**Problem**

A man wants to invest a sum of P50,000 in two investments. The first investment earns a rate of interest 4 times that of the second investment. In 3 years the first investment grows to P37,200. For 10 years, the second investment grows to P24,000.

- Find the sum invested in each rate of interest.

A. P35,000 and P15,000

B. P35,500 and P14,500

C. P30,000 and P20,000

D. P32,000 and P18,000

- Find the rate of interest of each.

A. 8% and 2%

B. 6% and 4%

C. 7% and 3%

D. 5% and 1%

**Solution**

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$x$ = 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:

$372 = x(1 + 12i)$

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

Second investment:

$240 = y(1 + 10i)$

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

$x + y = 50,000$

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

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

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

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

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

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

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

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

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 ]