Aptitude Discussion

Q. |
Yana and Gupta leave points $x$ and $y$ towards $y$ and $x$ respectively simultaneously and travel in the same route. After meeting each other on the way, Yana takes 4 hours to reach her destination, while Gupta takes 9 hours to reach his destination. If the speed of Yana is 48 km/hr, what is the speed of Gupta? |

✖ A. |
72 kmph |

✖ B. |
39 mph |

✔ C. |
32 kmph |

✖ D. |
None of these |

**Solution:**

Option(**C**) is correct

Yana and Gupta travel for the same amount of time till the time they meet between $x$ and $y$.

So, the distance covered by them will be the same as the ratio of their speeds. Let the time that they have taken to meet each other be $x$ hours from the time they have started.

Therefore, the cover the entire distance, Yana would take $x+4$ hours and Gupta will take $x+9$ hours.*Ratio of time taken* Yana : Gupta :: $x+4:x+9$

*$⇒$ Ratio of speeds* of Yana : Gupta :: $x+9:x+4$ or $1:\dfrac{x+4}{x+9}$

By the time Yana and Gupta meet, Yana would have travelled $48\times x$ kms. After meeting, this is the distance that Gupta takes 9 hours to cover.

Hence, Gupta's speed \(=\dfrac{48x}{9}\) km/hr.

But we know that the ratio of Yana's and Gupta's speeds are $1:\dfrac{x+4}{x+9}$

Therefore,

\(48:\dfrac{48x}{9}::1:\dfrac{x+4}{x+9}\)

\(\dfrac{x}{9}=\dfrac{x+4}{x+9}\)

\(\Rightarrow x^2+9x=9x+36\)

\(\Rightarrow x^2=36\text{ or }x=6\) hours.

Hence, speed of Gupta

\(=\dfrac{48x}{9}\)

\(=\dfrac{48\times 6}{9}\)

= 32 kmph

**Edit:** For formula based very short alternative solution, check comment by **Chirag Goyal.**

**Ritik Mittal**

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**Vineet**

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there r 2 buses which shuttle bet a hiill stastion at the top and a town at the bottom of the hill.they both start at the same time -one from bottom and the other from the top.they both travel along the same routeand cross each othr AT A CERTAIN POINT .after they cross,8 times the time taken by the bus ging down hill is equal to the difference bet the time taken by the two buses to reah their destinations from theri meeting point.

1/ the diference bet the time taken by them to reach their respectiv destinations from the meeting point s 16hrs.how many hrs after they start do they meet?

2/ the speed of the bus from the bottom of hill to the top is 10 km/.and it takes 8 hr more than the bus from top to complete its trip. what is the distance bet the top and bottom of hill

1. Busses meet after 6hrs. of their journey

2. Distance between their journey is 120km

Good question

I got relation between their speed's

If X = speed of bus going down the hill

& Y = speed of bus going to top hill

X = 3Y

Solution is lengthy to write.

**Vineet**

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the distance between kanpur and mumbai is 750 km. Two trains simultneously leave from kanpur and mumbai.After their meeting the train trav twrds mum reaches after 9 hrs and that trav twrds kanpur after hrs.find speed of train running twrds mum

one correction... that trav twrds kanpur after 4 hrs

Speed of Train running from Kanpur to Mumbai = 50 km/hr.

Speed of Train running from Mumbai to Kanpur = 75 km/hr.

thankx chirag for the previous ques...pls do explain this one with a sol

solution please..................................

**Chirag Goyal**

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Shortcut

$\dfrac{\text{Speed of A}}{\text{Speed of B}} $$= \sqrt{\dfrac{\text{Time taken by B after meeting}}{\text{Time taken by A after Meeting}}}$

Shortcut

if 'A' is the Speed of 'Gupta' & 'B' is the Speed of 'Yana'

'a' is time taken by 'Gupta' after meeting

'b' is time taken by 'Yana' after meeting

then

$$\dfrac{A}{B}=\sqrt{\dfrac{b}{a}}$$

God one, gets the answer. But how did you arrive at this result?

You can Derive this equation

X.............................Z.....................Y

Let 'Z' is the point of meeting between 'X' & 'Y'

if 'A' is the Speed of 'Yana' & 'B' is the Speed of 'Gupta'

Then Yana & Gupta will meet at 'Z' in same Time

i.e.

$$\dfrac{XZ}{A}=\dfrac{YZ}{B}$$

$$\dfrac{XZ}{YZ}=\dfrac{B}{A}\ \ \ \ \ \ \ \ (i)$$

After Meeting, Yana will cover Distance ZY in 'a' time with speed 'A' and Gupta Will Cover Distance ZX in 'b' time with speed 'B'

i.e.

$$\dfrac{ZY}{A}=a\ \ \ \ \ \ \ (ii)$$

And

$$\dfrac{ZY}{B}=b\ \ \ \ \ \ \ (iii)$$

Solving (i), (ii) & (iii) We got

$$\dfrac{A}{B}=\sqrt{\dfrac{b}{a}}$$

Correction in the third Equation

$$\dfrac{ZY}{B}=b\ \ \ \ \ \ \ (iii)$$

Replace 'ZY' with 'ZX'

Thank You.

Sorry Friends another Correction in First Equation

$$\dfrac{XZ}{YZ}=\dfrac{B}{A}\ \ \ \ \ \ \ \ (i)$$

Replace above equation with the below one

$$\dfrac{XZ}{YZ}=\dfrac{A}{B}\ \ \ \ \ \ \ \ (i)$$

Thank You.

**Anujpachauri**

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A simple solution is:

let say point $d$ is the point where $x$ and $y$ are met.

And distance from $x$ to $d$ is $a$ and $d$ to y is $l-a$;

and then,

$a/48=l-i-a/v$; -------- (1)

Where $v$ is speed of Gupta and

and $l-a/4=48$ -------- (2)

and $a/9=v$ -------- (3)

Upon solving these equations, we get,

$v=32$

That is the final answer.

Two persons X and Y start simultaneously from P and Q for Q and P respectively. After reaching their destinations, they head back towards their respective cities. For the first time they met on their journey at 22Km from Q. Next time,i.e., on return journey they met at a distance of 10Km from city P. Assuming constant speed throughout journey, find distance PQ