You stated your battery is 12V 7Ah battery.
LED (actually the bar light uses 4) is 12-14V (that's the drop), and they say can use up to 10W. Let's do the math:
$$
I = \frac{P}{E} = \frac{10W}{12V} = 833 mA \\
$$
No sense in calculating at 14V, you're limited to 12V.
$$
3 \cdot 0.833A = 2.499A
$$
So, BEST case, 2.8 hours. But you really need to see the curve for your specific battery, and it's derating for thermal, etc, to see if it can realistically output 7Ah-- manufacturers (for marketing purposes) usually put the 'best case' information on the battery, but only the datasheet for it will provide the truth.
I'd expect 2 hours and change. If you put an ammeter on it when you get it, and see how much current it's actually drawing, that will let you get a better idea.
Best Answer
All else being equal, yes. Of course, all things are never equal.
First, the lower-power unit will probably be physically smaller than the 50 watt, and this will include the light-emitting area. If you look directly at the LED while it operates, there is a good chance that the lower-power unit will appear brighter, since it is emitting light from a smaller area. However, the appearance is irrelevant. What counts is the total light emitted, not the power density per unit area.
Second, (again assuming a physically smaller LED) heat sinking is likely to be more effective on the higher-power LED. At the same power it is likely that the low power LED will run hotter, and this will reduce its efficiency somewhat.
Finally, the power rating for an LED refers to the total input power, not the visible output. So the two LEDs may not use the same process, and may have different intrinsic efficiencies.