1955 AHSME Problems/Problem 6

A merchant buys a number of oranges at $3$ for $10$ cents and an equal number at $5$ for $20$ cents. To "break even" he must sell all at:

$\textbf{(A)}\ \text{8 for 30 cents}\qquad\textbf{(B)}\ \text{3 for 11 cents}\qquad\textbf{(C)}\ \text{5 for 18 cents}\\ \textbf{(D)}\ \text{11 for 40 cents}\qquad\textbf{(E)}\ \text{13 for 50 cents}$

Solution

Since we are buying at $3$ for $10$ cents and $5$ for $20$ cents, let's assume that together, we are buying 15 oranges. That means that we are getting a total of $30$ oranges for $(10\times5) + (20\times3)$ cents. That comes to a total of $30$ oranges for $110$ cents. $110/30$ = $11/3$ . This leads us to $3$ for $11$ cents which is $\boxed{C}$ and we are done.

-Brudder

1955 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
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

1955 AHSME Problems/Problem 6

Solution

See Also