1959 AHSME Problems/Problem 17

Problem 17

If $y=a+\frac{b}{x}$, where $a$ and $b$ are constants, and if $y=1$ when $x=-1$, and $y=5$ when $x=-5$, then $a+b$ equals: $\textbf{(A)}\ -1 \qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$

Solution

Plugging in the x and y values, we obtain the following system of equations: \[\begin {cases} 1 = a - b \\ 5 = a - \frac{b}{5} \end {cases}\] We can then subtract the equations to obtain the equation $4 = 0.8b$, which works out to $b = 5$.

Plugging that in to the original system of equations: \[\begin {cases} 1 = a - 5 \\ 5 = a - \frac{5}{5} \end {cases}\] It is easy to see that $a = 6$, and that $a+b$ is $\boxed{\textbf{(E) } 11}$.

See also

1959 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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. AMC logo.png