Question

A plane left 30 mins late than its scheduled time and in order to reach the destination 1500km away in time it had to increase its speed by 100kmper hr from usual speed. Fund its usual speed

Answer 1

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$\underline{\underline{\huge{\textbf{Solution:}}}}$

Given,
A plane left 30 mins late than its scheduled time

It's Destination is 1500 km away

To reach it, it increased it's speed by 100 kmph from usual speed

Usual Speed = ?

$\underline{\large{\textsf{Step-1:}}}$
Assume the Usual Speed of the train as a Variable

Let Usual Speed of the plane be x kmph

$\underline{\large{\textsf{Step-2:}}}$
Express the Increased Speed of the Plane in terms of Usual speed.

Increased Speed = (Usual Speed + 250) kmph

$\implies$ Increased Speed = (x+250) kmph

$\underline{\large{\textsf{Step-3:}}}$
Find the relationship between the time taken by plane to reach the Destination with usual speed and time taken to reach the Destination with Increased Speed.

Let the time taken by plane to reach the Destination with Usual speed be $t_{u}$

Let the time taken by plane to reach the Destination with Increased speed be $t_{i}$

Given that,
The plane is $\textsf{30 min \textbf{Late}}$ than it's scheduled time

$\implies t_{u} > t_{i}$

$\implies t_{u} - t_{i} = 30\: min$

$\underline{\large{\textsf{Step-4:}}}$
Express $t_{u} - t_{i}$ in Hours
(As speed is considered in kmph)

$\bigstar \textsf{1 Hour = 60 Minutes}$

$\implies 1\: Min = \frac{1}{60}\: Hr$

$\implies t_{u} - t_{i} = 30 \times \frac{1}{60}\: Hr$

$\implies t_{u} - t_{i} = \dfrac{1}{2}\: Hr$ -----(1)

$\underline{\large{\textsf{Step-5:}}}$
Express time taken by train to reach the Destination with usual speed $t_{u}$ and time taken by train to reach the Destination with increased speed $t_{i}$ in terms of Usual speed'x'

We have,
$\bigstar \mathbf{time\: taken = \dfrac{Distance}{Speed}}$

$\implies t_{u} = \dfrac{1500}{x}$

$\implies t_{i} = \dfrac{1500}{x+250}$

$\underline{\large{\textsf{Step-6:}}}$
Substitute $t_{u}$ and $t_{i}$ in
(1) and solve to obtain an equation in One variable.

$\implies \dfrac{1500}{x} - \dfrac{1500}{x+250} = \dfrac{1}{2}\: Hr$ -----(1)

$\implies \dfrac{1500(x+250)-1500x}{x(x+250)} = \dfrac{1}{2}\: Hr$

$\implies \dfrac{1500x+375000-1500x}{x^{2}+250x} = \dfrac{1}{2}\: Hr$

$\implies \dfrac{375000}{x^{2}+250x} = \dfrac{1}{2}\: Hr$

Cross Multiply

$\implies 2(375000) = 1(x^{2} +250x)$

$\implies 750000 = x^{2} +250x$

$\implies x^{2} +250x-750000 = 0$

$\underline{\large{\textsf{Step-7:}}}$
Factorise the equation to get Value of Variable

$\implies x^{2} +250x-750000 = 0$

$\textsf{x^{2} \times (-750000) = -75000x^{2}}$
$\textsf{-750000x^{2} = 1000x \times -750x}$
$\textsf{1000x - 750x = 250x}}$

$\implies x^{2} +1000x - 750x - 750000 = 0$

$\implies x(x +1000)- 750(x+1000)= 0$

$\implies (x-750)(x +1000)= 0$

$\implies x = 750\: kmph$

While equating (x+1000) to zero,
A negative value is obtained.

$\therefore x \neq 0$
As Speed can't be negative.

$\textsf{Usual Speed of the Plane = \underline{\textbf{750\: kmph}}}$

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Answer 2

Solution :-

Let the original speed of train be x km/hr
New speed = (x + 100) km/hr

We know that,
Time = Distance / Speed

Given : A plane left 30 minutes or 1/2 hours later than the scheduled time.

According to the question,

=> 1500/x - 1500/(x + 100) = 1/2
=> (1500x + 15000 - 1500x)/x(x + 100) = 1/2
=> 2(15000) = x(x + 100)
=> 30000 = x² + 100x
=> x² + 100x - 30000 = 0
=> x² + 600x - 500x - 30000 = 0
=> x(x + 600) - 500(x + 600) = 0
=> (x - 500) (x + 600) = 0
=> x = 500 or x = - 600

∴ x ≠ - 600 (Because speed can't be negative)


Hence,
Its usual speed = 500 km/hr