Two positive numbers are in the ratio 3:4. The sum of thrice the first number and twice the second is 68. Find the smaller number.

Let the two positive numbers be $3x$ and $4x$, where $x$ is a common factor.

"The sum of thrice the first number and twice the second is 68."
Thrice the first number is $3(3x) = 9x$.
Twice the second number is $2(4x) = 8x$.
Their sum is 68:

The first number is $3x = 3(4) = 12$.
The second number is $4x = 4(4) = 16$.

Comparing 12 and 16, the smaller number is 12.

The final answer is 12.