Question

At a certain time in a deer park, the number of heads and the number of legs of deers and human visitors were counted and it was found there were $39$ heads & $132$ legs. If the number of deers and human visitors in the park are ${}^{\prime }{d}^{\prime }$ and ${}^{\prime }{h}^{\prime }$ respectively, Then find $\left|\frac{d-h}{3}\right|$.

Answer 1

Solution:

No. of legs of a deer $=4$

No. of legs of a human$=2$

No. of head of a deer $=1$

No. of head of a human $=1$

If $d$ no. of deers and $h$ no. of humans are there

So, total number of heads,

$d+h=39$

$\Rightarrow h=39-d$ ………$\left(1\right)$

and, total number of legs,

$4d+2h=132$

$\Rightarrow 4d+2(39-d)=132$ [from $\left(1\right)$]

$\Rightarrow 4d-2d+78=132$

$\Rightarrow 2d=54$

$\Rightarrow d=27$

So, from $\left(1\right)$,

$h=39-27=12$

Now,

$\left|\frac{d-h}{3}\right|=\left|\frac{27-12}{3}\right|=\left|\frac{15}{3}\right|=5$.

Hence, the required value is $5$