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(A) Let the initial number of bushes be $x$.The population changes over three years as follows:Year $1$: Increase of $10 \% \implies x 	imes (1 + 0.1

Vedclass · Accepted Answer

$25000$

Vedclass · Answer

(A) Let the initial number of bushes be $x$.The population changes over three years as follows:Year $1$: Increase of $10 \% \implies x 	imes (1 + 0.10) = 1.1x$Year $2$: Increase of $8 \% \implies 1.1x 	imes (1 + 0.08) = 1.1x 	imes 1.08$Year $3$: Decrease of $10 \% \implies (1.1x 	imes 1.08) 	imes (1 - 0.10) = 1.1x 	imes 1.08 	imes 0.9$Given that the final population is $26730$,we have:$x 	imes 1.1 	imes 1.08 	imes 0.9 = 26730$$x 	imes (\frac{11}{10}) 	imes (\frac{108}{100}) 	imes (\frac{9}{10}) = 26730$$x 	imes \frac{10692}{1000} = 26730$$x = \frac{26730 	imes 1000}{10692}$$x = 25000$Thus,the number of bushes in the beginning was $25000$.

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