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There are 5 locks and 5 keys and each of the 5 keys

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There are 5 locks and 5 keys and each of the 5 keys [#permalink]

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There are 5 locks and 5 keys and each of the 5 keys matches each of the 5 locks. What is the minimum and the maximum trial numbers of attempts needed to confirm that each of the 5 keys matches each of the 5 locks?
A. 5,15
B. 4,15
C. 5,10
D. 4,10
E. 5,20


pls. explain in detail.
[Reveal] Spoiler: OA
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Re: There are 5 locks and 5 keys and each of the 5 keys [#permalink]

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New post 10 Dec 2015, 11:20
robu wrote:
There are 5 locks and 5 keys and each of the 5 keys matches each of the 5 locks. What is the minimum and the maximum trial numbers of attempts needed to confirm that each of the 5 keys matches each of the 5 locks?
A. 5,15
B. 4,15
C. 5,10
D. 4,10
E. 5,20


pls. explain in detail.

hi
this is a derangement question. Basically we need to de-arrange the 4 keys. if all 4 keys goes in correct lock then the fifth key will automatically go into the correct lock.
so the minimum number is 4.

for the maximum number consider cases

so if you want to de arrange 1 thing you can do it in 0 ways. Incase of 2, you can do it in 1 way. put 1 wrong key in wrong lock. In case of 3 you can do it in 2 ways such that no key goes into the correct lock. similarly for 4 its 9 ways. now the 10 one will be the one where the keys of 4 lock matches the correct lock. since 4 are matched the 5th key will be matched itself.

hope it helps !!!
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Re: There are 5 locks and 5 keys and each of the 5 keys [#permalink]

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New post 10 Dec 2015, 14:27
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Assume you have locks 1-5 and keys A-E.

Minimum: assume you are lucky to find the correct key/combo on the first try. So, 1 -> A, 2 -> B, 3 ->C, and 4 -> D, then 5 must match with E. Therefore, you only need to try 4 combos at a minimum.

Maximum: assume that it takes as many guesses as possible. So, with the first key you try A, B, C, and D with no success, therefore E must be the match (so 4 attempts). For key 2 you no longer have E available so you try A, B, and C, with no success, therefore D must be the match (3 attempts). And so on for key 3 (2 attempts) and key 4 (1 attempt). Key 5 matches with the remaining lock for a total of 4 + 3 + 2 + 1 = 10 attempts.
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Re: There are 5 locks and 5 keys and each of the 5 keys [#permalink]

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New post 20 Dec 2015, 17:30
very good question.
I thought what are the actual number of tries that we need to "unlock" all the lockers, and got into the trap 5-15.
of course, if we do not need to unlock the lockers, but just simply to check..
minimum # of attempts is 4.
maximum is 10, as explained above.

very tricky one.
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Re: There are 5 locks and 5 keys and each of the 5 keys [#permalink]

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New post 10 Jul 2017, 15:44
I don't know guys, maybe you've already got this GMAT mindset, the way of thinking, I am still on my way
the description clearly asks
Quote:
There are 5 locks and 5 keys and each of the 5 keys matches each of the 5 locks. What is the minimum and the maximum trial numbers of attempts needed to confirm that each of the 5 keys matches each of the 5 locks?

So basically the first part is an assumption, that we need to confirm, otherwise there wouldn't be a question, right?
The second part asks about total min/max number of attempt to confirm the assumption, that all 5 keys match all 5 locks
Let's assume that the keys are A, B, C, D and F. And locks are 1, 2, 3, 4, and 5.

So If A matches 1, it doesn't mean by default that this gonna match 2, as this ideal cross-matching is the assumption that we need to check.
So to develop, I should try A for all 5 locks in order to confirm the assumption.
You want to check your understanding of the question, just don't read the responses.

I have certain problems with understanding this GMAT wording for such kind of questions. With math or geometry based question I have no such problem
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Re: There are 5 locks and 5 keys [#permalink]

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New post 09 Sep 2017, 23:03
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dhardubey wrote:
There are 5 locks and 5 keys and each of the 5 keys matches each of the 5 locks. What is the minimum and the maximum trial numbers of attempts needed to confirm that each of the 5 keys matches each of the 5 locks?
A. 5,15 B. 4,15 C. 5,10 D. 4,10 E. 5,2



Hi...

Pl post as per the rules of forum.
Give topic name as first few words of Q and search before posting.

Now as per Q..
Min..
All fit in correctly.
When first four have fit in, fifth is not required to be tested as it has to fit.
So 4
Max..
Choose the right key as last possibility.
First lock - 5th one Correct, so 4 attempts
Similarly others 3,2,1..
Total 4+3+2+1=10

D
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Re: There are 5 locks and 5 keys and each of the 5 keys [#permalink]

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New post 10 Jan 2018, 18:42
Hi All,

This is a poorly-worded question, so you might want to consider studying with practice materials that are better "designed'.' That having been said, the 'intent' of this question is probably that there are 5 locks and 5 keys - and each of the keys opens exactly one of the 5 locks. We're asked for the least/most number of attempts that it would take to determine the proper 'pairing' of each key to each lock.

To start, you should notice that the answer choices are all relatively small, so you can probably 'brute force' the solution - just 'map out' how the attempts would have to go (without need of any complex math).

Let's call the locks: A, B, C, D and E
Let's call the (matching) keys: a, b, c, d and e

IF.... we 'luck out' and manage to place each key with each lock on the first try, there would be...
a in A = 1st attempt
b in B = 2nd attempt
c in C = 3rd attempt
d in D = 4th attempt
Based on what we're told, with just one lock and one key left, there'd be no reason to make an attempt - that key would have to fit that lock. Thus, the LEAST number of attempts would be 4. Eliminate Answers A, C and E.

In that same way, we can now determine what would happen if we were 'unlucky' and took the maximum number of tries to open each lock.....
a in B/C/D/E = 4 attempts... and then we'd know that a would have to 'match' A.
b in C/D/E = 3 attempts... and then we'd know that b would have 'match' B.
c in D/E = 2 attempts... and then we'd know that c would have to 'match' C.
d in E = 1 attempt... and then we'd know that d would have to 'match' D.
That would leave just e in just E, which would not require an additional attempt.
Thus, the MOST number of attempts would be 10.

Final Answer:
[Reveal] Spoiler:
D


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Re: There are 5 locks and 5 keys and each of the 5 keys   [#permalink] 10 Jan 2018, 18:42
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