Question

A businessman bought some items for rs 7500 . he kept 5 items for himself and sold the remaining items at a profit os rs 25 per item. from the amount received 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,
[tex] \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} \\ x=\frac{5+115}{2} \text{ or }x=\frac{5-115}{2} \\ [/tex]
[tex]x=\frac{120}{2} \text{ or }x=\frac{-110}{2} \\ x=60 \text{ or }x=-55 \\[/tex]

The number 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

Answer 2

Answer:

Step-by-step explanation:

No. of items bought = x

Price per item = y

Total amount paid = x * y = xy = 7500

y = 7500 / x

Total items sold = (x - 5)

Price of each item = (y + 30)

Total revenue received = (x - 5) * (y + 30)

Since he can buy 4 more items from his revenue

(x - 5)(y + 30) - 7500 = 4 * y

(x - 5)[(7500/x) + 30] - 7500 = 4y

(x - 5)[(7500 + 30x) / x ] - 7500 = 4y

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

7500x + 30x² - 37500 - 150x - 7500x = 4xy

30x² + 7500x - 150x - 7500x - 37500 = 4(7500)

30x² - 150x - 37500 - 30000 = 0

30x² - 150x - 67500 = 0

30 (x² - 5x- 2250) = 0

x² - 5x - 2250 = 0

x = -45 , +50 (Solving quadratic equation)

Since cost cannot be negative, so x = 50 which is the number of items bought.

The price per item = 7500/50 = 150