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: Think about what the total of the grid must satisfy.
The grid has 3 rows and 4 columns sharing the same average x. Adding up all 3 row sums covers every cell exactly once, and each row sums to $4x$, so the grid total is $3\times4x=12x$: a multiple of 12.

Step 2: Work out the total available.
The four fixed numbers are 1, 15, 9 and 14, and eight of the nine numbers 2, 3, 4, 5, 7, 10, 11, 12, 13 also get placed. Adding all thirteen numbers:
\[ 1+2+3+4+5+7+9+10+11+12+13+14+15=106 \]
Since only eight of the nine free numbers are used, the real grid total is $106$ minus whichever one number is left out.

Step 3: Test each option for x directly.
Instead of solving algebraically, check each answer choice by seeing whether $106$ minus some single number from $\{2,3,4,5,7,10,11,12,13\}$ can equal $12x$:
$x=6 \Rightarrow 12x=72$; we would need to drop $106-72=34$, which is not in our list. Rejected.
$x=7 \Rightarrow 12x=84$; we would need to drop $106-84=22$, not in our list. Rejected.
$x=8 \Rightarrow 12x=96$; we would need to drop $106-96=10$, and 10 IS in our list. This works.
$x=9 \Rightarrow 12x=108$; this already exceeds 106, so it cannot be reached by dropping a number, since dropping only reduces the total further. Rejected.

Step 4: Final Answer.
Only $x=8$ survives this test, matching option 3, achieved by dropping the number 10 from the nine choices.
\[ \boxed{x=8} \]