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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?

  1. p=q
  2. p<q
  3. p>q
  4. $p = \frac {q}{2}$
asked in Quantitative Aptitude by (7.9k points) 115 216 353
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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 $x \times \dfrac{21}{20}\times \dfrac{9}{10}$  = $x\times \dfrac{189}{200} \color{red}{\hspace{0.5cm}i.e <x}$

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

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