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Maths Puzzle
There is a 50m long army platoon marching ahead. The last person in the platoon wants to give a letter to the first person leading the platoon. So while the platoon is marching he runs ahead, reaches the first person and hands over the letter to him and without stopping he runs and comes back to his original position.
In the mean time the whole platoon has moved ahead by 50m.
The question is how much distance did the last person cover in that time. Assuming that he ran the whole distance with uniform speed.
Read Solution (Total 4)
-
- It is given that the platoon and the last person moved with
uniform speed. Also, they both moved for the identical
amount of time. Hence, the ratio of the distance they
covered - while person moving forward and backword - are
equal.
Let's assume that when the last person reached the first
person, the platoon moved X meters forward.
Thus, while moving forward the last person moved (50+X)
meters whereas the platoon moved X meters.
Similarly, while moving back the last person moved [50-(50-
X)] X meters whereas the platoon moved (50-X) meters.
Now, as the ratios are equal,
(50+X)/X = X/(50-X)
(50+X)*(50-X) = X*X
Solving, X=35.355 meters
Thus, total distance covered by the last person
= (50+X) + X
= 2*X + 50
= 2*(35.355) + 50
= 120.71 meters
Note that at first glance, one might think that the total
distance covered by the last person is 100 meters, as he
ran the total lenght of the platoon (50 meters) twice.
TRUE, but that's the relative distance covered by the last
person i.e. assuming that the platoon is stationary. - 11 years agoHelpfull: Yes(7) No(0)
- Here -----------1F1L
here is the platoon the distance b/w first and last person is =50 m
let when last person meet first person platoon move x distance
so the situation is 2F1f---
when the platoon move 50 m ahead it means last person is at initial position of
first person
As last person run with uniform speed so ratio of speed first and last person remain same while last person moving forward and backward .
til the time last person handover letter to first person
first person move=x
last person move = 50 +x
when the last person come back
first person travel in that interval= 50-x
last travel during backward journey= 50-(50-x)=x
As both travel all in same time so,
x/50+x= 50-x/x
2(x square)= 2500
x square = 1250
x= 35.5
total distance travel by last man =x+50+x=2x+50=70+50=120m
- 13 years agoHelpfull: Yes(3) No(6)
- 120.71 metres
- 14 years agoHelpfull: Yes(2) No(2)
- 120 M
- 15 years agoHelpfull: Yes(0) No(3)
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