# Singles – 2

## 17 thoughts on “Singles – 2”

1. Girish |

sr m not able to visualize why did u subtract 9…. i got the answer by traditional method though… but .. also understood they cover 3 times the length of the pool… finding it difficult to visualize 3*13 – 9

• Ratish |

When A reaches point Q, he has covered (L+9) metres and we also know that when A reaches point Q, he has covered 3*13=39 metres.
Hence, L=30
Q
|————————————————.————-|

2. Girish, Ratish has already explained it perfectly ðŸ™‚ Just to reiterate, the thing is that on the second meeting A has travelled L + 9 (since he has returned 9 m), where L is the length of the pool (this is something you would need even in the traditional method!), and hence L + 9 must be 39 giving us L = 30.

regards
J

3. Girish |

got it sir… thank you ratish….

4. Harsh Vardhan |

Sir what about if the question is Two robots M and B start from the opposite ends A and B of a linear track respectively and keep running between the ends for infinite time. They meet first time at point 60m from A. If AB=100, which is point of 4th meeting?

• For more than 2 meetings, we would best be served by actually manually doing it ðŸ™‚ As in just roughly use your fingers to track the approx position of the two.

regards
J

• seema |

answer to this question would be 20 metres from A? right or wrong?

5. Here the distance of 1st meet and the 2nd meet of two ends are mentioned:

Sir do you plan to write something on this:

Two boats start towards each other, from the two points exactly opposite of each other on the opposite banks of a river, simultaneously. They meet at a distance of 410 m from one of the banks and continue sailing further till they reach the opposite banks. They take rest for 1 hr each and start off the return journey taking the same route. Now they meet at a distance of 230 m from the same bank. Find the distance between the two banks. (Assume that river water is still.

The only difference in the details you can see, Please see if you can write and include in TSD.

• This is actually the same kind of problem, as the rest time of 1 hour is same for both….we can ignore it ðŸ™‚ I guess the answer might be something like 730 m…since 410 * 3 + 230 = 1460 will get the first guy back to his start point.

regards
J

• Scrabbler how did you get 730?
I used 410/(d-410) = (2d – 230)/ (d+230)

d= 730

• chess player |

Hi J,
Can u please explain how did we get the value 430 in this sum?(which you have computed).
would be very grateful.

• It is a typo, it should have been 410. Rectified now!

regards
J

6. How do we figure out the shortcut in this case. Long method will be ratio of speeds and the below equations:

410/Sa = d-410/Sb

2d-230 /Sa = d+230/Sb

• I just did (430 * 3 + 230) / 2. Same bank so plus. Figure out the logic for yourself….very gut feel thing.

regards
J

7. Afreen |

sir why are we multiplying 13*3? like the total distance has been covered thrice right,but then why 13*3? its just not clicking to me

• See the figure carefully and imagine them moving, see the total distance covered till the second meeting. If you cannot visualise then you have to go with equations.

regards
J