, 2019

Barry’s 12 coins include at least one of each of the 5 coins of different values.

The total value of these 5 coins is $\$2.00+ \$1.00+\$0.25+ \$0.10+ \$0.05=\$3.40$.

Barry has the smallest total amount of money that he could have if each of his remaining $12-5=7$ coins has value $0.05 (the smallest possible value of a coin).

Thus, the smallest total amount of money that Barry could have is $$\text{3.40+7 0.05= 3.40+ 0.35= 3.75.}$$

Solution 1

The mean of the numbers $20,30,40$ is $\dfrac{20+30+40}{3}=\dfrac{90}{3}=30$.

Since each of the given answers has three numbers, then for the mean to equal 30, the sum of the three numbers must also equal $30\times3=90$.

Of the given answers, only (D) has numbers whose sum is 90 ($23+30+37=90$).

Solution 2

Since 20 is 10 less than 30 and 40 is 10 more than 30, the mean of the numbers $20,30,40$ is 30.

In each of the given answers, 30 is the middle number in the list of three numbers.

Thus, for the mean of the three numbers to equal 30, the first and last numbers must be equal “distances” away from 30 (with one number being less than 30 and the other greater than 30).

Looking at answer (D), 23 is 7 less than 30 and 37 is 7 more than 30 and so the mean of these three numbers is 30. (We may check that this isn’t the case for each of the other four answers.)