Difference between revisions of "2015 AMC 8 Problems/Problem 23"
(→Solution) |
(→Solution) |
||
Line 11: | Line 11: | ||
==Solution== | ==Solution== | ||
− | The numbers have a sum of <math>6+5+12+4+8=35</math>, which averages to <math>7</math>, which means <math>A, B, C, D, E</math> will have values <math>5, 6, 7, 8, 9</math>, respectively. Now it's process of elimination: Cup <math>A</math> will have a sum of <math>5</math>, so putting a <math>3.5</math> slip in the cup will leave <math>5-3.5=1.5</math>; However, all of our slips are bigger than <math>1.5</math>, so this is impossible. Cup <math>B</math> has a sum of <math>6</math>, but we are told that it already has a <math>3</math> slip, leaving <math>6-3=3</math>, which is too small for the <math>3.5</math> slip. Cup <math>C</math> is a little bit trickier, but still manageable. It must have a value of <math>7</math>, so adding the <math>3.5</math> slip leaves room for <math>7-3.5=3.5</math>. This looks good at first, as we do have slips smaller than that, but upon closer inspection, we see that no slip fits exactly, and the smallest sum of two slips is <math>2+2=4</math>, which is too big, so this case is also impossible. Cup <math>E</math> has a sum of <math>9</math>, but we are told it already has a <math>2</math> slip, so we are left with <math>9-2=7</math>, which is identical to the Cup C case, and thus also impossible. With all other choices removed, we are left with the answer: Cup <math>\boxed{\textbf{(D)}~D}</math> | + | The numbers have a sum of <math>6+5+12+4+8=35</math>, which averages to <math>7</math>, which means <math>A, B, C, D, |
+ | |||
+ | E</math> will have values <math>5, 6, 7, 8, 9</math>, respectively. Now it's process of elimination: Cup <math>A</math> | ||
+ | |||
+ | will have a sum of <math>5</math>, so putting a <math>3.5</math> slip in the cup will leave <math>5-3.5=1.5</math>; However, | ||
+ | |||
+ | all of our slips are bigger than <math>1.5</math>, so this is impossible. Cup <math>B</math> has a sum of <math>6</math>, but we are told that it already has a <math>3</math> slip, leaving <math>6-3=3</math>, which is too small for the | ||
+ | |||
+ | <math>3.5</math> slip. Cup <math>C</math> is a little bit trickier, but still manageable. It must have a value of <math>7</math>, so adding the <math>3.5</math> slip leaves room for <math>7-3.5=3.5</math>. This looks good at first, as we | ||
+ | |||
+ | do have slips smaller than that, but upon closer inspection, we see that no slip fits | ||
+ | |||
+ | exactly, and the smallest sum of two slips is <math>2+2=4</math>, which is too big, so this case is | ||
+ | |||
+ | also impossible. Cup <math>E</math> has a sum of <math>9</math>, but we are told it already has a <math>2</math> slip, so we | ||
+ | |||
+ | are left with <math>9-2=7</math>, which is identical to the Cup C case, and thus also impossible. | ||
+ | |||
+ | With all other choices removed, we are left with the answer: Cup <math>\boxed{\textbf{(D)}~D}</math>. | ||
==See Also== | ==See Also== |
Revision as of 22:45, 3 November 2019
Tom has twelve slips of paper which he wants to put into five cups labeled , , , , . He wants the sum of the numbers on the slips in each cup to be an integer. Furthermore, he wants the five integers to be consecutive and increasing from to . The numbers on the papers are and . If a slip with goes into cup and a slip with goes into cup , then the slip with must go into what cup?
Solution
The numbers have a sum of , which averages to , which means $A, B, C, D,
E$ (Error compiling LaTeX. ! Missing $ inserted.) will have values , respectively. Now it's process of elimination: Cup
will have a sum of , so putting a slip in the cup will leave ; However,
all of our slips are bigger than , so this is impossible. Cup has a sum of , but we are told that it already has a slip, leaving , which is too small for the
slip. Cup is a little bit trickier, but still manageable. It must have a value of , so adding the slip leaves room for . This looks good at first, as we
do have slips smaller than that, but upon closer inspection, we see that no slip fits
exactly, and the smallest sum of two slips is , which is too big, so this case is
also impossible. Cup has a sum of , but we are told it already has a slip, so we
are left with , which is identical to the Cup C case, and thus also impossible.
With all other choices removed, we are left with the answer: Cup .
See Also
2015 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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.