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Sets - confusing OA

by 4meonly » Fri Nov 14, 2008 6:03 am
A tester can pass a certain test only when all the three examiners pass him or her. A total of 30 testers took the test. If examiner A passed 20 testers, B passed 17 testers and C passed 15 testers, at least how many testers passed the test?
[spoiler]
OA 0
I get 2[/spoiler]
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by Tryingmybest » Fri Nov 14, 2008 7:24 am
Let us go from low to high passes in this case as this deals with minimum testers passed.

C passed 15 - Take first 15 pass

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 to 30
y y y y y y y y y y y y y y y NO

Now tester B passed 17 say he passed the last 17

1 to13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
N y y y y y y y y y y y y y y y y y y


NOW FROM ABOVE TRY TO GUESS HOW MAY C AND B PASSED

14 15

Well we know A passed 20 testers , since 20 testers could be grouped other than testers 14 15 meaning for example A could have passed 1 2 3 4 5 6 7 8 9 10 11 17 18 19 20 21 22 23 24 25.

We have 0 common testers answer 0
Last edited by Tryingmybest on Fri Nov 14, 2008 8:10 am, edited 4 times in total.
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by stop@800 » Fri Nov 14, 2008 7:30 am
In first attempt I also got 2
but here is the answer

Total 30

A passed 20
B passed 17
C passed 15

A: 0
B: 2
C: 0
AB: 10
AC: 10
BC: 5
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by srisl11 » Fri Nov 14, 2008 7:37 am
We need to find the minimum number of testers passed by all three (A, B and C)
B passed 17 testers and C passed 15 testers
So B and C have passed atleast 2 testers in common
If B and C have passed 2 testers in common then B has 15 and C has 13 different testers .(testers passed by either B or C but not both)

A has passed 20 testers ...Now A can pass 10 testers which B has
passed and 10 different testers which C has passed.
or A could have passed 15 testers which B has passed and 5 testers which C has passed....
(we can try out other combinations too)

From the above mentioned combinations we can say that 0 is the minimum number of testers passsed by all three A, B and C

Hope you understd....
(hope I didn't give a explanation which only I can understand )
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by gmat009 » Fri Nov 14, 2008 8:02 am
stop@800 wrote:In first attempt I also got 2
but here is the answer

Total 30

A passed 20
B passed 17
C passed 15

A: 0
B: 2
C: 0
AB: 10
AC: 10
BC: 5
Stop, can you plz. explain how you got AB,AC,BC
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by logitech » Fri Nov 14, 2008 9:44 am
stop@800 wrote:In first attempt I also got 2
but here is the answer

Total 30

A passed 20
B passed 17
C passed 15

A: 0
B: 2
C: 0
AB: 10
AC: 10
BC: 5
A: 0
B: 4
C: 0
AB: 9
AC: 11
BC: 4

?

What happened to 30 students ?
LGTCH
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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by EricLien9122 » Fri Nov 14, 2008 10:47 am
pass Not pass
A: 20 10
B: 17 13
C: 15 15

So, we know at least 15 teachers failed the test. Since the question didn't tell us any relationship between A, B and C, and asked about how many at least passed (minimum value), we have to assume the worst possible situation.

C failed 15, A and B failed 15 together, therefore answer is 0.

If this question is asking about at most (maximum value), we can assume the best possible situation.

A+B+C failed 15 together, and 15 passed.


I hope this makes sense, please correct me if I made a mistake.
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