A man covered a certain distance at some speed. Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. The distance (in km) is:

A. 35
B. 36
C. 37
D. 40

• Let distance = x km and usual rate = y kmph.
  X/Y-(X/Y+3)=40----I
  X/(Y-2)-X/Y=40-----II
  Then,
  On dividing (i) by (ii), we get: x = 40.
• 13 years agoHelpfull: Yes(6) No(6)
• (v+3)*(t-2/3)=vt
  
  (v-2)*(t+2/3)=vt
  
  salve equations and u get s 17.5km
• 13 years agoHelpfull: Yes(4) No(6)
• speed up by 3 km/hr time reduce by 40 min=2/3hrs...
  speed slow by 2km/hr time increse by 40 min =2/3hrs...
  both the distnace travel are same...
  
  
  let v is normal speed t time taken to cover that distance so the distance is distance=v*t;
  now according to speed up v'=v+3 and t'=t-2/3
  
  and speed slow v''=v-2 and t''=t+2/3...
  
  butv''*t''=v'*t'=v*t
  
  now given below two eqution are formed...
  (v+3)*(t-2/3)=vt
  
  (v-2)*(t+2/3)=vt
  
  salve eqn find v and t so distance =v*t;;;=17.5
  
• 13 years agoHelpfull: Yes(1) No(5)