A man walking at 3/4th of the speed, reaches his office late by 2 hours. What is the usual time?

a. 5 hours
b. 3 hours
c. 6 hours
d. 12 hours

• speed=1/time
  3/4th speed=4/3 time
  
  take usual time x
  so (4/3)*x-x=2hrs
  x=6hrs
• 8 years agoHelpfull: Yes(14) No(1)
• at 3/4th of speed he is late by' 2hrs'
  x-3/4(x)=2
  x=8
  so 8-2=6hrs(since 2 hrs late)
• 8 years agoHelpfull: Yes(12) No(6)
• ans)6hr
  explanation:-
  speed:distance/time
  assume speed = v,distance=d,time=t
  man walking at 3/4th of the speed=3/4v
  and he reaches his office late by 2 hr=t+2
  general v=d/t t=d/v--------equation 1
  but here t+2=d/(3/4)v ------------equation 2
  
  solve the 2 equations
  t/(t+2)=3/4
  t=6hr
  
  
• 8 years agoHelpfull: Yes(10) No(2)
• S1:S2=1:(3/4)
  Since speed and time are inversely proportional. So t1:t2=3/4:1=3:4
  Now,diff in time is 2 hrs. So,
  4x-3x=2
  X=2
  So usual time is 3x=6
• 8 years agoHelpfull: Yes(3) No(0)
• let speed be x and current speed=(3/4)x
  current time =t+2
  distance will be same in both cases
  distance = speed * time
  x*t= (3/4)*x(t+2)
  on solving t=6 hours.
• 8 years agoHelpfull: Yes(2) No(0)
• 3/4 =2hr
  1/4=x
  3/4 × 2= 1/4 × x
  X= 3/4 *2*1/4
  X=6
• 8 years agoHelpfull: Yes(2) No(0)
• please send me Techmahindra model papers @ lallipaskl94@gmail.com
• 8 years agoHelpfull: Yes(0) No(7)
• Speed=d/t,so t=d/speed ,d/(3/4*speed)=d/speed +2 on solving 4/3*d/speed-d/speed=2,d/speed =6 hours
• 8 years agoHelpfull: Yes(0) No(0)
• ans c ;d=v*t=3v/4*(t+2);t=6
• 8 years agoHelpfull: Yes(0) No(0)
• ans c ;d=v*t=3v/4*(t+2);t=6
• 8 years agoHelpfull: Yes(0) No(0)
• 'x' speed->y hours
  (3/4)*'x' speed->(y+2) hours
  speed is inversely proportional to time so
  x/((3/4)*x) =(y+2)/y
  4/3=(y+2)/y
  4y=3y+6
  y=6
  
• 8 years agoHelpfull: Yes(0) No(0)
• He decrease the speed 1/3. so if 1/3=2 hours then 1=2*3=6 hours.
• 8 years agoHelpfull: Yes(0) No(0)
• Speed Time
  Original s t
  Given 3s/4 t-2
  
  here distance is same. hence,
  3s/4 * t-2 = s*t
  solving this we get t = 2.
• 8 years agoHelpfull: Yes(0) No(0)
• formula-numerator of the fraction*time/difference between numerator and denaminator
  3*2/1=6 hours
• 7 years agoHelpfull: Yes(0) No(0)
• 6 hours.
  we know v1t1=v2t2; v1=3/4(v),v2=v,t1=t+2,t2=t
  by substituting these values in above formula ,we get the answer as 6hrs.
• 6 years agoHelpfull: Yes(0) No(0)