Answer:
1. So Chris has 10c coins and 20c coins and they're the same amount, I'll ignore the c for the calculations but keep in mind that these are cents.
Let x be the number of coins
10x = 20x
And well the 20 cents come to 60 cents more than the 10 cent coins which means the total amount given by the number of coins of 20 cents is 60 cents more ( + 60) than the amount given by the number of coins of 10 cents. Thus,
20x = 10x + 60
x = 6
So, 6 x 0.10 = 0.60$ and 6 x 0.20 = 1.20$
0.60$ + 1.20$ = 1.80$
2. Michael has 4 coins of 5 cents so, and 4 20 cents coins, 5 50 cents coins, one 10 cents coin.
We don't wanna use the 50 cents they would quickly reach the max number.
We want to use the smallest amounts of coins definitely, so
0.10(1) + 0.5(2) + 0.20(4) = 1$
So pretty much 7.
I didn't use 4 times the 0.5 because that would leave a gap of 0.10$ that can't be covered by any other coin so I replaced the 0.10$ and two of the 0.5$ coins by a single 0.20$ coin.