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

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Shlok02

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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 = 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 = 2/3 (4T/3)

C = 8T/9
T: C = 9:8

Using a common multipler
T = 9x, C = 8x,
A = 12x, N = 16x

T + C < min 3 digit number with distinct digits = 102
T + C < 102
17x < 102
x < 6

A + N = 28x
(A + N ) max = 28 * 5 = 140

Ans. 140 (c)

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It is given that the ratio of :
Trainees to Assistant = 3/4
Trainees to Nurses = 3/4 * Ratio of Trainees to Assistant = 3/4*3/4 = 9/16
Assistant to Coordinators = 2* Ratio of Trainees to Assistant = 2*3/4 = 3/2

Ratio of Trainees to Coordinators =
T/A = 3/4 (Let T = 3y, hence A = 4y)
T/N = 9/16 (N = 3y*16/9 = 16y/3)
A/C = 3/2 (From above, A= 4y, C = 8y/3
T/C = ?
Here by substitution - I get the ratio of T/C = 3y*3/8y = 9/8

The total number of Trainees + Coordinators < 102 (smallest 3 digit number with distinct digit)
Hence 9y+ 8y <102
17y<102
y< 6

Since as per question, y is a positive integer, the maximum value of y is 5.

Maximum possible value of Assistants and Nurse = 16y/3+4y = 16y+12y = 28y = 28*5 = 140

Hence answer is Option C - 140

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Bunuel

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Trainees = 3x
Assistants = 4x

The ratio of trainees to nurses is 3/4 of 3:4,

so: T : N = 9 : 16

Since T = 3x, nurses = 16x/3, so x must be a multiple of 3.

Let x = 3k.

Then:
Trainees = 9k
Assistants = 12k
Nurses = 16k

Also, assistants : coordinators = 2 * (3:4) = 3:2

So if assistants = 12k, coordinators = 8k.

Now: T + C = 9k + 8k = 17k

The smallest 3-digit number with distinct digits is 102.

So: 17k < 102 k < 6

Maximum k = 5.

Then: A + N = 12k + 16k = 28k

= 28 * 5 = 140

Option C

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T:A:N:C
3:4
then include other info, so T/N = 3/4 x 3/4 = 9/16
A/C = 3/4 x 2 = 6/4 = 3/2

T+C<102

You have some decimals so good to find a ratio where all are integers
T:A:N:C
9:12:16:8

9+8 = 17
102/17 = 6(but we must use 5 because its less than 102)
So the common factor is 5.
12x5 = 60
16x5 = 80
total = 140(final answer)

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Let us denote trainee as T, assistant as A, nurse as N, and coordinator as C.
T:A = 3:4
T:N=(3/4)*(T:A)=(3/4)*(3:4)=9:16
A:C=2*(T:A)=2*(3:4)=6:4

Rewriting T:A as 9:12 and A:C as 12:8, we get
T:A:N:C = 9:12:16:8

Let us consider T as 9x, A as 12x, N as 16x, C as 8x.
Smallest three-digit positive integer with distinct digits=102
T+C<102
9x+8x<102
17x<102
x<6
Therefore, maximum x=5 as x has to be an integer (9x, 12x, 16x, 8x have to be integers).

Thus, maximum possible combined number of assistants and nurses = 12x+16x = 28x = 28*5 = 140

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working out the ratios we get :-
trainees : Assistants: nurses: coordinators = 9:12:16:8

also 9x + 8x < 102 (smallest 3 digit number with distinct digits)
x< 6
x is a positive integer so lets take x =5 to get the maximum value of assistants + nurses = 12*5 + 16*5=140

Bunuel

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I got answer C, 140.
For three (or greater) part ratio problems, we need to find a common multiple to connect the separate ratio parts.

First translate the given information:
t/a = 3/4
t/n = 9/16
a/c = 3/2

t : a : n : c = 9 : 12 : 16 : 8

We are next told that the combined number of trainees and coordinators is less than 102 (smallest three-digit positive integer with distinct digits). This creates the equation 17x (where x is the multiplier of the above ratio) < 102.
This yields x is less than or equal to 5 since we are dealing with people, there is an integer constraint.

Now, the question is asking for the maximum possible combined number of assistants and nurses, which is 28x.
Take x = 7. 28 * 5 = 140, answer choice C.

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Here is the breakdown
T:A = 3:4
T:N = (3/4) x (3/4) = 9:16
A:C = 2 of T/A = 2 of 3/4 = 3/2 = 3:2 = 12:8
Therefore, T:A:N:C = 9:12:16:8
Now, T+C < The smallest three-digit positive integer with distinct numbers
100 x
101 x
102 (right)
T + C < 102
Here, T + C = 9 + 8 = 17
The sum will be the multiples of 17
17 x 5 = 85
17 x 6 = 102
Since 102 is equal, T+C = 85 < 102
Therefore, T = (85/17)*9 = 5*9 = 45, C = 8*5 = 40
Hence, A + N = 12*5 + 16*5 = 60 + 80 = 140

Answer is C

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Lets say we have t, a, n, c volunteers

Ratios given
t/a = 3/4
t/n = (3/4)*(t/a) = 9/16
a/c = 2*(t/a) = 3/2
combined this gives => t:a:n:c = 9:12:16:8

given combined t+c < 102 what is max a+n = ?

