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

Varun Khullar · 09:18 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

12345678910
4,7,... are eliminated
1235689
12368..
1238
123..left
..and so on.. answer is 140..

selebratinglife

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

Now I did it by some sort of calculated guess. I just kept on subtracting the number of terms before the number until it is eventually of the form 3n+1. Once it gets to that form we are done!
Now lets take the number 119, number of terms before it (of the form 3k+1) =40
So 119 will take the place of 119-40=79 after round 1, which is of the form 3k+1. So it's eliminated.
Now for 48, after the first round 48 will take the place of 48-16=32.
After second round,32-11=21,after 3rd,21-7=14, after 4th,14-5=9,after 5th,9-3=6, after 6th, 6-2=4. Now we are done.
Now before calculating for 93, lets calculate for 140. After 1st round 140 will take the position of 140-47=93. So it would always be one step ahead of 93. So our required answer is option e).

Before 150 gets condensed to a single round, there are 11 rounds of elimination. So W.r.t choice b), it can be safely eliminated.

crazyganesh

take first few numbers of the series

in the first round
1 4 7 10 13 (difference between each term is 3)
second
2 6 11 15 (differences are 4,5,4,5,...)
third
3 9 16(differences are 6,7,6,7,...)
etc

Differences between the first terms: 1, 1, 2, 3, 4, 5...

Elimination follows the equations...
1 + 3n
2 + 9n ; 6 + 9n
3 + 13n ; 9 + 13n
5 + 17n ; 13 + 17n
8 + 21n ; 18 + 21n
12 + 25n ; 24 + 25n
and so on...

Now,
48 = 9 + 13n
119 = 2 + 9n
38 = 2 + 9n

so these 3 are def not the answers.
Not able to pin down either 93 or 140 in this...

nikhilkumarverm

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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
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Solution

Everytime, we are supposed to group the entire series in lumps of three, and remove the first element from each group containing three or less elements.

Thus we have ,

i - n - e
1 - 150 - 1
2 - 100 - 2
3 - 67 - 3
4 - 44 - 5
5 - 29 - 8
6 - 19 - 12
7 - 12 - 17
8 - 8 - 23
9 - 5 - 30
10 - 3 - 38
11 - 2 - 47
12 - 1 - ?? ( < 57 )

where i is iteration number
n is number of elements
e is the first element of the series

The last element has to be less than 57 following the above sequence, and has to be chosen as 48. there must be a better method to do this and am just sharing my vague solution that worked.

ans (b) 48

implex

One more solution

after first step left=100
after second step=66
after 3rd step=44
after 4th step =29
after 5th =19
after sixth=12
after 7th=8
after 8th =5
after 9th =3
after tenth 2
after 11th =1

so the winner is 1 after 11th round
it was 2nd after 10th round
and 3rd after 9th round
and fifth after round 8
8th after round 7
12th after round 6
18th after round 5
27th after round 4
41 after round 3

and now it is easy to check it is 140 ..

harrison_hill

hi everybody
this is my first post.
i think answer is 140.

ist cancelled qqad devil=1+3d

2nd cancelled qqad devil= 2 + 4+6 +4 +6 till

3rd cancelled qqad devil= 3 +6+8+6+8-------

4th cancelled qqad devil= 5+9+12+9+12----

all the first four options cancell out leaving the last option

Varun Khullar

Quote:

Originally Posted by selebratinglife

Now I did it by some sort of calculated guess. I just kept on subtracting the number of terms before the number until it is eventually of the form 3n+1. Once it gets to that form we are done!
Now lets take the number 119, number of terms before it (of the form 3k+1) =40
So 119 will take the place of 119-40=79 after round 1, which is of the form 3k+1. So it's eliminated.
Now for 48, after the first round 48 will take the place of 48-16=32.
After second round,32-11=21,after 3rd,21-7=14, after 4th,14-5=9,after 5th,9-3=6, after 6th, 6-2=4. Now we are done.
Now before calculating for 93, lets calculate for 140. After 1st round 140 will take the position of 140-47=93. So it would always be one step ahead of 93. So our required answer is option e).

Before 150 gets condensed to a single round, there are 11 rounds of elimination. So W.r.t choice b), it can be safely eliminated.

nice one selebrating.. the series is
1, 2, 3, 5, 8, 12, 18, 27, 41, 62, 93, 140, 210, 315

isha_pisces26

According to me the answer comes out to be (e) 140

This is how:-

In the various rounds, the numbers eliminated are of the form
round 1: 1+3n
round 2: 2+4n / 2+9n
round 3: 3+6n / 3+14n
round 4: 5+9n / 5+19n
round 5: 8+13n / 8+24n
round 6: 12+18n / 12+29n
......

Now considering the answer choices:
(a) 93 = 3+6n where n=15
(b) 48 = 12+18n where n=2
(c) 119 = 5+19n where n=6
(d) 38 = 2+9n where n=4
(e) 140 does not have any such conversion

Hence answer is (e) 140

vipul.prakash

Round# -- no of terms eliminated ---- no of terms left
1 ------------50 -------------------------100
2 ------------34 ------------------------- 66
3 ------------22 ------------------------- 44
4 ------------15 ------------------------- 29
5 ------------10 ------------------------- 19
6 ------------ 7 -------------------------- 12
7 ------------ 4 -------------------------- 8
8 ------------ 3 -------------------------- 5
9 ------------ 2 -------------------------- 3

now when 3 terms are left the last term(3rd) will remain till end... working backwards... the last term(3rd) was 5th in round8... 8th in round7... 12th 1 round earlier... but not 19th.. instead 18th in earlier round(round 5)...
NOW similarly we work backwards... and find the 18th term in round 5 was which number term 1 round earlier....

it wont take much calc... to come at final ans as 140 !!!

vipul.prakash

the 140th term wud become 93rd, 62nd, 41st, 27th, 18th, 12th, 8th, 5th, 3rd and hence last term remaining..