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# GMAT Club Olympics 2024 (Day 6): To be allowed to participate in the

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Re: GMAT Club Olympics 2024 (Day 6): To be allowed to participate in the [#permalink]
­To be allowed to participate in the Olympic Games, an athlete must pass doping tests conducted by three different agencies. Out of 50 athletes, 40 passed the test by the first agency, 35 by the second agency, and 30 by the third agency. If all athletes passed the test by at least one agency, which of the following represent the lowest and highest numbers of athletes who might have passed the test by all three agencies?

A. 0 and 25
B. 0 and 30
C. 5 and 25
D. 5 and 27
E. 5 and 30­

Explanation:
Total athletes = 50
Passed by 1st agency = 40
2nd agency = 35
3rd agency = 30

The maximum:
3rd agency passed the least number of athletes, 30, hence, that is the highest number who might have passed all 3 tests, assuming these 30 passed the test by the 1st and the 2nd agency

The minimum:
1st agency did not pass 10 athletes. (50-40)
2nd agency did not pass 15 athletes   (50-35)
and the 3rd did not pass 20 athletes (50-30)
Assuming, each agency rejected different people, and no one was rejected by more than one agency, the total atheletes who did not pass were 50-10-15-20 = 5
Therefore, at least 5 athletes passed all three agencies.
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Re: GMAT Club Olympics 2024 (Day 6): To be allowed to participate in the [#permalink]
For max case, there has to be overlap like an eclipse,
Highest= min (40,35,30)=30

for Min, lets find the compliment of passes(rather failures)

Test 1=50-40=10
Test 2=50-35=15
Test 3=50-30=20
Sum=10+15+20=45
Hence out of 50, 45 are failures, since all have passed by atleast one agency, Min=50-45=5

E)
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Re: GMAT Club Olympics 2024 (Day 6): To be allowed to participate in the [#permalink]
Bunuel wrote:
­To be allowed to participate in the Olympic Games, an athlete must pass doping tests conducted by three different agencies. Out of 50 athletes, 40 passed the test by the first agency, 35 by the second agency, and 30 by the third agency. If all athletes passed the test by at least one agency, which of the following represent the lowest and highest numbers of athletes who might have passed the test by all three agencies?

A. 0 and 25
B. 0 and 30
C. 5 and 25
D. 5 and 27
E. 5 and 30­

 This question was provided by GMAT Club for the GMAT Club Olympics Competition Win over \$30,000 in prizes such as Courses, Tests, Private Tutoring, and more

­

­In three set prob, The max overlap will be given by the smallest circle in the venn diagram as it can be included in both of the circles to give the max possible overlap.
To calculate min overlap consider this. Take 35 overlap in only the whole 50 pool and see the leftout from the 50 pool i.e. 50-35=15.
then take 40 overlap in only the whole 50 pool and see the leftout from 50 i.e. 50-40-10.
Now see this the lowest 30 will be adjusted in the space left out by 35 and 40 overlap i.e 15+10=25 but what cannot be adjusted will be 30-25=5 that has to be the min overlap for the condition to hold that the all athletes have to pass the test by atleast one agency.

IMO min 5 and max 30 E Answer.
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Re: GMAT Club Olympics 2024 (Day 6): To be allowed to participate in the [#permalink]
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To be allowed to participate in the Olympic Games, an athlete must pass doping tests conducted by three different agencies. Out of 50 athletes, 40 passed the test by the first agency, 35 by the second agency, and 30 by the third agency. If all athletes passed the test by at least one agency, which of the following represent the lowest and highest numbers of athletes who might have passed the test by all three agencies?­

Consider this diagram (attached)

We know that a+b+c+d+e+f+g = 50 (Since all athletes passed the test by atleast one agency)

Also,
a+d+g+e = 40, (number of atheletes passing the test by the first agency)
b+d+g+f  = 35, (number of atheletes passing the test by the second agency)
c+e+g+f = 30, (number of atheletes passing the test by the third agency)

If you add all of these, we get

(a+b+c+d+e+f+g) + (d+e+f) + 2g = 105

So (d+e+f) + 2g = 55

Now we need to find min and max value,

If g is max, (d+e+f) should be min, so let us assume that (d+e+f) is 0
Doing so will result in g being 27.5
But we know that we can't have half athelete, so g's max value will be 27.
And (d+e+f) must be atleast 1

This will lead to only one option (5, 27), luckyy!

But we will try to find the min value for g as well.

So (d+e+f) + 2g = 55

And (a+b+c+d+e+f+g) = 50
So (a+b+c) + (d+e+f) + g = 50

Substituting the value of (d+e+f) = 55 - 2g, we get

g = (a+b+c) + 5

Hence g's minimum value must be 5.

Hence answer (D) 5 and 27

­
Attachments

IB5go.png [ 15.49 KiB | Viewed 2145 times ]

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Re: GMAT Club Olympics 2024 (Day 6): To be allowed to participate in the [#permalink]
­1st agency = 40
2nd agency = 35
3rd agency = 30

Maximum = 30
Because 30 is the minimum of athletes that passe test by any agency. To get the maximum no. we have to ssume that these passed the tests for other 2 too.

