FUN SHUNN

From: rency raghavan <rencykr@yahoo.co.uk>
Subject: Re: //Dil Se Desi// BS4U - 59
To: "KARAMJEET ARNEJA" <aap_ka_karam@yahoo.com>, "Raja" <tumhara.raja@yahoo.com>
Cc: dilsedesigroup@yahoogroups.com
Date: Saturday, 3 July, 2010, 11:31 AM

 

Another approach that works to the correct solution

The car makes three forward journey and two return journey. 

A little thought would show that the three forward journeys must be equal and let it be x

similarly both the return journey are of equal distance, be it y

So we have 3x - 2y = 20.......... ..eqn 1

The distance walked by each family is 20-x

lets take the third family who got in the car at last.

they were walking while the car made two forward and two return trips. 

considering their walking speed which is 1/5 th of the car, the distance walked is (2x+2y)/5

Equating the distance walked, we have (2x +2y)/5 = 20-x

ie 7x+2y =100........ .. eqn 2

adding 1 and 2 

10x = 120

or x=12 mile

So each family traveled in car for 12 miles and walked for 8 miles which makes their total travel time 2 hrs and 36 min.

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From: KARAMJEET ARNEJA <aap_ka_karam@ yahoo.com>
To: Raja <tumhara.raja@ yahoo.com>
Cc: dilsedesigroup@ yahoogroups. com
Sent: Sat, 3 July, 2010 0:31:11
Subject: Re: //Dil Se Desi// BS4U - 59

Wow! wow!! wow!!!

 

What an original ingenious approach. A little long but a 'you-simply- can't-go- wrong' one.

 

And your answer (naturally then) is absolutely correct.

 

I had it solved by 2 different approaches based eqauating time for three different journeys but both of them (short though) were way more complex than yours which uses the very basic concepts that every one can understand. And your explanation is simply flawless and adequately detailed. Sum it up in two words? "very impressive". Can't say it enough. So I won't. :)

 

I would have loved to see inputs from Rency and Shashi but looks like thay are enjoying a little summer time off too.

 

Alright then, let me see how soon can I get around to the 60th serving of BS4U.

 

Regards

Karamjeet

 

 

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From: Raja <tumhara.raja@ yahoo.com>
To: KARAMJEET ARNEJA <aap_ka_karam@ yahoo.com>
Cc: dilsedesigroup@ yahoogroups. com
Sent: Thu, July 1, 2010 1:15:08 PM
Subject: Re: //Dil Se Desi// BS4U - 59

 

Dear Karam, It felt good to see your BS4U after a long gap. Hope your assignment is successfully complete. Here is my solution to BS4U 59. It is really fantastically interesting puzzle, I really enjoyed it. Let car pick up R family first from farmhouse 'A', and leaves R at the point D. Then the car returns from D to point B where during this period M & S walked up to. From B car picked up another family say M and left at point E where R also reached walking (since they have to walk rest of the distance together) From E car returned to pick up family S who have walked from B to point  C. The car  picked up family S and took to destination F where during this period families R & M has reached walking. So, we get starting point A, then B,C,D,E, on the route and final destination F. Let AB = a miles, BC = b miles, CD = c miles, DE = d miles, and finally EF = e miles. MS walked 'a' miles, from A to B, time taken = a/4 hour (since walking speed is 4mph) This time is equal to time taken by car picking up R from A and leaving at D + returning empty car from D to Point B to pick up family M from there.= (AB+BC+CD)/20 + (DC+CB)/20 So,  a/4 = (a+b+c)/20 + (c+b)/20 = (a+2b+2c)/20 or     2a = b+c......... ......... ......... ......... ......... ......... ......... ......... ...[1] Similarly, while S walked from B to C (b miles), taking time b/4 hour, car picked up M from B and leaving them at point E + empty car returned from E to C to pick up family S. So, b/4 = (BC+CD+DE)/20 + (ED+DC)/20 or   b/4 = (b+c+d)/20 + (d+c)/20 = (b+2c+2d)/20 or   2b = c+d......... ......... ......... ......... ......... ......... ......... ......... .....[2] Also, when car left R at D and moved from D to B picked up M from B and left M at E where during this period R has reached. Therefore, R walked distance DC = d miles in time d/4 hour, while empty car moved back from D to B to pick up M + taking M from B to E. So, d/4 = (DC+CB)/20 + (BC+CD+DE)/20 = {(c+b) + (b+c+d)}/20 or ,  5d =  2b+2c+d or ,  2d = b+c......... ......... ......... ......... ......... ......... ......... ......... ....[3] Similarly, families R&M walked from E to F, time taken e/4. This time is equal to the time taken by empty car moving back from E to C, picking up family S  from there and moving with S from C to F. Therefore, e/4 = (ED+DC)/20 + (CD+DE+EF)/20 or, e/4 = {(d+c)+(c+d+ e)/20 = (2c+2d+e)/20 or,  2e = c+d......... ......... ......... ......... ......... ......... ......... ......... .....[4] From [1], [2], [3], & [4], a = b = c = d = e........... ......... ......... ......... ......... ......... ......... .....[5] Which mean a+b+c+d+e = 5a = 20miles (it is given in the puzzle). So a = 4 miles....... ......... ......... ......... ......... ......... ......... ......... ...[6] Then total time taken by the three families to reach F from A will be time taken by M & S from A to B + time taken by M from B to C + time taken by S from C to F. Therefore, Total time = a/4 + a/4 + (a+a+a)/20, put a = 4 from [6], Then, total time = 2hours 36 minutes. As they started from A at 12 Noon, therefore they reached F at 2:36 PM.   Regards, Raja --- On Tue, 29/6/10, KARAMJEET ARNEJA <aap_ka_karam@ yahoo.com> wrote: From: KARAMJEET ARNEJA <aap_ka_karam@ yahoo.com> Subject: //Dil Se Desi// BS4U - 59 To: dilsedesigroup@ yahoogroups. com Cc: "raj lohar" <tumhara.raja@ yahoo.com>, "rency raghavan" <rencykr@yahoo. co.uk>, "Shashikant Daptardar" <daptardar_sd@ yahoo.com> Date: Tuesday, 29 June, 2010, 8:39 PM Three families – Rohit's, Mohit's and Sohit's went on a short fun trip to a farmhouse in a big rented van that could carry all the 12 members (4 for each family). Unfortunately the van would not start on their way back. The farmhouse was located near a less frequented road and the location was not cell-phone friendly either – you know - no signal strength.  They had to reach back to the city limit (20 miles from the farmhouse) where they had their individual cars parked. Fortunately a small car stopped by and the driver volunteered to help them. Given the size of the car it could do 20 mph in that zone and carry only four persons at a time. The driver proposed that he could make three trips and drop one by one but all of them expressed desire to reach at the same time even if it meant walking all those 20 miles at the rate of 4 mph. Rohan (Rohit's wonder kid) suggested that using an appropriate combination of walking at 4mph and riding at 20mph, all the three families can reach the destination at the same time.   So the driver picked up one family while others continued walking. The driver dropped the first family at a distance pre-calculated by Rohan (to complete the rest of the journey walking) and drove back to pick up the second family (while the third family kept walking) and dropped them at again Rohan's directed distance (from where the second family would walk to the destination) and drove back and picked up the third family and drove them all the way to the destination and lo and behold as speculated by Rohan, all the three families did reach at the same time.   So friends, if the three families and started at 12 noon, what exact time did they make it to the destination?   Good luck and have fun.   Regards Karamjeet