1996 AJHSME Problems/Problem 2

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

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