Jason borrowed $18,000 in two loans. One loan charged 11% simple interest and the other charged 6% simple interest. After 1 yr, Jason paid a total of $1380. Find the amount borrowed in each loan.​

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Answer:

The principle amounts borrowed by Jason are [tex]$\$ 6000$[/tex] and [tex]$\$ 12000$[/tex] .

Step-by-step explanation:

Given

Jason borrowed [tex]\$18,000[/tex] in two loans.

Step 1 of 2

Let [tex]$x$[/tex], and [tex]$(\$ 18,000-x)$[/tex] be the principle amount borrowed by Jason.

Let [tex]$P_{1}=x$[/tex], and

[tex]$P_{2}=(\$ 18000-x)$[/tex].

Let [tex]$I_{1}$[/tex]be the simple interest paid on loan 1 , and [tex]$I_{2}$[/tex] be the simple interest paid on loan 2 .

[tex]I_{1}+I_{2}=\$ 1380[/tex]

[tex]r_{1}=11 \%[/tex] (interest rate of loan 1)

[tex]$r_{2}=6 \%$[/tex] (interest rate of loan 2)

Time in years for both the loan is 1 year.

Substitute the formula of simple interest.

[tex]$P_{1} r_{1} t+P_{2} r_{2} t=\$ 1380$[/tex]

Step 2 of 2

Substitute the values

[tex](x)(0.11)(1)+(\$ 18,000-x)(0.06)(1)=\$ 1380 \\[/tex]

[tex]&0.11 x+\$ 1080-0.06 x=\$ 1380[/tex]

[tex]&0.05 x+\$ 1080=\$ 1380[/tex]

[tex]&0.05 x=\$ 1380-\$ 1080 \\[/tex]

[tex]&0.05 x=\$ 300 \\[/tex]

[tex]&x=\$ 6000[/tex]  

Substitute the value [tex]$\$ 6000$[/tex] in [tex]$(\$ 18,000-x)$[/tex] to find the other principle amount.

[tex]\$ 18,000-\$ 6000=12,000[/tex]

The principle amounts borrowed by Jason are [tex]$\$ 6000$[/tex] and [tex]$\$ 12000$[/tex].

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