



Re: Riddler's DI Bash
by Satyam Gadha - Friday, 11 April 2008, 10:40 AM

@tanu shivnani

Che Tanu

Its nice to see your doubt on my assumption. Honestly,I took more than 10 mins to solve this question.
And when I finished that I realized there is no rocket science involved smile.
However, I will try my best to explain you.
Now, As you said you have understood the previous explanation till assumption B+F+D+E=23.
The maximum and minimum heights that could be accommodated with avalable size of beds are 6.5 and 3.5 respectively.
(Before I go further with my explanation, I will try to explain you one basic funda.
Whenever I see an addition like this in which you are supposed to determine the values of all unknowns,
I try to take the closer values for all of them during the hit and trial method.
e.g if A+B+C+D = 24, then I will take A=B=C=D = 6)
Based on the above funda, I tried to put the value of B and D as 6 feets each.
Now for E and F following values can go and sum must come as 23.
 E= 3.5  F= 7.5(can fit in the bigger room but will not fit any bed)
 E= 4    F= 7  (can fit in the bigger room but will not fit any bed)
 E= 4.5 F = 6.5 (can fit the bigger room and will fit in the biggest bed available)

I hope the explanation above helps and leaves no doubts.

Thanks for your doubts. Please do raise the doubts wherever you can, may be they wont help you sometimes but they may help the guy whom you asked.

Thanks again Tanu.




Re: Riddler's DI Bash
by Mou Sukoshi - Sunday, 13 April 2008, 10:18 PM

Botanical Prob

1>Whenever a plant of resistant variety is crossed with that of healthy variety, it results in the healthy plant also becoming resistant. So, % of resistant variety shd increase & that of healthy plants shd decrease when there's a cross betwn those two. Hence (C)


2> In stage 1, % of infected plants decrease. This can happen only when cross happens betwn resistant and infected plants take place. So, there wont be any change in no of healthy plants. Hence (D)

3>  Taking initially 100%=100.

No of healthy plants initially = 48 = No of healthy plants in stage 1

No of infected plants initially = 32

No of infected plants in stage 1 = 22 (In stage 1, cross is betwn infected & resistant plants. Initially infected plants die. No of healthy plants remain constant.)

No of inected plants in stage 2 = 40 (Total no of plants remain constant bcoz healthy plants become infected)

Diff in no of infected plants = 40 - 32 = 8

Since, all nos are multiples of 2, to get the minm no of plants, we can divide by 2. Hence (B) 4

Re: Riddler's DI Bash
by sahil madhani - Tuesday, 15 April 2008, 12:53 AM
  can u explain y option 2.a &  3.a  in 2nd riddle
Re: Riddler's DI Bash
by Pirates Are Back In Town - Sunday, 20 April 2008, 08:08 PM
  Hey Satyam,

I got a real good reason and a good way to enhance my vocab ....

Remember a good vocabulary makes you look smarter. I think that should be more motivating than the reason "Clearing the exam”. So this fundamental shift between the approaches makes quite a big difference in your results

A girl I really liked once ... said I was a Centaur , I nodded and agreed obviously not knowing what it meant , which by the way is a mythological creature with a goats lower body and a human upper. I never quite understood why she called me a centaur..... Point is a poor vocabulary can easily land you in trouble, more trouble than any exam can offer.

She also used words like Condescending and Derogatory very regularly that is when I got genuinely interested in building a vocabulary. It is surprising because for years my father insisted that i try and build a vocabulary, and even if I learn a word a day I can learn 365 words per yr ... , but then all of us know motivations differ. and whats better than a girl...... smile

1.Vocabulary books : A great way to start your vocabulary building process , most of these books follow the Etymological Approach , spiced up with lots of interesting story about words and thier history. I’m sure most CAT aspirants have heard of "Word Power Made easy" by Norman Lewis. Ideal for beginners with its small chapters and emphasis on basics.

For those of you who have finished WPME  go for Contextual Vocab Building...

2. Contextual Vocabulary building: This is when we learn words while reading; we try and guess the meaning of the words from its surrounding words or the context

For example the word Baldric is a deliciously rare word ....even I do not what it is ... .. But if you are reading “After the heated battle, Aragorn put the sword back into the baldric" the meaning of the word should not be too difficult to guess.

