A man covers a certain distance on a toy train. Had the train moved 4 km/hr faster, it would have taken 30 minutes less. If it moved 2 km/hr slower, it would taken 20 minutes more. Find the distance.

প্রশ্নে বলা হচ্ছে, একজন লোক টয় ট্রেনে একটি নির্দষ্ট দূরত্ব অতিক্রম করে। যদি ট্রেনটি 4 মিনিট/ঘণ্টা বেশি গতিতে ছুটত তবে 30 মিনিট কম লাগত। যদি 2 কিমি/ঘণ্টা কম গতিতে চলত তবে 20 মিনিট বেশি লাগে । দূরত্ব বের করুন।

The distance be x km and initial speed by y km/hr.

Original time taken $\frac{Dis\tan ce }{Speed } = \frac{x}{y}$

According to the first condition , $\frac{x}{y} - \frac{x}{y+4} = \frac{30}{60} \because \left[30 minutes = \frac{30}{60} hours\right]$

$\Rightarrow \frac{xy + 4x -xy}{y \left(y+4\right)} =\frac{1}{2}$

$\Rightarrow \frac{4x }{y \left(y +4\right)}= \frac{1}{2}$

$\Rightarrow 8x = y^2 + 4y$

$\therefore = \frac{y^2 + 4y}{8} . . . . . . . .. . . \left(i\right)$

Again, $\frac{x}{y-2} - \frac{x}{y} = \frac{20}{60}$

$\Rightarrow \frac{xy - xy + 2x}{y \left(y -2\right)} = \frac{1}{3}$

$\Rightarrow 6x = y^2 - 2y$

$\therefore x = \frac{y^{2- 2y} }{6} . . . . . .. . . . .. . \left(ii\right)$

From equation (i) & (ii) we get have

$\frac{y^2 + 4y}{8} = \frac{y^2 - 2y}{6}$

$\Rightarrow 6y^2 + 24y = 8y^2 - 16y$

$\Rightarrow 8y^2 - 6y^2 = 24 y + 16y \Rightarrow 2y^{2 } = 40y$

∴ $y = \frac{40}{2} = 20$

Putting the value of y = 20 in equation (i)

$x = \frac{y^2 + 4y}{8} = \frac{{\left(20\right)}^2 + \left(4 \times 20\right)}{8} = \frac{480}{8} = 60 km \left(answer\right)$