Question

A father and his son walked 240 meters. During this trip, the father made 100 fewer steps than did his son. Find the distance each one of them covers with one step, if the father’s step is 20 cm greater than his son’s step.

Answer 1

Hello!

The answer is:

The father's step is 80 cm

The son's step is 60 cm.

Why?

To solve the problem, we need to write equations in order to create a relation between the father's and son's steps, and the distance covered by them.

We know thay they covered 240 meters, and the father's step is 20 cm  (0.2m)greater than his son's step, so, we can write the following relation:

$NumberOfSteps_{Son}=\frac{240m}{x}$

and,

$NumberOfSteps_{Father}=\frac{240m}{x+0.2m}$

Now, if the know that the father made 100 fewer steps than his son, we have:

$NumberOfSteps_{Son}-100=\frac{240m}{x+0.2m}$

Then, substituting the first equation into the second equation, we have:

$\frac{240m}{x}-\frac{240m}{x+0.2m}=100\\\\\frac{240m(x+0.2m)-240mx}{x(x+0.2m)}=100\\\\240mx+0.2m*240m-240mx=(x)(x+0.2m)*100\\\\48m^{2}=100*(x^{2}+0.2mx)=100x^{2} +20mx\\\\48m^{2}=100x^{2}+20mx\\\\100x^{2}+20mx-48m^{2}$

We have a quadratic equation:

$100x^{2}+20mx-48m^{2}=0$

Where,

$a=100\\b=20\\c=-48$

So, solving it applying the quadratic formula, we have:

$\frac{-b+-\sqrt{b^{2} -4ac} }{2a}$

$\frac{-20+-\sqrt{20^{2} -4*100*-48} }{2*100}=\frac{-20+-\sqrt{400+19200} }{200}\\\\\frac{-20+-\sqrt{400+19200} }{200}=\frac{-20+-\sqrt{19600} }{200}\\\\\frac{-20+-\sqrt{19600} }{200}=\frac{-20+-(140)}{200}\\\\x_{1}=\frac{-20+140}{200}=0.6\\\\x_{2}=\frac{-20-140}{200}=-0.8$

Then, since distance cannot be negative, we need to take the positive value, so, the son's step is 0.6m or 60 cm.

Now, calculating the father's step, we have:

$FatherStep=SonStep+20cm\\\\FatherStep=60cm+20cm=80cm$

Hence, we have that the father's step is 80 cm.

Have a nice day!