Step 1: Fix angle reference and note the cycle length.
Take \(\theta=0^\circ\) at TDC between exhaust and intake (the valve-overlap TDC). One full 4-stroke cycle spans \(720^\circ\) of crank rotation and ends at the next same TDC (\(720^\circ\)).
Step 2: Map the inlet valve open interval.
Inlet opens \(10^\circ\) BTDC \(\Rightarrow\) at \(-10^\circ\) (i.e., \(710^\circ\) in the current cycle) and closes \(40^\circ\) ABDC of the intake stroke.
BDC of intake is at \(180^\circ\); hence close at \(180^\circ+40^\circ=220^\circ\).
Therefore, inlet is open on \([710^\circ,720^\circ]\cup[0^\circ,220^\circ]\).
Step 3: Map the exhaust valve open interval.
Exhaust opens \(25^\circ\) BBDC of the power stroke. Power stroke is \(360^\circ \to 540^\circ\); hence open at \(540^\circ-25^\circ=515^\circ\).
Exhaust closes \(15^\circ\) ATDC of the next TDC, i.e., at \(720^\circ+15^\circ\). Within our cycle this appears as it being still open on \([0^\circ,15^\circ]\) after \(720^\circ\).
Therefore, exhaust is open on \([515^\circ,720^\circ]\cup[0^\circ,15^\circ]\).
Step 4: Find when either valve is open and complement.
Union of open intervals:
\[
[0,220]\ \cup\ [515,720].
\]
Hence both valves are closed only on
\[
[220,515],
\]
whose length is \(515-220=295^\circ\).
Step 5: Convert to percentage over the \(720^\circ\) cycle.
\[
\text{Percentage closed}=\frac{295}{720}\times 100
=40.972%\ \approx\ 40.97%.
\]
Final Answer:
\[
\boxed{40.97%}
\]
Ravi had _________ younger brother who taught at _________ university. He was widely regarded as _________ honorable man.
Select the option with the correct sequence of articles to fill in the blanks.