1965 AHSME Problems/Problem 29

Problem

Of $28$ students taking at least one subject the number taking Mathematics and English only equals the number taking Mathematics only. No student takes English only or History only, and six students take Mathematics and History, but not English. The number taking English and History only is five times the number taking all three subjects. If the number taking all three subjects is even and non-zero, the number taking English and Mathematics only is:

$\textbf{(A)}\ 5 \qquad \textbf{(B) }\ 6 \qquad \textbf{(C) }\ 7 \qquad \textbf{(D) }\ 8 \qquad \textbf{(E) }\ 9$

Solution

The number of students taking all three subjects cannot be greater than or equal to $4$ , because then there would be $20$ or more students taking English and History only, and we know that $6$ students are taking Mathematics and History only. These three values sum to a number greater than $28$ , the total number of students. Thus, because we know that the number of students taking all three subjects is a non-zero even number, there must be $2$ of them. Thus, $10$ people must be taking only English and History. With the knowledge of $6$ students taking only Mathematics and History, we know the classes of $18$ students so far, so we have $10$ students left who are either taking Mathematics and English only or Mathematics only. Because these two categories have an equal number of students, the number of students taking English and Mathematics only is $\boxed{\textbf{(A) }5}$ .

1965 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 28	Followed by Problem 30
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1965 AHSME Problems/Problem 29

Problem

Solution

See Also