Question

A merchant bought some items for rupees 600, keeping 10 items for himself, he sold
remaining items at a profit of rupees 5 per item. From the amount received in this deal he
could buy 15 more items. Find the original price of each item.​

Answer 1

The original price of each item is Rs. 10

• Given,
• A merchant bought some items for rupees 600, keeping 10 items for himself
• Let the price of each item be = x
• Number of items bought for Rs. 600 = 600/x
• keeping 10 items for himself, he sold  remaining items
• ⇒Items sold = 600/x -10 = (600-10x)/x
• a profit of rupees 5 per item
• ⇒Selling price = Rs (x+5)
• From the amount received in this deal he  could buy 15 more items
• From given, we have,
• (x+5) (600-10x)/x = 600+15x
• (x+5) (600-10x) = x (600+15x)
• 5 (x+5) (120-2x) = 5x (120+3x)
• (x+5) (120-2x) = x (120+3x)
• upon solving, we get,
• -5x^2-10x+600 = 0
• x^2+2x-120 = 0
• (x+12)(x-10)=0
• x = -12, 10
• Therefore, the original price of each item is Rs. 10

Answer 2

The original price of each item is Rs. 26.5.

Step-by-step explanation:

Let the original price of each item is Rs. x.

So, the number of items that the merchant bought is $\frac{600}{x}$.

Now, keeping 10 items for himself, he sold the remaining items at a profit of Rs. 5 per item.

Now, the amount received in this deal he could buy 15 more items.

Hence, we can write the equation from the above conditions as

$(\frac{600}{x}- 10)(x + 5) = 15x$

⇒ (600 - 10x)(x + 5) = 15x²

⇒ (120 - 2x)(x + 5) = 3x²

⇒ 120x + 600 - 2x² - 10x = 3x²

⇒ 5x² - 110x - 600 = 0

⇒ x² - 22x - 120 = 0

Using quadratic formula, the value of x will be = $\frac{-(-22) + \sqrt{(- 22)^{2} - 4(1)(- 120)}}{2(1)}$

= Rs. 26.5 {Neglecting the negative root as x can not be negative}

Therefore, the original price of each item is Rs. 26.5. (Answer)