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Quadratic Equations — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsQuadratic Equations4 Marks
Question
While boarding an aeroplane, a passenger got hurt. The pilot showing promptness and concern, made arrangements to hospitalise the injured and so the plane started late by $30$ minutes to reach the destination, $1500\ km$ away in time, the pilot increased the speed by $100\ km/hr$. Find the original speed/hour of the plane.
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Answer

Let the original speed of the plane be $x\ km/hr$
Increased speed of the plane $= (x + 100) km/hr$
Total Distance $= 1500\ km,$
We know that, $\text{Time}=\frac{\text{Distance}}{\text{Speed}}$
Time taken to reach the destination at original speed $=\text{t}_1=\frac{1500}{\text{x}}\text{ hr}$
Time taken to reach the destination at increasing speed $=\text{t}_2=\frac{1500}{\text{x}+100}\text{ hr}$
Acording to the question, $t_1 - t_2 = 30\ min$
$\Rightarrow\frac{1500}{\text{x}}-\frac{1500}{\text{x}+100}=\frac{30}{60}$
$\Rightarrow\frac{1500\text{(x}+100)-1500\text{x}}{\text{x}(\text{x}+100)}= \frac{1}{2}$
$\Rightarrow\frac{1500\text{x}+150000-1500\text{x}}{\text{x}^2+100\text{x}}=\frac{1}{2}$
$\Rightarrow\frac{150000}{\text{x}^2+100\text{x}}=\frac{1}{2}$
$\Rightarrow 300000 = x^2 + 100x$
$\Rightarrow x^2 + 100x - 300000 = 0$
$\Rightarrow x^2 + 600x - 500x - 300000 = 0$
$\Rightarrow x(x - 600) - 500(x + 600) = 0$
$\Rightarrow (x - 500 )( x + 600) = 0$
$\Rightarrow x - 500 = 0$ or $x = -600 = 0$
$\Rightarrow x = 500$ or $x = -600$
Since, speed cannot be negative.
Thus, the orginal speed/hour of the plane is $500\ km/hr.$

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