Aptitude Discussion

Q. |
Mukesh, Suresh and Dinesh travel from Delhi to Mathura to attend Janmashtmi Utsav. They have a bike which can carry only two riders at a time as per traffic rules. Bike can be driven only by Mukesh. Mathura is 300 km from Delhi. All of them walk at 15 km/h. All of them start their journey from Delhi simultaneously and are required to reach Mathura at the same time. If the speed of bike is 60km/h, then what is the shortest possible time in which all three can reach Mathura at the same time? |

✖ A. |
\(\dfrac{58}{7}\) hr |

✔ B. |
\(\dfrac{65}{7}\) hr |

✖ C. |
$10$ hr |

✖ D. |
None of these |

**Solution:**

Option(**B**) is correct

Mukesh starts from Delhi (say $A$).

He has to take off the other two (say Dinesh) on his bike, take him up to a certain point (say $C$) drop him there and return for Suresh.

Meanwhile, Suresh starts walking.

Suresh and Mukesh meet at (say $B$) Mukesh picks up Suresh at $B$ and turn towards Mathura.

All of them arrive together at Mathura (say $D$).

$A$------------ $B$ ------------- $C$ ------------- $D$

As Mukesh drives at 60km/h and Suresh (as well as Dinesh) walk at 15 km/h. So Mukesh will travel 4 times the distance $(60/15=4)$ Suresh has travelled in the same time.

Here Mukesh has travelled $=AC+CB$, while Suresh has travelled $=AB$ distance in the same time.

$\text{Distance by Mukesh} $ $= 4 \times \text{(Distance by Suresh)}$

$ \Rightarrow AC+CB=4(AB)$

$\Rightarrow (AB+BC)+CB=4(AB)$

$\Rightarrow BC+CB = 3(AB)$

$\Rightarrow BC=\dfrac{3}{2}(AB)$ -------- $(1)$

Similarly by the time Dinesh reaches point $D$ from $C$ walking, Mukesh and Suresh reach $D$ riding bike.

Here also distance travelled by bike $(=BD)$ is 4 times the distance travelled on foot $(=CD)$

$\Rightarrow CB+(BD)=4(CD)$

$\Rightarrow BC+(BC+CD)=4(CD)$

$\Rightarrow 2(BC)= 3(CD)$

$\Rightarrow CD= \dfrac{2}{3}(BC)$ -------- $(2)$

Now, it is given that total distance is given as $300 \text{ km}$

$\Rightarrow AB+BC+CD=300$

Using values from, equations $(1)$ and $(2)$,

$\Rightarrow AB+\dfrac{3}{2}(AB)+\dfrac{2}{3}(BC)=300$

$\Rightarrow AB+\dfrac{3}{2}(AB)+\dfrac{2}{3}\times \dfrac{3}{2}(AB)=300$

$\Rightarrow AB+\dfrac{3}{2}(AB)+AB=300$

$\Rightarrow \dfrac{7}{2}(AB)=300$

$\Rightarrow AB=\dfrac{600}{7}$

So, $BC$

$=\dfrac{3}{2}(AB)$

$=\dfrac{3}{2}\times \dfrac{600}{7}$

$=\dfrac{900}{7}$

Similarly, $CD$

$=\dfrac{2}{3}(BC)$

$=\dfrac{2}{3} \times \dfrac{900}{7}$

$=\dfrac{600}{7}$

Now, total time $(T)$ can be calculated using total time bike is used, since bike has been used for the whole journey.

$\Rightarrow T = \dfrac{AC+CB+BD}{60}$

$\Rightarrow T = \dfrac{(AB+BC)+CB+(BC+CD)}{60}$

$\Rightarrow T = \dfrac{AB+3BC+CD}{60}$

$\Rightarrow T = \dfrac{\frac{600}{7}+(3 \times \frac{900}{7}) +\frac{600}{7}}{60}$

$\Rightarrow T = \dfrac{600+3 \times 900 + 600}{7 \times 60}$

$\Rightarrow T = \dfrac{\bf{65}}{\bf{7}}$

**Edit:** Thank you **Chirag Goyal** for pointing out the error, and providing the alternate calculation. Solution has been updated and correct answer choice has been changed from Option C to Option B.

**Chirag Goyal**

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Thank you Chirag for bringing it to my attention. I've updated the solution.

**Chirag Goyal**

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Hi Guys, Correct Option is (B)

A------------ B ------------- C ------------- D

Here is the explanation that why Distance AB & CD will always be the same irrelevant to Speed of Bike and Pedestrian.

Proof: Let the ratio of Bike Speed to Walk Speed = x : y

Assume Mukesh along with Suresh rides with speed 'x' from Delhi at point A towards Mathura at point D

For the shortest time condition, Mukesh has to drop suresh Somewhere between A & D (say C) meanwhile Dinesh is walking with a speed 'y'.

Now time taken by Biker from A to C and back to B to pick up 'Dinesh'

is same as of time taken by Dinesh from A to B.

i.e.

(AB + BC + CB) / x = AB / y here BC = CB

solving above we got, BC / AB = (x - y) / 2y

Flashback, when Mukesh has Dropped Suresh at point C

Now Time Taken by Suresh to cover distance of CD must be same as of time taken by Mukesh to go back to pick up Dinesh at point B and to reach Mathura and meets with Suresh at point D at the same time.

i.e. CD / y = (CB + BC + CD) / x here BC = CB

so BC / CD = (x - y) / 2y

Acc. to Ques. x = 60km/hr ; y = 15km/hr

putting altogether we got 3AB = 3CD = 2BC

AD = 300 km (given)

hence

AB = BC = (600/7)km ; BC = (900/7)km

total Journey of Mukesh is AB + BC + CB + BC + CD

= (600 + 900 + 900 + 900 + 600)/7 km with speed of 60 km/hr

Shortest Time = Distance Traveled / Bike Speed = 3900/7*60 = 65/7 hr

Option (B) 65/7 hr is the Correct Answer.

There is an error, Please correct this

'AB = BC = (600/7)km'

To this

'AB = CD = (600/7)km'.

Thank you.

**Sheldon**

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Kudos, this is one great explanation. I have seen this question before someplace else, but could not understand the solution there as it was incorrectly solved. This solution appears to be correct one.

Thank you very much for the excellent job.

To $\bf{Admin\ Solution}$

I Believe you $\text{missed a key piece of solution}$

Take a look at that when $Dinesh$ is Dropped by $Mukesh$ at Point $C$

& $Mukesh$ turns towards $B$ to pick up $Suresh$ Meanwhile $Dinesh$ had covered some part of $CD$ while $Mukesh's$ Journey over $CB$ alone.

I strongly suggest you to make an improvement.

Thank You.