we start with a logical reasoning problem

**Problem 1**

A, B, C, D and E decide to run a race. Before the race, five predictions of the outcome are made: (1) ECBAD (as in A first, C second, etc), (2) DAECB, (3) DCBAE, (4) BDAEC, (5) DCBEA. No prediction was completely correct but two of them correctly predicted the placements of exactly two of the runners. The remaining three predictions were totally wrong. What was the actual outcome of the race?

A, B, C, D and E decide to run a race. Before the race, five predictions of the outcome are made: (1) ECBAD (as in A first, C second, etc), (2) DAECB, (3) DCBAE, (4) BDAEC, (5) DCBEA. No prediction was completely correct but two of them correctly predicted the placements of exactly two of the runners. The remaining three predictions were totally wrong. What was the actual outcome of the race?

## 3 comments:

1 arranged all the results in rows keeping alignment in tact..

2 as we know tht only two places of two races is correct and rest all are completely wrong we have to search for the rows where there is a unique element..

for eg if we check for the 1st and the 5th result..in the first column we get

E and B as distinct which no other results have..so we know that E or B can be our potential first place..simillarly we carry out for all the coulumns..

3.we will obtain sevral results and from there on we wil have to put the sequence and check for consistency...

hence one gets the correct answer.. BAEDC.

Since C,B are common in same places in 3 predictions, thus they must be false. Hence somewhat true predictions would be DAECB and BDAEC.

DAECB

BDAEC

If we take A,E in 1st and B,C in 2nd we find D to be 4th which is also not predicted by other 3 wrong predictions

ANS:- BAEDC

Nice quest bro :)

BAEDC.. and as explained by others..

Achilles's methodology helps breaking the problem too..:-)

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