A merchant visited 3 fairs. At the first, he doubled his money and spent $30; at the second he tripled his money and spent $54; at the third, he quadrupled his money and spent $72, and then had $48 left. How much money had he at the start?

$29

29x2=58
58-30=28
28x3=84
84-54=30
30x4=120
120-72=48

$29 no cap

Let's break down the information provided step by step to find the initial amount of money the merchant had.

1. At the first fair, the merchant doubled his money and spent $30.
Let's assume the initial amount of money the merchant had is "x".
After doubling his money, the merchant had 2x.
He spent $30, so he had (2x - 30) left.

2. At the second fair, the merchant tripled his money and spent $54.
The amount he had left after the first fair was (2x - 30).
After tripling his money, the merchant had 3 * (2x - 30) = 6x - 90.
He spent $54, so he had (6x - 90 - 54) left.

3. At the third fair, the merchant quadrupled his money and spent $72.
The amount he had left after the second fair was (6x - 90 - 54).
After quadrupling his money, the merchant had 4 * (6x - 90 - 54) = 24x - 540 - 216.
He had $48 left after spending at the third fair, so we can set up the equation: 24x - 540 - 216 = 48.

4. Simplify the equation and solve for x:
24x - 540 - 216 = 48
Combine like terms: 24x - 756 = 48
Add 756 to both sides: 24x = 804
Divide both sides by 24: x = 33.5

Therefore, the merchant had $33.5 at the start.

To solve this problem, we can work backward by figuring out how much money the merchant had after each fair. Let's go step by step:

1. In the third fair, the merchant quadrupled his money and spent $72, leaving him with $48. This means that before spending $72, he had $48 + $72 = $120.

2. Now let's move on to the second fair. The merchant tripled his money and spent $54, so before spending $54, he had $120 + $54 = $174.

3. Finally, in the first fair, the merchant doubled his money and spent $30, so before spending $30, he had $174 + $30 = $204.

Therefore, the merchant had $204 at the start.