How can two friends with different walking and cycling speeds cover 20 miles sharing just one bicycle between them and reach the destination at same time?
Two friends Arun and Bijen are to travel a distance of 20 miles starting at the same point at the same time. They have only one bicycle to share between them. They have agreed that one who rides the bicycle would drop it on the way and walk on while the other walking behind would reach the bicycle and starts riding it.
None of them would waste even a second on the way. Arun walks at the speed of 4 miles/hr but rides at the speed of 10 miles/hr, while Bijen walks at the speed of 5 miles/hr and rides at the speed of 8 miles/hr.
Can you devise an ingenious scheme by which the two friends would reach the 20 mile point at the same time?
Recommended time to solve: 20 minutes.
Hint: Think simply.
Solution to the Sharing a bicycle puzzle
True. We'll move directly to the heart of the puzzle and explore the relations between the travel times of the two friends. Time is the most important parameter in the puzzle because it's stated that both will take the same time to cover 20 miles.
This fact determines values of other parameters.
To derive the time equation, let's assume in the simplest scenario, Arun travels distance X walking taking time TAW, and rest of the distance Y cycling taking time TAR. For Bijen time to cover distance X cycling be TBR and time to cover distance Y walking be TBW.
The following shows the schematic.
Arun starts walking first with Bijen starting on the cycle simultaneously. At a point distant X miles from the starting point, Bijen drops the cycle and starts walking the rest distance Y to the finishing point. Both reaches the destination at the same time.
As time T to travel a distance D equals D divided by Speed S, the times in hours would be,
TAW = X/4
TAR = Y/10
TBR = X/8, and,
TBW = Y/5.
But, TAW + TAR = X/4 + Y/10 = TBR + TBW = X/8 + Y/5,
Or, (X/4 - Y/5) = (X/8 - Y/10) = 1/2(X/4 - Y/5).
Math then ascertains that,
(X/4 - Y/5)=0,
Or, X/4 = Y/5,
Or, X/Y = 4/5.
Add 1 to both sides,
(X + Y)/Y = 9/5,
Or, 20/Y = 9/5.
So finally a breakthrough result,
Y= 100/9 miles, and,
X = 80/9 miles.
Success so soon! If Arun walks first with Bijen riding the cycle, at a distance of 80/9 miles Bijen would leave the Cycle behind and start walking. Reaching this exchange point Arun would use the cycle to cover the rest 100/9 miles and reach the end point simultaneously with Bijen.
This is the first solution that has been forced by the conditions that the two friends would share a single cycle between them and they would reach the end point at the same time.
But in addition to these necessary conditions, the critical element has been the ratio 4/5 of both pair of speeds.
Because of this equality of the ratio only we could derive the breakthrough relation, X/Y = 4/5 and then determine the value of X and Y.
Question is: Can there be any solution other than this solution with a single cycle exchange point?
As an answer we would tell you to divide X into four equal parts each 20/9 miles long and divide Y into five equal parts each also 20/9 miles long. Effectively, you would divide then the whole 20 miles into 9 equal parts.
Now with Arun first riding the segment 1, he will again ride on alternate segment 3, 5, 7 and 9, a total of 5 segments of total length 100/9 miles and walk the rest 80/9 miles in four segments.
When Arun rides Bijen walks.
So in this scheme Bijen will walk 100/9 miles and will ride 80/9 miles.
Portion that Arun and Bijen walk as well as ride are exactly same as before, only difference being Arun starts riding first as if reversing the first solution from the end point.
This equivalence of the new scheme with the first successful scheme would ascertain that again both friends would reach destination at the same time.
Following shows the schematic for this second solution.
Again we would raise the same question: can there be any more solution?
In fact there would be infinite more solutions.
Just divide each 20/9 miles long segment into nine other segments and you would get the third solution.
And so on.
But that is theoretical. Practically the third solution onward would have segment length too small to be called feasible ones.
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