Any reading material is a good source for building contextual vocabulary; In fact spending a few minutes with the newspaper should give you 20+ good words everyday.

3. Word lists: Possibly the most mundane(Syn: prosaic .. I dont remember .. but I guess I am write ... ) of learning techniques, except when you have limited time and the exam is looming large right in front of you.

But perhaps the most important thing is to have a burning desire for words, for words define the world around you and your command over them gives you a distinct advantage in better understanding /communicating with the world.

At Last ... Keep using the new words you learn ... otherwise time spent in  learning new words would actually be the time killed .....

So Why dont we start a using these words .. when we Write a post in the forum .. let us delve into the sea of Words ... ...

Imagine you have to give a woman a compliment, does beautiful convey all that you want to say
or "alluring, angelic, appealing, beauteous, bewitching, charming, classy, comely, cute, dazzling, delicate, delightful, divine, elegant, enticing, exquisite, fair, fascinating, gorgeous, graceful, grand, lovely, magnificent, marvelous, pulchritudinous, radiant, ravishing, refined, resplendent, slightly, splendid, statuesque, stunning, sublime"

Does the trick better? .....  if the girl is aspiring for CAT I can bet

it works Dud e ... try it ..

Happy Hunting ... (Words, not Gals).... smile
Re: Riddler's DI Bash
by Satyam Gadha - Monday, 21 April 2008, 08:34 AM

@Pirates Are Back In Town

Che Tanuj

It was an awesome post and i would like to say that I realy enjoyed it. It was more motivating than any other manuscript I have ever read for building vaocabulay.
The third para of your post is very much eye opener. I always tried to prepare for Vocab to get through exams..else I never had or have interest in building my vocab.
My circle of friends is not that encouraging for vocab. As I ma working in a software firm... I do know lots of technical words and do use it in my daily coversations.
And I have been reading a lot and lots of financial newspapers, books and magazines for last couple of years. So the words coming from finance world or economics world do not bother me much.
However, I dont understand all of them and sometimes I get the word but didn't get the meaning of sentence which is more imporatnt relatively. You know economics...

Now regarding the motivation for vocab, I guess I have no one to look at around me but soon I will find one way or the other.
An idea just clicked me... I will tell  you in the end of paragraph.

Now regarding the suggestions, you have given I would really like to rate them on the cale of 10.. if you dont mind, I would like to add my comments too here...

1. Vocabulary books : They are really good in their work but most of the time we dont use them in our day to day conversations. SO after the exams are over ,thesre is a good chance of loosing them.
   However if you try the same kind of books like "Word Power made Handy" again after finishing one.. same as repeating the old stuff in a new different format. The words will stay with you..but you really need to read them with your eyes stuck on them and keep referring them again and again.
 So the rating for this goes 6

2. Contextual Vocabulary building : Now, this option is great .. but you need to be really shameful in front of your colleagues and friends. (Dont ask me why...) The kind of humiliation or pity which I see in their eyes makes me feel bad for my vocab preparation.
   They think that I am doing this only for some competitive exam. And if I try to use it in my day to day sentences they again make fun out of me. Its not that easy. Again you have to write each word or underline them as you are reading the context contents. Till the time I will get the word meanings after reading the content.. I , generally, loose the sentence which has the underlined word or may be I loose interest caz I have to read the same para again to understand the usage of the underlined word.
   However a good hard work can pay better than the first option and hence the rating would go as 8.

3. Word Lists: No comments.. what you said is completely right. Rating would be 3 

Now my idea , to be in touch with the words you learn everyday, you can do following:
 1. Start a blog of your own, if you are really creative or read a lot of books and magazines or even newspapers.

 2. If you are not that eager to start a blog of yours.. start commenting on all the articles or blogs with more intutive words which you have learned in some othger blog or magzine before.
 3. Write a diary of yours, every night before you sleep. And use all the words which you have learned through out the day.
 4. Read as many articles or anything crap in all the sectiions of literature as you can.

And thanks Tanuj for your great post again. I promise, the next post will be more elegant.

Happy Hunting ... (Words, not Gals).... smile




Re: Riddler's DI Bash
by sMILING GadhUUU - Monday, 21 April 2008, 11:42 PM

Average Height of the group = 5.5

Total students =6

Total Height =33

Height of A + Height Of C= 10

Height of B+D+E+F=23



All Six Students were satisfied bo have his bed..

