Question

A plane left 30 min. later than the schedule time and in order to reach its destination 1500km away in time in has to increase its speed by 250 km /h from its usual speeds .find usual speed ?

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Answer 1

To find :- The usual speed = ?

Answer :- 750

Solution :-

Let the usual speed of the plane be x km/hr. Then,

Time taken to cover 1500 km with the usual speed =

$\frac{1500}{x} hrs$

Time taken to cover 1500km with the speed (x+250)km/hr =

$\frac{1500}{x + 250}$

$therefore \: \frac{1500}{x} = \frac{1500}{x + 250} + \frac{1}{2} \\ \\ \: \frac{1500}{x} - \frac{1500}{x + 250} = \frac{1}{2} \\ \\ \frac{1500 + 1500 \times 250 - 1500x}{x(x + 250)} = \frac{1}{2} \\ \\ \frac{1500 \times 250}{ {x}^{2} + 250 } = \frac{1}{2} \\ \\ = > 750000 = {x}^{2} + 250x \: \\ = > {x + 250x}^{2} - 750000 \: = 0\\ = > by \: using \: middle \: term \: split \: method \\ \\ = > {x}^{2} + 1000x - 750x - 750000 = 0 \\ \\ = > x(x + 1000) - 750(x + 1000) = 0 \\ \\ = > (x + 1000) \: (x - 750) = 0 \\ x = - 1000 \: or \: x = 750 \\ \\ hence \\ \: speed \: never \: be \: negative \\ \\ therefore \\ x = 750$

Hence the actual speed of the plane is 750km/hrs

Answer 2

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore Original\:speed=750\:km/h}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• In the given question informatio given about a plane left 30 min. later than the schedule time and in order to reach its destination 1500km away in time in has to increase its speed by 250 km /h from its usual speeds.

• We have to find Original speed.

$\underline \bold{Given : } \\ \implies Time(t) = 30 \: min \\ \\ \implies Distance(d) = 1500 \: km \\ \\ \implies Increased \: speed = 250 \: km/h \\ \\ \underline \bold{To \: Find : } \\ \implies Original \: speed = ?$

• According to given question :

$\bold{Let \: original \: speed = x \: km/h} \\ \implies Original \: time = \frac{Distance}{Speed} \\ \\ \implies Original \: time = \frac{1500}{x} - - - - - (1) \\ \\ \bold{For \: Increased \: speed : } \\ \implies Original\: time - 30 \: min = \frac{distance}{original \: speed + 250} \\ \\ \bold{Putting \: value \: of \: Original \: time \: from \: (1)} \\ \implies \frac{1500}{x} - \frac{1}{2} = \frac{1500}{x + 250} \\ \\ \implies \frac{1}{2} = \frac{1500}{x} - \frac{1500}{x + 250} \\ \\ \implies \frac{1}{2} = \frac{375000}{ {x}^{2} + 250x } \\ \\ \implies {x}^{2} + 250x = 750000 \\ \\ \implies {x}^{2} + 250x - 750000 = 0 \\ \\ \implies {x}^{2} + 1000x - 750x - 750000 = 0 \\ \\ \implies x(x + 1000) - 750(x + 1000) = 0 \\ \\ \implies (x - 750)(x + 1000) = 0 \\ \\ \bold{Note : Value \: of \: speed \: never \: be \: in \: negative} \\ \bold{\implies x = 750 \: km/h}$