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17 Jul 2026, 08:00
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Difficulty:
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Question Stats:
65% (02:17) correct
35%
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wrong
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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
A. 30
B. 34
C. 36
D. 40
E. 60
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M46-50
17 Jul 2026, 09:00
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GMAT Club Official Solution:
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 v be the number of vans and b be the number of buses.
Since the product of the number of vans and the number of buses equals the total number of students transported, we get:
vb = 3v + 14b
vb - 3v = 14b
v(b - 3) = 14b
v = 14b/(b - 3)
Since v must be an integer, we want to see clearly when b - 3 divides the numerator. To do that, rewrite the numerator as a multiple of b - 3 plus a remainder:
v = 14b/(b - 3) =
= (14b - 42 + 42)/(b - 3) =
= [14(b - 3) + 42]/(b - 3) =
= 14 + 42/(b - 3)
Since v must be an integer, 42/(b - 3) must also be an integer. Therefore, b - 3 must divide 42. Also, b must be greater than 3, because if b is 1 or 2, v is negative, and if b is 3, v is undefined.
Since b < 15, we get b - 3 < 12.
The positive divisors of 42 that are less than 12 are:
1, 2, 3, 6, and 7
Now find the possible totals:
If b - 3 = 1, then b = 4 and v = 56, so v + b = 60.
If b - 3 = 2, then b = 5 and v = 35, so v + b = 40.
If b - 3 = 3, then b = 6 and v = 28, so v + b = 34.
If b - 3 = 6, then b = 9 and v = 21, so v + b = 30.
If b - 3 = 7, then b = 10 and v = 20, so v + b = 30.
Thus, the possible total numbers of vehicles are:
30, 34, 40, and 60
Therefore, 36 could NOT be the total number of vehicles used.
Answer: C.
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 v be the number of vans and b be the number of buses.
Since the product of the number of vans and the number of buses equals the total number of students transported, we get:
vb = 3v + 14b
vb - 3v = 14b
v(b - 3) = 14b
v = 14b/(b - 3)
Since v must be an integer, we want to see clearly when b - 3 divides the numerator. To do that, rewrite the numerator as a multiple of b - 3 plus a remainder:
v = 14b/(b - 3) =
= (14b - 42 + 42)/(b - 3) =
= [14(b - 3) + 42]/(b - 3) =
= 14 + 42/(b - 3)
Since v must be an integer, 42/(b - 3) must also be an integer. Therefore, b - 3 must divide 42. Also, b must be greater than 3, because if b is 1 or 2, v is negative, and if b is 3, v is undefined.
Since b < 15, we get b - 3 < 12.
The positive divisors of 42 that are less than 12 are:
1, 2, 3, 6, and 7
Now find the possible totals:
If b - 3 = 1, then b = 4 and v = 56, so v + b = 60.
If b - 3 = 2, then b = 5 and v = 35, so v + b = 40.
If b - 3 = 3, then b = 6 and v = 28, so v + b = 34.
If b - 3 = 6, then b = 9 and v = 21, so v + b = 30.
If b - 3 = 7, then b = 10 and v = 20, so v + b = 30.
Thus, the possible total numbers of vehicles are:
30, 34, 40, and 60
Therefore, 36 could NOT be the total number of vehicles used.
Answer: C.
General Discussion
17 Jul 2026, 08:42
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Given
V - 3 students per vehicle - x vans = 3x students
B - 14 students per vehicle - y buses = 14y students
let Total students be T = 3x + 14y
but also given than xy = T
therefore the equation will be xy = 3x + 14y
given that y<15 (we know that x and y should be positive integers)
so xy - 3x = 14y
x = 14y/(y-3) = 2*7*y / (y-3)
for x to be integer, lets substitute values of y from 4 to 14 (as y<15), and y>4 because anything less than 4 will result in negative integer for x which is not possible
y= 4 => x = 28*2 = 56, total number of vehicles = 60 (E eliminated)
y= 5 => x = 35, V = 40 (D eliminated)
y=6 => x = 28, V = 34 (B eliminated)
y=7 => x=49/2 not an integer so eliminate
y=8 => x= 2*7*8/5 again not an integer
y=9 => x = 21, V = 30 (A eliminated)
so left over C = 36, and the answer
V - 3 students per vehicle - x vans = 3x students
B - 14 students per vehicle - y buses = 14y students
let Total students be T = 3x + 14y
but also given than xy = T
therefore the equation will be xy = 3x + 14y
given that y<15 (we know that x and y should be positive integers)
so xy - 3x = 14y
x = 14y/(y-3) = 2*7*y / (y-3)
for x to be integer, lets substitute values of y from 4 to 14 (as y<15), and y>4 because anything less than 4 will result in negative integer for x which is not possible
y= 4 => x = 28*2 = 56, total number of vehicles = 60 (E eliminated)
y= 5 => x = 35, V = 40 (D eliminated)
y=6 => x = 28, V = 34 (B eliminated)
y=7 => x=49/2 not an integer so eliminate
y=8 => x= 2*7*8/5 again not an integer
y=9 => x = 21, V = 30 (A eliminated)
so left over C = 36, and the answer
