We need to determine two (positive) relatively prime numbers such that their sum is 150, each is less than 100, and their difference is a multiple of 7.
Let the sought numbers be $a$ and $b$.

$a+b=150$, if $a=75+x, b=75-x$. Their difference, $a-b=2 x$. This is a multiple of 7 if $x=7 m$ (and $m>0$). Therefore,

$$
a=75+7 m, \quad b=75-7 m .
$$

If $a$ and $b$ are relatively prime, $m$ cannot be an odd number.

However, $75+7 m<100$ if $7 m<25$, that is, $m=1,2,3$.

Of these values of $m$, only 2 can be considered, so

$$
a=75+14=89, \quad b=75-14=61
$$

These are indeed relatively prime. (Each is a prime number.)

Klara Kalocsa (Lea Raskai's daughter, 5th grade, Budapest V.).