Difference between revisions of "2015 AMC 8 Problems/Problem 23"
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==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 2 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 2 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 11:42, 30 November 2015
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 
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 2 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. 
