in Quantitative Aptitude edited by
227 views
1 vote
1 vote

A motorbike leaves point $\text{A}$ at $1$ pm and moves towards point $\text{B}$ at a uniform speed. A car leaves point $\text{B}$ at $2$ pm and moves towards point $\text{A}$ at a uniform speed which is double that of the motorbike. They meet at $3:40$ pm at a point which is $168$ km away from $\text{A}$. What is the distance, in km, between $\text{A}$ and $\text{B}?$

  1. $364$
  2. $378$
  3. $380$
  4. $388$
in Quantitative Aptitude edited by
13.4k points
227 views

1 Answer

1 vote
1 vote

Let the distance between $A$ and $B$ be $D$ km, and the speed of the motorbike be $x$ km /hour. Then the speed of the car is $2x$ km/hour.



Let's say they met at point $C$.



We know that,  $\text{Speed} = \dfrac{\text{Distance} }{\text{Time}}$

For motorbike:

$x = \dfrac{168}{\frac{8}{3}} \quad \left[\because 2 \;\text{hours}\; 40 \;\text{minutes} = 2 + \frac{40}{60} = 2 + \frac{2}{3} = \frac{8}{3} \;\text{hours}\right]$

$\Rightarrow \boxed{x = 63}$ km/hours

For car:

$2x = \dfrac{D-168}{\frac{5}{3}}$

$\Rightarrow2(63) = \dfrac{3(D-168)}{5} \quad \left[\because 1\;\text{hours}\; 40 \;\text{minutes} = 1 + \frac{40}{60} = 1 + \frac{2}{3} = \frac{5}{3}\;\text{hours} \right]$

$\Rightarrow 42\times5 = D-168$

$\Rightarrow 168 + 210 = D$

$\Rightarrow \boxed{D = 378\;\text{km}}$

$\therefore$ The distance between $A$ and $B$ is $378\;\text{km}.$

Correct Answer $:\text{B}$

edited by
10.3k points
Answer:

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true