A trader bought 200 exercise books at N300 each. He sold z books at N360 each, while the remaining books were sold at N10 less than the cost price per book. If he made a total profit of N9,000, find the value of z:

The trader bought 200 exercise books at N300 each.

$$\text{Total Cost} = 200 \times \text{N}300 = \text{N}60,000$$

The trader sold $z$ books at N360 each.

$$\text{Revenue}_1 = z \times \text{N}360 = 360z$$

The number of remaining books is $200 - z$.
The selling price for these books is N10 less than the cost price (N300).
Selling price for remaining books = N300 - N10 = N290.

$$\text{Revenue}_2 = (200 - z) \times \text{N}290$$

Total Profit = (Revenue from all sales) - (Total Cost)
Total Profit = $\text{Revenue}_1 + \text{Revenue}_2 - \text{Total Cost}$
Given total profit is N9,000.

$$9000 = 360z + 290(200 - z) - 60000$$

First, distribute 290 into the parenthesis:

$$9000 = 360z + (290 \times 200) - (290 \times z) - 60000$$ $$9000 = 360z + 58000 - 290z - 60000$$

Combine the terms with $z$:

$$360z - 290z = 70z$$

Combine the constant terms:

$$58000 - 60000 = -2000$$

Substitute these back into the equation:

$$9000 = 70z - 2000$$

Add 2000 to both sides of the equation:

$$9000 + 2000 = 70z$$ $$11000 = 70z$$

Divide both sides by 70 to find $z$:

$$z = \frac{11000}{70}$$ $$z = \frac{1100}{7}$$