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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...")
 
m (formatted answer)
 
(One intermediate revision by one other user not shown)
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Plugging that in to the original system of equations:
 
Plugging that in to the original system of equations:
 
<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 <math>\boxed{\textbf{(E) } 11}</math>.
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== See also ==
 +
{{AHSME 50p box|year=1959|num-b=16|num-a=18}}
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{{MAA Notice}}

Latest revision as of 11:29, 21 July 2024

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 (Problems • Answer Key • Resources)
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

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