Total= 40+ 35 + 30 = 105
Minimum = 105 - 50 -50 = 5

Ans E
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Re: GMAT Club Olympics 2024 (Day 6): To be allowed to participate in the [#permalink]
Consider test passed cases 30,35,40

Max all common is 30.

Min all common is 0.

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Re: GMAT Club Olympics 2024 (Day 6): To be allowed to participate in the [#permalink]
Explanation: Minimum: 10 didn't pass the first one, 15 the second, and 20 the third one, if these are all different people, it means 45 people didn't pass all three tests.
Maximum: 30 people passed the third one, and are also included in the first and second test.
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Re: GMAT Club Olympics 2024 (Day 6): To be allowed to participate in the [#permalink]
Bunuel wrote:
­To be allowed to participate in the Olympic Games, an athlete must pass doping tests conducted by three different agencies. Out of 50 athletes, 40 passed the test by the first agency, 35 by the second agency, and 30 by the third agency. If all athletes passed the test by at least one agency, which of the following represent the lowest and highest numbers of athletes who might have passed the test by all three agencies?

A. 0 and 25
B. 0 and 30
C. 5 and 25
D. 5 and 27
E. 5 and 30­

 This question was provided by GMAT Club for the GMAT Club Olympics Competition Win over \$30,000 in prizes such as Courses, Tests, Private Tutoring, and more

­

­total is 50
1st agency is 40 , 2nd 35 and 3rd 30
maximum can be 30 and least can be 5
option E is correct
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Re: GMAT Club Olympics 2024 (Day 6): To be allowed to participate in the [#permalink]
Given: To be allowed to participate in the Olympic Games, an athlete must pass doping tests conducted by three different agencies. Out of 50 athletes, 40 passed the test by the first agency, 35 by the second agency, and 30 by the third agency.

Asked: If all athletes passed the test by at least one agency, which of the following represent the lowest and highest numbers of athletes who might have passed the test by all three agencies?

For highest numbers of athletes who might have passed the test by all three agencies : -
Attachment:

Screenshot 2024-07-15 at 9.05.50 PM.png [ 38.48 KiB | Viewed 1788 times ]

The highest numbers of athletes who might have passed the test by all three agencies = 25

For lowest numbers of athletes who might have passed the test by all three agencies : -
Attachment:

Screenshot 2024-07-15 at 9.05.59 PM.png [ 39.97 KiB | Viewed 1788 times ]

The lowest numbers of athletes who might have passed the test by all three agencies = 5

IMO C­
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Re: GMAT Club Olympics 2024 (Day 6): To be allowed to participate in the [#permalink]
­To find the maximum no of athletes who passed test from all three agencies would be the the maximum overlap
so, bigger circle is of 50 athletes who passed atleast 1 test, another of 35 athletes inside 50 athletes circle & last circle of 30 athletes inside 40 athletes circle.
maximum would be 30 athletes

to find minimum, let's find the no of athletes who failed the test, minimum overlap
1. Athletes who failed test 1: 10
2. Athletes who failed test 2: 15
3. Athletes who failed test 3: 20

Sum is 45, & this could be the unique no of athelets
So 50-45 =5

IMO E
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Re: GMAT Club Olympics 2024 (Day 6): To be allowed to participate in the [#permalink]
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Let a,b,c be the number of participants passing from one, two and all three agencies respectively. According to the data given

a+b+c = 50 (all athletes pass from at least one agency)
a+2b+3c = 105

Reducing the above two equations gives
b +2c = 55

Maximum value of c can be 27 as can be seen above with b=1

Now minimum value of C cannot be zero as that would make b=55, but the total number of athletes only sum to 50.
By hit and trial, we find that c has to be a minimum of 5 for b=45 to sum up to 50 with a=0

Therefore, D

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Re: GMAT Club Olympics 2024 (Day 6): To be allowed to participate in the [#permalink]
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Bunuel wrote:
­To be allowed to participate in the Olympic Games, an athlete must pass doping tests conducted by three different agencies. Out of 50 athletes, 40 passed the test by the first agency, 35 by the second agency, and 30 by the third agency. If all athletes passed the test by at least one agency, which of the following represent the lowest and highest numbers of athletes who might have passed the test by all three agencies?

