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GMAT Club World Cup 2026 (Day 8): At a volunteer clinic, every

1 2 3 4 5

remdelectus

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15 Jul 2026, 22:38

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let Trainees-t, assistants-a, nurses-n, coordinators-c
Ratios; t:a=3:4 then a=4/3t
ratio of trainees to nurses is 3/4 of trainees to assistants
t/n=3/4(t/a)
because t/a=3/4
t/n=3/4*3/4=9/16
therefore t:n=9:16,
n=16/9t
so, a:c=3:2, c=2/3a=2/3*4/3t=8/9t
since the denominators are 3 and 9, let t=9k
a=12k, n=16k, c=8k
the smallest 3 digit +ve integers with distinct digits is 102
since t+c<102 so 9k+8k=17k<102
17k/17<102/17, k<6 therefore max integer for k=5
Max assistants and nurses
a+n=12k+16k=28k
k=5
a+n=12(5)+16(5)=140
ANS: C. 140

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OPTION C
Given
Ratio of T:A=3:4
Ratio of T:N=3/4*3:4= 9:16
Ratio of A:C= 2*T:A= 2*3:4= 3:2
and T+C<smallest three digit unique +ve no = 102

now we need to make the ratios consistent so

C:A
A:T
T:N
=( 2:3 )*4
( 4:3)*3
(9:16) = 8:12:9:16
C:A:T:N
now T+C<102 = 9K+8K<102, 17K<102, K<6
since we need to find the max no of assistants and nurses lets take max k so 5
12*5+16*5=140 ANS

Bunuel

At a volunteer clinic, every volunteer is assigned exactly one of four roles: trainee, assistant, nurse, or coordinator. The ratio of trainees to assistants is 3 : 4. The ratio of trainees to nurses is 3/4 of the ratio of trainees to assistants and the ratio of assistants to coordinators is twice the ratio of trainees to assistants. If the combined number of trainees and coordinators is less than the smallest three-digit positive integer with distinct digits, what is the maximum possible combined number of assistants and nurses?

A. 112
B. 128
C. 140
D. 168
E. 196

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T:a=3x:4x
T:n=9x:16x
A:c=6x:4x

T:a:n:c=9x:12x:16x:8x

T+c<102
17x<102
X<6

Maximum required so x=5

A+n=140

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T:A = 3x:4x

T:N = (3/4) (T:A)
(3/4) (3x/4x) = T:N = 9x/16x

A:C = 2T/A = 6x/4x

multiply eq 1 with 3
we get, T:A = 9x:12x
and we have T:N = 9x:16x

now multiply eq 3 with 2,

A:C = 12x:8x

so now we have, T:A:N:C = 9x:12x:16x:8x

told that T+C < smallest three digit distinct positive integer

smallest three digit distinct positive integer = 102

T+C <102
9x+8x < 102
17x<102
x< 6

we need max value of A+N = 28x
for that x needs to be maximun and max value of x =5

so 28(5) =140

choice C

Bunuel

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16 Jul 2026, 00:15

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Trainee (T): Assistants is 3:4,

T: Nurse(N) is (3/4)*(3/4), so 9:16

A: Cordinator (C) is (9/16)*2 so 3:2. if we set a random variable x for each of the roles as T:A:N:C we can do

3x:4x:(16/3)x:(8/3)x then mutliply times 3 to get rid of the fraction so 9x:12x:16x:8x as the ratios

the 3x:4x comes from the ratio of T:A then I use the 3x from T in the T:N ratio so N is 3x*(16/9) and lastly I used the A:C ratio and pulled the 4x, so C is 4x*(2/3).

T+C = 9x + 8x = 17x which has to be less than 102, the smallest 3 digit number with distinct digits so we want to maximize 17j<102 which gives you 5. If j is 5, A =60 and N = 80. A+N =140. C

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T:A 3:4
T:N 3/4X3/4= 9:16
A:C 2X3/4= 3:2
From T:A we can rewrite it as 3k:4k
From T:N 9:16 so N= T.(16/9)= 3k(16/9)= 16k/3 meaning k must be a multiple of 3 meaning we can let k be 3m
T= 9m, A=12m, N=16m
For A:C 3:2. C=A.2/3= 12m.2/3= 8m
Going back to the constraints the least 3-digit integer with unique digits is 102
We need T+C<102
9m+8m<102
17m<102
m<6
Since m is a positive integer the max value is 5 which works giving T+C=85
So A+N= 12X5+16X5= 140
Ans C

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The first thing we need to do is find the common ratios across all 4 categories.

