CAT 2008: Quantitative Questions a Day 134 -Till end -> The Discussions

Aarav · 08:04 PM

Hello All,

This will be the last thread for QQAD 2008. From next Monday, we will have 2 questions a day in Monday - Thursday period, one in the morning at 5:30 (as is happening now), and next at 1:30 PM. Friday's QQAD will be a challenging one and hence we will restrict ourselves to just 1 problem that day, and Saturday also we will solve 1 problem. Thus, QQAD will approx. (with -4% deviation) see 54 more problems in this thread ending Nov. 7.

The students this season have been exemplary in their discipline to be regular in QQAD

and I'm expecting no lesser response from you in the coming 5-6 weeks.

Keep walking, Good Luck

Just to Add ... 2 more Practice Tests will be delivered to you, one on 14/10 and the last on 04/11.

implex

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Quantitative Question # 134
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The 150 Quant devils of QQAD are given individual numbers from 1 to 150, and a contest happens in multiple rounds to select the Ultimate QQAD devil. The elimination follows a weirdo pattern. In the 1st round starting from first devil, every 3rd devil is eliminated i.e. 1st, 4th, 7th, .... This repeats again from the first numbered (among the remaining) devil in the next round (leaving 3, 5, 8, 9, ...). This process is carried out repeatedly until there is only the winner left. What is the number of the Ultimate QQAD Devil?

(a) 93 (b) 48 (c) 119 (d) 38 (e) 140

implex · 08:26 AM

Quote:

Originally Posted by implex

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Quantitative Question # 134
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The 150 Quant devils of QQAD are given individual numbers from 1 to 150, and a contest happens in multiple rounds to select the Ultimate QQAD devil. The elimination follows a weirdo pattern. In the 1st round starting from first devil, every 3rd devil is eliminated i.e. 1st, 4th, 7th, .... This repeats again from the first numbered (among the remaining) devil in the next round (leaving 3, 5, 8, 9, ...). This process is carried out repeatedly until there is only the winner left. What is the number of the Ultimate QQAD Devil?

(a) 93 (b) 48 (c) 119 (d) 38 (e) 140

The algo for the problem is simple, but doing it manually seems a bit tough

Replace the terms at 3k+1 position by a high number say 200
now sort the array again
repeat the process. Till there is only one term left..

or we can use pointers and much much easier

Let me think how it can be done manually

after first round all 3k+1 terms are gone
after second round all terms of form n-[(n+1)/3]=3k+1 will be deleted

option c and D will be deleted after 2nd round

after third round all terms of the form n-[(n+1)/3]-[(n-[(n+1)/3]+1)/3]=3k+1 will be deleted
93-31-[64/3]=93-31-20=42 safe
48-16-11=27 safe
140-47-31=62 safe

so left is 93 and 48 140

after fourth round all terms of the form
n-[(n+1)/3]-[(n-[(n+1)/3]+1)/3]-n-[(n-[(n+1)/3]-[(n-[(n+1)/3]+1)/3]+1)/3]=3k+1 will be deleted

fine till here!!

crap ahead!!

implex

Can someone check by making a program

I am too lazy, use the algo i have given, its pretty obvious, but use pointers, else it will be a mess.

implex

Let me explain my process again
First round all 3k+1 terms are gone

Now every remaining term will shift to the left, by a value which is equal to the number of terms before it which got deleted.
For round 1 it is [(n+1)/3]

so new positions of every number is n-[(n+1)/3]

similarly after round 2 new position will
be n-[(n+1)/3]-[(n-[(n+1)/3]+1)/3]

and for every round

now we note that all the terms of the form 3k+1 are deleted
so we do option hunting

and land up at 48

I hope I am right

P.S: Though i gave the algorithm, I didn't program, there are high chances I have done a mistake. Kindly bear with me

anil.iitkgp · 08:15 AM

ok i have written a program in python..
hope i will get sum mark.. lol

import sys

l=range(1,int(sys.argv[1])+1)

while len(l)!=1:
_l =[]
j=0
while j<len(l):
_l.append(l[j])
j+=3
for item in _l:
l.remove(item)

print l

answer is coming as 140,
run this program on terminal as python filename.py 150

dazed

Quote:

Originally Posted by anil.iitkgp

ok i have written a program in python..
hope i will get sum mark.. lol

import sys

l=range(1,int(sys.argv[1])+1)

while len(l)!=1:
_l =[]
j=0
while j<len(l):
_l.append(l[j])
j+=3
for item in _l:
l.remove(item)

print l

answer is coming as 140,
run this program on terminal as python filename.py 150

I used a trial analysis
Am getting the ans as C-119

Here is how:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 (sample set 1/10)
-1 4 7 10 13 (round 1)
2 3 5 6 8 9 11 12 14 15
- 2 6 11 15(round 2)
3 5 8 9 12 14
- 3 9 (round 3)
5 8 12 14
- 5 14(round 4)
ans: 12 *10=120(winner)

since it is approximation - take the next closest ans - 119.

Here is another trial

1 2 3 4 5 6 7 8 9 10 (sample set 1/15)
2 3 5 6 8 9
3 5 8 9
8*15=120

implex

Yet another attempt

After round 1
the options turn into
62,32,79,25,93

now see this 140 becomes 93 so 140 will die after 93( Thanks bhaskar for this logic). 79 and 25 will die in round 2

so we are left with 32 and 93
now 32 becomes 21 and 93 becomes 62

again
21 becomes 14 and 62 becomes 41
14 becomes 9 and 41 becomes 27
9 dies in round 3 remember?

so left is 27

which is 140

ah finally

HYgoldking

The 150 Quant devils of QQAD are given individual numbers from 1 to 150, and a contest happens in multiple rounds to select the Ultimate QQAD devil. The elimination follows a weirdo pattern. In the 1st round starting from first devil, every 3rd devil is eliminated i.e. 1st, 4th, 7th, .... This repeats again from the first numbered (among the remaining) devil in the next round (leaving 3, 5, 8, 9, ...). This process is carried out repeatedly until there is only the winner left. What is the number of the Ultimate QQAD Devil?

(a) 93 (b) 48 (c) 119 (d) 38 (e) 140

After each step first numbers are 2,3,5,8,12,17 ... 38, ... 93 ... 138

So answer should be 38 or 93

first number is 38, after 9 steps
first number is 93, after 14 steps

anil.iitkgp · 08:55 AM

thanks implex,
just rephrasing your solution so others can understand it.

lets say answers may b 93,48,119,140

after one round of elimination they will get left shifted, which depends on number of numbers divisible by 3 before them
num (numbers divisible by 3 before 93)=31, so 93 will become 62(93-31)...
like this 48,119,140 will become 32,79,25,93
now remove all the numbers from list which gives 1 as remainder when divided by 3
now proceed as above, untill 1 option is left.