Question

walking at a speed of 20% more than original speed,a person requires 6 min less than normal speed.find the time required usually?​

Answer 1

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Answer 2

given the 20% increase in speed with reduced time of 6min lesser than usual find the original speed

Explanation:

1. let the original speed be 'u' km/min and let the time required while this speed be 't' min.
2. let the increased speed be 'v' km/min and the time required while this speed be 'T' min.
3. then,                              $v= u + 0.2u = 1.2u$

                                               $T= t - 6$

    4.let the distance covered be 'x' then,                                                                                      

                                     $x = ut$           ( original speed)

                                     $x = vT$           (  increased speed)

    5. from point 3. we get,        $ut = 1.2u (t - 6)$

    6. solving the above eq. we get $t = 36 min$ . hence the original time required was 36 mins.