There are 4 categories respectively:

T, A, N, C

T:A = 3:4

T:N or T/N = 3/4th of T/A or 3/4[T/A] or 3/4 X 3/4 = 9/16

For A:C = Twice T:A or 2 X 3/4 = 6/4 or 3/2

So far if we were to equalize the others, then we'd have to assign some common 'x' number of persons to all the categories and then equalize.

Let's say T = 3x.
Then, T/N = 9/16 or 3x / N = 9/16
16 X 3x = 9N
N = 16x / 3

Similarly with A/C:
A = 4x

4x / C = 3/2
C = 8x / 3

So we have 3x : 4x : 16x/3 : 8x/3
And if we equalize all of these by cancelling the denominators and multiplying by 3, then we have the ratio of
T : A : N : C = 9x : 12x : 16x : 8x

Now from the Question, we know that T + C is less than the the smallest three digit integer with distinct numbers, i.e., 102
T + C < 102
9x + 8x < 102
17x < 102

Therefore, x < 6.
The maximum value that it can then assume is 5.

So calculating the maximum possible value for A + N = 12x + 16 = 60 + 80 = 140 --> C

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T/A=3/4 , T/N= 3/4(T/A)
So A/N = 3/4
So T:A:N: = 9: 12: 16
A/C = 2(T/A)= 3/2 = 12/8
So, T:A:N:C = 9:12:16:8

Now, T+C <102, Max(A+N) =?
9x+8x< 102
17x < 102 => x<6
so max would be x=5
12x+16x = 60 +80 = 140.
Answer is C

Bunuel

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may be somewhere like option A ) 112

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Ans C - 140

T/A = 3/4, so T = 3A/4 ... (1)
T/N = 3/4 * 3/4 = 9/16 ... (2)
A/C = 6/4, so C = 4A/6 ... (3)
Smallest 3 digit positive integer = 102
Given, T+C < 102
Now, from (1) and (3) above
3A/4 + 4A/6 < 102
A < 102*12/17
ie A < 72

For A+N, we will use (1) and (2) where,
3A/4 = 9N/16
i.e. N = 4A/3
So, A + 4A/3 = 7A/3

We know A<72,
then 7A/3 will be less than 7*72/3 i.e. 168
Hence, available value in options is 140 (ans C)

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Trainees (T) / Assistants (A) / Nurses (N) / Coordinator (C)
T : A = 3:4
T : N = (3/4) of T : A -----> (3/4)*(3/4) = 9/16
A : C = 2*(T : A) -------> (3/4)*2= 3/2
T+C< 102 (three-digit positive integer with distinct digits )
What is Max A+N=?

T : A : N= 9:12:16
T : A : N : C= 9:12:16:8
K = Ratio multiplier
Total volunteer = 9k+12k+16k+8k=45 K
T+C = 17K
T+C<102 ------> 17K<102 ------> K< 102/17 ------> K<6
Max A+N = 12k+16k -----> (12*5) + (16*5) =140
Ans. C
IMO

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Ans choice C: Given T:A = 3/4 ; T:N = 3/4 of T:A ; A:C = 2 x (T/A) = 6/4 ; T+C<102 (Smallest 3 digit integer with unique digits)

Combining all the Ratios we get T:A:N:C = 9:12:16:8

Since T + C = 17K < 102 (K is a constant)

K<6 ; K = 5,4,3,2,1
In order to get Max (A + N) , we need to consider Max K = 5;

Therefore Max A + N = 5 (12+16) = 140.

Bunuel

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Answer is (C) 140

Here's the information that we are given:
Volunteers can be put into one out of the four available groups trainee (t), assistant (a), nurse (n), co-ordinator (c)
The ratio t/a = 3/4
The ratio t/n is 3/4 times t/a, hence t/n = (3/4)*(3/4) = 9/16
The ratio a/c is twice that of t/a, hence a/c = 2*(3/4) = 6/4
Combined number of t and c is less than the smallest three digit positive integer with unique digits ==> the smallest three digit positive integer is 100. But here, 0 repeats. So we go to 101. Here 1 repeats. Thus, we go to 102 (which is what we need)
Therefore, t+c < 102

Now, let's try and find the common ratio of all 4 numbers.
t/a = 3/4 ==> t = (3/4)a
thus, a/t= 4/3 = 12/9
t/n = 9/16 ===> t = (9/16)n
Thus, a:t:n = 12:9:16
Also, a/c = 6/4 = 12/8
Thus, a:t:n:c = 12:9:16:8
Let x be the common factor of the ratio.
Thus,
a = 12x
t = 9x
n = 16x
c = 8x

