Question

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A businessman bought some items for Rs 600.Keeping 10 items for himself he sold he sold the remaining items at a profit of Rs 5 per item. From the amount received in his deal he could he could buy 15 more items. Find the original price of each item.

Answer 1

ANSWER:

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Total cost price of some items =Rs 600.(Given)

Let the original cost price of one item be= Rs x

Therefore,

$number \: of \: items \: purchased = \frac{600}{x} \\ \\ number \: of \: items \: sold \: after \: keeping \: 10 \: items = \frac{600}{x} - 10 \\ \\ selling \: price \: of \: each \: item = Rs (x + 5) \\ \\ total \: selling \: price \: of \: all \: items = Rs(x + 5)( \frac{600}{x} - 10) \\ \\ profit = selling \: price - cost \: price \\ \\ therefore \\ net \: profit \: made \: in \: the \: deal \\ = Rs(x + 5)( \frac{600}{x} - 10) - 600 \\ \\ but \: profit \: is \: equal \: to \: cost \: price \: of \: 15 \: items \: = Rs15x \\ = (x + 5)( \frac{600}{x} - 10) - 600 = 15x \\ \\ therefore \\ x( \frac{600}{x} - 10) + 5( \frac{600}{x} - 10) - 600 = 15x \\ \\ therefore \\ 600 - 10x + \frac{3000}{x} - 50 - 600 - 15x = 0 \\ \\ - 10x + \frac{3000}{x} - 50 - 15x = 0 \\ \\ - 25x + \frac{3000}{x} - 50 = 0 \\ \\ - 25 {x}^{2} + 3000 - 50x = 0.........(multiplying \: both \: the \: sides \: by \: x) \\ \\ {x}^{2} - 120 + 2x = 0.........(dividing \: both \: sides \: by \: - 25) \\ \\ {x}^{2} + 2x - 120 = 0 \\ \\ {x}^{2} + 12x - 10x - 120 = 0 \\ \\ x(x + 12) - 10(x + 12) = 0 \\ \\ (x + 12)(x - 10) = 0 \\ \\ therefore \\ x + 12 = 0 \: or \: x - 10 = 0 \\ x = - 12 \: or \: x = 10$
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x≠ -12 as cost price cannot be negative .


Therefore,
x=10.
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Hence,
the original cost of one(each)item is Rs 10.

Answer 2

Step by step explanation on attachment

Given

Total price of all items = 600

Number of itms he keep with himself after selling some items = 10

Profit he earned from each item = 5