After traveling 108 km, a cyclist observed that he would have required 3 hours less if he could have traveled at a speed 3 km/hr more. At what speed did he travel?

প্রশ্নে বলা হচ্ছে যে, Mr. Sakil প্রতিদিন একটি নির্দিষ্ট সময়ে অফিসের উদ্দেশ্যে বাসা থেকে বের হন। তিনি যদি ঘণ্টায় 5 কিলোমিটার বেগে হাঁটেন, তবে তাঁর 7 মিনিট দেরি হয়। অন্যদিকে ঘণ্টায় 6 কিলোমিটার বেগে হাঁটলে তিনি 5 মিনিট আগেই অফিসে পৌঁছেন। আপনাকে অফিস থেকে বাসার দূরত্ব বের করতে হবে। 

Let. distance of office from the house be x km.

According to question,

$\frac{x}{5}-\frac{7}{60}=\frac{x}{6}+\frac{5}{60}$ [Minutes is converted by hour]

$$\begin{gathered} \Rightarrow \frac{x}{5}-\frac{x}{6}=\frac{5}{60}+\frac{7}{60} \\ \Rightarrow \frac{6x-5x}{30}=\frac{5+7}{60} \\ \Rightarrow \frac{x}{30}=\frac{1}{5} \\ \Rightarrow x=\frac{30}{5} \\ \therefore x=6 km \end{gathered}$$

Let, 1^st day sail=y hours and 2^nd day= 7-y

According to the first condition, y-(7-Y)=3

y=5

So, 1^st day sail = 5 & 2^nd day said = 5-3= 2

Again, speed of 2^nd day and speed of 1^st day = x-5

According to the first condition, 

$$\begin{gathered} 2\times x+5\left(x-5\right)=59 \\ \Rightarrow 7x=59+25=84 \\ \therefore x=\frac{84}{7}=12 \\ {2}^{nd} day speed = 12 kmph \mathit{a}\mathit{n}\mathit{s}\mathbf{.} \end{gathered}$$

Let,

The original speed = x km/h

ATQ, 

108/x - 108/x + 3 = 3

=> 108x + 324 - 108x/x² + 3x = 3

=> 3x² + 9x - 324 = 0

=> 3(x² + 3x - 108) = 0

=> x² + 3x - 108 = 0

=> x² + 12x - 9x - 108 = 0

=> (x + 12)(x - 9) = 0

=> x = 9

Therefore, original speed 9.

প্রশ্নে বলা হচ্ছে যে, 108 কিলোমিটার যাওয়ার পরে একজন সাইকেল আরোহী বুঝতে পারলেন যে তিনি যদি 3 km/hr বেশি বেগে যেতেন, তবে তার 3 ঘণ্টা কম সময় লাগতো। সাইকেল আরোহীর গতিবেগ কত ? 

$$\begin{gathered} Let, the speed be x km/hr. \\ Since dis\tan ce is 108 km, time=\frac{108}{x} hrs \\ When speed is increased by 3 km/hr, speed is=(x+3) km/hr \\ \therefore The required time=\frac{108}{x+3}hrs \\ According to the question, \\ \frac{108}{x}-\frac{108}{x+3}=3 \\ \Rightarrow \frac{108\left(x+3\right)-108x}{x\left(x+3\right)}=3 \\ \Rightarrow \frac{108x+324-108x}{x^2+3x} \\ \Rightarrow 324=3x^2-9x \\ \Rightarrow 324-3x^2-9x=0 \\ \Rightarrow -3x^2-9x+324=0 \\ \Rightarrow -3\left(x^2+3x-108\right)=0 \\ \Rightarrow x^2+3x-108=0 \\ \Rightarrow x^2+12x-9x-108=0 \\ \Rightarrow x\left(x+12\right)-9\left(x+12\right) \\ \Rightarrow \left(x+12\right)\left(x-9\right)=0 \\ So either, x-9=0 \\ or, x+12=0 \\ if, x+12=0 \\ then, x=-12 \\ \therefore The speed 9km [Since velocity x\ne -12] \\ He traveled at a speed of 9 km/hr. \mathbf{\left(}\mathit{a}\mathit{n}\mathit{s}\mathbf{\right)} \end{gathered}$$

Let, he took x hours at 60 mph.

$\therefore$He took (5-x) hours at 80 mph.

According to question

60x+80(5-x)=340

=60x+400-80x=340

=20x=400-340=60

$$\begin{gathered} \therefore x=\frac{60}{20}=3 \\ \therefore Jane took 3 hours at 60 mph.\left(answer\right) \end{gathered}$$