You invested $4000, part at 5% and the remainder at 9% annual interest. At the end of the year, the total interest from these investments was $311. How much was invested at each rate?

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You invested $4000, part at 5% and the remainder at 9% annual interest. At the end of the year, the total interest from these investments was $311. How much was invested at each rate?

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Answer 1 · 2016-09-03T09:14:41

$1225$ $1225$ at $5 %$ $5%$ and $2775$ $2775$ at $9 %$ $9%$

Explanation:

Let the part invested at $5 %$ $5%$ be $x$ $x$ and the part invested at $9 %$ $9%$ be $y$ $y$

So we can write $x + y = 4000$ $x+y=4000$ and $\frac{5}{100} \times x + \frac{9}{100} \times y = 311$ $5/100timesx+9/100timesy=311$

or

$5 x + 9 y = 31100$ $5x+9y=31100$

Multiplying both sides of $x + y = 4000$ $x+y=4000$ by $5$ $5$

We get

$5 x + 5 y = 20000$ $5x+5y=20000$

Subtracting $5 x + 5 y = 20000$ $5x+5y=20000$ from $5 x + 9 y = 31100$ $5x+9y=31100$

We get

$5 x + 9 y - 5 x - 5 y = 31100 - 20000$ $5x+9y-5x-5y=31100-20000$

or

$4 y = 11100$ $4y=11100$

or

$y = \frac{11100}{4}$ $y=11100/4$

or

$y = 2775$ $y=2775$ ------------------------Ans $1$ $1$

So plugging the value $y = 2775$ $y=2775$ in the equation $x + y = 4000$ $x+y=4000$

we get

$x + 2775 = 4000$ $x+2775=4000$

or

$x = 4000 - 2775$ $x=4000-2775$

or

$x = 1225$ $x=1225$ ---------------------------Ans $2$ $2$

Answer 2 · 2016-09-03T10:41:32

Improved my method by cutting out s step.

$2775 was invested at 9%
$1225 was invested at 5%

Explanation:

$A very different approach!$ $color(red)("A very different approach!")$

$The explaining takes longer than the actual calculation$ $color(red)("The explaining takes longer than the actual calculation")$

$Determine the proportion dinvested at 9%$ $color(blue)("Determine the proportion dinvested at 9%")$

Suppose all the money was invested at 5% then the income would be $\frac{5}{100} \times $ 4000 = $ 200$ $5/100xx$4000 = $200$

Suppose all the money was invested at 9% then the income would be $\frac{9}{100} \times $ 4000 = $ 360$ $9/100xx$4000 = $360$

Consider this transition of total interest received by varying the amount deposited in each account.

This can be modelled by modelling just one account. If all of the money is in the 9% account then none is in the 5% account. If all the money is in the 5% account then there is none in the 9% account. So one account infers directly how much is in the other as the funds available is fixed at $4000

The result is a straight line graph where the gradient is the change in interest depending on how much is in each account.

Tont B

$m = \frac{360 - 200}{$4000}$ $m= (360-200)/("$4000")$

The equation of this graph will be:

$y = \frac{360 - 200}{4000} x + 200$ $y=(360-200)/4000 x+200$

$y = \frac{1}{25} x + 200$ $y=1/25x+200$

We are told that the target interest is $311.

Set $y = $ 311$ $y=$311$ giving

Dropping the $ sign for now

$311 = \frac{1}{25} x + 200$ $311=1/25x+200$

Subtract 200 from both sides

$111 = \frac{1}{25} x$ $111=1/25x$
Multiply both sides by 25

$x = $ 2775 \leftarrow$ $x=$2775 larr$ sum deposited in the 9% account

Thus the amount of the principle sum in the 5% account is:

$$ 4000 - $ 2775 = $ 1225$ $$4000-$2775 = $1225$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Check$ $color(blue)("Check")$

$For the 5% account$ $color(brown)("For the 5% account")$

$\frac{5}{100} \times $ 1225 = $ 61.25$ $5/100xx$1225 = $61.25$ interest

$For the 9% account$ $color(brown)("For the 9% account")$

$\frac{9}{100} \times $ 2775 = $ 249.75$ $9/100xx$2775 =$249.75$ interest

$$ 249.75$ $$249.75$
$ul($color(white)(2)61.25) larr " Add"$
$$311.00$

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You invested $4000, part at 5% and the remainder at 9% annual interest. At the end of the year, the total interest from these investments was $311. How much was invested at each rate?

2 Answers

Explanation:

Explanation:

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