Question

The time taken by a person to cover 150km was 2hr 60min more than the time taken in the return journey. If he returned at a speed of 10km per hour more than the speed while going. Find the speed per hour in each direction

Answer 1

Answer:

The time should be 2hr 30mins more, not 2hr 60min. Here the correct answer is below

Step-by-step explanation:

Let the speed be x and time be t1 and t2

So, Speed of departing journey =

$x = \frac{d}{t} \\ x = \frac{150}{t + 2.5} \\ t + 2.5 = \frac{150}{x} ............ \: i$

Now, Speed of return journey =

$x = \frac{d}{t} \\ x + 10= \frac{150}{t} \\ t = \frac{150}{x + 10}.............ii$

Using (ii) in (i), we get,

$\frac{150}{x + 10} + 2.5 = \frac{150}{x} \\ \frac{150}{x} - \frac{150}{x + 10} = 2.5 \\ \frac{150(x + 10) - 150(x)}{(x)(x + 10)} = 2.5 \\ \frac{150x + 1500 - 150x}{ {x}^{2} + 10x } = 2.5\\ 1500 = 2.5( {x}^{2} + 10x) \\ \frac{1500}{2.5} = {x}^{2} + 10x \\ {x}^{2} + 10x - 600 = 0 \\ using \: quadratic \: formula \: we \: get \\ {b }^{2} - 4ac \: = {10}^{2} - 4(1)( - 600) = 100 + 2400 = 2500 = d \\ x = \frac{ - b + \sqrt{d} }{2a} \: or \: \frac{ - b - \sqrt{d} }{2a} \\ x = \frac{ - (10) + \sqrt{2500} }{2(1)} \: or \: \frac{ - (10) - \sqrt{2500} }{2(1)} \\ x = \frac{ - 10 + 50}{2} \: or \: \frac{ - 10 - 50}{2} \\ x = \frac{40}{2} \: or \: \frac{ - 60}{2} \\ x = 20 \: or \: - 30 \\ x = - 30 \: is \: neglected \: because \: speed \: cannot \: be \: negative \\ so \: speed \: of \: going= 20km |hr \\ \\ speed \: of \: return = x + 10 = 20 + 10 = 30km |hr$

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