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.Join The Discussion
Comments ( 6 )
Related Questions on Speed Time and Distance
A. 48 min.
B. 60 min.
C. 42 min.
D. 62 min.
E. 66 min.
A. 262.4 km
B. 260 km
C. 283.33 km
D. 275 km
E. None of these
A. 4 hours
B. 4 hours 30 min.
C. 4 hours 45 min.
D. 5 hours
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