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

Question

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

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

Bunuel · Accepted Answer

GMAT Club Official Explanation:

Let's start by minimizing the number of athletes who might have passed the test by all three agencies:

The least number of athletes who could have passed both the first and the second tests can be calculated as follows: (Test 1) + (Test 2) - (Total) = 40 + 35 - 50 = 25. This scenario occurs when all athletes are passed either by the first agency (40), the second agency (35), or both (25), meaning no athlete is failed by both the first and the second agencies.

Using the same logic, the least number of athletes who could have passed the first and the second, AND the third tests can be calculated as: (Test 1 and Test 2) + (Test 3) - (Total) = 25 + 30 - 50 = 5.

Or consider this. After the first step, we'd have two groups of athletes: the first group consists of 25 athletes who were passed by both the first and the second agencies, and the second group consists of the remaining 25 athletes who were passed by either the first or the second agency but NOT by both. The third agency passed 30 athletes. To minimize those passed by all three, the third agency could have passed all 25 from the second group, who were NOT passed by both the first and the second agencies. We are then left with 5 athletes from the 30 passed by the third agency, who must fall into the first group (those passed by both the first and the second agencies). Thus, these 5 athletes are those who were passed by all three agencies.)

Therefore, the least number of athletes who must have passed all three agencies is 5.

_______________________

As for the maximum number of athletes who might have passed the test by all three agencies:

Since the third agency passed the lowest number of athletes, 30, the number of athletes passed by all three agencies cannot possibly be more than that.

Let's check whether 30 is possible.

If all three agencies passed 30 athletes, we'd be left with 20 athletes to accommodate. The first agency, which passed 40 athletes, could have passed 10 (40 - 30 = 10) athletes out of those 20 athletes, leaving 10 athletes to accommodate. However, the second agency, which passed 35 athletes, could have passed only 5 (35 - 30 = 5) athletes out of those 10 athletes, leaving 5 athletes who were passed by neither of the agencies. Since we are given that all athletes passed the test by at least one agency, this scenario is not possible.

Essentially, if x is the maximum number of athletes that could have been passed by all three agencies, then:

40 - x would be the number left from the first agency to accommodate other athletes, 
 35 - x would be the number left from the second agency to accommodate other athletes, 
 30 - x would be the number left from the third agency to accommodate other athletes.

Thus, we need such a maximum x that: x + (40 - x) + (35 - x) + (30 - x) >= 50. This gives x <= 27.5. Since x must be an integer, the maximum x is 27.

Hence, the maximum number of athletes who might have passed the test by all three agencies is 27.

Answer: D.

meet459 · Answer

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

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

busygmatbee1290 · Answer

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.

Sof22 · Answer

Let's begin with the highest numbers of athletes who might have passed the test by all three agencies.
If 30 students might have passed the test by all three agencies, the total number will be less than 50;  30+10+5 = 45 
If 27 students might have passed the test by all three agencies, the total number will be equal to 50;  27+13+8+3 = 51  (assumed that no athlete might have passed the test by two agencies)
If we suppose that one athlete might have passed the test by two agencies,  27+3+13+7=50 
Therefore, the highest number is 27.

The lowest number is 5.
Firstly, because D is the only option that has 27
Secondly, because if it is 5,
when  40+10=50  (5 in all 3 agencies, 25 in the 1st and the 3rd, 20 in the first and the 2nd)
If we choose 0, we number is greater than 50

Therefore, the correct answer is D. 5 and 27

0ExtraTerrestrial · Answer

Let's denote the number of athletes who passed only one test from any agency out of the three- A, the number of athletes who passed exactly two tests each from one of the three agencies- B, and the number of athletes who passed tests from all three agencies- C, such that sets of employees considered in the three countings are disjoint.

It follows that
 A+B+C = 50 ........(i)

And

A+2B+3C= 40+35+30= 105 ........(ii)

Subtracting (i) from (ii) gives,

B+2C= 55 ........(iii)

or
 ​​​​​​​B+C+C= 55

which can again be put in (i) to obtain

​​​​​​​C-A= 5 ........(iv)

We need to both minimise and maximise C.
1. Since C cannot be negative, the minimum value for C from (iv) follows to be C= 5, let's check if B and A agree to the input, A comes out as zero and B as 45, since 45+5+0 = 50 (i) and 0+2(45)+3(5)= 105 (ii), this case is valid.
2. B+2C is an odd number, with 2C even, this means B is odd, here to maximise C we start by putting B= 1, which gives C= 27 and A= 22, since 22+1+27= 50 (i) and 22+2(1)+3(27)= 105 (ii), this case is valid.

So the minimum and maximum values for C are 5 and 27 respectively. Thus D is correct.

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