Aptitude Discussion

Q. |
$A$ and $B$ start running simultaneously. $A$ runs from point $P$ to point $Q$ and $B$ from point $Q$ to point $P$. $A$'s speed is 6/5 of $B$'s speed. After crossing $B$, if $A$ takes 5/2 hr to reach $Q$, how much time does $B$ take to reach $P$ after crossing $A$? |

✖ A. |
3 hr 6 min |

✖ B. |
3 hr 16 min |

✖ C. |
3 hr 26 min |

✔ D. |
3 hr 36 min |

**Solution:**

Option(**D**) is correct

$A->......................................<-B$

\(\begin{align*} \dfrac{V_A}{V_B}=\sqrt{\left(\dfrac{t_B}{t_A}\right)}\\ \Rightarrow \left(\dfrac{6}{5}\right)^2=\dfrac{t_B}{t_A}\\ t_B=\dfrac{36}{25}\times \dfrac{5}{2}\\ \end{align*}\)

= 3.6 hour

= **3 hr 36 min**

**LINZ T MATHEW**

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**Anurag**

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can anyone please explain the solution in detail

the solution would be like this

let A and b meet at point X, distance between them is say D. so distance a has to travel is X , while distance for B to travel is D-X, therefore time taken by A should be equal to time taken by B.

which can be expressed as

X/Sa = D-X/Sb ........(1)

we know Sa/Sb=6/5......(2)

so X/D-X = 6/5 ........(3)

remaining time taken by A to travel (D-X) distance is

Ta = D-X/Sa = 5/2 ......(4)

which is given

so remaining diatance to be covered by B is X which should be covered in Tb min

Tb= X/Sb

now substitute the value of 2 and 3 in equation 4 , you would be getting the equation as

(Sa/Sb)=(Tb/Ta)^0.5