If i decrease my speed by 20% of my original speed. I reach my office 7 min. late. What is my usual time taken to reach the office?

• Its 28seconds.
  we know that distance=speed*time.
  here distance to office is constant, then speed will be inversely proportional to time.let actual speed be s, then speed at now will be (4s)/5 i.e., s-(20% of s).and if t is the actual time taken, then time taken now will be (t+7).
  now product of speed and time is constant, that implies
  s*t=(4s/5)*(t+7)
  5t=4t+28
  t=28
• 12 years agoHelpfull: Yes(21) No(1)
• Here let original speed be x. (1)
  speed decreased by 20% mean 0.8x. (2)
  time taken b t (3)
  and after speed decreased, time= t-7min. (4)
  Using 1-2 = 3-4, will get 0.2x= 7min.
  so x=(7min/0.2)
  =35min
• 12 years agoHelpfull: Yes(3) No(7)
• vt=.8v(t-7)
  t=28
  
• 12 years agoHelpfull: Yes(3) No(0)
• d=s*t1
  d=(s-(20/100)s)*t2=0.8t2
  s*t1=0.8s*t2
  t1=0.8t2
  since t2-t1=7min
  sub t1, t2-0.8t2=7min
  t2=7min/0.2=70/2min=35min
  t1=35min-7min=28min
• 12 years agoHelpfull: Yes(2) No(0)
• if a man walks with (x/y) of his usual speed and he takes 't' hours to cover a certain distance then the time to cover the same distance with usual speed
  
  =xt/(y-x)here x=1,y=5 and t=7mins
  =7/4
  =1.75 mins
  hence usual time he takes to reach office is 1.75 mins.. pls confirm if this was an option...!!!
  
• 12 years agoHelpfull: Yes(0) No(12)
• sorry.. there's a correction in the explanation that-
  he takes 't' hours more time to cover a distance.....
  
  answer is otherwise correct to my expectations..!!!
• 12 years agoHelpfull: Yes(0) No(0)