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805+ (Hard)|   Word Problems|         
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GMAT Club World Cup 2026 (Day 10): A school used only vans and buses 17 Jul 2026, 23:36
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A school used only vans and buses to transport students on a field trip. Each van carried exactly 3 students, and each bus carried exactly 14 students. The number of vans used, when multiplied by the number of buses used, was equal to the total number of students transported. If fewer than 15 buses were used, which of the following could NOT be the total number of vehicles used?

A. 30
B. 34
C. 36
D. 40
E. 60

let total vans be x and buses be y
total students carried
3x+14y
y<15 ,
target which cannot be value of x+y
also given
x*y = 3x+14y
x(y-3) = 14y
x = 14y / (y-3)

y has to be at least 4
upon checking with values
y at 4
14*4/ ( 4-3) = 56 vans and 4 bus ; 60

similarly check with values other y = 5,6,7,8,9
y= 5 ; 70/ 2= 35 ; 40

y = 6 ; 14*6 / 3 ; 28+6 ; 34

All values of total is possible except ; 36

OPTION C 36 is correct
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GMAT Club World Cup 2026 (Day 10): A school used only vans and buses 17 Jul 2026, 23:57
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1 van = 3 students
1 bus= 14 students.

say number of van = x. so total students in van = 3x
number of bus = y. so total students in bus= 14y

xy = 3x+14y

y<15

find = x+y

now y cant be less than 4. as soon as it goes below 4, value of x becomes negative or zero. and thats not possible.

so range for y: 3<y<15

xy - 3x = 14y
x(y-3) = 14y
x = 14y/ y-3

y-3 must divide 14y to get value of x.

y= 4 , x=56 . x+y = 60
y=5, x=35, x+y= 40
y=6, x=28, x+y = 34
y=7 , not possible.
y=8, not possible.
y= 9, x= 21, x+y= 30

only choice C is not there.
Choice C
Bunuel
A school used only vans and buses to transport students on a field trip. Each van carried exactly 3 students, and each bus carried exactly 14 students. The number of vans used, when multiplied by the number of buses used, was equal to the total number of students transported. If fewer than 15 buses were used, which of the following could NOT be the total number of vehicles used?

A. 30
B. 34
C. 36
D. 40
E. 60


 


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GMAT Club World Cup 2026 (Day 10): A school used only vans and buses 18 Jul 2026, 00:36
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V = number of vans
B = number of buses

Each van carries 3 students. Each bus carries 14 students.

Total students = 3V + 14B

Given: V × B = total students

So,

-----> VB = 3V + 14B

Rearrange:

-----> VB - 3V - 14B = 0

Add 42 to both sides:

-----> VB - 3V - 14B + 42 = 42

Factor:

-----> (V - 14)(B - 3) = 42

Also, B < 15.

Now list the factor pairs of 42.

42 = 1 × 42
= 2 × 21
= 3 × 14
= 6 × 7

Since B < 15,

B - 3 can be:

1, 2, 3, 6, 7, 14

Now find the corresponding values of V and V + B.

B - 3 = 1

B = 4

V - 14 = 42

V = 56

Total vehicles = 60


B - 3 = 2

B = 5

V - 14 = 21

V = 35

Total vehicles = 40



B - 3 = 3

B = 6

V - 14 = 14

V = 28

Total vehicles = 34


B - 3 = 6

B = 9

V - 14 = 7

V = 21

Total vehicles = 30


B - 3 = 7

B = 10

V - 14 = 6

V = 20

Total vehicles = 30


B - 3 = 14

B = 17 (>15)

Not allowed.

Possible totals are: 30, 34, 40, 60

36 is not possible.

Answer: C
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GMAT Club World Cup 2026 (Day 10): A school used only vans and buses 18 Jul 2026, 01:17
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Given that:
there are certain number of buses and vans carrying students (let them be b and v respectively)
buses carry 14 students each ==> number of students is buses = 14b
vans carry 3 students each ==> no. of of students in vans = 3v
Also, b<15
lastly, v*b = s (total number of students)

Thus,
3v + 14b = vb
3v + 14b - vb = 0
v*(3-b) +14b = 0
v*(3-b) = -14b
v = -14b/(3-b)
v = 14b/(3-b)

Now, we know that v and b have to be positive integers only. Thus, we know that we have to use the concept of Diophantine equations here.

v = (14b - 42 + 42)/(b-3) ___ (we chose 42 as the addition/subtraction number because we want (b-3) in the numerator)
v = 14*(b-3)/(b-3) +42/(b-3)
v =. 14 + 42/(b-3)______(1)

we also know, b<15 ==> (b-3)<12

For v to be an integer, (b-3) = some factor of 42 less than 12 (from the above conditions) = {1, 2, 3, 6, 7}
Let's try the above values in equation 1
Optionb-3bvv+b
E145660
D253540
B362834
A692130
A7102030

