XAT 2011 — QA & DI Question 32

It can be observed that, irrespective of the value of x, mode of these numbers will be 4.

Now, the median of these numbers will depend on the value of x,

If x < 4 then the median of these seven numbers will be 4.

Now, as the mode is 4, the median cannot be 4.

(the question states that mean, median and mode are arranged in ascending order.)

Hence x cannot be less than 4.

Now,

If 4 < x ≤ 8

the median will be x & the mean will be,

$\frac{50+x}{7}$

Now $\frac{50+x}{7}$ > 7 and x ≤ 8

∴ 4, x and $\frac{50+x}{7}$ will form an AP only if $\frac{50+x}{7}$ > x

∴ $\frac{50+x}{7}$ - x = x - 4

∴ $\frac{50+x}{7}$ + 4 = 2x

∴ x = 6

Hence, x = 6 is a possible answer.

Now, if x > 8 then median will be 8 & mean will be 

$\frac{50+x}{7}$

Now, if x > 8 then $\frac{50+x}{7}$ is greater than 8.

∴ Increasing order of mean, median and mode will be,

4, 8, $\frac{50+x}{7}$

Now, they are in A.P.

∴ 8 - 4 = $\frac{50+x}{7}$ - 8

∴ 12 = $\frac{50+x}{7}$

∴ x = 12 × 7 – 50

∴ x = 34

Hence, sum of all possible values of x = 6 + 34 = 40

Hence, option (e).