@albiesriram said:
b? 3pi3/484?
@scrabbler said:
Bro thoda woh sin^-1 wala sum karke batao na.. mujhe uss ka answer toh aa gaya but by assuming a=b both C and D where giving the same answer. is there any other approach which would give a gud return on time investment?
@iLoveTorres said:Bro thoda woh sin^-1 wala sum karke batao na.. mujhe uss ka answer toh aa gaya but by assuming a=b both C and D where giving the same answer. is there any other approach which would give a gud return on time investment?
Put a = 2b....shayad teesra option aata hai?
Sin^-1 x = k then cos^-1 x = (pi/2 - k) types kuch try kiya...could simplify further by taking x = 1 maybe, not checked...
regards
scrabbler
Sin^-1 x = k then cos^-1 x = (pi/2 - k) types kuch try kiya...could simplify further by taking x = 1 maybe, not checked...
regards
scrabbler
@albiesriram said:
Please ignore, read it wrong 😞 no wonder answers weren't matching :banghead:
regards
scrabbler
@scrabbler said:Put a = 2b....shayad teesra option aata hai?Sin^-1 x = k then cos^-1 x = (pi/2 - k) types kuch try kiya...could simplify further by taking x = 1 maybe, not checked...regardsscrabbler
at x=1 you would get an value of the form a/0~undefined. that is the problem here.. mithai de diye saying (-1 inclusive)"x"(1 inclusive) but you cant take 1 here
@iLoveTorres said:at x=1 you would get an value of the form a/0~undefined. that is the problem here.. mithai de diye saying (-1 inclusive)"x"(1 inclusive) but you cant take 1 here
Sorry not 1....:splat: dimaag not working...I meant sin = cos i.e. 1/rt2, so that dono 45 deg 😲 my bad!!
regards
scrabbler
regards
scrabbler
"There are two trains on parallel tracks facing opposite directions. The starting and ending points of the trains form a rectangle 5m x 150m. Two men standing at the tail end of each train start moving towards each other along the line of the train at the same time when the trains start moving such that each train is travelling in the direction of the man present in the train. The speeds of the men are 18 Kmph and 36 Kmph respectively with respect to the corresponding trains in which they are moving. The speed of trains are 3 times the speed of the corresponding men. After how much time, does the distance between the two men become minimum, given that the tracks are separated by 5m?"
150 m to travel
total speed = 4(18+36) kmhr =3*5*4 =60 m/s
150/60 = 2.5s.??
TSD questions are good source of VA/RC practice.
TSD questions are good source of VA/RC practice.
A car which was driven in fog passed a man walking at 3 km/hr in the same direction. He could see the car for 4 min & upto a distance of 100 m.what was the speed of the car.
@iLoveTorres said:A car which was driven in fog passed a man walking at 3 km/hr in the same direction. He could see the car for 4 min & upto a distance of 100 m.what was the speed of the car.
4.5 kmph?
3 + 1.5
regards
scrabbler
3 + 1.5
regards
scrabbler
In 4 minutes the man would have travelled 3/15 km = 200m So total distaance traveled by car is 200+100 300 m/ 4min ;75m/min =4.5km/hr = too slow may be wrong.
@iLoveTorres said:A car which was driven in fog passed a man walking at 3 km/hr in the same direction. He could see the car for 4 min & upto a distance of 100 m.what was the speed of the car.
4.5??
@albiesriram said:In 4 minute the man would have travelled 3/15 km = 200m So total distaance traveled by car is 200+100 300 m/ 4min ;75m/min =4.5kn/hr = too slow may be wrong.
sorry for spamming....
The weather is like that...which led us to answer 4.5 
@iLoveTorres said:A car which was driven in fog passed a man walking at 3 km/hr in the same direction. He could see the car for 4 min & upto a distance of 100 m.what was the speed of the car.
4.5