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3v = 14b = bv

b < 15

Test each option, only 36 fails
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Answer: C) 36

v = number of vans, b = number of buses

v*b = 3v + 14b and b < 15
After rearranging we get: v = 14b/(b-3)
We don't want variables in both the numerator and the denominator, so we can try to rewrite 14b in a way that would cancel out the denominator, then split the terms.
x(b-3) = 14b
If x is 14, we get 14b - 42, which is 42 short of 14b, so: 14b = 14(b-3) + 42
v = 14(b-3)+42 divided by (b-3)
v = 14 + 42/(b-3)
So 42 must be divisible by b-3
If b must be less than 15, then b-3 must be less than 12.
The possible divisors of 42 are 1,2,3,6,7,14 -> we can stop here as we have found the first one above 12
The possible values of b are therefore 4, 5, 6, 9, and 10. We plug these into the equation v = 14 + 42/(b-3)
The possible values of v are therefore 30, 34, 40, 60
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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?
vb=3v+14b
vb-3v=14b
v=14b/(b-3)
Fewer than 15 buses were used. As from equation we can see b must be more than 3.
When b=4, v=56, Total=60
When b=5, v=35, Total=40
When b=6, v=28, Total=34
When b=9, v=21, Total=30

We didn't get only option C) 36.
Not possible values of vehicles used= 36

C
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Say vans V and buses b will form the equation as per the given question that:
vb = 3v + 14b
or we can say vb - 3v = 14b
v(b-3) = 14b
v= 14b/(b-3) = 14 + 42/(b-3)
adding b to both sides:
v + b = 14 + b + 42/(b-3)
b cannot be 3 and b -3 must be positive divisor of 42 which is less than 12 because b<15 so b can be either 1,2,6 and 7
So in the equation v = 14 + 42/(b-3) substituting values only 36 doesn't comes to be a value of v
so ans 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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Given that School uses only Van and buses to transport students on a field trip.

one Van carried exactly 3
one bus carried exactly 14

no of Vans* no of buses= total students transported.

Also, fewer than 15 buses were used
we have to determine which can't be number of total vehicles used among all possible options:
Let b buses and v vans were used.

each bus will carry 14, so b bus will carry 14b
each van will carry 3, so v van will carry 3v

So, total students transported will be 14b+3v

Given that v*b= 14b+3v
b=3v/(v-14) and b is less than 15 given already
3v/(v-14)<15
by re-arranging v= 14b/(b-3), given b<15 ( integer value only possible from 1 to 14 )

check by putting all possible values of b from 1 to 14
For b=1, negative value
b=2, negative,
b=3, not defined
b=4, v=56, possible and total no of vehicles= 60
b=5, v=35 , possible and total vehicle = 40
b=6, v=28, possible and total vehicle = 34
b=7, v= not integer, not possible
b=8, v= not integer, not possible
b=9, v=21, possible, total vehicle = 30
b=10, v=20, possible, total vehicle= 30
b=11, v= not integer, not possible
b=12, v= not integer, not possible
b=13, v= not integer, not possible
b=14, v= not integer, not possible
Possible values of total vehicles are 60, 40, 34, 30

36 is not possible among options

Answer-C
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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?

Let the number of vans and buses be v & b respectively.

vb = 3v + 14b

vb - 3v = 14b
v(b-3) = 14b
\(v = \frac{14b}{b-3} = \frac{14(b-3) + 42}{b-3} = 14 + 42/b-3\)

b < 15

b = 4; v = 14 + 42 = 56; b+v = 4 + 56 = 60
b = 5; v = 14 + 21 = 35; b+v = 5 + 35 = 40
b = 6; v = 14 + 14 = 28; b+v = 6 + 28 = 34
b = 9; v = 14 + 7 = 21; b+v = 9 + 21 = 30

Total vehicles used = {30, 34, 40, 60}

36 can not be the number of vehicles used.

IMO C
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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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Correct Answer C

Let v be vans and b be buses

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

We know from the equation that b>3 and from the question that b<15, which means 4<= b <= 14. v has to be an integer since it’s number of vans.

