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alexira [117]
3 years ago
7

$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 e

arned from the amount invested at 6% by $833.73, how much is invested at each rate?
Mathematics
1 answer:
ch4aika [34]3 years ago
8 0

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,<u> From Simple Interest method</u>

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

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