3 bed sizes 4,5 and 6. As all students were satisfied there heights cannot be more that 6+ 0.5 or less than 4-0.5

therefore max height possible =6.5

Min height possible =3.5

Now let us suppose the heights of the tallest person D be the max possible =6.5

and of F be the min possible = 3.5

Therefore Height of D+F =10

So Height of B+E = 13

But Both B and D are shorter that the tallest Student D who is 6.5,

So non of them can be more than 6.5

We give them the max heights possible for them

So B=E=6

Therefore B+E=12

The 1 feet height that cannot be accomodated among B or D , has to be given to any one of A , C,D or F.

but D already has his max height =6.5 , so no change possible.

We already know that A and C are 5 feet each., So no change possible.

Only possibility The height of F is more by 1 feet..

We increase the height of F by 1 feet and check.

3.5 + 1 = 4.5, It is still the minimum Height so, the final Heights are.

A= 5







FIRST ROOM can have only beds of sizes 4 and 5.

Clearly a student of Height more than 5.5 cannot stay in room 1. (So B,D,E cannot stay in Room 1)

Therefore Occupants of Room 1: A ,C and F

Occupants of Room 2: B, D and E


1) Roommate of E = Ans(a) D

2) Group in small Room = And(b)CAF

3)Students With same Height= A and C



x Number of Healthy Plants crossed with xNumber of Infected Plants = 0 healthy Plants, 2x Infected Plants

x Number of Infected Plants Crossed with x Number of Resistant Plants = 0 Infected Plants , x Resistant Plants

x Number of Healthy Plants crossed with x Number of Resistant Plants = 0 Healthy Plants, 2x Resistant Plants

Observe the graph.

Lets Assume that there are total 100 Plants.

h healthy, r resistant and i infected.

Now In any stage of the Experiment,Only 2 types of the plants are crossed, So the number of Plants of the 3rd Type would

remain the same at the end of the Stage,

So the Percentage of the 3rd Type of Plant which didnot participate in a stage would remain the same

Pecentage =( No of plants of the Type / Total Number of Plants) x 100

From the graph Percentage of Resistant Plant Remains Constant From Stage 1 to Stage 2 (22.66 %)

Also Percentage of Infected Plants remain constant from Stage 2 to Stage 3.(44.44 %)

So Resistant Plants DID NOT PARTICIPATE in Stage 2 experiment

And Infected Plants DID NOT PARTICIPATE in Stage 3 Experiment


Stage 3 = Healthy x Resistant

Stage 2 = Healthy x Infected

Stage 1= Resistant x Infected ( Only combination Left)

QUESTIONS 1 ,2 and 4 can be solved with this much work out only..

Question 1:Stage for crossing between Healthy and Resistant    Ans(3)Stage 3

Question 2 : Percentage of Change in NUMBER OF HEALTHY

PLANTS in Stage 1. Ans(d)As Only Resistant and Infected Plants

participated in Stage 1, No change for

Healthy Plants

Question 4: Stage 3, Crossing Between And(c)Resistant and Healthy.

Question 2:

INITIALLY THE PERCENTAGES OF r , i and h are 20,32 and 48 Percent Respectively.

Now as we have to find the minimal change in the value of i , from the start to the end fo the experiment.

Let us consider what Minimum values of the 3 types of plants , we can possibly have.

The Number of Plants being an integral quantity , We cannot have any number in fractions.

To simplyfy things divide the 3 respective percentages by there LCM that is 4,

Therefore we would get 5,8 and 12..

The number of Plants in the categories cannot be less than this..As then the Resistant ones would not remain an integer..

So now we have the minimum values for





Now consider the 1st Stage:

IN THE FIRST Stage Resistant plants are crossed with Infected plants.

As a result all the Infected plants that are crossed woutd die.

Suppose x infected plant are used in STAGE1:

Now consider any Fraction (a / b),

If the value of b is decreased, the value of fraction a / b would increase.

Similar increase happens here, In stage 1, the Healthy plants do not participate.

And there number remains the same as 12 ( As per our minimum assumption)

But the percentage of healthy plants increase from 48 % to 53 %.

