We can simplify the inequality by dividing both sides by 3:
$$\frac{60g}{3}>5$$
Simplifying further:
$$20g>5$$
Dividing both sides by 20:
$$g>\frac{1}{4}$$
Now we can check each set of numbers:
A. None of these numbers are greater than $\frac{1}{4}$, so none of them satisfy the inequality.
B. All of these numbers are greater than $\frac{1}{4}$, so they all satisfy the inequality.
C. Two of these numbers are greater than $\frac{1}{4}$ (3 and 4), but 2 is not, so only two of them satisfy the inequality.
D. All of these numbers are greater than $\frac{1}{4}$, so they all satisfy the inequality.
Therefore, the set of numbers in which all the numbers satisfy the inequality is $\boxed{\textbf{(B)}\ 3, 4, \text{ and }5}$.
In which of the following sets of numbers are all the numbers solutions to 60g>3×5?
A. 1, 2, and 3
B. 3, 4, and 5
C. 2, 3, and 4
D. 4, 5, and 6
1 answer