A shopkeeper buys some books for £80. If he had bought 4 more books for the same amount, each book would have cost £1 less. Find the number of books he bought?

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Let b = the number of books he bought for £80

Let c = the cost of each book

Then, b*c = £80 {Eqn 1}

And (b+4) * (c-£1) = £80 {Eqn 2}

Two equations, two unknowns. So you can find a unique solution. And you can do this several ways. I'll use substitution. For convenience, I'll drop the £ sign.

Rewrite Eqn 1 as b = 80/c

Then substitute into Eqn 2: (80/c + 4) * (c- 1) = 80

Now multiply out using FOIL method: (80/c * c) + (80/c * -1) + (4c) + (4 * -1) = 80

Rewrite: 80 - 80/c + 4c - 4 = 80

Rearrange: -80/c + 4c = 4

Divide both sides of equation by 4: -20/c + c = 1

Rearrange: -20/c + c - 1 = 0 {Eqn 3}

Get a common denominator of c: (-20 + c^{2} - c) / c = 0 {Eqn 4}

Notice in Eqn 4 that the numerator must be equal to zero in order for the equation to hold. So, we can ignore the c in the denominator

Rewrite: c^{2}-c-20 = 0 {Eqn 5}

Now factor Eqn 5: (c-5)(c+4) = 0. The solution is c = 5 or c = -4. Since c is a cost, it cannot be negative, so the only solution is 5

The cost of each book originally is £5

Substituting back into Eqn 1, the number of books bought by the shopkeeper is
16

Check your work

Originally, he bought 16 books @ £5/each = £80

If he bought 4 more, he would have bought 20. Each book would cost £4 (ie, one £ less). Again, he would have paid 20* £4 = £80, which is the same amount.

One point about the wording of the problem. The problem should have made it clear that all books cost the same amount.