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Quiz Results (Posted on 2004-09-02) Difficulty: 3 of 5
Five students each answered five questions on an quiz consisting of two multiple-choice questions (A, B or C) and three True-False questions. They answered the questions as follows:
Student Q1 Q2 Q3 Q4 Q5 
Alex    A  A  T  T  T 
Bert    B  B  T  F  T 
Carl    A  B  T  T  F 
Dave    B  C  T  T  F 
Eddy    C  A  F  T  T 
No two students got the same number of correct answers. Who got the most correct answers?

See The Solution Submitted by Brian Smith    
Rating: 3.2222 (9 votes)

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Solution solution using binary | Comment 11 of 30 |
there are only 2 sets of possibilities for their individual scores: 5 4 3 2 1 or 4 3 2 1 0
i)assuming 5 4 3 2 1
only one of them has the set of all correct choices. if this were alex, (let 1=correct answer 0=wrong answer)
 
TABLE 1
alex=perfect score
a a t t t
alex 1 1 1 1 1 = 5
bert 0 0 1 0 1 = 2
carl 1 0 1 1 0 = 3
dave 0 0 1 1 0 = 2
eddy 0 1 0 1 1 = 3

we can see from here that the scores are
5 2 3 2 3, or not all scores are unique
therefore alex's answers must NOT be the set of all correct choices

doing the same thing with all the other four, none of them gives the others unique scores:


TABLE 2
bert=perfect score
b b t f t
alex 0 0 1 0 1 = 2
bert 1 1 1 1 1 = 5
carl 0 1 1 0 0 = 2
dave 1 0 1 0 0 = 2
eddy 0 0 0 0 1 = 1

scores: 2 5 2 2 1

--

TABLE 3
carl=perfect score
a b t t f
alex 1 0 1 1 0 = 3
bert 0 1 1 0 0 = 2
carl 1 1 1 1 1 = 5
dave 0 0 1 1 1 = 3
eddy 0 0 0 1 0 = 1

scores: 3 2 5 3 1

--

TABLE 4
dave=perfect score
b c t t f
alex 0 0 1 1 0 = 2
bert 1 0 1 0 0 = 2
carl 0 0 1 1 1 = 3
dave 1 1 1 1 1 = 5
eddy 0 0 0 1 0 = 1

scores: 2 2 3 5 1

--

TABLE 5
eddy=perfect score
c a f t t
alex 0 1 0 1 1 = 3
bert 0 0 0 0 1 = 1
carl 0 0 0 1 0 = 1
dave 0 0 0 1 0 = 1
eddy 1 1 1 1 1 = 5

scores: 3 1 1 1 5

therefore, the only possibility of scores would be 4 3 2 1 0.


ii) since 4 3 2 1 0 is the correct set of scores, then one of them (and only one) must get 4 correct and also, only one of them must get 0


A)assuming carl to have 4 correct answers
this means:
we would need only to change one of his
answers to get the set of correct choices.

carl=a b t t f.

ex.if carl got item 5 wrong, then correct set of answers must be a b t t t.

the catch is this: Looking at table 3 reprinted below,


TABLE 3
carl=perfect score
a b t t f
alex 1 0 1 1 0 = 3
bert 0 1 1 0 0 = 2
carl 1 1 1 1 1 = 5
dave 0 0 1 1 1 = 3
eddy 0 0 0 1 0 = 1

scores: 3 2 5 3 1

the only possible item carl can get a wrong must be item 4, since this will make eddy's score=0
(we want one of them to get 0 since one has to be 0)

so, then the correct choices must be a b t f f
but checking the others' scores agains this
(a b t f f), we get:
alex=2, bert=3, carl=4, dave=4, eddy=2
which are not unique scores. so then carl=4 must be false

applying the same priciple to the other tables:

B)assuming alex=4

reproducing TABLE 1 above:

a a t t t
alex 1 1 1 1 1 = 5
bert 0 0 1 0 1 = 2
carl 1 0 1 1 0 = 3
dave 0 0 1 1 0 = 2
eddy 0 1 0 1 1 = 3

scores: 5 2 3 2 3

alex=4 must not be correct, because to give alex a 4, he must have gotten only 1 item incorrect. but in this case, the minimum score is 2, so that if we change only one item for alex to get a 4, the least score we can produce from the others will be a 1 (since 2-1=1). nobody then will get a 0.

so alex=4 is clearly false.


C) assuming dave=4

TABLE 4
dave=perfect score
b c t t f
alex 0 0 1 1 0 = 2
bert 1 0 1 0 0 = 2
carl 0 0 1 1 1 = 3
dave 1 1 1 1 1 = 5
eddy 0 0 0 1 0 = 1

scores: 2 2 3 5 1

examining table 4 above, only by changing item 4 to a f will give us a score of 0 (which is for eddy)
so then the correct set of choices, to give dave a 4, must be b c t f f
checking the other person's answers against this will give us the scores:

(checklist= b c t f f)
alex=1
bert=3
carl=2
dave=4
eddy=0

this is clearly a unique set of scores therefore dave=4 must be true. so then the correct answers must be:

b c t f f


but then, HOW ARE WE SURE that this is the only solution to the problem?

Then we have to examine the last 2 possibilities, one where bert=4 (table 2) and another where eddy=4 (table 5). If we apply the same principles above to tables 2 and 5, we will find out that both are also false. only the case DAVE=4 gives us a unique set of scores :)





Edited on September 12, 2004, 4:36 pm
  Posted by i_wish on 2004-09-05 18:32:39

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