From the above table we can safely say (C) 36 is the only option which is not the value of v+b. Hence, that is our answer
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GMAT Club World Cup 2026 (Day 10): A school used only vans and buses 18 Jul 2026, 02:05
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v=No. of vans
b=No. of buses
Each van carries 3 students
each bus carries 14 students
students=3v+14b
vb-3v-14b=0
(v-14)(b-3)=42
we know fewer than 15 buses were used so y<15
Let x=b-3
x must be a positive divisor of 42
so: x <(1,2,3,6,7,14,21,42)
since b=x+3<15 meaning x<(1,2,3,6,7)
b-3. b. v-14=42/b-3. v. v+b
1. 4. 42 56. 60
2 5. 21 35 40
3 6 14 28 34
6. 9. 7 21 30
7 10 6 20. 30

the possible total number of vehicles are: 30-A, 34-B, 40-D, 60-E
ANS: C.36
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GMAT Club World Cup 2026 (Day 10): A school used only vans and buses 18 Jul 2026, 02:17
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3v+14b = number of students transported was

vb=3v+14b

rearrage vb-3v-14b=0

now it needs factoring

vb-3v-14b+42=42

(v-14)(b-3)=42

now here its a bit of consideration, so we are told its less than 15 buses so b-3<12, so make a table;
v-14, b-3, v, b total vehicles
total vehicles should be the options and see which ones are possible and which ones are not

36 is not - final
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GMAT Club World Cup 2026 (Day 10): A school used only vans and buses 18 Jul 2026, 02:21
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Total no. of students= 3V+14B
Also, Total number of students= V*B
B<15
3V+14B=VB
V=(14B)/(B-3)
Value of B must be more than three. Otherwise number of Vans will be negative which is not possible.
B...............V................(B+V)
4..............56...............60 (E)
5...............35..............40 (D)
6..............28...............34 (B)
7..............Fraction.....Not possible
8..............Fraction.....Not Possible
9..............21................30 (A)

We got all options except (C)36.

C. 36
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GMAT Club World Cup 2026 (Day 10): A school used only vans and buses 18 Jul 2026, 02:52
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Equation will become - 3V+14B=V*B
Where v stands for vans and B stands for buses
Upon simplying we get - B = 3v/v-14
And given that b<15 so finally we get v>=18 thus V*B>=34
Thus A is invalid and our answer
Bunuel
A school used only vans and buses to transport students on a field trip. Each van carried exactly 3 students, and each bus carried exactly 14 students. The number of vans used, when multiplied by the number of buses used, was equal to the total number of students transported. If fewer than 15 buses were used, which of the following could NOT be the total number of vehicles used?

A. 30
B. 34
C. 36
D. 40
E. 60


 


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GMAT Club World Cup 2026 (Day 10): A school used only vans and buses 18 Jul 2026, 03:08
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3V+14B=VB
VB-3V=14B
V(B-3)=14B
V=14B/B-3
Given that B<15 and that B is not equal to 3 we can test cases starting from B=4 all the way to 14
For the first one we get
(14x4)/(4-3)= 56 (56+4=60)
(14x5)/5-3= 35 (35+5)=40
(14x6)6-3=28 (28+6)=34
.
.(14x9)/9-3= 12 (21+9)=30
.
.
.(13x14)/13-3= Not integer
(14x14)/14-3= Not integer
From the above only 36 is missing
Ans C
Bunuel
A school used only vans and buses to transport students on a field trip. Each van carried exactly 3 students, and each bus carried exactly 14 students. The number of vans used, when multiplied by the number of buses used, was equal to the total number of students transported. If fewer than 15 buses were used, which of the following could NOT be the total number of vehicles used?

A. 30
B. 34
C. 36
D. 40
E. 60


 


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GMAT Club World Cup 2026 (Day 10): A school used only vans and buses 18 Jul 2026, 03:23
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It is given in the question steam that each van carries 3 students exactly and each bus carries 14 students exactly, hence the total number of students = 3v + 14b
also we know that b<15
Given,
v*b = 3v + 14b
v = 14b/(b-3)
Now we need to see that b-3 will be an integer for the term v to be an integer and also a factor of 14

if b-3 = 1, then b=4 and hence v= 56 --> v+b = 56+4 = 60 (option E and hence it is possible)
if b-3 = 2, then b=5 and hence v= 35 --> v+b = 35+5 = 40 (option D and hence it is possible)
if b-3 = 3, then b=6 and hence v= 28 --> v+b = 28+6 = 34 (option B and hence it is possible)
if b-3 = 4, then b=7 and hence v = 24.5 not an integer ignore
if b-3 = 5, then b=8 and hence v is again not an integer, ignore
if b-3 = 6, then b=9 and again v is not an integer, ignore
if b-3 = 7, then b=10 and hence v = 20 --> v+b = 30 (option A and hence it is possible)

