#  > The FaaDoO Engineer's Lounge >  > Chit-Chat/Share your interests, hobbies etc/FaaDoO Engineers ke Kisse!! >  >  F! Contest - Master-Mind of the DAY: Day 7

## Sakshi Dutta

*Hey FaaDoOs!!

This is the Seventh thread for master mind of the day...

**Are you    planning to become an MBA? Or wanna take the GRE route? Or are you tired    of studying and wanna jump into a job straight after your   engineering??

Whatever be your interest, one thing is for sure – you are going to come    across a lot of ‘aptitude’ questions along your way to realizing your    dreams!

More details HERE 

Match your wits against one daily puzzler of a question! And who knows    if you are the first one to crack it, you might just be the winner of a    Rs.100 mobile recharge!!

Here's the First Question for DAY 7 of Master mind of the day--*

*Q1.) Rahul likes statistics and so he calculated the average height of his friends as X. He also calculated the average of the average heights of all the possible pair of his friends (two friends taken at a time) as Y.
Later, he calculates the average of the average heights of all possible triplets (three friends taken at a time) as Z.
Find the relationship among  X,Y,Z ?* 

*Correct Answer will win FREE Mobile Recharge worth Rs.100!!!

Entries will be accepted till 7PM, 5th Dec. 2011!*





  Similar Threads: F! Contest: Master-Mind of the DAY F! Contest - Master-Mind of the DAY: Day 8 F! Contest - Master-Mind of the DAY: Day 6 F! Contest - Master-Mind of the DAY: Day 4 F! Contest - Master-Mind of the DAY: Day 2

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## Sagar Agarwal

ans 1 ==  for X:Y:Z     is  1:2:3

---------- Post added at 06:17 PM ---------- Previous post was at 06:16 PM ----------

or ans 1 == X:Y:Z   is  1:2:3

---------- Post added at 06:18 PM ---------- Previous post was at 06:17 PM ----------




> or ans 1 == X:Y:Z   is  1:2:3


i wanted to write

or ans 1 ==== X:Y:Z   is  3:2:1

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## yrvsmurthy

the answer for the first question of day7 isy>x>z

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## abhinavgupta

answer for question of day 7 is  X=Y=Z

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## somesh.km1

can nt say anything any relation do not holds true for all possible cases

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## gaddamraghavareddy

X,Y,Z are equal.
X=Y=Z

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## wasimakram

ans fr the 1st question 
x=y=z

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## vikas1306

Hi, my answer is Z (> or =) Y(> or =) X i.e. Z can be equal to as well as greater than Y and Y can be equal to as well as greater than X depending upon the values.

---------- Post added at 10:13 AM ---------- Previous post was at 10:10 AM ----------




> Hi, my answer is Z (> or =) Y(> or =) X i.e. *Z can be equal to or greater than Y and Y can be equal to or greater than X depending upon the values*.


I want rephrase my earlier post as shown in bold.

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## ashwanisingh835

I think the relation will be
   X>Y=Z
MEANS X is the greatest and Y and Z are equal

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## sharan0619

Ans.1)x=y=z.....

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## sharan0619

But it should be given that the number of students in the class should be a multiple of 6,only then the averages of pairs and triplets can be taken

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## Rajatswm

Equal : X=Y=Z

solution:
let there be n students with heights a1, a2, a3.... an.
so average = 1/n{(sum of all heights)}

now for avg heights of pair: 
there will be nC2 pairs like (a1+a2)/2 , (a1+a3)/2 and so on.. so every students height was added (n-1) times. 
so avg of pair =(1/nC2)*(n-1)*{sum of all heights/2) = 1/n {sum of all heights}

for triplets:
there will be total nC3 triplets like... (a1+a2+a3)/3, (a1+a2+a4)/3 and so on.. by observation you can see that the sum of all triplet is equal to
 (n-1)*(n-2)/3{sum of all heights)
hence avg: (1/nC3)*sum = 1/n {sum of all heights}

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## akhil691

Question 1): X=Y=Z

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## sreekanthzipsy

ans is x=y=zex:three students 5 ,5.1, 5.2    average=5.1aver of pairs  1st 2 studs are 5.05 and 2nd 2 studs are 5.15 and total average is 5.1 aver of triplets3 students is 5.1(x=y=z)...........

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## prasanjeet roy

According to me*                        Z<Y<X*

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## ashwanisingh835

X=y=z is the correct answer

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## rohitmitsgwalior

The Correct Answer Is 
x=y=z

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## rider

x=y=z, all the three averages are equal

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## Sakshi Dutta

> Equal : X=Y=Z
> 
> solution:
> let there be n students with heights a1, a2, a3.... an.
> so average = 1/n{(sum of all heights)}
> 
> now for avg heights of pair: 
> there will be nC2 pairs like (a1+a2)/2 , (a1+a3)/2 and so on.. so every students height was added (n-1) times. 
> so avg of pair =(1/nC2)*(n-1)*{sum of all heights/2) = 1/n {sum of all heights}
> ...


Very well explained..!!  :): 


---------- Post added at 06:28 PM ---------- Previous post was at 06:28 PM ----------

*Correct Answer: X=Y=Z !!

 @abhinavgupta   takes home a cool Rs.100 Mobile Recharge!!

Well done everyone who got the correct answer.. 

Thread closed..Please continue to today's thread...


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