Difference between revisions of "2002 AMC 10B Problems/Problem 25"
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Let <math>S</math> be the sum of the numbers. Let <math>x</math> be the number of integers, and the mean be <math>y</math>. We have the equation <math>\frac{S}{x} = y</math>. We also have <math>\frac{S+15}{x+1} = y+2</math> and <math>\frac{S+16}{x+2} = y+1</math>. From the first equation, we have <math>S = xy</math>. We can multiply both sides by the denominators of the second and third equations to yield <math>S+15 = xy+2x+y+2</math> and <math>S+16 = xy+x+2y+2</math>. Since <math>S = xy</math>, we can substitute that into the equations and subtract by <math>xy</math> to get <math>15 = 2x+y+2</math> or <math>13 = 2x+y</math>. We also get <math>x+2y = 14</math>, and adding the two equations up gives <math>3x+3y = 27</math>, or <math>x+y = 9</math>. Subtracting <math>x+y=9</math> from <math>2x+y=13</math> gives <math>x = 4</math> and <math>y = 5</math>. | Let <math>S</math> be the sum of the numbers. Let <math>x</math> be the number of integers, and the mean be <math>y</math>. We have the equation <math>\frac{S}{x} = y</math>. We also have <math>\frac{S+15}{x+1} = y+2</math> and <math>\frac{S+16}{x+2} = y+1</math>. From the first equation, we have <math>S = xy</math>. We can multiply both sides by the denominators of the second and third equations to yield <math>S+15 = xy+2x+y+2</math> and <math>S+16 = xy+x+2y+2</math>. Since <math>S = xy</math>, we can substitute that into the equations and subtract by <math>xy</math> to get <math>15 = 2x+y+2</math> or <math>13 = 2x+y</math>. We also get <math>x+2y = 14</math>, and adding the two equations up gives <math>3x+3y = 27</math>, or <math>x+y = 9</math>. Subtracting <math>x+y=9</math> from <math>2x+y=13</math> gives <math>x = 4</math> and <math>y = 5</math>. | ||
− | We can check that this is the correct answer by having <math>4, 9, 6, | + | We can check that this is the correct answer by having <math>4, 9, 6,</math> and <math>1</math> as our numbers. The sum is 20, and the mean is 5. Adding 15 gives that the sum is 35, and the number of integers is 5, so the mean is 2, an increase of 2. Adding 1 to the integers gives that the sum is 36, and the number of numbers is 6, for a mean of 6, a decrease of 1. |
We can also prove that since the sum is 20, the number of integers is 4, and the mean is 5, we have <math>\frac{20}{4} = 5</math>, which is true. The second equation gives <math>\frac{35}{5} = 7 = 5+2</math>, which satisfies the condition. We also have <math>\frac{36}{6} = 6 = 7-1</math>, which also satisfies the condition. | We can also prove that since the sum is 20, the number of integers is 4, and the mean is 5, we have <math>\frac{20}{4} = 5</math>, which is true. The second equation gives <math>\frac{35}{5} = 7 = 5+2</math>, which satisfies the condition. We also have <math>\frac{36}{6} = 6 = 7-1</math>, which also satisfies the condition. |
Latest revision as of 21:31, 22 July 2021
Problem
When is appended to a list of integers, the mean is increased by . When is appended to the enlarged list, the mean of the enlarged list is decreased by . How many integers were in the original list?
Solution 1
Let be the sum of the integers and be the number of elements in the list. Then we get the equations and . With a lot of algebra, the solution is found to be .
Solution 2
We let be the original number of elements in the set and we let be the original average of the terms of the original list. Then we have is the sum of all the elements of the list. So we have two equations: and Simplifying both equations and we get, Solving for and , we get and .
Solution 3
Warning: This solution will rarely ever work in any other case. However, seeing that you can so easily plug and chug in problem 25 it is funny to see this.
Plug and chug random numbers with the answer choices, starting with the choice of numbers. You see that if you have 4 5s and you add 15 to the set, the resulting mean will be 7; we can verify this with math adding in 1 to the set you result in the mean to be 6. Thus we conclude that 4 is the correct choice or
Video solution by Canadamath
https://www.youtube.com/watch?v=hYq8GgJ0it4
Solution 4
Let be the sum of the numbers. Let be the number of integers, and the mean be . We have the equation . We also have and . From the first equation, we have . We can multiply both sides by the denominators of the second and third equations to yield and . Since , we can substitute that into the equations and subtract by to get or . We also get , and adding the two equations up gives , or . Subtracting from gives and .
We can check that this is the correct answer by having and as our numbers. The sum is 20, and the mean is 5. Adding 15 gives that the sum is 35, and the number of integers is 5, so the mean is 2, an increase of 2. Adding 1 to the integers gives that the sum is 36, and the number of numbers is 6, for a mean of 6, a decrease of 1.
We can also prove that since the sum is 20, the number of integers is 4, and the mean is 5, we have , which is true. The second equation gives , which satisfies the condition. We also have , which also satisfies the condition.
Therefore, since we correctly solved it and checked our work multiple ways, we can be confident that the answer is .
See also
2002 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last problem | |
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 AMC 10 Problems and Solutions |
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