Now, we are given that, t+c < 102
Thus, 9x+8x < 102
17x < 102
x<6

Now, for the number of people to be whole numbers, x has to be an integer. For us to find the maximum number of a+n , x has to be the largest possible integer less than 6 i.e. 5
Thus a+n = 12x+16x = 12*5+16*5 = 60+80 = 140

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T:A = 3:4
T:N = 3/4 (3/4)= 9:16
A:C = 2(3/4) = 3:2

So, T:A:N = 9:12:16

If A is 12, then C should be 8

T:A:N:C = 9:12:16:8

Now, it is given that T + C < 102 (smallest 3 digit number with all unique digits)

9x + 8x = 17x < 102 , x<6

x has to be integer, so x <= 5

Max value of A+N = 12x + 16x = 28x = 28*5 = 140 (Max integer value of x can be 5)

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T/ A =3/4 ===> T= 3k
T/N = (3/4)^2 = 9/16====> since t =3k, N = 16k/3
so k must be divisible by 3
k = 3m
therefore t = 9m, n= 16m, a= 12m
a:c = 3/2
a = 12m
12m /c = 3/2
c = 8m
9m +8m <102
m<6 ==> m = 5
ans =140

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Trainee : Assitnt = 3 : 4

Trainee : Nurse = 3/4 of 3/4 = 9 : 16

Assist : Coordinator = twice 3/4 = 3 : 2

So combine:

T : A : N : C = 9 : 12 : 16 : 8

Trainees + coordinators = 9 + 8 = 17 parts.

Smallest 3-digit number with different digits = 102.

Largest multiple of 17 below 102 is 85, hence multiplier = 5.

Assistants + nurses = (12 + 16) × 5
= 28 × 5
= 140

Answer: C

Bunuel

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4 Roles
Trainees => T
Assistants => A
Nurses => N
Coordinators => C

Ratio of trainees to assistants = T : A = 3 : 4 = 9 : 12
Ratio of trainees to nurses = T : N = 3/4 (3/4) = 9 : 16
Ratio of assisstants to coordinators = A : C = 2 * (3/4) = 3 : 2 = 12 : 8

WIth the given ratios and tallying them we can say that
T : A : N : C = 9 : 12 : 16 : 8

Let us consider x is the common multiplier for the above ratios
=> T = 9x
=> A = 12x
=> N = 16x
=> C = 8x

We know that,
T + C < 102
=> 9x + 8x < 102
=> x < 6

For max possible combination of assisstant and nurses, x should be max value
=> x = 5
=> A = 12x = 12*5 = 60
=> N = 16x = 16*5 = 80

=> A + N = 140

C. 140

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T, A, N, C

T/A = 3/4 -> A = 4T/3
T/N = 3/4 * 3/4 = 9/16 -> N = 16T/9
A/C = 2 * 3/4 = 3/2 -> C = 2A/3 = 8T/9

ratio = 1:4/3:16/9:8/9
ratio = 9:12:16:8

smallest three-digit positive integer with distinct digits = 102

T+C < 102
9x+8x = 17x < 102
x < 6
x is at most 5

12x+16x = 28x = 28*5 = 140

IMO C

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trainee, assistant, nurse, coordinator

trainee:assistant = 3:4
trainee:nurse = 9:16
assistant:coordinator = 3:2

assistant = 4/3 * trainee
nurse = 16/9 * trainee
coordinator = 2/3 * assistant = 8/9 * trainee

least common multiple of 3 and 9: 9

9:9*4/3:9*16/9:9*8/9
9:12:16:8

trainee = 9k
assistant = 12k
nurse = 16k
coordinator = 8k

the smallest three-digit positive integer with distinct digits is 102

trainee + coordinator = 9k+8k = 17k
17k < 102 only if k<=5

if k=5, assistant + nurse = 12k+16k = 28k = 140

Answer C

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Given T:A = 3:4

Let T = 3K and A = 4K

Ratio of T:N is 3/4 of T:A

T:N = 9:16

Since T = 3K --> N = 16K/3

So, K must be a multiple of 3. Let K = 3M.

Then: T = 9M, A = 12M and N = 16M

The ratio of A:C is twice T:A:
--> A:C = 3:2

So, C = 8M

Given, T + C < 102
--> 9M + 8M < 102
--> 17M < 102
--> M < 6

So, the largest possible value is M = 5.

Therefore, A + N = 12M + 16M = 28M = 28*5 = 140

Answer: (C) 140