# 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)__

**1**

^{st}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.

**2**

^{nd}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 ( 5 )

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

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