Difference between revisions of "1996 AJHSME Problems/Problem 2"
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| − | + | ==Problem== | |
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| + | Jose, Thuy, and Kareem each start with the number 10. Jose subtracts 1 from the number 10, doubles his answer, and then adds 2. Thuy doubles the number 10, subtracts 1 from her answer, and then adds 2. Kareem subtracts 1 from the number 10, adds 2 to his number, and then doubles the result. Who gets the largest final answer? | ||
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| + | <math>\text{(A)}\ \text{Jose} \qquad \text{(B)}\ \text{Thuy} \qquad \text{(C)}\ \text{Kareem} \qquad \text{(D)}\ \text{Jose and Thuy} \qquad \text{(E)}\ \text{Thuy and Kareem}</math> | ||
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| + | ==Solution== | ||
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| + | Jose gets <math>10 - 1 = 9</math>, then <math>9 \cdot 2 = 18</math>, then <math>18 + 2 = 20</math>. | ||
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| + | Thuy gets <math>10 \cdot 2 = 20</math>, then <math>20 - 1 = 19</math>, and then <math>19 + 2 = 21</math>. | ||
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| + | Kareem gets <math>10 - 1 = 9</math>, then <math>9 + 2 = 11</math>, and then <math>11\cdot 2 = 22</math>. | ||
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| + | Thus, Kareem gets the highest number, and the answer is <math>\boxed{C}</math>. | ||
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== See also == | == See also == | ||
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* [[AJHSME Problems and Solutions]] | * [[AJHSME Problems and Solutions]] | ||
* [[Mathematics competition resources]] | * [[Mathematics competition resources]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 12:49, 23 October 2016
Problem
Jose, Thuy, and Kareem each start with the number 10. Jose subtracts 1 from the number 10, doubles his answer, and then adds 2. Thuy doubles the number 10, subtracts 1 from her answer, and then adds 2. Kareem subtracts 1 from the number 10, adds 2 to his number, and then doubles the result. Who gets the largest final answer?
Solution
Jose gets 
, then 
, then 
.
Thuy gets 
, then 
, and then 
.
Kareem gets , then 
, and then 
.
Thus, Kareem gets the highest number, and the answer is 
.
See also
| 1996 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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. 
