Question

Two friends separated by a certain distance start walking towards each other. When they meet one of them has walked 20 meters more than the other. If the ratio of the distances that each has covered is 2: 3, find the distance that originally separated them.​

Answer 1

Answer:

$\boxed{\bf \: Distance\:between\:two\:friends= \: 100 \: m \: }$

Step-by-step explanation:

As it is given that, the ratio of the distances that each has covered is 2: 3.

Let assume that distance covered by first friend be 2x and other friend by 3x.

So, it means

$\sf \: Distance\:between\:two\:friends = 2x + 3x = 5x \: m$

Further given that, when they meet, one of them has walked 20 meters more than the other.

$\sf \: 3x - 2x = 20$

$\implies\sf \: x = 20$

So,

$\sf \: Distance\:between\:two\:friends$

$\sf \: = \: 5x$

$\sf \: = \: 5 \times 20$

$\sf \: = \: 100 \: m$

Hence,

$\implies\boxed{\bf \: Distance\:between\:two\:friends= \: 100 \: m \: }$

$\rule{190pt}{2pt}$

Additional Information

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$