Question

I Want Right Answer.

A plane left 30 minute late than its scheduled time and in order to reach the destination 1500 km away in time it had to increase its speed by 100 km/h from the usual speed. Find its usual speed.

Answer 1

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

Given,
The Plane left 30 minutes late than it's scheduled time

Distance to the destination is 1500 km

Plane's Speed was increased by 100 kmph from usual speed

Usual Speed = ?



$\underline{\Large{\textsf{Step-1:}}}$
Assume a Variable for Usual speed of the Plane

Let $\textsf{x kmph}$ be the usual Speed of the Plane

$\underline{\Large{\textsf{Step-2:}}}$
Note the Increased Speed of the Plane

Increased Speed of the Plane = (x+100) kmph



$\underline{\Large{\textsf{Step-3:}}}$
Note the Distance covered by the Plane

Distance covered the plane = Distance from source to Destination

Distance covered the plane = 1500 kmph



$\underline{\Large{\textsf{Step-4:}}}$
Note the time taken to reach the destination with usual speed and Increased speed.

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

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

$\implies t_{i}= \dfrac{1500}{x+100}$



$\underline{\Large{\textsf{Step-5:}}}$
Note the relation between the time taken by plane to reach the Destination with both usual speed and Increased speed

Let the time taken to reach the destination with usual speed and with increased speed be $\textbf{ t_{u} }$ and $\textbf{ t_{i} }$ respectively.

Given, the Plane is 30 min Late than it's scheduled time.

$\bigstar\: \: \boxed{\mathbf{speed \propto \: \frac{1}{time}}}$

If speed is more, time taken will be less.
If Speed is less , time taken will be more.

$\therefore$ Time taken to reach the destination with usual speed will be more than the time taken to reach the destination with increased speed.

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

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



$\underline{\Large{\textsf{Step-6:}}}$
Expressing $t_{u}-t_{i}$ in Hours

$t_{u}-t_{i}$ has to be expressed in Hours as the the Speed is in kmph.

We have,
$\bigstar\: \: \boxed{\mathbf{1\: Hour = 60\: minutes}}$

$\implies 1\: min = \dfrac{1}{60}\: Hr$

$\implies 30\: min = \dfrac{30}{60}\: Hr$

$\implies 30\: min = \dfrac{1}{2}\: Hr$

$\therefore\: t_{u}-t_{i}= \dfrac{1}{2}\: Hr$



$\underline{\Large{\textsf{Step-7:}}}$
Substitute $t_{u}$ and $t_{i}$ to obtain an equation in one variable.

$\implies \dfrac{1500}{x}-\frac{1500}{x+100} = \dfrac{1}{2}$

$\implies \dfrac{1500(x+100)-1500x}{x(x+100)}=\dfrac{1}{2}$

$\implies \dfrac{1500x+150000-1500x}{x(x+100)}=\dfrac{1}{2}$

$\implies \dfrac{1500x+150000-1500x}{x(x+100)}=\dfrac{1}{2}$

$\implies \dfrac{150000}{x(x+100)}=\dfrac{1}{2}$

Cross Multiply

$\implies 2(150000) = x(x+100)$

$\implies 300000 = x^{2}+100x$

$\implies x^{2}+100x-300000=0$



$\underline{\Large{\textsf{Step-8:}}}$
Factorise the Equation to get value of the variable

$\implies x^{2}+100x-300000=0$

$\textsf{-3000 x^{2} = 600x \times -500x}$
$\textsf{600x -500x = 100x}$

$\implies x^{2}+600x-500x-300000 = 0$

$\implies x(x+600)-500(x+600)=0$

$\implies (x+600)(x-500)=0$

$\implies x-500 = 0$

$\implies x =500\: kmph$


While equating (x+600) to zero,
a negative value is obtained.

$x \neq -650$



$\blacksquare \: \: \textsf{The Usual Speed of the Plane = \underline{\Large{\textbf{500\: kmph}}}}$

Answer 2

Let, the usual time taken by the aeroplane = x km/hr

- Distance to the destination = 1500 km

Case (i) :

Speed = Distance / Time = (1500 / x) Hrs

Case (ii) :

Time taken by the aeroplane = (x - 1/2) Hrs

Distance to the destination = 1500 km

Speed = Distance / Time

= 1500 / (x - 1/2) Hrs

Increased speed = 250 km/hr

⇒ [1500 / (x - 1/2)] - [1500 / x] = 250

⇒ 1/(2x2 - x) = 1/6

⇒ 2x2 - x = 6

⇒ (x - 2)(2x + 3) = 0

⇒ x = 2 or -3/2

Since, the time can not be negative, The usual time taken by the aeroplane = 2 hrs and the usual speed = (1500 / 2) = 750 km/hr