This is because of the decrease in the total number of plants by x ( The infected plants that die during Stage 1).

Let us find out would would be the value of x ,

h initially = 5

h at the end of stage 1=5


( 5 / ( 25-x) )*100 =53.33

x = 2.5 , Therefore the change in Number of Infected Plants at the end of STAGE 1= 8- 2.5 = 6.5

Similarly in stage 2 the value of h plants decreases from 53.33 to 33.33 % , A decrease of 20 %.

This decrease in the number of h , actually is getting converted to infected Plants.

53.33 percent of h at the end of stage 1 = 12,Therefore 33.33 % of h at the end of stage 2 = 7.5 (As total plants in stage 1

and 2 are same)

Let us see the change in values of h , i and r because of the x infected plants that die.

Therefore the change in Number of Infected Plants at the end of STAGE 2 = 6.5 +4.5

               INITIALLY                    STAGE 1                        STAGE 2

r                 5                                 5                                    5

i                  8                               6.5                                  11

h                12                               12                                 7.5

Therefore, Minimum change possible Ans(a) 3.


Eliminate the noise, by removing all the Positions for the 4 values 2,3,6,5 that are not possible.

Start with the Second Guess as we know all four positions are incorrect.

1 2 3 4

2 x x -

3 - x x

6 x x

5 x x -

Check the positions one after the other with the responses so that they follow the results,

If any response does not satisfy, leave it and try the next combination

Finally you would oblige for 6352..

How many digit,s placements can be determined.. Ans(d) All 4 digits

How many numbers are possible Ans(a) Only 1.

Let there be Light----------------->

Re: Riddler's DI Bash
by sMILING GadhUUU - Monday, 21 April 2008, 11:43 PM

And the Tollywood Question…

Man it rocked the Heart out of Me, My Lungs were gasping for Breath , My braining crying it's lobes out, back straining and Legs Trying to catch sub unconcious sleep.

With all this i only have my understanding and workout..

TG SIRG Pleaseeeeeee look into this problem doesnt look like it has an answer..


It is clear from the question that all the 4 Members of the family made guesses from 1 to 4.

So There are 3 Possibilites for the Guesses.    

(1) The member correctly guesses the actor and also guesses the correct position.

(2) The actor guessed , Does come amongst the 1st 4 of the final List, But the Positon guessed in incorrect.

(3) The Actor guessed does not appear in the 1st 4 of the final list.

It appears that the No of correct entries point to the number of actors Who were guessed correctly to be amongst the TOP 4 at the same time the position guessed was also correct.

Secondly the Wrongly guessed point to the Number of Actors guessed that could not even make it to the TOP 4 of the final List.


First, We have to find out the 4 actors that could make it to the TOP 4,

Then we can work on there positions.

Rishu , Mona Spicy made 1 wrong guesses each..

So 1 actor out of the 4 guessed by them could not make it to the top 4.

But on the other way 3 actors guessed by each of them were among the TOP 4.

Out of the 3 actors that Spicy guessed correctly, She could not guess the position of Any one correctly.

At this point we consider only the Actors incorrectly guessed by the 4 family members as we can eliminate them and decide on the final 4.

So Mrs Sharma had 2 wrong guess,

Rishu , Mona and Spicy 1 Each.

If we would try to find the Final 4 They have to be any three out of FAED.

We can try out the combinations out of FAE, AED,EDA,FED, And add 1 actor such

That we have 2 in the 4 guessed by Mrs Sharma, And 3 each in the lists of Rishu, Mona and Spicy.

We would find that DCFE satisfies the 4 conditions,

2,3,3,3 actors out of this list of Four (DCFE) Respectively are there in the guesses of

Mrs S, Rishu Mona and Spicy,

Now consider the correct guessed made by the 4 family members,

                 Mrs Sharma              Rishu         Mona          Spicy

F                                                 2                3              1  

E                                                4                 1              3

D                    4                           1                                4

C                    3                                            4


Now Spicy had no correct guesses, All the options of F,E and D guessed by spicy would be incorrect, We would strike them of


                       1                2                    3                         4

F                      X                                     

E                                                             X

D                                                                                      X




Mrs Sharma made one correct guess, So either D is 4 Or C IS 3,

But D is not 4 .So C IS 3.

