A starts from a place at 11.00 A.M. and travels at a speed of 4 kmph, B starts at 1.00 P.M. and travels with speeds of 1 kmph for 1 hour, 2 kmph for the next 1 hour, 3 kmph for the next 1 hour and so on. At what time will B catch up with A?

a) 9.24 b)9.32 c)9.48 d)none

• At 1:00 pm the dif'n between A & B = 8 km
  after 2:00 pm ................... = 11 km (as B's speed
  is 1 and A's 4km, then eqv speed=(4-1)=3km)
  After 3:00....................... = 13km (as B's speed 2km)
  After 4:00....................... = 14km
  after 5:00....................... = 14km (A's speed= B's
  speed)
  after 6:00....................... = 13km
  after 7:00....................... = 11km
  after 8:00....................... = 8km
  after 9:00....................... = 4km
  and now the eqv speed is= (9-4) =5km/hr;
  and the remaning distance is 4km;
  then, time=(60*4)/5=48 min;
  then the meeting time is=9:00+48min=9:48 pm
• 8 years agoHelpfull: Yes(4) No(0)
• ans is 9.26(approx)........................
• 11 years agoHelpfull: Yes(0) No(5)
• none, If we consider that B goes for 9 hr then it crosses 45 km in 9 hrs where as A crosses 8+36 =44 km and after this B travels with 10km/hr and A travels with a Speed of 4km/hr .
  Therefore B catch up with A within 8-9 hrs..
• 8 years agoHelpfull: Yes(0) No(0)
• A starts at 11.00 am with 4kmph speed that means he covers 8 km in 2 hrs i.e., at 1 pm.Now at 1 pm b starts,his speed increases by 1 kmph for every hour.there fore they both can catch up at speeds for 7 hrs 4+4+4+4+4+4+4=28 for A
  
  1+2+3+4+5+6+7=28 for B
  But A has a head start of 2 hrs this can be covered in the next hour but again A covers 4hrs in this 1 hr.In this way it can be worked out and the answer is 9.48hrs.
• 5 years agoHelpfull: Yes(0) No(0)