Question

A retiree receives $5120 a year interest from $40,000 placed in two bonds, one paying 14% and the other 12%. How much is invested in each bond? 3)

Answer 1

Answer:

Investment in first bond is $ 16,000

In second bond is $ 24,000

Step-by-step explanation:

Let x be the amount invested in first bond,

Since, the total invested amount = $ 40,000,

So, the total amount invested in second bond = ( 40,000 - x ) dollars,

Given,

In first bond,

Interest rate = 14%,

While in second bond,

Interest rate = 12%,

Thus, the total interest from both bonds in a year,

I = 14% of x + 12% of ( 40,000 - x )

$=\frac{14x}{100}+\frac{12(40000-x)}{100}$

$=0.14x+0.12(40000-x)$

According to the question,

I = $ 5120,

$\implies 0.14x+0.12(40000-x)=5120$

$0.02x+4800=5120$

$0.02x=320$

$\implies x = 16000$

Hence, the amount invested in first bond = $ 16,000,

The amount invested in second bond = $ 40,000 - $ 16,000 = $ 24,000