A particle returns to the starting post after 10s. If the rate of change of velocity during the motion is constant in magnitude Then it’s location after 7 seconds will be the same as that after-? Now my super-efficient and productive brain (#sarcasm) can figure out that it’s 3 seconds, since it has got a constant velocity and 10-7=3. But I doubt future questions like this will be that easy. Is there anyway to solve it ‘properly’?

No, you are given that the acceleration is constant. The velocity is then a linear function of time and the position is quadratic. The equation for the velocity and position when the acceleration is constant is $$v=v_0+at\\\s=s_0+v_0t+\frac 12at^2$$ You should be able to derive these by integration. $s_0$ is the position at $t=0$ and $v_0$ is the velocity at $t=0$. We are given $s(10)=s(0)=s_0$. By taking the derivative you can show that $v(5)=0$, so $v_0=-5a$ and plugging in you can find that your intuition is correct.