Question

$13,957 is invested, part at 7% and the rest at 6%. If the interest earned from the amount invested at 7% exceeds the interest earned from the amount invested at 6% by $833.73, how much is invested at each rate?

Answer 1

Answer:

The Amount invested at 7% interest is $12,855

The Amount invested at 6% interest = $1,102  

Step-by-step explanation:

Given as :

The Total money invested = $13,957

Let The money invested at 7% = $p_1$  = $A

And The money invested at 6% = $p_2$ = $13957 - $A

Let The interest earn at 7% = $I_1$

And The interest earn at 6% = $I_2$

$I_1$ -  $I_2$ = $833.73

Let The time period = 1 year

Now, From Simple Interest method

Simple Interest = $\dfrac{\textrm principal\times \textrm rate\times \textrm time}{100}$

Or,  $I_1$ = $\dfrac{\textrm p_1\times \textrm 7\times \textrm 1}{100}$

Or,  $I_1$ = $\dfrac{\textrm A\times \textrm 7\times \textrm 1}{100}$

And

$I_2$ = $\dfrac{\textrm p_2\times \textrm 6\times \textrm 1}{100}$

Or,  $I_2$ = $\dfrac{\textrm (13,957 - A)\times \textrm 6\times \textrm 1}{100}$

∵  $I_1$ -  $I_2$ = $833.73

So, $\dfrac{\textrm A\times \textrm 7\times \textrm 1}{100}$ -  $\dfrac{\textrm (13,957 - A)\times \textrm 6\times \textrm 1}{100}$ = $833.73

Or, 7 A - 6 (13,957 - A) = $833.73 × 100

Or, 7 A - $83,742 + 6 A = $83373

Or, 13 A = $83373 + $83742

Or, 13 A = $167,115

∴ A = $\dfrac{167115}{13}$

i.e A = $12,855

So, The Amount invested at 7% interest = A = $12,855

And The Amount invested at 6% interest = ($13,957 - A) = $13,957 - $12,855

I.e The Amount invested at 6% interest = $1,102

Hence,The Amount invested at 7% interest is $12,855

And The Amount invested at 6% interest = $1,102   . Answer