Examveda
Examveda

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

Join The Discussion

Comments ( 6 )

  1. Bomri Bagra
    Bomri Bagra :
    2 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

  2. Nazmul Hossain
    Nazmul Hossain :
    4 years ago

    4x/3-x=16

  3. Taibur Rahman
    Taibur Rahman :
    5 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

  4. Rj Al-Amin
    Rj Al-Amin :
    5 years ago

    wrong at 1st line of 1st solution

  5. Mishu Dhar
    Mishu Dhar :
    6 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)

  6. Veenark73 Kshirasagar
    Veenark73 Kshirasagar :
    6 years ago

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

Related Questions on Speed Time and Distance