Question

A trader bought a number of articles for 1200.
Keeping 10 items for himself, he sold each of the
rest at 2 more than what he paid for it, thus
getting a profit of 60 on the whole transaction.
Find the number of articles purchased.​

Answer 1

Answer:

100

Step-by-step explanation:

Let the number of articles purchased = n

Cost of each article = $\frac{1200}{n}$

He sold (n - 10) articles at ($\frac{1200}{n}$ + 2)

Total sale price = (n - 10) X ($\frac{1200}{n}$ + 2)

Profit = 60 = Total sale price - Purchase cost

So,

60 = [(n - 10) X ($\frac{1200}{n}$ + 2)] - 1200

60 = 1200 - $\frac{12000}{n}$ + 2n - 20 - 1200

60 = - $\frac{12000}{n}$ + 2n - 20

60n = 12000 + 2n² - 20n

2n² - 80n + 12000 = 0

n² - 40n + 6000 = 0

n² + 60n - 100n + 6000 = 0

(n + 60) X (n - 100) = 0

n = - 60 OR n = 100

Since the number of articles cannot be negative

n = 100