From this we can say we will never see the value of v+b to be equal to 36 i.e option C
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GMAT Club World Cup 2026 (Day 10): A school used only vans and buses 18 Jul 2026, 03:39
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From the statement, we can get this equation
vb = 3v + 14b
vb - 3v = 14b
v (b-3) = 14b
v = 14b / (b-3)
v = 14 + 42 / (b-3)

Since v must be integer -> 42/(b-3) = should be integer as well

Stated b<15 -> then b-3 < 12
- then factor of 42 that match with b-3 < 12 should be, b-3 = [ 1, 2, 3, 6, 7 ]
- then b = [4, 5, 6, 9, 10]
- then v
1. v = 14+42=56; b+v = 4+56=60
2. v = 14+21=35; b+v = 5+35=40
3. v = 14+14=28; b+v = 6+28=34
4. v = 14+7=21; b+v = 9+21=30
5. v = 14+6=20; b+v = 10+20=30

So, the answer which is NOT the total number b+v is C.36

Bunuel
A school used only vans and buses to transport students on a field trip. Each van carried exactly 3 students, and each bus carried exactly 14 students. The number of vans used, when multiplied by the number of buses used, was equal to the total number of students transported. If fewer than 15 buses were used, which of the following could NOT be the total number of vehicles used?

A. 30
B. 34
C. 36
D. 40
E. 60


 


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GMAT Club World Cup 2026 (Day 10): A school used only vans and buses 18 Jul 2026, 03:53
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Bunuel
A school used only vans and buses to transport students on a field trip. Each van carried exactly 3 students, and each bus carried exactly 14 students. The number of vans used, when multiplied by the number of buses used, was equal to the total number of students transported. If fewer than 15 buses were used, which of the following could NOT be the total number of vehicles used?

A. 30
B. 34
C. 36
D. 40
E. 60


 


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3v + 14b = vb

v = 14b/b-3

= (14b + 42 - 42) / (b-3)


= 14 + 42/(b-3)

Hence, b - 3 should be a divisor or 42

42 = 1* 2 * 3 * 7

b-3 = 1

b = 4

v = 56

t = 60

b - 3 = 2

b = 5

v = 14 + 21 = 35

t = 30

b - 3 = 3

b = 6

v = 14 + 7 = 28

t = 34

b - 3 = 6

b = 9

v = 14 + 7 = 21

t = 30

b - 3 = 7

b = 10

v = 14 + 6 = 20

t = 30

So, 36 is not possible.

Option C
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GMAT Club World Cup 2026 (Day 10): A school used only vans and buses 18 Jul 2026, 04:00
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Total number of students = 3v+14b
Product of vans and buses = vb

So 3v+14b = vb
Solving, we get v = 14b/(b-3)
Rewriting, v= 14 + 42/(b-3) --this is eqn 1

For v to be integer, b-3 should be a positive factor of 42.
Since b< 15, then b-3<12. So possible values of b-3 are 1, 2, 3, 6, 7. Which gives possible b values as 4, 5, 6, 9, 10
For these values, we use eqn 1 to get respective values of b. Which are: 56, 35, 28, 21, 20 respectively. Summing, we get, v+b as 60, 40, 34, 30, 30

So 36 is NOT a possible value
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GMAT Club World Cup 2026 (Day 10): A school used only vans and buses 18 Jul 2026, 04:25
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Lets say v is number of vans and b in number of bus

Given => 3v + 14b = vb and b < 15

so we have vb - 3v - 14b = 0
we need possible values of v & b
One thing if we add 42 on both sides of equation we can simplify it

vb - 3v - 14b + 42 = 42
v(b-3) - 14 (b-3) = 42
(b-3)(v-14) = 42

So we need possible factor of 42 remembering b < 15.

factors (b-3) (v-14) = (42,1) (21,2) (14,3) (7,6) (6,7) So if calculate b and v for all possible factors we see that 36 is the only not possible value.

Ans - C
Bunuel
A school used only vans and buses to transport students on a field trip. Each van carried exactly 3 students, and each bus carried exactly 14 students. The number of vans used, when multiplied by the number of buses used, was equal to the total number of students transported. If fewer than 15 buses were used, which of the following could NOT be the total number of vehicles used?

