Difference between revisions of "2020 AMC 8 Problems/Problem 20"
(→Video Solution) |
|||
Line 45: | Line 45: | ||
If Tree 4 now has a height of <math>11</math>, then Tree 5 would need to have height <math>22</math>, but in that case <math>S</math> would equal <math>88</math>, which is not <math>1</math> more than a multiple of <math>5</math>. So we instead take Tree 4 to have height <math>44</math>. Then the sum of the heights of the first 4 trees is <math>22+11+22+44 = 99</math>, so using a height of <math>22</math> for Tree 5 gives <math>S=121</math>, which is <math>1</math> more than a multiple of <math>5</math> (whereas <math>88</math> gives <math>S = 187</math>, which is not). Thus the average height of the trees is <math>\frac{121}{5} = \boxed{\textbf{(B) }24.2}</math> meters. | If Tree 4 now has a height of <math>11</math>, then Tree 5 would need to have height <math>22</math>, but in that case <math>S</math> would equal <math>88</math>, which is not <math>1</math> more than a multiple of <math>5</math>. So we instead take Tree 4 to have height <math>44</math>. Then the sum of the heights of the first 4 trees is <math>22+11+22+44 = 99</math>, so using a height of <math>22</math> for Tree 5 gives <math>S=121</math>, which is <math>1</math> more than a multiple of <math>5</math> (whereas <math>88</math> gives <math>S = 187</math>, which is not). Thus the average height of the trees is <math>\frac{121}{5} = \boxed{\textbf{(B) }24.2}</math> meters. | ||
+ | |||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/uCV9-WpaIjw | ||
+ | |||
+ | ~savannahsolver | ||
==Video Solution== | ==Video Solution== |
Revision as of 18:22, 26 February 2021
Contents
Problem 20
A scientist walking through a forest recorded as integers the heights of trees standing in a row. She observed that each tree was either twice as tall or half as tall as the one to its right. Unfortunately some of her data was lost when rain fell on her notebook. Her notes are shown below, with blanks indicating the missing numbers. Based on her observations, the scientist was able to reconstruct the lost data. What was the average height of the trees, in meters?
Solution 1
We will show that ,
,
,
, and
meters are the heights of the trees from left to right. We are given that all tree heights are integers, so since Tree 2 has height
meters, we can deduce that Trees 1 and 3 both have a height of
meters. There are now three possible cases for the heights of Trees 4 and 5 (in order for them to be integers), namely heights of
and
,
and
, or
and
. Checking each of these, in the first case, the average is
meters, which doesn't end in
as the problem requires. Therefore, we consider the other cases. With
and
, the average is
meters, which again does not end in
, but with
and
, the average is
meters, which does. Consequently, the answer is
.
Solution 2
Notice the average height of the trees ends with therefore the sum of all five heights of the trees must end with
. (
*
=
)
We already know Tree
is
meters tall. Both Tree
and Tree
must
meters tall - since neither can be
.
Once again, apply our observation for solving for the Tree
's height. Tree
can't be
meters for the sum of the five tree heights to still end with
. Therefore, the Tree
is
meters tall.
Now the Tree
can either be
or
. Find the average height for both cases of Tree
. Doing this, we realize the Tree
must be
for the average height to end with
and that the average height is
.
Solution 3
As in Solution 1, we shall show that the heights of the trees are ,
,
,
, and
meters. Let
be the sum of the heights, so that the average height will be
meters. We note that
, so in order for
to end in
,
must be one more than a multiple of
. Moreover, as all the heights are integers, the heights of Tree 1 and Tree 3 are both
meters. At this point, our table looks as follows:
If Tree 4 now has a height of , then Tree 5 would need to have height
, but in that case
would equal
, which is not
more than a multiple of
. So we instead take Tree 4 to have height
. Then the sum of the heights of the first 4 trees is
, so using a height of
for Tree 5 gives
, which is
more than a multiple of
(whereas
gives
, which is not). Thus the average height of the trees is
meters.
Video Solution by WhyMath
~savannahsolver
Video Solution
See also
2020 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.