Question

DIRECTIONS for questions 45 and 46: Read the information below and answer the question that follows.
It is possible to arrange eight of the nine numbers 2, 3, 4, 5, 7, 10, 11, 12, 13 in the vacant squares of the 3 by 4 array shown below so that the arithmetic average of the numbers in each row and column is the same integer.

1			15
	9
		14

45. The arithmetic average is:

• A. 6
• B. 7
• C. 8
• D. 9

Hint:

Every row and column shares the same average x, so the whole grid's total must be a multiple of 12, since there are 3 rows of 4 cells each.

Answer 1

The Correct Option is C

Step 1: Use the fact that every row and column has the same average.
Let the common average be x. The array has 3 rows of 4 cells each, so each row sums to \(4x\), and the whole grid total (summing over the 3 rows) is \(3\times4x=12x\). Checking with the columns gives the same total: each column sums to \(3x\), and with 4 columns the total is \(4\times3x=12x\) too. So the grid total must be \(12x\), a multiple of 12.

Step 2: Add up the numbers involved.
The four numbers already fixed in the grid are 1, 15, 9 and 14. The nine numbers to pick eight from are 2, 3, 4, 5, 7, 10, 11, 12, 13. Adding all thirteen numbers together:
\[ 1+15+9+14+2+3+4+5+7+10+11+12+13 = 106 \]

Step 3: Account for the one number left out.
106 is not a multiple of 12 (\(106=12\times8+10\)), so one of the nine free numbers must be dropped to bring the total down to a multiple of 12. Dropping 10 gives:
\[ 106-10=96=12\times8 \]
96 is a multiple of 12, with \(x=8\).

Step 4: Check that a valid arrangement exists for x = 8.
Using the eight numbers 2, 3, 4, 5, 7, 11, 12 and 13 (leaving out 10), the grid can indeed be completed so every row sums to 32 and every column sums to 24, for example:

1	13	3	15
12	9	7	4
11	2	14	5

Every row here adds to 32 (four times 8) and every column adds to 24 (three times 8), confirming the average of 8 is achievable, which rules out 6, 7 and 9 (options 1, 2 and 4).

Final Answer:
\[ \boxed{x=8} \]