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A bank loaned out
$14,000, part of it at the rate of
5% annual interest, and the rest at
3% annual interest. The total interest earned for both loans was
$440.00. How much was loaned at each rate?

A bank loaned out $\$ 14,000$ , part of it at the rate of $5 \%$ annual interest, and the rest at $3 \%$ annual interest. The total interest earned for both loans was $\$ 440.00$ . How much was loaned at each rate?

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Add and subtract money amounts: word problems

Full solution

Q. A bank loaned out

\$ 14,000

, part of it at the rate of

5 \%

annual interest, and the rest at

3 \%

annual interest. The total interest earned for both loans was

\$ 440.00

. How much was loaned at each rate?

Denote Loaned Amounts: Let's denote the amount loaned at $5$ % as $x$ and the amount loaned at $3$ % as $y$ . We have two unknowns and will need two equations to solve for them. $\newline$ The first equation comes from the fact that the total amount loaned out is $\$14,000$ : $\newline$ $x + y = 14,000$
Equation for Total Amount: The second equation comes from the total interest earned, which is $\$440$ . The interest earned from the amount loaned at $5$ \% is $0.05x$ , and the interest earned from the amount loaned at $3$ \% is $0.03y$ . The sum of these interests is the total interest: $\newline$ $0.05x + 0.03y = 440$
Equation for Total Interest: Now we have a system of two equations with two unknowns: $\newline$ $1$ ) $x + y = 14,000$ $\newline$ $2$ ) $0.05x + 0.03y = 440$ $\newline$ We can solve this system using substitution or elimination. Let's use the substitution method. From the first equation, we can express $y$ in terms of $x$ : $\newline$ $y = 14,000 - x$
Solve Using Substitution: Substitute $y = 14,000 - x$ into the second equation: $\newline$ $0.05x + 0.03(14,000 - x) = 440$ $\newline$ Now, let's distribute the $0.03$ into the parentheses: $\newline$ $0.05x + 420 - 0.03x = 440$
Substitute and Simplify: Combine like terms: $\newline$ $0.02x + 420 = 440$ $\newline$ Now, subtract $420$ from both sides to isolate the term with $x$ : $\newline$ $0.02x = 440 - 420$ $\newline$ $0.02x = 20$
Isolate Term with x: Divide both sides by $0.02$ to solve for $x$ : $\newline$ $x = \frac{20}{0.02}$ $\newline$ $x = 1,000$ $\newline$ So, $\$1,000$ was loaned at $5\%$ interest.
Solve for x: Now, substitute $x$ back into the equation $y = 14,000 - x$ to find $y$ : $\newline$ $y = 14,000 - 1,000$ $\newline$ $y = 13,000$ $\newline$ So, $\$13,000$ was loaned at $3\%$ interest.

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