This year's Veterans Day parade featured a novelty limousine that sadly was filled with only a few remaining veterans. Two kids were trying to estimate the length of this cool limo. But the limo was moving, albeit steadily. One of the boys decided to walk alongside the moving limo from its back to the front in parade's direction. He made 30 steps. Then he turned back and walked alongside the limo in the opposite direction. He made 15 steps. Assuming that each step is 2 feet, can we figure out limo's length?
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7 comments:
d, the distance the limo traveled while the boy walked
L, the length of the limo
60 feet – d = L
30’ + d = L
60 – d = 30 +d
60’-30’ = 2d = 30’
d=15’
L=45 feet
Gads, my algebra is terribly rusty!
One additional assumption: the boy's pace was also very steady.
With that, we introduce three variables - limo's length (X), limo's speed (L) and boy's speed (B). The speeds would be in feet per second. Then we have these equations:
X/(L+B)=30
X/(B-L)=60
==>
X=30*L+30*B=60*B-60*L
90*L=30*B
L=B/3
X=40*B
What this tells us is that we need to know either the speed of a limo or the speed of the boy in order to know the length. Let's say the the boy had a watch or a good internal sense of time, and he timed his steps precisely one step per second, making his speed 2 ft per second. With that, we get an 80 ft limo. If they boy walked a bit faster, he would be looking at the longest limo in the world! http://www.limohead.co.uk/top-10-longest-super-stretch-limousines-in-the-world-ever/
Let l be the length of the limo.
We assume that the boy walks at constant pace (in both directions) and the limo drives at a slow and constant pace also (and it must be slower than the boy's pace or else he would never catch front of the limo). We do not need to know how fast the boy walks nor how slowly the limo drives.
At t=0, the boy is at the back end of the limo. He and the limo proceed forward at their respective rates until the boy reaches the front.
Let t=t1 represent the time when the boy is at the front of the limo and turns around.
He has covered 2 ft/step * 30 steps = 60 feet.
Let d1 represent the distance the limo travels in this time period (which is the distance its rear end travels in this time period).
d1 + l = 60 feet.
The limo continues moving forward at its slow pace and the boy now walks in the opposite direction at the same pace as before until he reaches the back end of the limo.
Let t=t2 represent the time when the boy reaches the back end of the limo.
He has covered 2 ft/step * 15 steps = 30 feet.
Let d2 represent the distance the limo travels in this time interval.
Since rate x time = distance, and the boy covered 60 feet in the first interval from t = 0 to t = t1, and only 30 feet in the second interval from t = t1 to t = t2, the first time interval is twice as long as the second, i.e. (t1 - 0) = 2*(t2 - t1). So the limo must also cover twice the distance in the first time interval as the second time interval, i.e. d1 = 2*d2.
The boy and the rear of the limo are at the same point at t = 0 and at t = t2, so d1 + d2 = 60ft - 30 ft.
Combining this we get
2*d2 + d2 = 30ft
3*d2 = 30ft
d2 = 10ft
d1 = 20ft
l = 60ft - d1 = 60ft - 20ft = 40 feet
@Maria: Congratualations on not only your little one but on creating puzzles that are not the easiest thing in the world to solve. It took me awhile to "see" this one; I'm not sure I got it yet.
Let the length of the care = x in steps.
Let the movement towards the boy of the car = y in steps.
15 + y = x
Let the length of the car (in steps) be x
Let the amount the car moves in the same direction as the boy = y
30 - y = x
Add the two conditions together to get
2x = 45
x = 22.5
So the car's length = 22.5 steps or 45 feet.
I sure await the answer to this one: I'm not at all sure it is right.
Well, you all were very close and kj got it absolutely right. 40 feet.
assume that x is the length of the limo in feet
y is how much limo moves (in feet) when the boy does 1 step
then when boy moves in the direction of limo's motion:
x + 30y = 30 *2 = 60
when boys moves in the opposite direction:
x = 15 *2 + 15y = 30 + 15y
substitute 30 + 15y instead of x in the first equation:
30 + 15y + 30y = 60
45y = 30
y = 2/3
x = 30 + 15y = 30 + 10 = 40 feet
Puzzle point for everyone who dared to tackle this one.
This is a remarkable puzzle. No matter how you solve this (I now have my own method), at some point you have to make a guess or rather an unusual observation that is really not based in standard mathematics. The key to it is to realize that either the time is twice what it is one way when compared to the other, or the distance covered by the limo is twice going the same direction as the boy as the limo's distance when the boy goes in the opposite direction.
I don't think I've ever seen a problem quite like this.
dang...dang...dang! need to seriously look at that limosine company, driver policy and practice...seem like the
victims could not opened the door from the inside...driver may have had the children safety aka drunk passengers
lock-on...mmmm, so only the skinny people got out??? dry them tears...time to sue...driver not shaken playin stupid...
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