Question

The following sequence of numbers is arranged in increasing order: 1, x, x, x, y, y, 9, 16, 18. Given that the mean and median are equal, and are also equal to twice the mode, the value of y is

• A. 5
• B. 6
• C. 7
• D. 8

Answer 1

The correct option is D 8

No's are 1, x, x, x, y, y, 9, 16, 18

$\because$ has largest frequency so Mode = x

Now given, mean = Median = 2 Mode

Mean = y = 2x ...(i)

Again Mean = $\frac{3x+2y+44}{9}$

$2x=\frac{3x+2y+44}{9}$

$\Rightarrow 15x-2y=44$ ...(ii)

So, by (i) and (ii) x = 4 and y = 8.