(9+8)x < 102 => x < 6

so x = 5 => (12+16)*5 = 28*5 = 140

Ans - C

Bunuel

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As given in the question
The ratio of T: A=3:4
T/N=(3/4)*(T/A)= 9/16
A/C= 2(T/A)=3/2
T:A:N:C= 9:12:16:8
T+C<102=> 17x<102=> x<6
Xmax=5
Sum of assistants and nurses maximum= 28*x=28*5=140

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T=3/4A, T=3/4*3/4N=9/16N, A=3/2C, C=2/3*4/3*9/16N=1/2N
T+C<102, 3/4 A+2/3 A<102 so A<72,
T+C<102, 9/16 N+1/2 N= 17/16 N<102 so N<96
Combined above two A+N < 168. So max possible is 140.

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

We have been given - T + C is < 102 (smallest 3 digit possible number with distinct digit)
We need to find the max possible for A + C

Now T:A = 3:4 and A:C = 3:2 Therefore, T:A:C = 9:12:8
We get T:C = 9:8
let the variable be x so 9x + 8x = 17x. The max 2 digit multiple of 17 is 85 (It can't be a three digit because the next multiple of 17 is 102. The required number is lesser than 102)
So now we have T = 45 and C = 40
Basis the same we can find A = 60.

Now, A + N = 60 + N
T:N = 9:16 so if T is 45 then N is 80

So A + N = 60 + 80 = 140.
Answer is C - 140.

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From the question, we've 4 pieces of information.
1. t/a = 3/4 -> t = 3a/4
2. t/n = 3/4 * t/a -> a/n = 3/4 -> a = 3n/4
3. a/c = 2 * t/a -> c = a^2/2t
4. t + c < 102

Finding the maximum a+n.

3a/4 + a^2/2t < 102
3a/4 + a^2/(2*3a/4) < 102
3a/4 + 4a/6 < 102
9a+8a < 102*12
So, a < 102*12/17

Next to n, 3n/4 < 102*12/17
n < 4*102*3/17
n < 102*12/17

So, max a+n < 2*102*12/17, which is an integer.

2448/17 ~ 140 -> Choice C.

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Good question, I solved by ratios simultaneously
Started writing ratios beneath and bringing them on same scale

T : A : N : C
3k:4k:C
9k: A :16k
T :3k:2k

So we bring them to same scale
Which becomes - 9k:12k:16k:8k

T+C<102 so 9k+8k <102
Thus k<6
So max integer is 5
Thus - A+N = 12k + 16k = 28k and 28*5 = 140
Option C

Bunuel

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C, 140

'smallest three-digit integer with distinct digits' is 102, not 100, not 101, so T + C < 102

T/A = 3/4. T/N is three quarters of that, so 9/16. A/C is double it so 3/2.

everything back to T: A = 4T/3, N = 16T/9, C = 8T/9. The ninths force T = 9m, so T = 9m, A = 12m, N = 16m, C = 8m

T + C = 17m < 102 puts m at 5 max (17 times 6 is exactly 102, and we need strictly less). A + N = 28m = 140

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

Satisfying all say,
T=9x, A=12x, N=16x, C=8x

& 9x+8x<102
.:17x<102
.: x<6
.: x={1,2,3,4,5}
.: A+N=12x+16x=28x
.: Maximum possible no of A+N=28*5=140

.: C is the correct answer

Bunuel

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Given,
Trainees(T) : Assistants (A) = 3:4
T =3x
A = 4x

Trainees (T) : Nurses (N) = 3T/4A
T/N = (3*3)/(4*4)
T/N = 9/16
So, T : N = 9k : 16k
Since T = 3x, taking k = x/3,
N = 16x/3

So, x must be teh multiple of 3.
Taking x = 3p

T = 9p, A = 12 p, N = 16p

Assitants (A) : Coordinator (C) = 2(T:A) = 3/2

C = 8 p

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

Given,
T+C < 102
9p+8p = 17p <102
p<6

MAximum integer = 5

So, A + N = 12p + 16p = 28 p = 28 * 5 = 140

Answer : C (140)

Bunuel

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T = trainees
A = assistants
N = nurses
C = coordinators

Given T : A = 3 : 4

The ratio of T to N is 3/4 of the ratio of T to A.

T : N = (3/4) × (3/4) = 9 : 16

Also, A : C = 2 × (3 : 4) = 6 : 4 = 3 : 2

combining all the ratios.

From:

T : A = 3 : 4

and

A : C = 3 : 2

Make A the same.

LCM of 4 and 3 is 12.

Multiply:

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

T : A : C = 9 : 12 : 8

Now match T with T : N = 9 : 16.

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

Let the common multiplier be k.

Then,
T = 9k
A = 12k
N = 16k
C = 8k

Given: T + C < 100

----> 9k + 8k < 100

----> 17k < 100

The largest integer value of k is: k = 5

Now find the maximum value of A + N.

A + N = 12k + 16k = 28k = 28 × 5 = 140

Answer: C

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Trainee: 3x
assistant : 4x
nurse:n
coordinator=c

(3x/n)=3/4 * (3x/4x) ==> n= 16x/3

4x/c= 2* (3x/4x) ==> c=8x/3

3x +8x/3 <123 ==> x<=21

max 4x+16x/3 ==> 196

Bunuel

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