2 men at same tym start walking towards each other from A n B 72 kms apart. sp of A is 4kmph. Sp of B is 2 kmph in 1st hr,2.5 in 2nd, 3 in rd. n so on… when will they meet
i in 7 hrs ii at 35 kms from A iii in 10 hrs iv midway

• After the first hour, they are
  4+2=6 km closer to each other. After the second hour, they are
  4+2.5=6.5 km closer to each other, and so on. The reduction of distance per hour is therefore an arithmetic progression:
  6, 6.5, 7, ...
  with initial value a=6 and common difference d=0.5.
  
  The arithmetic series (adding these distances) can be expressed in terms of hours, h, as
  S(h) = ah + dh(h-1)/2
  = 6h + 0.5(h^2-h)/2
  = 6h + (h^2-h)/4
  and they meet when this sum equals 72:
  6h + (h^2-h)/4 = 72
  24h + h^2 - h = 288
  h^2 + 23h - 288 = 0
  (h - 9) (h + 32) = 0
  
  We have no interest in a negative solution, so the positive one h=9 is the one we want. They meet after 9 hours.
  
  That doesn't match either of the answers that are times, so let's check the distance from A, which at a steady 4 km/h is clearly 36km, half the total distance.
  
  They meet midway (answer iv).
• 12 years agoHelpfull: Yes(34) No(4)
• Yup.. answer is (iv).
  Distance travelled together in each hour is 6+6.5+7+...
  this must be equal to 72.
  solving this we get total time travelled is 9 hrs.
  that is 'A' has travelled 9*4 = 36.
  so they meet halfway.
• 8 years agoHelpfull: Yes(1) No(1)
• In one hour two men complete 4+2=6km/h
  and in 2nd h 4+2.5=6.5 , in 3rd h-->7 and so on
  obtained series is in AP so
  by s=a+d(n-1)/2
  sh=ah+dh(h-1)/2
  here a=6 and d=0.5=1/2
  so
  6h+0.5h(h-1)/2=72
  h^2-23h-288=0
  by simplification h=9.
  
  a can reach 36km that is halfway in 9 hours.
• 7 years agoHelpfull: Yes(0) No(1)