A. 30
B. 34
C. 36
D. 40
E. 60


 


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GMAT Club World Cup 2026 (Day 10): A school used only vans and buses 18 Jul 2026, 04:28
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let #vans - v
#buses - b

vb=#students =3v+14b
v=14b/(b-3)

v+b==> b(b+11)/(b-3)
since v+b>0 && b<15 ==> 4<= b <=14
b=4 ==> v+b=60
b=5 ++. 40
b=6 ==> 34
b=9 ==> 30
hence ans 36
Bunuel
A school used only vans and buses to transport students on a field trip. Each van carried exactly 3 students, and each bus carried exactly 14 students. The number of vans used, when multiplied by the number of buses used, was equal to the total number of students transported. If fewer than 15 buses were used, which of the following could NOT be the total number of vehicles used?

A. 30
B. 34
C. 36
D. 40
E. 60


 


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GMAT Club World Cup 2026 (Day 10): A school used only vans and buses 18 Jul 2026, 05:18
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School -> Vans and buses to transport students

Let x be the number of vans and y be the number of buses
Also,

y < 15

Van carries 3 students
Bus carries 14 students

Total number of students = 3*x + 14*y = xy
=> 3x+14y = xy
=> x(3 - y) = -14y
=> x = 14y/(y - 3)

Now we also know that y < 15 and here y >= 4 since x cant be negative
=> 4<= y < 15

By substituting different values of y and finding possible values of x + y
We come to the conclusion that,
36 vehicles not possible

=> C. 36
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GMAT Club World Cup 2026 (Day 10): A school used only vans and buses 18 Jul 2026, 05:24
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IMO: C
vb = 3v + 14 b
vb - 3v -14b = 0
vb - 3v -14b +42 = 42
(v-14)(b-3)=42
b<15 ---. b-3 = positive divisor of 42
so the only possible totals are 30, 34, 40, 60
36 does not occur
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LibrarianTurtle
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GMAT Club World Cup 2026 (Day 10): A school used only vans and buses 18 Jul 2026, 05:58
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It is a tough one.

\(3V+14B=VB\)
\(B<15\)
\(14B=VB-3V ---->14B=V(B-3) ----\frac{(14B)}{(B-3) }= V\)
Here is the tough part. We can rewrite this equation otherwise we have to do number crunching which takes time.
\(\frac{14B}{(B-3)}------->\frac{14(B-3)+42}{(B-3)} -----> 14 + \frac{42}{(B-3)}\)
B-3 should divide 42 and B<15
B=4 V=56 Total=60
B=5 V=35 Total=40
B=6 V=28 Total=34
B=9 V=21 Total=30
Total can't be 36
If we try to do other wise we get V>17.5 and B<15 which takes quite lots of number crunching
IMO
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GMAT Club World Cup 2026 (Day 10): A school used only vans and buses 18 Jul 2026, 06:03
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let vans = v, buses = b

given:

3v + 14b = vb

We can say

vb − 3v − 14b = 0

(v − 14)(b − 3) = 42

also b < 15



v + b = total vehicles

factor pairs of 42:

1×42 (15,45) total = 60

2×21 (16,24) total = 40

3×14 (17,17) total = 34

6×7 (20,10) total = 30

7×6(21,9) total = 30

14×3 (28,6) = 34

21×2 (35,5) total = 40

42×1 (56,4) totl = 60

36 is not there

ansis c
Bunuel
A school used only vans and buses to transport students on a field trip. Each van carried exactly 3 students, and each bus carried exactly 14 students. The number of vans used, when multiplied by the number of buses used, was equal to the total number of students transported. If fewer than 15 buses were used, which of the following could NOT be the total number of vehicles used?

A. 30
B. 34
C. 36
D. 40
E. 60


 


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GMAT Club World Cup 2026 (Day 10): A school used only vans and buses 18 Jul 2026, 06:10
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Let the number of vans = v
The number of buses = b
v*b = 3v + 14b
v*(b-3) = 14b
v = 14b/(b-3)
Lets test b = 1,2,3.....9
b = 1, 14/-2, v = -7 (not an answer)
b = 2, 28/(2-3), v = -28 (not an answer)
b = 3, 42/0, v = 0 (not answer)
b = 4, 56/1, v = 56, total vehicles = 4 + 56 (answer choice E). Eliminate.
b = 5, 70/2, v = 35, total vehicles = 5+35 (answer choice D). Eliminate.
b = 6, 84/3, v = 28, total vehicles = 6+28 (answer choice B). Eliminate
b = 7, 98/4, v = not an integer (out-of-scope)
b= 9, 126/6, v = 21, total vehicles = 9+21 (answer choice A). Eliminate
The more we do for "b", the number of vehicles will decline.

Therefore, only 36 is not possible. Answer C.
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