# Logic puzzle about travel times and distances

Each of the four person – Pawan, Rahul , SGR, Ankit –travelled at a different uniform speed out of 10 km/hr, 15 km/hr,20km/hr, and 25 km/hr for different time durations- 30 hours, 40 hours, 50 hours and 60 hours – not necessary in that order. It is also known that-

1. Rahul travelled for a shorter distance than Ankit and a longer Distance than SGR, who travelled a longer distance than pawan.

2. The person who travelled the longest distance neither travelled at the highest speed nor travelled for the longest duration.

3. The person who travelled a distance of 750 km is not the one who travelled the longest distance.

4. No two persons travelled the same distance.

What is the name of the Person who travelled for the shortest period?

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I think the (recreational-mathematics) and (homework) tags are inherently contradictory. Which of them is right? What have you tried? –  Henning Makholm Sep 13 '12 at 13:08

Set up a $4\times 4$ table of the possible distances that may have been covered: $$\qquad\ \ \ 10\qquad \ \ \ 15 \qquad\ \ 20 \qquad\ \ \ 25$$ $$30 \qquad 300\qquad 450 \qquad\ 600 \qquad\ 750$$ $$40 \qquad 400\qquad 600 \qquad\ 800 \qquad 1000$$ $$50 \qquad 500\qquad 750 \qquad 1000 \qquad 1250$$ $$60 \qquad 600\qquad 900 \qquad 1200 \qquad 1500$$ The upper row gives the possible speends, and the left column gives the possible times. The 16 products are the possible distances, and the 4 distances that have actually been covered must lie in different rows and in different columns. Now start to eliminate.

By (2), the longest distance is not in the right column or in the bottom row, so it can be at most $1000$. By (2) and (4) we can eliminate three entries in the last column, as well as the entry $1200=20\cdot 60$ in the bottom row. This leaves only one possible time for the person with speed $25$, namely $30$ hours with covered distance $25\cdot 30=750$.

Eliminate the other three entries in the row of time $30$, as well as the entry $15\cdot 50=750$. We are left with the table: $$\qquad\ \ \ 10\qquad \ \ \ 15 \qquad\ \ 20 \qquad\ \ \ 25$$ $$30 \qquad - \qquad\ - \qquad\ -\ \qquad\ 750$$ $$40 \qquad 400\qquad 600 \qquad\ 800 \qquad -$$ $$50 \qquad 500\qquad -\qquad 1000 \qquad\ -$$ $$60 \qquad 600\qquad 900 \qquad\ -\ \qquad\ -$$ Now show that it is not possible to find a solution which contains the distance $20\cdot 40=800$.

It follows that the person with speed $20$ must have travelled for time $50$. The rest is now easy, and rule (1) then gives you the names to go with the distances.

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Thanks for the explanation. –  user221287 Sep 13 '12 at 16:03

1 will be last: you need to match the speeds and times, then this will let you put the names to them.

From 3, 750km can come from 50*15 or from 30*25. If it is 50*15, the 60 will have to go with 10 to not be the longest (clue 2). Then 30 can't go with 25 (clue 4) so it is 30*20=600, which duplicates 60*10-contradiction. So 750=30*25.

I'll let you finish.

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Got the complete solution by trying, Thanks. –  user221287 Sep 13 '12 at 13:33