TITLE: Burning a rope to count time
QUESTION [10 upvotes]: A rope burns irregularly in 16 minutes and costs 32 rupees, while a second rope burns also irregularly in 7 minutes and costs 14 rupees. Both can be lit only at one end and can be turned off and lit again as many times we want, until they are completely burned. In what way can we count 1 minute by using such ropes and how much will it cost? (Multiple ropes can be used but we are looking for the combination that will cost less).

My attempt: We light the two ropes simultaneously so when the second is fully burnt, 7 minutes have passed, so the remaining from the first one will be burnt in 9 minutes. Then we light a second of 7 mins and this way we have one of 2 minutes. Following the same procedure twice, we can have another one of 2 minutes. Then we light another of 7 minutes, along with the two of 2 minutes one after the other, and we have one of 3 minutes left (7-2-2). Then with the one of 3 minutes and another one of 2 minutes we can count 1 minute. 
...but this would be a disaster, cost-wise :)
Any other ideas are most welcome!!!

REPLY [5 votes]: New answer - 116 rupees
So, apparently we can get even cheaper than 120 rupees, this time by buying only a single 16-minute rope, and six 7-minute ropes (cost is $1 \times 32 + 6 \times 14 = 116$). This is how you do it:

Create a 2-minute rope with the 16-minute rope and two 7-minute ropes (remaining ropes: $4 \times 7, 1 \times 2$)
Use the 2-minute rope to turn two 7-minute ropes into 5-minute ropes (remaining ropes: $2 \times 7, 2 \times 5$)
Use one of the 5-minute ropes to turn the remaining 7-minute ropes into 2-minute ropes (remaining ropes: $1 \times 5, 2 \times 2$)
Use the two 2-minute ropes to reduce the 5-minute rope to the 1-minute rope you really wanted all along

I no longer dare to say this might be optimal.
Old answer - 120 rupees
The cheapest I have been able to work out so far is 120 rupees: 2 16-minute ropes and 4 7-minute ropes (cost is $2 \times 32 + 4 \times 14 = 120$). You do this as follows:

Light the two 16-minute ropes and a 7-minute rope to create two 9-minute ropes (remaining ropes: $2 \times 9$, $3 \times 7$)
Use a 9-minute rope and a 7-minute rope to create a 2-minute rope (remaining ropes: $1 \times 9$, $2 \times 7$, $1 \times 2$)
Light the two 7-minute ropes and the 2-minute rope to create two 5-minute ropes (remaining ropes: $1 \times 9$, $2 \times 5$)
Use the 9-minute rope and one of the 5-minute ropes to create a 4-minute rope (remaining ropes: $1 \times 5$, $1 \times 4$)
Use your 5-minute rope and 4-minute rope to create the desired 1-minute rope

I'm reasonably confident this is optimal, but I have no idea how to conclusively prove it.