6. a tech company bought a number of computers for its employees: 6 costing $1600 each, 4 costing $1200…

6. a tech company bought a number of computers for its employees: 6 costing $1600 each, 4 costing $1200 each, and $y$ costing $900 each, where $y$ is a positive odd integer. if the median price for all the computers purchased by the company was $1200, what is the greatest possible value of $y$?

Answer

Explanation:

Step1: Count total computers

Total number of computers = $6 + 4 + y = 10 + y$

Step2: Define median position

For $n=10+y$ items, median is at position $\frac{n+1}{2}=\frac{11+y}{2}$ when sorted.

Step3: Set median condition

Sorted prices: $y$ items of $$900$, 4 items of $$1200$, 6 items of $$1600$. For median to be $$1200$, the median position must fall within the $$1200$ group. So the number of items before the $$1200$ group ($y$) must be less than the median position, and the cumulative count up to $$1200$ must be at least the median position: $y < \frac{11+y}{2} \leq y + 4$ First solve left inequality: $2y < 11 + y$ $y < 11$

Step4: Apply odd integer rule

$y$ is a positive odd integer, so the greatest odd integer less than 11 is 9. Verify: total computers = $10+9=19$, median position = $\frac{19+1}{2}=10$. The first 9 items are $$900$, items 10-13 are $$1200$, so the 10th item is $$1200$, which matches the median.