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 dollars. The absolute difference between the amounts Ashley and Betty had to spend was
dollars. The absolute difference between the amounts Betty and Carlos had was
dollars, between Carlos and Dick was
dollars, between Dick and Elgin was
dollars, and between Elgin and Ashley was
dollars. How many dollars did Elgin have?
Solution
Working backwards, if , then
. Since
is a positive integer,
.
Since , we know that
. But if
, which is the smallest possible "plus" value, then
, which is too much money.
Hence, . But since
must be a positive integer, that leaves only two possibilities:
or
, which correspond with
and
.
Concentrating only on , we have
leading to
, which leads to
, which leads to
. Thus far we have given out
dollars. This means that Dick must have
dollars. However, the difference between Carlos and Dick is not
dollars.
Thus, the right answer must be . Verifying, if
, then
,
, which leads to
. Thus far, we have given out
dollars, leaving
dollars for Dick. Dick does indeed have
dollars more than Carlos, and
dollars more than Elgin.
See also
1997 AHSME (Problems • Answer Key • Resources) | ||
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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