Nisha bought a number of oranges at 2 for a r | vedclass

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(A) Let the number of oranges bought of each type be $60$ (the $LCM$ of $2$ and $3$ is $6$,so $60$ is a convenient multiple).Cost Price $(C.P.)$ of th

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$6$

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(A) Let the number of oranges bought of each type be $60$ (the $LCM$ of $2$ and $3$ is $6$,so $60$ is a convenient multiple).Cost Price $(C.P.)$ of the first type of $60$ oranges: Since $2$ oranges cost $1$ rupee,$60$ oranges cost $60 / 2 = 30$ rupees.Cost Price $(C.P.)$ of the second type of $60$ oranges: Since $3$ oranges cost $1$ rupee,$60$ oranges cost $60 / 3 = 20$ rupees.Total $C.P.$ for $120$ oranges $(60 + 60)$ $= 30 + 20 = 50$ rupees.To make a profit of $20 \%$,the total Selling Price $(S.P.)$ should be: $S.P. = C.P. + (20 \% 	ext{ of } C.P.) = 50 + (0.20 	imes 50) = 50 + 10 = 60$ rupees.This $S.P.$ of $60$ rupees is for $120$ oranges.Therefore,the $S.P.$ for $1$ dozen ($12$ oranges) $= (60 / 120) 	imes 12 = 0.5 	imes 12 = 6$ rupees.

Nisha bought a number of oranges at $2$ for a rupee and an equal number at $3$ for a rupee. To make a profit of $20 \%$ she should sell a dozen for (in $Rs.$)

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