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07 Aug 2026, 05:33
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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
07 Aug 2026, 05:33
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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:
\(vb = 3v + 14b\)
\(vb - 3v = 14b\)
\(v(b - 3) = 14b\)
\(v = \frac{14b}{b - 3}\)
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:
\(v = \frac{14b}{b - 3} =\)
\(= \frac{14b - 42 + 42}{b - 3} =\)
\(= \frac{14(b - 3) + 42}{b - 3} =\)
\(= 14 + \frac{42}{b - 3}\)
Since \(v\) must be an integer, \(\frac{42}{b - 3}\) 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:
\(vb = 3v + 14b\)
\(vb - 3v = 14b\)
\(v(b - 3) = 14b\)
\(v = \frac{14b}{b - 3}\)
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:
\(v = \frac{14b}{b - 3} =\)
\(= \frac{14b - 42 + 42}{b - 3} =\)
\(= \frac{14(b - 3) + 42}{b - 3} =\)
\(= 14 + \frac{42}{b - 3}\)
Since \(v\) must be an integer, \(\frac{42}{b - 3}\) 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
