1993 AMC 8 Problem 12

Attempt Problem 12 of the 1993 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1993 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

12.

If each of the three operation signs, +,+, −,-, ×,\times, is used exactly once in one of the blanks in the expression
5 x‾ 4 x‾ 6 x‾ 35 \, \underline{\phantom{x}} \, 4 \, \underline{\phantom{x}} \, 6 \, \underline{\phantom{x}} \, 3
then the value of the result could equal
99
1010
1515
1616
1919
Answer: E
Concepts:order of operationscasework
Small hint:
Multiplication happens before addition and subtraction, so where you place ×\times matters most
Big hint:
Work through each way of assigning +,+, −,-, ×\times to the three blanks and compute the result
Solution:
The six arrangements give 5×4+6−3=23,5 \times 4 + 6 - 3 = 23, 5×4−6+3=17,5 \times 4 - 6 + 3 = 17, 5+4×6−3=26,5 + 4 \times 6 - 3 = 26, 5−4×6+3=−16,5 - 4 \times 6 + 3 = -16, 5+4−6×3=−9,5 + 4 - 6 \times 3 = -9, and 5−4+6×3=19.5 - 4 + 6 \times 3 = 19.
The only value among the choices is 19.19.
Thus, the correct answer is E .
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