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Doctors Medicines MCQs Q&A

S1: While crossing a busy road we should obey the policeman on duty.
P: We should always cross the road at the zebra crossing.
Q: We must look to the signal lights and cross the road only when the road is clear.
R: If there are no signal lights at the crossing, we should look to the right, then to the left and again to the right before crossing the road.
S: If the road is not clear we should wait.
S6: We should never run while crossing a road.

The Proper sequence should be:

S1: While crossing a busy road we should obey the policeman on duty.
P: We should always cross the road at the zebra crossing.
Q: We must look to the signal lights and cross the road only when the road is clear.
R: If there are no signal lights at the crossing, we should look to the right, then to the left and again to the right before crossing the road.
S: If the road is not clear we should wait.
S6: We should never run while crossing a road.

The Proper sequence should be: Correct Answer QRPS

We must look to the signal lights and cross the road only when the road is clear. If there are no signal lights at the crossing, we should look to the right, then to the left and again to the right before crossing the road. We should always cross the road at the zebra crossing. If the road is not clear we should wait.

Related Questions

If a + b + c + d = 4, then find the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$     + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$     + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$     + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$     is?
If a + b + c + d = 4, then the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$     + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$     + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$     + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$     is?
A thief is spotted by a policeman from a distance of 400 meters. When the policeman started running after the thief, the thief also started running. Assuming the speed of the thief to be 8 km/h and that of the policeman is 10 km/h, how far will the thief have run before he is overtaken by the policeman?
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\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaqGjbGaaeOzaiaabckacaqG4bGaaeiOaiabgUcaRiaabckadaWc % aaWdaeaapeGaaGymaaWdaeaapeGaamiEaiabgkHiTiaaigdacaaIZa % aaaiabg2da9iaabckacaaIXaGaaGymaiaacYcacaqGGcGaaeiDaiaa % bIgacaqGLbGaaeOBaiaabckacaqGMbGaaeyAaiaab6gacaqGKbGaae % iOaiaabckacaqG0bGaaeiAaiaabwgacaqGGcGaaeODaiaabggacaqG % SbGaaeyDaiaabwgacaqGGcGaae4BaiaabAgacaqGGcWaaeWaa8aaba % WdbiaabIhacqGHsislcaaIXaGaaG4maaGaayjkaiaawMcaa8aadaah % aaWcbeqaa8qacaaI1aaaaOGaaeiOaiabgUcaRiaabckadaWcaaWdae % aapeGaaGymaaWdaeaapeWaaeWaa8aabaWdbiaadIhacqGHsislcaaI % XaGaaGymaaGaayjkaiaawMcaa8aadaahaaWcbeqaa8qacaaI1aaaaa % aakiaacckacaGGUaaaaa!70B8! {\rm{If\;x\;}} + {\rm{\;}}\frac{1}{{x - 13}} = {\rm{\;}}11,{\rm{\;then\;find\;\;the\;value\;of\;}}{\left( {{\rm{x}} - 13} \right)^5}{\rm{\;}} + {\rm{\;}}\frac{1}{{{{\left( {x - 11} \right)}^5}}}\;.