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 <math>10</math>. Jose subtracts <math>1</math> from the number <math>10</math>, doubles his answer, and then adds <math>2</math>. Thuy doubles the number <math>10</math>, subtracts <math>1</math> from her answer, and then adds <math>2</math>. Kareem subtracts <math>1</math> from the number <math>10</math>, adds <math>2</math> 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 ==
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{{AJHSME box|year=1996|num-b=1|num-a=3}}
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* [[AJHSME]]
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* [[AJHSME Problems and Solutions]]
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* [[Mathematics competition resources]]
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{{MAA Notice}}

Latest revision as of 13:16, 22 December 2024

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?

$\text{(A)}\ \text{Jose} \qquad \text{(B)}\ \text{Thuy} \qquad \text{(C)}\ \text{Kareem} \qquad \text{(D)}\ \text{Jose and Thuy} \qquad \text{(E)}\ \text{Thuy and Kareem}$

Solution

Jose gets $10 - 1 = 9$, then $9 \cdot 2 = 18$, then $18 + 2 = 20$.

Thuy gets $10 \cdot 2 = 20$, then $20 - 1 = 19$, and then $19 + 2 = 21$.

Kareem gets $10 - 1 = 9$, then $9 + 2 = 11$, and then $11\cdot 2 = 22$.

Thus, Kareem gets the highest number, and the answer is $\boxed{C}$.


See also

1996 AJHSME (ProblemsAnswer KeyResources)
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png