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 $\boxed{C}$ .

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.

Page

Toolbox

Search

1996 AJHSME Problems/Problem 2

Problem

Solution

See also