# Math Help - Simple interest question

1. ## Simple interest question

A company invests $50000, some into an account paying 5.2% per annum simple interest and the remainder in an account paying 4.5% simple interest per annum. After 2 years the$50000 had grown to $54976. How much went into each account? What is the formula for this type of question? I worked it out by trial and error, but I know that this isnt the right way. Thanks 2. Originally Posted by Local_lunatic A company invests$50000, some into an account paying 5.2% per annum simple interest and the remainder in an account paying 4.5% simple interest per annum. After 2 years the $50000 had grown to$54976. How much went into each account?

What is the formula for this type of question? I worked it out by trial and error, but I know that this isnt the right way.

Thanks
Recall that Simple Interest = Principle * Rate * Time

The interest gained here is, of course, 54976 - 50000 = 4976

So now we set up our simple interest equations. we have two accounts, so we must have a sum of the two simple interests giving 4976

Thus we have: $Principle_1 * Rate_1 * Time + Principle_2 * Rate_2 * Time = 4976$

so let's say the first account is the one that gives 5.2% interest. And let's say we invest $x$ dollars into this account. what do we have remaining? well, $(50000 - x)$ dollars, of course.

so our equation becomes:

$x*5.2 \% * 2 + (50000 - x)*4.5 \% *2 = 4796$

or $2(0.052)x + 2(0.045)(50000 - x) = 4796$

now solve for $x$, that will be the amount you invested into the first account. then subtract that answer from 50000 to find how much you invested into the second account

3. Originally Posted by Jhevon
Recall that Simple Interest = Principle * Rate * Time

The interest gained here is, of course, 54976 - 50000 = 4976

So now we set up our simple interest equations. we have two accounts, so we must have a sum of the two simple interests giving 4976

Thus we have: $Principle_1 * Rate_1 * Time + Principle_2 * Rate_2 * Time = 4976$

so let's say the first account is the one that gives 5.2% interest. And let's say we invest $x$ dollars into this account. what do we have remaining? well, $(50000 - x)$ dollars, of course.

so our equation becomes:

$x*5.2 \% * 2 + (50000 - x)*4.5 \% *2 = 4796$

or $2(0.052)x + 2(0.045)(50000 - x) = 4796$

now solve for $x$, that will be the amount you invested into the first account. then subtract that answer from 50000 to find how much you invested into the second account
That's basically where I managed to get to except that I simplified it down a bit more.

$0.104x + 0.09(50000 - x) = 4796$

or $0.052x + 0.045(50000 - x) =2398$

How would I go about solving it for x though?

4. Originally Posted by Local_lunatic
That's basically where I managed to get to except that I simplified it down a bit more.

$0.104x + 0.09(50000 - x) = 4796$

or $0.052x + 0.045(50000 - x) =2398$

How would I go about solving it for x though?

do you know how to expand brackets?

$2(0.052)x + 2(0.045)(50000 - x) = 4796$

$\Rightarrow 0.104x + 0.09(50000 - x) = 4796$

now we expand the brackets for the second term. we take the 0.09 and multiply each term in the brackets one at a time

$\Rightarrow 0.104x + 0.09(50000) + 0.09(-x) = 4796$

$\Rightarrow 0.104x + 4500 - 0.09x = 4796$

$\Rightarrow 0.014x + 4500 = 4796$

Can you take it from here?

5. So from the beginning

$2(0.052)x + 2(0.045)(50000-x) = 4976$

$0.104x + 0.09(50000-x) = 4976$

$0.104x + 4500 - 0.09x = 4976$

$0.104x-0.09x = 4976 - 4500$

$0.014x = 476$

$476 / 0.014 = 34000$

$50000 - 34000 = 16000$

Account 1(x) =$34000 Account 2 =$16000

Here's another one.

Junior invests $4000, some into an account paying 6.5%pa simple interest and the remainder into an account paying 5.8%pa simple interest. After 18months the$4000 has grown to $4391.60. How much did Junior put into each account? $1.5(0.065)x + 1.5(0.058)(4000-x) = 381.6$ $0.0975x + 0.087(4000-x) = 381.6$ $0.0975x + 348 - 0.087x = 381.6$ $0.095x - 0.087x = 381.6 - 348$ $0.0105x = 33.6$ $33.6 / 0.0105 = 3200$ $4000-3200 = 800$ Account 1(x) =$3200
Account2 = $800 Have I done these right? 6. Originally Posted by Local_lunatic So from the beginning $2(0.052)x + 2(0.045)(50000-x) = 4976$ $0.104x + 0.09(50000-x) = 4976$ $0.104x + 4500 - 0.09x = 4976$ $0.104x-0.09x = 4976 - 4500$ $0.014x = 476$ $476 / 0.014 = 34000$ $50000 - 34000 = 16000$ Account 1(x) =$34000
Account 2 =$16000 Here's another one. Junior invests$4000, some into an account paying 6.5%pa simple interest and the remainder into an account paying 5.8%pa simple interest. After 18months the $4000 has grown to$4391.60. How much did Junior put into each account?

$1.5(0.065)x + 1.5(0.058)(4000-x) = 381.6$

$0.0975x + 0.087(4000-x) = 381.6$

$0.0975x + 348 - 0.087x = 381.6$

$0.095x - 0.087x = 381.6 - 348$

$0.0105x = 33.6$

$33.6 / 0.0105 = 3200$

$4000-3200 = 800$

Account 1(x) = $3200 Account2 =$800

Have I done these right?
it should be $1.5(0.065)x + 1.5(0.058)(4000-x) = \color {red}391.6$

but you're setup looks correct. work out the answer with the corrected RHS

7. My bad, the question was meant to be $4381.60 instead of$4391.60. Thanks for helping me understand it