Question on Time, Speed and Distance

Question

A child while going to school reduces his speed to 4/5th of his actual speed and reaches 15 minutes late. Find his actual speed.

My attempt:

Speed is in the ratio = 5:4

Distance is constant.

Therefore, time is in the ratio = 4:5

Therefore, the actual time taken to reach school = 60 minuites.

From here, I am not able to find the actual speed. Please help regarding this. Thankyou.

Answer 1

You're right, there's not enough data to calculate the speed.

The same distance travelled $$s=vt$$ can be covered in the time longer by $15\,\text{min}$ at $\frac 45$ of the original speed: $$s=\frac45v(t+15\,\text{min})$$ so $$vt=\frac45v(t+15\,\text{min})$$ and $v$ disappears, as well as $s$.

You can only find, as you did, $t$: $$\frac54t=t+15\,\text{min}$$ $$\frac14t=15\,\text{min}$$ $$t=4\cdot 15\,\text{min} = 60\,\text{min}$$

You can get arbitrary $v$ by choosing appropriate $s=v\cdot{1\,\text{h}}.$

Answer 2

We have

• $\frac {D} v=t $
• $\frac {D} {\frac{4v}5}=t+15 $

then

• $\frac54 t=t+15 \implies t=60$ min

and thus we can only deduce that

• $v=\frac {D} {60}$ km/min $=D$ km/h

Answer 3

Let the actual speed $100x$ m/minute and he normally takes $y$ minutes

So, the distance $100xy$ m

Now $100xy=80x(y+15)\implies5y=4(y+15)\iff y=60$

So, any finite value of $x$ will satisfy the given condition.