Is the average (arithmetic mean) height of Porthos, Athos, and Aramis

#1 Porthos is 10 centimetres taller than Athos, and 9 centimeters taller than Aramis.
let P = x
so Athos = x-10
Aramis = x-9
so avg ; x+x-9+x-10/3 ; 3x-19/3
median ; x-9
value of x is not know ; insufficient
#2
P = 190
but relation and value with other 2 are not given; insufficient
from 1 & 2
x=190
so
we get
551>543
sufficient
IMO C

Archit3110 The median x-9 has to be a positive number because it's height of a person. Also for any value of x (here x>9) we can compare mean and median of the list. Please correct me if I'm wrong.

Porthos = p, Athos = t, Aramis = r;

(1) Porthos is 10 centimetres taller than Athos, and 9 centimeters taller than Aramis. sufic.

\(p=t+10…p=r+9…average=sum/num=\frac{p+t+r}{3}=\frac{p+p-10+p-9}{3}…median=p-9\)
\(Question:\frac{p+p-10+p-9}{3}>p-9…3p-19>3p-27…19<27?…Answer:yes\)

(2) Porthos height is 190 centimetres insufic.

Answer (A)

Using Statement 1, your working out was spot on, the only thing is you walked past the answer, you actually had enough info in your solution to determine that A was sufficient alone.

As you correctly mentioned, x-9 is the median, in addition to this, the average height can be expressed as (3x-19)/3 (as you all correctly noted). We can rewrite this average height as x - 19/3

But the thing you walked past was that these two values are indeed comparable (also bear in mind that heights can never be negative).

We can compare the median of x-9 to the average of x-19/3. 19/3 is approx 6.33

X-9 is clearly going to be LESS than x-6.33, therefore, the average is indeed greater than the median and (1) is sufficient alone. The key point is, it does NOT matter what the value of 'x' is, x will always be positive, and the other heights can never be negative either, so x must be 9 at the very least.

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