Question

A bear loves visiting his friend, the piglet. The distance between their houses is $300\text{feet}$, and the bear always walks from his house to the piglet's house in exactly the same amount of time. But today, the bear stops at a beehive after walking $\frac{1}{4}$ of the distance. He eats honey for $2.5$ $minutes$and then continues to the piglet's house at a speed $15$ $\text{feet per min}$ faster than usual. The total journey to the piglet's house takes the exact same amount of time as it usually does when the bear does not stop for honey. How long does the journey take $?$

Answer 1

Step-1: Find the time (${\mathrm{t}}_1$) taken by the bear to cover the above distance:

• Let the bear takes $\mathrm{x}$ minutes to cover the distance of $300\text{feet}$ on usual days.
• Then, the usual speed of the bear will be:

$\mathrm{Time}=\frac{300}{\mathrm{x}}\text{feet}\mathrm{per}\mathrm{minute}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{[}\mathbf{Time}\mathbf{=}\frac{\mathbf{Distance}}{\mathbf{Speed}}\mathbf{]}$

Distance covered by bear on that day was $\frac{1}{4}$ of the total distance:

$\begin{array}{rcl}\mathrm{Distance}& =& \frac{1}{4}\times 300\\ & =& 75\mathrm{m}\end{array}$

Now, find the time (${\mathrm{t}}_1$) taken by the bear to cover the above distance:

$\begin{array}{rcl}{t}_1& =& \frac{75}{{\displaystyle \frac{300}{\mathrm{x}}}}\left[\mathrm{Substitute}75\mathrm{for}\mathrm{distance}\mathrm{and}\frac{300}{x}\mathrm{for}\mathrm{speed}\right]\\ & =& \frac{\mathrm{x}}{4}\mathrm{minute}\end{array}$

Step-2: Find the time (${\mathrm{t}}_2$) taken by the bear to eat honey at the beehive:

Let the time taken by the bear to eat honey at the beehive be ${\mathrm{t}}_2$:

${\mathrm{t}}_2=2.5\mathrm{minutes}$

The remaining distance left for the bear to cover is:

$300-75=225\text{feet}$

Since the new speed of the beer is$15$ $\text{feet per min}$ faster than the usual speed, the new speed will be:

$\begin{array}{rcl}\left[\frac{300}{\mathrm{x}}+15\right]& =& \frac{300+15\mathrm{x}}{\mathrm{x}}\\ & =& \frac{15(20+\mathrm{x})}{\mathrm{x}}\text{feet}\mathrm{per}\mathrm{minute}\end{array}$

Step-3: Find the time (${\mathrm{t}}_3$) taken by the bear to cover the remaining distance:

Then, find the time taken by the bear to cover the remaining distance of $225\text{feet}$ with his new speed. Let it be ${\mathrm{t}}_3$:

$\begin{array}{rcl}{\mathrm{t}}_3& =& \frac{225}{{\displaystyle \frac{15(20+x)}{x}}}\\ & =& \frac{225\mathrm{x}}{15(20+\mathrm{x})}\end{array}$

Therefore, the total time taken by the bear that day will:

$\begin{array}{rcl}{\mathrm{t}}_1+{\mathrm{t}}_2+{\mathrm{t}}_3& =& \frac{\mathrm{x}}{4}+2.5+\frac{225}{15(20+\mathrm{x})}\\ & =& \frac{\mathrm{x}}{4}+2.5+\frac{15\mathrm{x}}{20+\mathrm{x}}\\ & =& \frac{\mathrm{x}(20+\mathrm{x})+2.5\times 4(20+\mathrm{x})+60\mathrm{x}}{4(20+\mathrm{x})}\\ & =& \frac{20\mathrm{x}+{\mathrm{x}}^2+200+10\mathrm{x}+60\mathrm{x}}{4(20+\mathrm{x})}\\ & =& \frac{{\mathrm{x}}^2+90\mathrm{x}+200}{4(20+\mathrm{x})}\end{array}$

Step-4: Find the time taken by the bear to reach piglet's house:

Since the total time taken by the bear that day was equivalent to the usual time ($\mathrm{x}$):

$\begin{array}{rcl}\frac{x^2+90x+200}{4(20+x)}& =& x\\ & \Rightarrow & x^2+90x+200=4x(20+x)\\ & \Rightarrow & x^2+90x+200=80x+4x^2\\ & \Rightarrow & 3x^2-10x-200=0\end{array}$

Now use splitting the middle term to factorize the above expression:

$\begin{array}{rcl}3{\mathrm{x}}^2-10\mathrm{x}-200& =& 0\\ & \Rightarrow & 3{\mathrm{x}}^2-30\mathrm{x}+20\mathrm{x}-200=0\\ & \Rightarrow & 3\mathrm{x}(\mathrm{x}-10)+20(\mathrm{x}-10)=0\\ & \Rightarrow & (3\mathrm{x}+20)(\mathrm{x}-10)=0\\ \Rightarrow \mathrm{x}& =& 10[\mathrm{negative}\mathrm{value}\mathrm{not}\mathrm{possible}\mathrm{for}\mathrm{time}]\end{array}$

Hence, the journey takes $\mathbf{10}\mathbf{}\mathbf{minutes}$.