Difference between revisions of "1995 AHSME Problems/Problem 25"
(→Solution) |
|||
| Line 7: | Line 7: | ||
Let <math>a</math> be the smallest element, so <math>a+18</math> is the largest element. Since the mode is <math>8</math>, at least two of the five numbers must be <math>8</math>. The last number we denote as <math>b</math>. | Let <math>a</math> be the smallest element, so <math>a+18</math> is the largest element. Since the mode is <math>8</math>, at least two of the five numbers must be <math>8</math>. The last number we denote as <math>b</math>. | ||
| − | Then their average is <math>\frac{a + (8) + (8) + b + (a+18)}5 = 12 \Longrightarrow 2a + b = 26</math>. Clearly <math>a \le 8</math>. Also we have <math>b \le a + 18 \Longrightarrow 26-2a \le a + 18 \Longrightarrow 8/3 < 3 \le a</math>. Thus there are a maximum of <math>6</math> values of <math>a</math> which corresponds to <math>6</math> values of <math>b</math>; listing shows that all such values work. | + | Then their average is <math>\frac{a + (8) + (8) + b + (a+18)}5 = 12 \Longrightarrow 2a + b = 26</math>. Clearly <math>a \le 8</math>. Also we have <math>b \le a + 18 \Longrightarrow 26-2a \le a + 18 \Longrightarrow 8/3 < 3 \le a</math>. Thus there are a maximum of <math>6</math> values of <math>a</math> which corresponds to <math>6</math> values of <math>b</math>; listing shows that all such values work. The answer is <math>\boxed{B}</math>. |
== See also == | == See also == | ||
Latest revision as of 20:59, 10 February 2019
Problem
A list of five positive integers has mean 
and range 
. The mode and median are both 
. How many different values are possible for the second largest element of the list?
Solution
Let 
be the smallest element, so 
is the largest element. Since the mode is , at least two of the five numbers must be . The last number we denote as 
.
Then their average is 
. Clearly 
. Also we have 
. Thus there are a maximum of 
values of which corresponds to values of ; listing shows that all such values work. The answer is 
.
See also
| 1995 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Problem 26 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
