Question

A man walked 12 km at a certain rate and then 6 km further at a rate of 12 km/hr faster. If he
had walked the whole distance at the faster rate, his time would have been 20 minutes less.
How long did it really take him to walk the 18 km?​

Answer 1

Solution :-

Let us assume that, initial speed of man is x km/h.

Case 1) :- walked 12km at x km/h and next 6 km at (x + 12)km/h .

So,

→ Total time = (Distance/speed) = [(12/x) + 6/(x + 12)] hours.

Case 2) :- walked whole distance of 18km at the rate (x + 12) km/h.

So,

→ Total time = (Distance/speed) = 18/(x + 12)] hours.

Given that, he save 20 minutes in case (2) than case(1).

A/q,

→ [(12/x) + 6/(x + 12)] - 18/(x + 12) = 20 minutes

→ [{12(x + 12) + 6x} /x(x+12)] - 18/(x + 12) = 20/60 hours.

→ [(12x + 144 + 6x)/x(x+12)] - 18/(x + 12) = 1/3

→ [(18x + 144)/x(x+12)] - 18/(x + 12) = 1/3

→ (18x + 144 - 18x) / (x² + 12x) = 1/3

→ 144 * 3 = x² + 12x

→ x² + 12x - 432 = 0

→ x = [-12 ± √{(12)² - 4*(-432)}] / 2 = [-12 ± √(144 + 1728)]/2

→ x = [-12 ± √(1872]/2 = [-12 ± 12√13]/2

→ x = [-12 + 12√13]/2 or, [-12 - 12√13]/2

since speed can't be in negative.

Therefore,

→ x = [-12 + 12√13]/2 = 12(√13 - 1)/2 = 6(√13 - 1)

→ x ≈ 15.6 km/h.

Hence,

→ Actual time by man = [(12/x) + 6/(x + 12)] = (12/15.6) + (6/27.6) = 0.98 ≈ 1 Hour.(Approx). (Ans.)

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