Let

$x$ = sum invested at greater interest

$y$ = sum invested at lesser interest

$i$ = interest of $y$

$4i$ = interest of $x$

$F = P(1 + rt)$

First investment:

F = P37,200, P = x, r = 4i, and t = 3 yrs

$372 = x(1 + 12i)$
$x = \dfrac{37,200}{1 + 12i}$

Second investment:

F = P24,000, P = y, r = i, and t = 10 yrs

$240 = y(1 + 10i)$
$y = \dfrac{24,000}{1 + 10i}$

$x + y = 50,000$

$\dfrac{37,200}{1 + 12i} + \dfrac{24,000}{1 + 10i} = 50,000$

$\dfrac{372}{1 + 12i} + \dfrac{240}{1 + 10i} = 500$

$\dfrac{372(1 + 10i) + 240(1 + 12i)}{(1 + 12i)(1 + 10i)} = 500$

$372(1 + 10i) + 240(1 + 12i) = 500(1 + 12i)(1 + 10i)$

$(372 + 3,720i) + (240 + 2,880i) = 500(1 + 22i + 120i^2)$

$612 + 6,600i = 500 + 11,000i + 60,000i^2$

$60,000i^2 + 4,400i - 112 = 0$

$i = 0.02 ~ \text{and} ~ -0.093$

Use i = 0.02 and 4i = 0.08

$x = \dfrac{37,200}{1 + 12(0.02)} = \text{P}30,000$

$y = \dfrac{24,000}{1 + 10(0.02)} = \text{P}20,000$

Answer for Part 1: [ C ]

Answer for part 2: [ A ]