Using,
b=4, we get v=56; total vehicles is 60 (E)
b=5, we get v=35; total vehicles is 40 (D)
b=6, we get v=28; total vehicles is 34 (B)
b=7, we get v=21; total vehicles is 28
b=10, we get v=20; total vehicles is 30 (A)

The only option is C that does not fall within the range.
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Here's my solution for this question
Attachments

Screenshot_2026-07-17-22-18-42-74_40deb401b9ffe8e1df2f1cc5ba480b12.jpg
Screenshot_2026-07-17-22-18-42-74_40deb401b9ffe8e1df2f1cc5ba480b12.jpg [ 792.01 KiB | Viewed 948 times ]

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Be v no of vans & b be no of buses
van carries 3 students and bus carries 14 students
vb = 3v+14b
rearrange: vb-3v -14b = 0
14*3 = 42
add 42 to factorize
(v-14) (b-3) = 42
since fewer than 15 buses were used b<15 and b-3 can only be divisor of 42
that are at most 11:
1,2,3,6,7
this gives:
buses vans total

4 56 60
5 35 40
6 28 34
9 21 30
10 20 30

so possible are: 30,34,40 & 60
Answer: C. 36
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v b
3 14
b<15
vb = 3v + 14b
v = 14b/b-3 = 14 + 42/b-3(I am trying to isolate b for better comparison)
now, b and v are natural numbers, so 42/b-3 should be an integer
b,v,t(b+v)
4,60,64
5,35,40
6,28,34
9,21,30
10,20,30
17,17,34
24,16,40
45,15,60
so the only it can't take is 36.
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vb=3b+14b
simon's favorite factoring trick
vb-3b-14b+42--(v-14)(b-3)=42
fewer than 15 buses means b<15--(b-3)<12

v-14+b-13=f1+f2
v+b-17=f1+f2
total vehicles=17+(factors pairs of 420)


Pair (1,42)= total=17+1++42=60
pair(2,21)=17+2+21=40
pair (3,41) total=17+3+14=34
pair (6,7) total=17+6+7=30
36 is impossible
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vans =v, buses = b
total buses <15
v * b = 3v+14b
vb- 3v = 14b
v(b-3)= 14b
v= 14b/b-3
for b=4 v=56 b t =60
b=5 v=35 t = 40
b=6 v = 28 t=34
b=9 , v v=21 t = 30

except 36 all totals valid option C
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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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Hopefully there is a faster way to solve this
Vans = V, Buses = B
Given condition- B< 15
from the Q, We get the equation
V x B = 3V + 14B
Additionally, at LHS, the result must be greater than 3V to get a positive value of B
Hence B >= 4
Thus we can say 4 <=B <=14
Now taking different values of B,

at B = 5,
5v = 3v + 70
v = 35 and B + V = 40. Exclude D

at B = 6
3V = 14*6
v = 28 and B + V = 34. Exclude B

At B = 5
v = 14*4 = 56
B + V = 60. Exclude E

B = 11-14 and 7-8 gives non integer values for V

at B = 10,
7V = 140, V= 20
B + V =3-. Exclude A

Only value not found is 36
Answer (C) = 36
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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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vb = 3v + 14b
v = 14b/b-3
14b should be divided by b-3.
at b =4, we get total =60
similarly the option we don't get is C)36
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IMO C

Let no of van be x, no of bus =y
Then as per Q, 3x+14y= xy
or, x= 14y/y-3
Hence y>3 and Y<15
values of y = 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4
For only few values of y we get x as a whole no (since x cant be fraction)
yxx+y
102030
92130
62834
53540
45660
Hence only no which cant be x+y = C. 36
Hope this helps
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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Students = S
Van = V
Bus = B

As per question stem,
3V + 14B = S = VB
V=14*B/(B-3) and B<15

Number of Vans V is integer,
possible values of B = 4,5,6,9,10

For
B= 4; V= 56 : V+B = 60 Choice E eliminated
B= 5 then V+B = 40 Choice D eliminated
B=6 then V+B = 34 Choice B eliminated
B=9 then V+B = 37
B=10 then V+B = 30 Choice A eliminated
Correct answer is C = 36. 36 could not be the total number of vehicles used

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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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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van -> 3, bus -> 14
let #van be x and #bus be y (# = no. of)
xy = 3x + 14y
y<15

on solving x in terms of y
x = 14y/y-3
y>3
y<15
so 15>y>3
for y=4, x = 56, total vehicles T = 56+4 = 60
similarly for others
y=5, x=35, T=40
y=6, x=28, T=34
y=9, x=21, T=30
y=10, x=20, T=30
(note that we have to check for all the integers from 15 to 3, these were the cases which gave me valid solution)
only option C, which says 36 is not possible.
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