Progressions - The problem of the pill's weight

Let the number of the flask containing the 30-mg pills be $n$ . We know that the total number of pills weighed is $\dfrac {15(16)}2 = 120$ . The weight of these 120 pills is 2540 mg, which means that:

$\begin{aligned} 120\times 20 + n(30-20) & = 2540 \\ 2400 + 10n & = 2540 \\ 10n & = 140 \\ \implies n & = \boxed{14} \end{aligned}$

Let the number of 20-mg pills weighed be $x$ , and the number of 30-mg pills weighed be $y$ . The total number of pills weighed is $\displaystyle \sum_{n=1}^{15} n = \dfrac{15(16)}{2} = 120$ . We can then form two equations:

$x+y=120 \implies x=120-y\implies$ Eq.(1)

$20x+30y=2540\implies$ Eq.(2)

Substitute Eq.(1) into Eq.(2):

$20(120-y)+30y=2540\\ 2400-20y+30y=2540\\ 10y=140\\ y=\boxed{14}$