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Speed Time and Distance
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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:

Answer & Solution
Correct Answer: Option A
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.
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9 Comments
Abhishek Mohapatra
Abhishek Mohapatra 1 year ago
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.

Temp Ml
Temp Ml 2 years ago
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
Zaffar Bhat
Zaffar Bhat 2 years ago
d/(3/4 )s- d/s=16
D/s=48 ans
Bomri Bagra
Bomri Bagra 4 years ago
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
Nazmul Hossain
Nazmul Hossain 7 years ago
4x/3-x=16
Taibur Rahman
Taibur Rahman 7 years ago
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
Rj Al-Amin
Rj Al-Amin 7 years ago
wrong at 1st line of 1st solution
Mishu Dhar
Mishu Dhar 8 years ago
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)
Veenark73 Kshirasagar
Veenark73 Kshirasagar 8 years ago
Y we have go take 3/4 as 4/3