Solution: Suppose there is x grams of salt in the water, which originally has x 1, and after the first pouring, it has x2, and so on.

x 1= 100*80%=80

X2=60*80%=48 after the first pouring.

At this time, the concentration of brine is 48/ 100=48%.

The remaining x3 after the second pouring = 60 * 48% = 28.8.

At this time, the concentration of brine is 28.8/ 100=28.8%.

After the third pouring, the residual X4=60*28.8%= 10.368.

At this time, the concentration of brine is10.368/100 =10.368%.

Answer: The concentration of brine in the last cup is 10.368%.