Difference between revisions of "1999 AMC 8 Problems/Problem 11"

Line 1: Line 1:
Problem 11
+
==Problem==
  
Each of the five numbers 1, 4, 7, 10, and 13 is placed in one of the five squares so that the sum of the three numbers in the horizontal row equals the sum of the three numbers in the vertical column. The largest possible value for the horizontal or vertical sum is
+
Each of the five numbers 1,4,7,10, and 13 is placed in one of the five squares so that the sum of the three numbers in the horizontal row equals the sum of the three numbers in the vertical column. The largest possible value for the horizontal or vertical sum is
 +
(A) 20 (B) 21 (C) 22 (D) 24 (E) 30
  
 +
==splution==
  
 +
(D) 24: The largest sum occurs when 13 is placed in the center. This sum is 13 + 10 + 1 = 13 + 7 + 4 = 24.
 +
Note: Two other common sums, 18 and 21, are possible.
  
<math>[asy]
+
==see also==
draw((0,0)--(3,0)--(3,1)--(0,1)--cycle);
 
draw((1,-1)--(2,-1)--(2,2)--(1,2)--cycle);[/asy]</math>
 
  
 
+
{{AMC8 box|year=1999|num-b=10|num-a=12}}
The answer is obviously (D) because you have to put the biggest number (13) in the middle. Then, in order to make it equal, you put the next biggest number (10) on the top, and then you put 7 on the left side. Since the difference for the numbers are all 3, you have to put the next biggest one (4) on the horizontal row, and then you put the (1) on the vertical row, and get letter (D) as your answer.
 

Revision as of 15:13, 4 November 2012

Problem

Each of the five numbers 1,4,7,10, and 13 is placed in one of the five squares so that the sum of the three numbers in the horizontal row equals the sum of the three numbers in the vertical column. The largest possible value for the horizontal or vertical sum is

(A) 20 (B) 21 (C) 22 (D) 24 (E) 30

splution

(D) 24: The largest sum occurs when 13 is placed in the center. This sum is 13 + 10 + 1 = 13 + 7 + 4 = 24. Note: Two other common sums, 18 and 21, are possible.

see also

1999 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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