Difference between revisions of "1997 AHSME Problems/Problem 18"
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<math>70 + 2m = 110</math> | <math>70 + 2m = 110</math> | ||
− | <math>m = 20</math>, giving answer <math>\boxed{ | + | <math>m = 20</math>, giving answer <math>\boxed{E}</math>. |
The original list is <math>\{10, 16, 20, 32, 32\}</math>, with mean <math>\frac{10 + 16 + 20 + 32 + 32}{5} = 22</math> and median <math>20</math> and mode <math>32</math>. | The original list is <math>\{10, 16, 20, 32, 32\}</math>, with mean <math>\frac{10 + 16 + 20 + 32 + 32}{5} = 22</math> and median <math>20</math> and mode <math>32</math>. | ||
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== See also == | == See also == | ||
{{AHSME box|year=1997|num-b=17|num-a=19}} | {{AHSME box|year=1997|num-b=17|num-a=19}} | ||
+ | {{MAA Notice}} |
Latest revision as of 14:13, 5 July 2013
Problem
A list of integers has mode and mean
. The smallest number in the list is
. The median
of the list is a member of the list. If the list member
were replaced by
, the mean and median of the new list would be
and
, respectively. If were
instead replaced by
, the median of the new list would be
. What is
?
Solution
Let there be integers on the list. The list of
integers has mean
, so the sum of the integers is
.
Replacing with
will increase the sum of the list from
to
.
The new mean of the list is , so the new sum of the list is also
.
Thus, we get , leading to
numbers on the list.
If there are numbers on the list with mode
and smallest number
, then the list is
Since replacing with
gives a new median of
, and
must be on the list of
integers since
is odd,
, and the list is now
The sum of the numbers on this list is , so we get:
, giving answer
.
The original list is , with mean
and median
and mode
.
The second list is , with mean
and median
.
The third list is with median
.
See also
1997 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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.