1961 AHSME Problems/Problem 24

Problem 24

Thirty-one books are arranged from left to right in order of increasing prices. The price of each book differs by $\textdollar{2}$ from that of each adjacent book. For the price of the book at the extreme right a customer can buy the middle book and the adjacent one. Then:

$\textbf{(A)}\ \text{The adjacent book referred to is at the left of the middle book}\qquad \\ \textbf{(B)}\ \text{The middle book sells for \textdollar 36} \qquad \\ \textbf{(C)}\ \text{The cheapest book sells for \textdollar4} \qquad \\ \textbf{(D)}\ \text{The most expensive book sells for \textdollar64 } \qquad \\ \textbf{(E)}\ \text{None of these is correct }$

Solution

Let the leftmost book cost $c$ dollars. The 16th book (the middle book) costs $c+30$ dollars, and the 31st book costs $c+60$ dollars.

If the book to the left of the middle book is used, the equation to solve for the price of the leftmost book is $c+30+c+29=c+60$ . Solving yields $c=1$ , so the price of the middle book is $\textdollar{31}$ and the price of the rightmost book is $\textdollar{61}$ .

If the book to the right of the middle book is used, the equation to solve for the price of the leftmost book is $c+30+c+31=c+60$ . Solving yields $c=-1$ , which can not happen.

Thus, the answer is $\boxed{\textbf{(A)}}$ .

1961 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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

1961 AHSME Problems/Problem 24

Problem 24

Solution

See Also