                        1                2                    3                         4

F                      X                                     X

E                                                             X

D                                                             X                        X

C                     X                X               Correct                    X


One of Monas Guess F is 3, C is 4  ,E is 1 is correct.

F cannot be 3 and C cannot be 4, So E is 1.

                        1                2                       3                     4

F                      X                                        X

E                  Correct           X                      X                     X

D                     X                                        X                      X

C                     X                X               Correct                    X


One of Rishus F at 2,   E at 4 or D at 1 is correct

E cannot be 4, D cannot be 1 So F is 2.


                        1                2                       3                     4

F                      X            Correct                 X                     X

E                  Correct           X                      X                     X

D                     X                X                       X                     smile 

C                     X                X               Correct                    X


The only place left for D is 4,but there is a contradiction,As spicy had guessed D to be 4 ,It could not be correct ...

My brain could not find a better way so switching of my brain for now..


Re: Riddler's DI Bash
by sMILING GadhUUU - Monday, 21 April 2008, 11:44 PM


But I do believe if this question( Apna Tollywood )would have been correct , It would have been easy to solve it by options , But I think somewhere there is something missing , Because if we consider the second Question

Who, out of the following, could not make into the first four?  

(a) Dev
(b) Farukh
(c) Ganesh
(d) Chitwan
(e) Bobby

If we can find The one actor among these five who could not make it to the first 4, the rest 4 did make it to the first 4.

There are only 5 combinations to try DFGC,FGCB,GCBD,CBDF,BDFG..

But I could not find if any of them could satify the conditions implied on the guesses of the 4 Family members.

HAH,, Guys it feels great to be back to my basic.. After a break of almost a month.

Feels like at home….

And by the Way satt , I do have a very limited vocabulary but do take my contribution to yours and pirates list of Words to describe Girls--- Almanac….

Keep up the good work Pirates , and if any one of you is a maniac like me and derives interest out of blasphemy , I think you would really relish a book called

The Complete Idiots Guide to Verbal Self Defence.. The words would make you fall in love with it……..

Let there be light------------------>

Re: Riddler's DI Bash
by AMIT RAJ - Friday, 25 April 2008, 07:10 PM

Hi Smiling G,

Plz shorten your name.


Re: Riddler's DI Bash
by omkar patwardhan - Tuesday, 24 June 2008, 12:13 PM

hey no one sems to have replied here for so far.

i got terribly stuck in the well bagles pico riddle.

the no. of digigts which can be easily determined and confirmed is unanimously i think 4.

but the no. of numbers ( well is this alliteration......or the sound is repeated hwat do u call this figure of sppech?)is confusing to me. someone please enlighten me.

Re: Riddler's DI Bash
by omkar patwardhan - Tuesday, 24 June 2008, 12:44 PM

About the roomies.

first one goes e according to me coz not sufficient data to solve it.

second one goes also e coz both b and c are equally likely.

third one i  disagree i think no one will have the same height.

for the plant thing.

i go with option c for the first and last question.

frankly, the remaining two about plant can anyone help me why ther eis no change in healthy plant percentage i did not understand the posts supporting it.

Re: Riddler's DI Bash
by Ramkrishna Roy - Tuesday, 24 June 2008, 01:11 PM
  1.A 2.B, 3.D
Re: Riddler's DI Bash
by Total Donkey - Friday, 27 June 2008, 04:06 AM
  @DI riddler

really awesome collection of puzzles, the simplicity of the problems was the best part (though i spent my time to appreciate the problems)... really worth doing at 4 in the morning     

btw, has the last question about the tollywood got some discrepancy or I got the meaning of correct and wrong guesses wrongly

I took them as
correct guesses- number of guesses rightly done and at the right place
wrong guesses - number of people guessed who are not among the top 4 itself

Kindly clarify

Re: Riddler's DI Bash
by shubham singh - Wednesday, 2 July 2008, 02:11 PM
  very nice puzzles and di questions tg.please put more such type of prbms on ur site..thanks a ton nyways 4 the help u r providing to the cat
Re: Riddler's DI Bash
by Rahul Garg - Monday, 7 July 2008, 07:43 PM

Hi Smiling G.

I wanna ask you one thing about the explanation in first set of questions...

If we take A=5.5 and C=4.5, still A+C=10 holds good and now the answer for the question 3 would be hard to determine.

