A businessman bought some items for ₹600, keeping 10 items for himself he sold the
remaining items at a profit of ₹5 per item. From the amount received in this deal he
could buy 15 more items find the original price of each item.
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Answer:
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}
x
600
.
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(
x
600
−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)}
2(1)
−(−22)+
(−22)
2
−4(1)(−120)
= Rs. 2The 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}
x
600
.
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(
x
600
−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)}
2(1)
−(−22)+
(−22)
2
−4(1)(−120)
= 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) 6.5 {Neglecting the negative root as x can not be negative}
Therefore, the original price of each item is Rs. 10.
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