Should it not be 135-y+135-2y-x…..wher x being the distance c walks during the time taken by A to cover (135-y)….pls correct me if i am wrong…or do we assume c stays at starting point waiting…

Hi J,
Please help me with this problem – One day I started from A for B at 12 noon. My friend started from B for A at 2 p.m. We met on the way at 4:05 p.m. and reached our destinations at exactly the same time. What time was it?

The times till the meeting are 245 and 125 minutes. Since they reach at the same time, we can prove that the time taken after the meeting will be rt (245 * 125) = 175 minutes => the time will be 175 min past 4:05 = 7 pm.

Practice , more than anything else, Pranita. And awareness that such solutions/methods exist in the first place! When solving sums in practice, try to look for any key, any hint, any lever you can use to get a quicker solution. In time it will become second nature…

Hi J, I have a problem in understanding the question from simcat-5 that i took. The que goes like this:
Everyday, Deepak rides a bicycle on a road from point A to point B at a constant speed. Buses ply to and fro between points A and B. Everyday, Deepak, the bus going from A to B and the bus going from B to A meet at a roadside restaurant. Today Deepak started 30 mins later than his usual time. As a result he met a bus going from B to A at a distance 5x short of the restaurant 5 mins after the usual time. Similarly he meets the bus going from A to B at a distance 8x short of the restaurant 10 mins earlier than his usual time. What is the ratio of speeds of Deepak, the bus going from A to B and the bus goint from B to A?

What i dont understand is the meeting of the bus bus going from A to B and Deepak. How come they meet 8x dist short of restaurant and that too, 10 mins before the usual time when deepak started 30 mins late. Shouldn’t they be meeting at a point past the restaurant at some time later than the usual time? Please help! Thanks!

Imagine that you go for a walk every day, and your dad starts driving after you an hour later. He picks you up at some point X. If one day you start half an hour later than normal, but you dad starts at his usual time, you would not have got as far as you usually do before he catches up, right (since you have less of a lead)? So you will meet before X, and at an earlier time than normal (think from the dad’s point of view!)

HI, I had a doubt in the first case where the time taken by A is 7 hrs and 48 mins. So when I was calculating, I took the situation as that C must be walking when A is coming back after dropping B, i.e, the relative speed of A and C to be considered. Can you tell me where this approach is wrong?

You’ll have to take two parts, one while A is dropping B in which he is moving in same direction as C, and the second when he is returning, and hence in the opposite direction. I suspect you could get the answer through that but it would be rather complicated.

Should it not be 135-y+135-2y-x…..wher x being the distance c walks during the time taken by A to cover (135-y)….pls correct me if i am wrong…or do we assume c stays at starting point waiting…

No, the distance C walks IN TOTAL becomes y. The same as that walked by B. Think about the symmetry of the problem!

regards

J

Hi J,

Please help me with this problem – One day I started from A for B at 12 noon. My friend started from B for A at 2 p.m. We met on the way at 4:05 p.m. and reached our destinations at exactly the same time. What time was it?

Regards ðŸ™‚

Harsh

The times till the meeting are 245 and 125 minutes. Since they reach at the same time, we can prove that the time taken after the meeting will be rt (245 * 125) = 175 minutes => the time will be 175 min past 4:05 = 7 pm.

regards

J

rt (245 * 125) is the ratio of the speeds of my friend & I..

so now either [245 * 5/7] or [125 * 7/5] = 175 minutes!

Thanks a lot ðŸ˜€

rt (245 * 125) = 175 Could you please elaborate.

It is an easy enough proof, try it ðŸ™‚

regards

J

Sir how to identify symmetry in questions like above??

Practice , more than anything else, Pranita. And awareness that such solutions/methods exist in the first place! When solving sums in practice, try to look for any key, any hint, any lever you can use to get a quicker solution. In time it will become second nature…

regards

J

Hi J, I have a problem in understanding the question from simcat-5 that i took. The que goes like this:

Everyday, Deepak rides a bicycle on a road from point A to point B at a constant speed. Buses ply to and fro between points A and B. Everyday, Deepak, the bus going from A to B and the bus going from B to A meet at a roadside restaurant. Today Deepak started 30 mins later than his usual time. As a result he met a bus going from B to A at a distance 5x short of the restaurant 5 mins after the usual time. Similarly he meets the bus going from A to B at a distance 8x short of the restaurant 10 mins earlier than his usual time. What is the ratio of speeds of Deepak, the bus going from A to B and the bus goint from B to A?

What i dont understand is the meeting of the bus bus going from A to B and Deepak. How come they meet 8x dist short of restaurant and that too, 10 mins before the usual time when deepak started 30 mins late. Shouldn’t they be meeting at a point past the restaurant at some time later than the usual time? Please help! Thanks!

Imagine that you go for a walk every day, and your dad starts driving after you an hour later. He picks you up at some point X. If one day you start half an hour later than normal, but you dad starts at his usual time, you would not have got as far as you usually do before he catches up, right (since you have less of a lead)? So you will meet before X, and at an earlier time than normal (think from the dad’s point of view!)

regards

J

Can we have a shortcut as adding the ratios for no. of distinct points.

sorry posted in wrong thread

HI, I had a doubt in the first case where the time taken by A is 7 hrs and 48 mins. So when I was calculating, I took the situation as that C must be walking when A is coming back after dropping B, i.e, the relative speed of A and C to be considered. Can you tell me where this approach is wrong?

You’ll have to take two parts, one while A is dropping B in which he is moving in same direction as C, and the second when he is returning, and hence in the opposite direction. I suspect you could get the answer through that but it would be rather complicated.

regards

J