Difference between revisions of "1997 AHSME Problems/Problem 22"

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<math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10 </math>  
 
<math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10 </math>  
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==Solution==
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Working backwards, if <math>6 \le E \le 10</math>, then <math>6 \pm 11 \le A \le 10 \pm 11</math>.  Since <math>A</math> is a positive integer, <math>17 \le A \le 21</math>.
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Since <math>17 \le A \le 21</math>, we know that <math>17 \pm 19 \le B \le 21 \pm 19</math>.  But if <math>B=36</math>, which is the smallest possible "plus" value, then <math>E + A + B = 6 + 17 + 36 = 59</math>, which is too much money.
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Hence, <math>17 - 19 \le B \le 21 - 19</math>.  But since <math>B</math> must be a positive integer, that leaves only two possibilities:  <math>B = 1</math> or <math>B=2</math>, which correspond with <math>E = 9</math> and <math>E = 10</math>.
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Concentrating only on <math>E=9</math>, we have <math>E=9</math> leading to <math>A = 9 + 11 = 20</math>, which leads to <math>B = 20 - 19 = 1</math>, which leads to <math>C = 1 + 7 = 8</math>.  Thus far we have given out <math>9 + 20 + 1 + 8 = 38</math> dollars.  This means that Dick must have <math>56 - 38 = 18</math> dollars.  However, the difference between Carlos and Dick is not <math>5</math> dollars.
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Thus, the right answer must be <math>\boxed{E}</math>.  Verifying, if <math>E = 10</math>, then <math>A = 10 + 11 = 21</math>, <math>B = 21 - 19 = 2</math>, which leads to <math>C = 2 + 7 = 9</math>.  Thus far, we have given out <math>10 + 21 + 2 + 9 = 42</math> dollars, leaving <math>56 - 42 = 14</math> dollars for Dick.  Dick does indeed have <math>5</math> dollars more than Carlos, and <math>4</math> dollars more than Elgin.
  
 
== See also ==
 
== See also ==
 
{{AHSME box|year=1997|num-b=21|num-a=23}}
 
{{AHSME box|year=1997|num-b=21|num-a=23}}
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Revision as of 19:39, 9 August 2011

Problem

Ashley, Betty, Carlos, Dick, and Elgin went shopping. Each had a whole number of dollars to spend, and together they had $56$ dollars. The absolute difference between the amounts Ashley and Betty had to spend was $19$ dollars. The absolute difference between the amounts Betty and Carlos had was $7$ dollars, between Carlos and Dick was $5$ dollars, between Dick and Elgin was $4$ dollars, and between Elgin and Ashley was $11$ dollars. How many dollars did Elgin have?

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$

Solution

Working backwards, if $6 \le E \le 10$, then $6 \pm 11 \le A \le 10 \pm 11$. Since $A$ is a positive integer, $17 \le A \le 21$.


Since $17 \le A \le 21$, we know that $17 \pm 19 \le B \le 21 \pm 19$. But if $B=36$, which is the smallest possible "plus" value, then $E + A + B = 6 + 17 + 36 = 59$, which is too much money.


Hence, $17 - 19 \le B \le 21 - 19$. But since $B$ must be a positive integer, that leaves only two possibilities: $B = 1$ or $B=2$, which correspond with $E = 9$ and $E = 10$.


Concentrating only on $E=9$, we have $E=9$ leading to $A = 9 + 11 = 20$, which leads to $B = 20 - 19 = 1$, which leads to $C = 1 + 7 = 8$. Thus far we have given out $9 + 20 + 1 + 8 = 38$ dollars. This means that Dick must have $56 - 38 = 18$ dollars. However, the difference between Carlos and Dick is not $5$ dollars.


Thus, the right answer must be $\boxed{E}$. Verifying, if $E = 10$, then $A = 10 + 11 = 21$, $B = 21 - 19 = 2$, which leads to $C = 2 + 7 = 9$. Thus far, we have given out $10 + 21 + 2 + 9 = 42$ dollars, leaving $56 - 42 = 14$ dollars for Dick. Dick does indeed have $5$ dollars more than Carlos, and $4$ dollars more than Elgin.

See also

1997 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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
All AHSME Problems and Solutions

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