Question

a businessman bought some otems for rs. 7500 . he kept 5 items for himself & sold remaining items at a profit of 25 rs. per item from the amount received in his deal he could buy 6 more items find the original price of each item.

Answer 1

Let the number of items bought be x.
Total cost of x item is Rs 7500.

(i) Then, price of each items $= \frac{\text{ Total cost} }{\text{ Number of items}}= \frac{7500}{x}$


Number of items sold after keeping 5 for himself = x-5

Profit per item = Rs 25
Then, total profit =  (Profit per item) × (Total number of items sold) =25(x-5)

With the profit amount he could buy 6 more items than the number of items he bought previously.
Therefore, from the total profit amount, he could buy (6+5)=11 items, because he sold only (x-5) items.

Then, total cost of 11 items = total profit =25(x-5)

(ii) Then, price of each items  $= \frac{\text{ Total cost} }{\text{ Number of items}}= \frac{25(x-5)}{11}$


Both (i) and (ii) represents the cost of each item, then the two must be equal.
According to problem,
$\frac{7500}{x}= \frac{25(x-5)}{11}$
$\text{Cross multiply} \\ 7500*11=25(x-5)*x \\ 825000=25x(x-5)$
$\text{Divide both sides by 25} \\ 3300=x(x-5) \\ 3300=x^2-5x \\ \text{Subtract 3300 from both sides:}$
$\\ 0=x^2-5x-3300 \\ x^2-5x-3300=0 \\ \text{Applying quadratic formula: }\\ x= \frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ x=\frac{-(-5)\pm\sqrt{(-5)^2-4(1)(-3300)}}{2(1)} \\ x= \frac{5\pm\sqrt{25+13200}}{2}$
$x=\frac{5\pm\sqrt{13225}}{2} \\ x=\frac{5\pm115}{2}$
[tex]x=\frac{5+115}{2} \text{ or }x=\frac{5-115}{2} \\ x=\frac{120}{2} \text{ or }x=\frac{-110}{2} \\ x=60 \text{ or }x=-55 \\[/tex]

The number of items cannot be negative, so we reject x=-55.

Therefore, the number of items bought=x=60.

Total cost of 60 items is Rs7500.
Cost of each item $= \frac{\text{ Total cost} }{\text{ Number of items}}= \frac{7500}{60}=\text{Rs}125$


Answer: Rs 125