A. 0 and 25
B. 0 and 30
C. 5 and 25
D. 5 and 27
E. 5 and 30­

 This question was provided by GMAT Club for the GMAT Club Olympics Competition Win over \$30,000 in prizes such as Courses, Tests, Private Tutoring, and more

­

Let’s go by highest number of people
Say 30 people is common for all, so we have 0 remaining in third, 5 in second and 10 in first. Considering remaining as unique total becomes 45 and hence cannot be answer
Next try 27, so we got 3 remaining from third, 8 from second and 13 from first. Now consider 1 common between third and first.
So total becomes 27+. 2 unique third + 1 common third and first + 8 unique second + 12 unique first = 50

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Re: GMAT Club Olympics 2024 (Day 6): To be allowed to participate in the [#permalink]
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­n(A) = 40
n(B) = 35
n(C) = 30

n(AnB - C) = b
n(CnB - A) = c
n(AnC - B) = a

n(AnBnC) = x

now,
50 = 40 + 35+ 30 -a -b -c -2x
55 = a + b + c + 2x ----(1)
For Max(x) :
a+b+c = 1 (Possible)
{x = 27} (possible because x=<min(A,B,C) i.e. 30)

For Min(x) :

40>=a+b+x ----(2)
35>=b+c+x ----(3)
30>=a+c+x ----(4)

also, from equation 1,
a+b+x = 55-c-x
put in eq (2)
c+x>=15
put in eq(3)&(4)
b<=20
a<=15
put these in eq(2) i.e. x<=40-a-b
implies, x>=40-20-15 i.e. {x>=5}
so, Min(x) = 5;

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Re: GMAT Club Olympics 2024 (Day 6): To be allowed to participate in the [#permalink]
Bunuel wrote:
­To be allowed to participate in the Olympic Games, an athlete must pass doping tests conducted by three different agencies. Out of 50 athletes, 40 passed the test by the first agency, 35 by the second agency, and 30 by the third agency. If all athletes passed the test by at least one agency, which of the following represent the lowest and highest numbers of athletes who might have passed the test by all three agencies?

A. 0 and 25
B. 0 and 30
C. 5 and 25
D. 5 and 27
E. 5 and 30­

 This question was provided by GMAT Club for the GMAT Club Olympics Competition Win over \$30,000 in prizes such as Courses, Tests, Private Tutoring, and more

­

­Drawing a venn diagram can be easier to visualise. We call people who passed the test by the first agency A, by the second agency B, and by the third agency C. We shall have 50 = 40 + 35 + 30 + (overlap of A, B and C) - (overlap of A and B) - (overlap of A and C) - (overlap of C and B) => 50 = 105 + (overlap of A, B and C) - [(overlap of A and B) + (overlap of A and C) + (overlap of C and B)]

So we shall have lowest number of athletes who might have passed by all 3 agencies when [(overlap of A and B) + (overlap of A and C) + (overlap of C and B)] is maximised and vice versa, we will have the highest number of athletes who might have passed by all 3 agencies when [(overlap of A and B) + (overlap of A and C) + (overlap of C and B)] is minimised. Since all athletes passed the test by at least one agency => minimum of overlap of A,B,C > 0 => 5 => try plug in answers into the diagram to find the highest number that fit => 30 => E

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Re: GMAT Club Olympics 2024 (Day 6): To be allowed to participate in the [#permalink]
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Re: GMAT Club Olympics 2024 (Day 6): To be allowed to participate in the [#permalink]
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To be allowed to participate in the Olympic Games, an athlete must pass doping tests conducted by three different agencies. Out of 50 athletes, 40 passed the test by the first agency, 35 by the second agency, and 30 by the third agency. If all athletes passed the test by at least one agency, which of the following represent the lowest and highest numbers of athletes who might have passed the test by all three agencies?

A. 0 and 25
B. 0 and 30
C. 5 and 25
D. 5 and 27
E. 5 and 30­

​​​​​​For highest numbers of athletes, we need to cover maximum number while maintaining the condition that all athletes passed the test by at least one agency.
1st agency -
40 passed, 10 failed
2nd agency -
total 35 athletes passed - 27 passed from previous 40, 8 passed from previous 10
3rd agency -
total 30 athletes passed - 27 passed from previous 27, 2 passed from last remaining, and 1 passed from any of the above.

There is only one answer that contains highest as 27.
So D. 5 and 27 is the correct choice.
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Re: GMAT Club Olympics 2024 (Day 6): To be allowed to participate in the [#permalink]
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Total number of athletes: n=50
Number of athletes passing each agency's test:
• First agency: A=40
• Second agency: B=35
• Third agency: C=30
Total count = 105. Hence, there are 55 additional "slots" where an athlete can be part of more than a single agency.
If we start with 0 athletes who were tested by all 3 and let x be the number of athletes tested by 2 agencies, we get (50-x) +2x = 105
Which gives x=55.
So, the minimum is not 0. The options show that the minimum in that case is 5.

Moving on to the maximum, we can now use the formula.
Total=A+B+C−(sum of EXACTLY 2−group overlaps)−2∗(all three)+Neither
50=105 - (exact 2) -2*(all three)
To maximise (all three) group, minimise (exactly 2) group.
(Exact 2) + 2(all 3) = 55
Minimum of Exact 2 group is 1 since 55 is not divisible by 2.
Hence, All 3 = 54/2 = 27.
Option D.

We can laso solve for minimum using the formula (Exact 2) + 2(all 3) = 55
We can see that the minimum value that works for (All Three) group is 5 which leaves 45 for (Exact 2) group and the equation holds. ­
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