Question

a businessman bought some items for rs 7500. he kept 5 items for himself and sold the remaining items at a profit of rs 30 per item from the amount received in this deal he could buy 4 more items find the original price for each item.​

Answer 1

_______________________


$\sf{\underline{Given:}}$


$\sf{No.\:of\:items\:bought}$ $=\:x$

$\sf{Price\:per\:item}$$=\:y$

$\sf{Total\:amount\:paid}$ $= x \times y = xy = 7500$

$y = \frac{7500}{x}$


$\sf{Total\:items\:sold}$$= (x - 5)$


$\sf{Price\:of\:each\:item}$ $= (y + 30)$


$\sf{Total\:revenue\:received}$ $= (x - 5) \times (y + 30)$


$\sf{\underline{Since\:he\:can\:buy,}}$

$\sf{4\:more\:items\:from\:his\:revenue.}$


$(x - 5)(y + 30) - 7500 = 4 \times y$


$(x - 5)( \frac{7500}{x} + 30) - 7500 = 4y$


$(x - 5)( \frac{7500 + 30x}{x} ) - 7500 = 4y$


$(x - 5)(7500 + 30x) - 7500x = 4xy$


$7500x + 30 x^{2} - 37500 - 150x - 7500x = 4xy$


${30x}^{2} + 7500x - 150x - 7500x - 37500 = 4(7500)$


${30x}^{2} - 150x - 37500 - 30000 = 0$


$30x^{2} - 150x - 67500 = 0$


$30( {x}^{2} - 5x - 2250) = 0$


${x}^{2} - 5x - 2250 = 0$


$\sf{\underline{Solving\:the\:quadratic\:equation}}$

$\sf{\underline{We\:get,}}$

$x = - 45 \: \: \: \: \: x = 50$


$\sf{\underline{Since\:the\:cost\:cannot\:be\:negative,}}$

$x = - 45$ is discarded.


$\sf{\underline{So,}}$

$x = 50$


$\sf{\underline{Therefore,}}$

$\sf{50\:items\:were\:bought.}$


$\sf{The\:price\:per\:item}$ $\frac{7500}{50} = 150$


_______________________