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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:

A. 48 min.

B. 60 min.

C. 42 min.

D. 62 min.

E. 66 min.

### 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.

1. 4x/3-x=16

2. 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

3. wrong at 1st line of 1st solution

4. 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)

5. Y we have go take 3/4 as 4/3

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