$A, B, C$ are three towns connected by straight roads from $A$ to $B$ , $B$ to $C$ and $C$ to $A$ . $A B = 5 k m, B C = 6 k m$ and $C A = 7 k m$ . Two cyclists start simultaneously form $A$ and go in different roads with same speed. They meet at $D$ . Then length of $B D$ is

1 K M

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2 K M

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3 K M

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4 K M

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Solution

The correct option is D. 4km

Since, both the cyclists have same speed, therefore, distance covered by them in particular time intervals will also be same.
Let the distance covered by both the cyclist in a particular time interval be $7 k m$ .
Therefore, one cyclist would have covered $A B$ and $2 k m$ of $B C$ as well and is now at point $E$ .
The second cyclist would have covered $A C$ and is now at point $C$ .
Therefore, distance between the two cyclists will be $C E = 6 - 2 = 4 k m$
Now, if cyclist one covers $2 k m$ then the other cyclists also cover $2 k m$ and thus meeting each other at $D$ .
Thus, distance covered by first cyclist from $B = 2 k m + 2 k m = 4 k m$
Hence, $B D = 4 k m$

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