Now C and F also have same height.

Please suggest me why this is not possible???

If any one else can suggest,

Re: Riddler's DI Bash
by sathyan sadagopan - Tuesday, 30 September 2008, 05:23 AM
  hi smilin gadhuu,
i think there s a small mistak in d explanation..u hav mentioned

Let us find out would would be the value of x ,

h initially = 5

h at the end of stage 1=5


( 5 / ( 25-x) )*100 =53.33

x = 2.5

but the initial value of h = 12
so it s (12/(25-x))*100=53.33

however d answer s correct value of x=2.5..

gr8 work smilin gadhuu good explanation..thank u

Re: Riddler's DI Bash
by Atul Mittal - Tuesday, 30 September 2008, 03:35 PM
  Hi all,
Hope this table could help you in getting the solution of second question
Please correct me if its wrong somewhere





Stage 2

Stage 3

Plant Types













































The ans are:- 1)C 2)d 3)C 4)a
Re: Riddler's DI Bash
by varun pandey - Wednesday, 1 October 2008, 10:29 AM

Answer to Tollywood problem:





There will be two sequences of ranking of actors which will satisfy the constraints in the question

Dev,Bobby,Ganesh,Chitwan and Dev,Ganesh,Chitwan,Bobby.

Please correct me if i am wrong.

Re: Riddler's DI Bash
by shantanu ajatshatru - Friday, 3 October 2008, 11:15 AM

1.a 2.b 3.d

1.c 2.d 3.d 4.c

1.d 2.b 3.d 4.e

Re: Riddler's DI Bash
by rashi agarwal - Tuesday, 7 October 2008, 03:01 PM
  first puzzle


Re: Riddler's DI Bash
by vasudha gupta - Wednesday, 8 October 2008, 04:46 PM
  fourth question:
1) d
2) b
3) a
4) e
Re: Riddler's DI Bash
by Balaji Kunjeti - Wednesday, 22 July 2009, 03:33 PM

(A) 1. a, 2.b, 3.d

(B) 1.c, 2.d, 3.b, 4.c

(C) 1.d, 2.a

(D) 1.d, 2.b, 3.d, 4.e

Re: Riddler's DI Bash
by Netra Mehta - Sunday, 13 September 2009, 01:39 PM
  Plz sum1 answer dis question..
its on data sufficiency..

The question below is followed by two statements, A and B. Answer the question using the following instructions:
Mark (1) if the question can be answered by using statement A alone but not by using statement B alone.
Mark (2) if the question can be answered by using statement B alone but not by using statement A alone.
Mark (3) if the question can be answered by using either of the statements alone.
Mark (4) if the question can be answered by using both the statements together but not by either of the statements alone.
Mark (5) if the question cannot be answered on the basis of the two statements.

Can we find a sequence of 1000 consecutive positive integers, in which there are exactly 5 prime numbers?

The sequence A! + 2, A! + 3, …, A!+ 1001 has no prime numbers, (A >200).
The sequence B!+1, B!+2, …, B!+1000 has exactly 10 prime numbers, (B > 200).


1) 1
2) 2
3) 3
4) 4
5) 5

Re: Riddler's DI Bash
by Netra Mehta - Sunday, 13 September 2009, 02:42 PM
  Heii Smilin G!!

In puzzle 1 y r u takin the height ig A & c both as 5
Cant it b 4.5 & 5.5??
Re: Riddler's DI Bash
by prasad navale - Friday, 18 September 2009, 10:28 PM

u say height of D as 6.5 feet and want him to get bed of 6 Feet??


by ur given data...available beds are of 4,5,6 feet means one can be shorter dan 4 feet bt not taller than 6 feet clearly

Re: Riddler's DI Bash
by gitanshu amola - Tuesday, 14 June 2011, 12:28 PM

dude i have follw the same approach but am confused in 1 of yr asumption that hw could d be 6.5 feet ?? there is no bed with 6+.. and at last they have said that all guys r satisfied with there beds !! hw could d will be satisfied ??? i thnk b,d,e, are 6 fit each. i am getting answers as (a,b,a)......

please explain a.s.a.p smile

Re: Riddler's DI Bash
by nw its ma turn to tame d cat .. - Tuesday, 14 June 2011, 10:42 PM