1,585 views

1 vote

At the end of the year $1998,$ Shepard bought none dozen goats. Henceforth, every year he added $p\%$ of the goats at the beginning of the year and sold $q\%$ of the goats at the end of the year where $p>0$ and $q>0.$ If Shepard had $9$ dozen goats at the end year $2002,$ after making the sales for that year, which of the following is true?

- $p=q$
- $p<q$
- $p>q$
- $p = \frac {q}{2}$

0 votes

Assuming the total no. of goats are $x$

- if percentage of adding (increase) i.e $p$ = $10\%$ & percentage of selling (decrese) percentage $q$ = $10\%$ , then at the end of the year , number of goats will be $x \times \dfrac{11}{10}\times \dfrac{9}{10}$ = $x\times \dfrac{99}{100}$ ⇒ which is always lesser than x.

$\color{red}{So, p = q \hspace{0.1cm}is \hspace{0.1cm}not\hspace{0.1cm} possible.}$

Now,

- if $p<q$

Assuming $p = 10\%$ & $q = 11.11\%$,

then at the end of the year number of goats will be $x \times \dfrac{11}{10}\times \dfrac{8}{9}$ = $x \times \dfrac{88}{90} \color{red}{ \hspace{0.5cm}i.e<x}$

$\color{red}{∴ p<q \hspace{0.1cm}is\hspace{0.1cm} not\hspace{0.1cm} possible.}$

- if $p>q$

Assuming $p = 11.11\%$ & $q = 10\%$,

then at the end of the year number of goats will be $x\times \dfrac{10}{9}\times \dfrac{9}{10}$ = $x \color{blue}{\hspace{0.5cm}i.e =x}$

$ {\color{blue} {∴ p>q\hspace{0.1cm} is \hspace{0.1cm}possible.}}$

- if $p>\dfrac{q}{2}$

Assuming $p = 5\%$ & $q = 10\%$,

then at the end of the year number of goats will be $x \times \dfrac{21}{20}\times \dfrac{9}{10}$ = $x\times \dfrac{189}{200} \color{red}{\hspace{0.5cm}i.e <x}$

$\color{red}{∴ p=\dfrac{q}{2}\hspace{0.1cm} is\hspace{0.1cm} not\hspace{0.1cm} possible.}$