Shortcut Method:

\(5 < x < 15\) and \(5 < y < 15\)

We have assumed that Perimeter or ABC > Perimeter of DEF
So, Maximum possible integer difference = \(x − y\)

Take Maximum value of \(x = 15\)
Take Minimum value of \(y = 5\)
integer difference = 10

Since, we cannot have 15 and 5 as the sides here, the difference of 10 is not possible.

So, what is the next possible integer value? It is 1 less than 10 = \(10 - 1 = 9\)

Hence, option C

I only considered integer values !! My bad !!

KarunMendiratta
what if the difference is greater than 9. if the unknown sides are 5.01 and 14.99 shall not A be the answer?

For example, with third sides of 14.99 and 5.01 , the difference would be 9.98 . However, the question asks for the greatest possible integer difference.



The perimeter of a triangle with sides 5 and 10 is $P=15+x$, where $\(5<x<15\)$.

The difference in perimeters is $\(\left|P_1-P_2\right|=\left|\left(15+x_1\right)-\left(15+x_2\right)\right|=\left|x_1-x_2\right|\)$.
To maximize this difference, we need to choose values for the two third sides, $\(x_1\)$ and $\(x_2\)$, that are as far apart as possible.
- $\(x_1\)$ must be a number very close to 15 , but not 15 itself. For example, $\(x_1=14.999 \ldots\)$
- $\(x_2\)$ must be a number very close to 5 , but not 5 itself. For example, $\(x_2=5.001 \ldots\)$

The difference $\(\left|x_1-x_2\right|\)$ will be a number that is less than 10 but can be arbitrarily close to 10 . The set of all possible differences is the interval $\((0,10)\)$.

The question asks for the greatest possible integer value in this interval. The integers in the interval $\((0,10)\)$ are $\(\{1,2,3,4,5,6,7,8,9\}\)$. The largest of these is 9 .

So, while the numerical difference can be a decimal like 9.99 , the greatest integer difference is 9 .

damn, I got the right calculation but missed the integer part