\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaqGjbGaaeOzaiaabckacaqG4bGaaeiOaiabgUcaRiaabckadaWc % aaWdaeaapeGaaGymaaWdaeaapeGaamiEaiabgkHiTiaaigdacaaIZa % aaaiabg2da9iaabckacaaIXaGaaGymaiaacYcacaqGGcGaaeiDaiaa % bIgacaqGLbGaaeOBaiaabckacaqGMbGaaeyAaiaab6gacaqGKbGaae % iOaiaabckacaqG0bGaaeiAaiaabwgacaqGGcGaaeODaiaabggacaqG % SbGaaeyDaiaabwgacaqGGcGaae4BaiaabAgacaqGGcWaaeWaa8aaba % WdbiaabIhacqGHsislcaaIXaGaaG4maaGaayjkaiaawMcaa8aadaah % aaWcbeqaa8qacaaI1aaaaOGaaeiOaiabgUcaRiaabckadaWcaaWdae % aapeGaaGymaaWdaeaapeWaaeWaa8aabaWdbiaadIhacqGHsislcaaI % XaGaaGymaaGaayjkaiaawMcaa8aadaahaaWcbeqaa8qacaaI1aaaaa % aakiaacckacaGGUaaaaa!70B8! {\rm{If\;x\;}} + {\rm{\;}}\frac{1}{{x - 13}} = {\rm{\;}}11,{\rm{\;then\;find\;\;the\;value\;of\;}}{\left( {{\rm{x}} - 13} \right)^5}{\rm{\;}} + {\rm{\;}}\frac{1}{{{{\left( {x - 11} \right)}^5}}}\;.\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaqGjbGaaeOzaiaabckacaqG4bGaaeiOaiabgUcaRiaabckadaWc % aaWdaeaapeGaaGymaaWdaeaapeGaamiEaiabgkHiTiaaigdacaaIZa % aaaiabg2da9iaabckacaaIXaGaaGymaiaacYcacaqGGcGaaeiDaiaa % bIgacaqGLbGaaeOBaiaabckacaqGMbGaaeyAaiaab6gacaqGKbGaae % iOaiaabckacaqG0bGaaeiAaiaabwgacaqGGcGaaeODaiaabggacaqG % SbGaaeyDaiaabwgacaqGGcGaae4BaiaabAgacaqGGcWaaeWaa8aaba % WdbiaabIhacqGHsislcaaIXaGaaG4maaGaayjkaiaawMcaa8aadaah % aaWcbeqaa8qacaaI1aaaaOGaaeiOaiabgUcaRiaabckadaWcaaWdae % aapeGaaGymaaWdaeaapeWaaeWaa8aabaWdbiaadIhacqGHsislcaaI % XaGaaGymaaGaayjkaiaawMcaa8aadaahaaWcbeqaa8qacaaI1aaaaa % aakiaacckacaGGUaaaaa!70B8! {\rm{If\;x\;}} + {\rm{\;}}\frac{1}{{x - 13}} = {\rm{\;}}11,{\rm{\;then\;find\;\;the\;value\;of\;}}{\left( {{\rm{x}} - 13} \right)^5}{\rm{\;}} + {\rm{\;}}\frac{1}{{{{\left( {x - 11} \right)}^5}}}\;.\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaqGjbGaaeOzaiaabckacaqG4bGaaeiOaiabgUcaRiaabckadaWc % aaWdaeaapeGaaGymaaWdaeaapeGaamiEaiabgkHiTiaaigdacaaIZa % aaaiabg2da9iaaigdacaaIXaGaaiilaiaabckacaqG0bGaaeiAaiaa % bwgacaqGUbGaaeiOaiaabAgacaqGPbGaaeOBaiaabsgacaqGGcGaae % iDaiaabIgacaqGLbGaaeiOaiaabAhacaqGHbGaaeiBaiaabwhacaqG % LbGaaeiOaiaab+gacaqGMbGaaeiOamaabmaapaqaa8qacaqG4bGaey % OeI0IaaGymaiaaiodaaiaawIcacaGLPaaapaWaaWbaaSqabeaapeGa % aGynaaaakiaabckacqGHRaWkcaqGGcWaaSaaa8aabaWdbiaaigdaa8 % aabaWdbmaabmaapaqaa8qacaWG4bGaeyOeI0IaaGymaiaaigdaaiaa % wIcacaGLPaaapaWaaWbaaSqabeaapeGaaGynaaaaaaGccaGGGcGaai % Olaaaa!6E72! {\rm{If\;x\;}} + {\rm{\;}}\frac{1}{{x - 13}} = 11,{\rm{\;then\;find\;the\;value\;of\;}}{\left( {{\rm{x}} - 13} \right)^5}{\rm{\;}} + {\rm{\;}}\frac{1}{{{{\left( {x - 11} \right)}^5}}}\;.\)If x + 1/(x - 13) = 11, then what will be the value of (x – 13)5 + 1/(x – 11)5?
The Hamiltonian of a particle is given by $$H = \frac{{{p^2}}}{{2m}} + V\left( {\left| {\overrightarrow {\bf{r}} } \right|} \right) + \phi \left( { + \left| {\overrightarrow {\bf{r}} } \right|} \right)\overrightarrow {\bf{L}} .\overrightarrow {\bf{S}} ,$$        where $$\overrightarrow {\bf{S}} $$ is the spin, $$V\left( {\left| {\overrightarrow {\bf{r}} } \right|} \right)$$  and $$\phi \left( {\left| {\overrightarrow {\bf{r}} } \right|} \right)$$  are potential functions and $$\overrightarrow {\bf{L}} \left( { = \overrightarrow {\bf{r}} \times \overrightarrow {\bf{p}} } \right)$$   is the angular momentum. The Hamiltonian does not commute with
A thief is noticed by a policeman from a distance of 200 m. The thief starts running and the policeman chases him. The thief and policeman run at the rate of 10 km/hr and 11 km/hr respectively. At what distance will the policeman catches the thief?