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Difference between revisions of "1959 AHSME Problems/Problem 17"

(Created page with "== Problem 17== If <math>y=a+\frac{b}{x}</math>, where <math>a</math> and <math>b</math> are constants, and if <math>y=1</math> when <math>x=-1</math>, and <math>y=5 </math> w...")
 
Line 10: Line 10:
 
<cmath> \begin {cases} 1 = a - 5 \\ 5 = a - \frac{5}{5} \end {cases}</cmath>
 
<cmath> \begin {cases} 1 = a - 5 \\ 5 = a - \frac{5}{5} \end {cases}</cmath>
 
It is easy to see that <math>a = 6</math>, and that <math>a+b</math> is (E) 11.
 
It is easy to see that <math>a = 6</math>, and that <math>a+b</math> is (E) 11.
 +
== See also ==
 +
{{AHSME 50p box|year=1959|num-b=16|num-a=18}}
 +
{{MAA Notice}}

Revision as of 12:49, 20 July 2020

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 (E) 11.

See also

1959 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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All AHSME Problems and Solutions

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