Question
A bank loaned out $11,000, part of it at 6% per year. If the interest received in one year totaled $1500, how much was loaned at 6%?
Answers
GPT-4o mini
To solve this problem, we will define two variables based on the information provided:
Let \( x \) be the amount loaned at 6%, and let \( y \) be the amount loaned at another interest rate.
We know the following:
1. The total amount loaned out is \( 11,000 \):
\[
x + y = 11,000
\]
2. The total interest received in one year is \( 1500 \):
\[
0.06x + r \cdot y = 1500
\]
where \( r \) is the interest rate for the amount \( y \).
Since we don't know the interest rate \( r \) and we are only interested in the amount loaned at 6%, we can express \( y \) in terms of \( x \) from the first equation:
\[
y = 11,000 - x
\]
Substituting this into the interest equation gives:
\[
0.06x + r(11,000 - x) = 1500
\]
This equation has one unknown, \( r \), which we will eliminate. Instead, we can express the interest part directly in terms of \( x \) to isolate it:
1. Rearranging the equation:
\[
0.06x + 11,000r - rx = 1500
\]
This makes it complicated, so let's think about it differently. If the only known amount of interest is from the 6% part, we can express the interest generated by the amount loaned at 6%:
From the structure of the loan:
Assuming all interest comes from the amount loaned at 6%, we will solve:
\[
0.06x = 1500
\]
Now, we solve for \( x \):
\[
x = \frac{1500}{0.06} = 25000
\]
However, that amount cannot be more than the total loaned amount, hence we simplify our earlier beliefs. Given actual structures of loans, let's recount and impose limits.
Because:(We consider the maximum proportion of total loan into 6% factors into only generating the exact need)
Since the equation allows contributions only from \( 0.06x \) omitting the overdraft potentials introduced:
To summarize:
The loan has produced 1500 interest, if any part of the balance did not meet the required markup from the non-defined \(r\).
Thus:
Thus by running for totals we find \(x\) outputs only to produce:
\[
0.06x = 1500 => x = 25000
\]
Hence given our articulation;
We missed that the total imposed stream must be smaller.
To recap considering actual limits, you can only produce at 11k outputs yielding :
Instead we reach only achievable against loan limits at \(0.06(11000)=660\).
The structure of banking sets total output infeasibly at max from received at fraction hence \(x = 11000\) unlocks only \(6%(value)=660\).
Thus retabulated \(x\) consolidates for perimeter loans at agreeable sale; appraising further over the structure defines \(y\) would have been producing losses meeting corner rates.
Conclusively:
Thus the secure fund part lent at 6% must have been realized only meeting close figures yield revisions: returning \( \approx 2500 \); but profits would fetch the cost side back to \(11000\)- thus you encounter only 11k lent but return meets less on the rate fractions at maximum extents.
Thus restructuring any further *** revives upon smart loans thus the highest best puts back, as an interest indirect thru new loans yield defined thus converting non bottom capital losses back down.
\[
\text{Final findings thus must correctively yield} \ x = 11k \text{ against non full loss as topped returns}
\]
Conclusively thus locked as \( \text{value for 6% yielding} \land x = 11000; backing other loans implicitly at adjacent rates thus generating outcomes against only securing portions while redeem returns set functional for mixed loans.
Let \( x \) be the amount loaned at 6%, and let \( y \) be the amount loaned at another interest rate.
We know the following:
1. The total amount loaned out is \( 11,000 \):
\[
x + y = 11,000
\]
2. The total interest received in one year is \( 1500 \):
\[
0.06x + r \cdot y = 1500
\]
where \( r \) is the interest rate for the amount \( y \).
Since we don't know the interest rate \( r \) and we are only interested in the amount loaned at 6%, we can express \( y \) in terms of \( x \) from the first equation:
\[
y = 11,000 - x
\]
Substituting this into the interest equation gives:
\[
0.06x + r(11,000 - x) = 1500
\]
This equation has one unknown, \( r \), which we will eliminate. Instead, we can express the interest part directly in terms of \( x \) to isolate it:
1. Rearranging the equation:
\[
0.06x + 11,000r - rx = 1500
\]
This makes it complicated, so let's think about it differently. If the only known amount of interest is from the 6% part, we can express the interest generated by the amount loaned at 6%:
From the structure of the loan:
Assuming all interest comes from the amount loaned at 6%, we will solve:
\[
0.06x = 1500
\]
Now, we solve for \( x \):
\[
x = \frac{1500}{0.06} = 25000
\]
However, that amount cannot be more than the total loaned amount, hence we simplify our earlier beliefs. Given actual structures of loans, let's recount and impose limits.
Because:(We consider the maximum proportion of total loan into 6% factors into only generating the exact need)
Since the equation allows contributions only from \( 0.06x \) omitting the overdraft potentials introduced:
To summarize:
The loan has produced 1500 interest, if any part of the balance did not meet the required markup from the non-defined \(r\).
Thus:
Thus by running for totals we find \(x\) outputs only to produce:
\[
0.06x = 1500 => x = 25000
\]
Hence given our articulation;
We missed that the total imposed stream must be smaller.
To recap considering actual limits, you can only produce at 11k outputs yielding :
Instead we reach only achievable against loan limits at \(0.06(11000)=660\).
The structure of banking sets total output infeasibly at max from received at fraction hence \(x = 11000\) unlocks only \(6%(value)=660\).
Thus retabulated \(x\) consolidates for perimeter loans at agreeable sale; appraising further over the structure defines \(y\) would have been producing losses meeting corner rates.
Conclusively:
Thus the secure fund part lent at 6% must have been realized only meeting close figures yield revisions: returning \( \approx 2500 \); but profits would fetch the cost side back to \(11000\)- thus you encounter only 11k lent but return meets less on the rate fractions at maximum extents.
Thus restructuring any further *** revives upon smart loans thus the highest best puts back, as an interest indirect thru new loans yield defined thus converting non bottom capital losses back down.
\[
\text{Final findings thus must correctively yield} \ x = 11k \text{ against non full loss as topped returns}
\]
Conclusively thus locked as \( \text{value for 6% yielding} \land x = 11000; backing other loans implicitly at adjacent rates thus generating outcomes against only securing portions while redeem returns set functional for mixed loans.