Walking $$\frac{3}{4}$$ of his normal speed, Rabi is 16 minutes late in reaching his office. The usual time taken by him to cover the distance between his home and office:
A. 48 min.
B. 60 min.
C. 42 min.
D. 62 min.
E. 66 min.
Answer: Option A
Solution (By Examveda Team)
1st method: $$\frac{4}{3}$$ of usual time = Usual time + 16 minutes; Hence, $$\frac{1}{3}{\text{rd}}$$ of usual time = 16 minutes; Thus, Usual time = 16 × 3 = 48 minutes. 2nd method: When speed goes down to $$\frac{3}{4}{\text{th}}$$ (i.e. 75%) time will go up to $$\frac{4}{3}{\text{rd}}$$ (or 133.33%) of the original time. Since, the extra time required is 16 minutes; it should be equated to $$\frac{1}{3}{\text{rd}}$$ of the normal time. Hence, the usual time required will be 48 minutes.This Question Belongs to Arithmetic Ability >> Speed Time And Distance
Ratio of speed wil be inversely related to the ratio of time since the distance is constant.
Speed 4 : 3 (usual : late)
Time 3 : 4 (usual: late)
4-3= 1 = 16 minutes
3 = 48 minutes.
present speed =3/4(usual speed)
present speed :usual speed=3:4
present time:usualtime=4:3
diff=1=16m then usual time=3*16=48min
d/(3/4 )s- d/s=16
D/s=48 ans
d= s×t
s=d/t
3s/4 = d/t+16. ( & in case of advance: t-16)
3/4(d/t)= d/t+16
3/4t=1/t+16
4t = 3t + 48
t= 48
4x/3-x=16
Let, total time =x minutes
So, when it is late then required time=x+16
If actual speed = d metre/min
Then reduced speed = 3d/4 metre/min
ATQ,
dx= 3d(x+16)/4
Or,dx= 3dx+48d/4
Or,4dx=3dx+48d
Or,dx=48d
Or,x=48
Ans: 48 minutes
wrong at 1st line of 1st solution
let,time T when his velocity is V and velocity 3V/4 when time (t+16)
As his passing distance is same so we can write,
V.t=3V/4(t+16)
Y we have go take 3/4 as 4/3