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just olya [345]
4 years ago
13

A man is on a 1/4 on a bridge. A train is coming the same direction he is going. The man can run across the bridge in the same d

irection and make in barely in time. He can also run backwards towards the train and also barely make it. How fast is the train going relative to the man.

Mathematics
1 answer:
Anon25 [30]4 years ago
4 0

Answer:

The train is twice fast going relative to the man.

Step-by-step explanation:

Consider the provided information.

Let the distance between the train and beginning of bridge is x and length of bridge is y.

A man is on a 1/4 on a bridge. Thus, the 1/4 of y is y/4.

The train is going x distance in time man runs y/4 distance.

Also if the train is going x + y in time man runs the distance 3y/4.

For better understanding refer the figure 1:

So, if train goes x+y-x distance in time man covers the distance 3y/4 - y/4

Now solve 3y/4 - y/4  = 2y/4

The train covers y distance in the time man runs 2y/4 = y/2

That means train covers 2 times of the distance cover by the man or the train goes twice as fast as man.

Hence, train is twice fast going relative to the man.

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\textit{area of a triangle}=\cfrac{1}{2}\cdot b\cdot h\qquad 
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<h3>Given</h3>
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<h3>Find</h3>
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<h3>Solution</h3>

For each of the y rows of squares, there are x segments at the top, plus another x segments at the bottom. The total number of horizontal segments is then

... horizontal segment count = (y +1)x

Likewise, for each of the x columns of squares, there are y segments to the left, plus another y segments to the right of the entire area. Then the total number of vertical segments is

... vertical segment count = (x+1)y

The total segment count is ...

... total segments = horizontal segments + vertical segments

.. = (y+1)x +(x+1)y

... total segments = 2xy +x +y

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<u>Check</u>

We know a square (1×1) has 4 segments surrounding it.

... count = 2·1·1 +1 +1 = 4 . . . . (correct)

We know the 3×3 window in the problem statement has 24 segments.

... count = 2·3·3 +3 +3 = 18 +3 + 3 = 24 . . . . (correct)

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3 years ago
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oksano4ka [1.4K]

Answer:

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