[2008] DI and VA/RC Question a Day by Arun Sharma's Mindworkzz

[implex]

Originally Posted by prakharc

Can anyone have 6 correct bets??? If 6 are correct then 7th will definetly be correct....So I think it'll be 0-5 correct bets??

oops yeah right !!
so assume 4 and 5 correct bets

blunder on my part !

[milindagrawal]

The first condition is that each person has minimum 1 correct bet.
So it should have been 1,2,3,4,5,6(here 6=7). But then its given that no person had all the correct bets.
Also, no person got all wrong. Which means 0,1,2,3,4,5 is also not possible.
Is the question contradictory in itself or am I missing something?
help!!

[implex]

Originally Posted by milindagrawal

The first condition is that each person has minimum 1 correct bet.
So it should have been 1,2,3,4,5,6(here 6=7). But then its given that no person had all the correct bet.
Also, no person got all wrong.
Is the question contradictory in itself or am I missing something?

milind each horse has one correct bet
and not each person !!

[shivam_01]

Originally Posted by implex
No two guys have same number of correct bets
max bet can eb 7 but not possible as given
so the number of correct bets are 0,1,2...6

Now clealry as one has 6 correct and other has 5 correct
so at max mismatch between these two is 3
hence one of C and e has got 6 other 5
hence H5 is the winner
similarly H6 stands second


1) b 2) C

rest coming !!

i didd nt get ur point how can 6 be corrrect for a person in case it is so 7th will also be correct please elaborate .. good question

[implex]

Originally Posted by shivam_01

i didd nt get ur point how can 6 be corrrect for a person in case it is so 7th will also be correct please elaborate .. good question

I thought their were 7 persons
usse sab gadbad ho gayi

[ankaj]

Originally Posted by milindagrawal

The first condition is that each person has minimum 1 correct bet.
So it should have been 1,2,3,4,5,6(here 6=7). But then its given that no person had all the correct bet.
Also, no person got all wrong.
Is the question contradictory in itself or am I missing something?

I guess question refers that every horse gets atleast one correct bet...a person can have all incorrect bets....

[ankaj]

Originally Posted by shivam_01

First one to post the answers on the thread
found it after many hardships any ways first of all thanks to PG and arun sharma

sequence is H6, H2, H3, H1, H5 , H4 , H7
so the answers are

1) Option c
2) Option a
3) Option b
4) Option c
5) Option a

lets c what is the take of others

@ Shivam @ Masoom....guys both of yours sol seems to satisfy the ranks of horses and given conditions....

does this mean that we can can't be determined as answers....
if so, arriving at 1 list is difficult [atleast i am missing some thing]..arriving at 2 would need some sharp logic....

[shivam_01]

i have found another set which satisfies the same condition
sequence is H7, H6, H3, H1, H2 , H4 , H5

so two sequences is H6, H2, H3, H1, H5 , H4 , H7
&
sequence is H7, H6, H3, H1, H2 , H4 , H5

finally i think
sequence is

Rank 1 H7 or H6
Rank 2 H6 or H2
Rank 3 H3
Rank 4 H1
Rank 5 H2 or H5
Rank 6 H4
Rank 7 H5 or H7

so the answers after modification are

1) Option e
2) Option e
3) Option e
4) Option e
5) Option a

[nishant_rungta]

My answers:

The right sequence is:

1. H7
2. H6
3. H3
4. H1
5. H2
6. H4
7. H5

ANSWERS:

1. D
2. C
3. A
4. B
5. A

Will edit the post in a few moments for the solution

PS EDIT:

I had a few issues which confused me. This is a very tough problem

First, we cannot have a guy who can make exactly 6 correct bets because one bet is automatically wrong this way. Hence, there is a guy who makes all 7 wrong bets and a guy who makes a maximum of 5 right bets.

These lines in the question were quite confusing this way:

(1) Every horse was bet upon correctly by at least 1 person &
(3) No person got all seven bets correct.

Now that we know one person made a maximum of 5 right bet, we need to find that guy who makes 5 right bets. Let us assume its A but then A doesn't give the requisite combination of 0,1,2,3,4,5 right bets. Then lets go to E. He gives us the maximum variety in the bets. Now we only need to shuffle 2 of E's bets such that we get a combination of 0,1,2,3,4,5 right bets. It is for sure that E makes 5 (which is maximum) right bets.

Now the question is how do we select 2 of E's bets and interchange them. Let us select those pair of bets which are interchanged in other guy's set of bets. Look at H3 and H1's position in E's set. They are at position 3 and 4 respectively. Now this pair of choice is at posiiton 4 and 3 respectively in F's set of bets. Now let us go about checking whether this satisfies all our conditions. But it doesn't.

So now let us hunt for another such combination. Take H5 and H7 from E's set of bets. They are at positions 1 and 7 respectively. Now this pair of horse position is at position 7 and 1 respectively in B's set of bets. Hence we can narrow down that this maybe the right set of bets when we interchange only the position of 5 and 7 in E's set of bets.

Hence, I can come down to the following set by tweaking the positions in E's set of bets:

H7
H6
H3
H1
H2
H4
H5

Now this combo gives us the following results:

A makes 1 right bet which is H3 position 3
B makes 2 right bets which are H7 and H5 at position 1 and 7 respectively
C makes 3 right bets
D makes 4 right bets
E makes 5 right bets
F makes 0 right bets

I hope this solution makes sense


Nishant

[v-factor]

Phewww....gud one
I got the sequence on 4th attempt.
7 6 3 1 2 4 5

Answers:-
1.d
2.c
3.a
4.b
5.a

Adding my approach(if u can call it an approach)
It can be deduced that the the no. of correct bets are 0,1,2,3,4,5 only as 6 correct bets automatically mean 7th is also right which violated the given condition.

Now I listed all possible contenders for different ranks eg.
1st-> H1, H7, H5,H6
2nd->H2, H3, H6, H7 and so on
Now starts the painful part
I started by assuming H1 as correct and coouldn't create the right combo.So next went with H5, H6 (taking care that I dont consider the already wrong scenarios). Finally some brute force rearrangement and stumbled upon the solution.