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lina2011 [118]
3 years ago
11

$6,000 dollars is invested in two different accounts earning 3% and 5% interest. at the end of one year, the two accounts earned

$220 in interest. how much money was invested at 5%?
Mathematics
2 answers:
Anettt [7]3 years ago
8 0

<u>Answer-</u>

<em>$2000</em><em> were invested at 5%</em>

<u>Solution-</u>

Let x amount of money was invested at 5% and (6000-x) amount was invested as 3%

We know that,

i=\dfrac{P\cdot r\cdot t}{100}

Putting the values,

i_1=\dfrac{P_1\cdot r_1\cdot t}{100}

i_1=\dfrac{x\cdot 5\cdot 1}{100}=\dfrac{5x}{100}

And

i_2=\dfrac{P_2\cdot r_2\cdot t}{100}

i_2=\dfrac{(6000-x)\cdot 3\cdot 1}{100}=\dfrac{3(6000-x)}{100}

According to the question,

\Rightarrow \dfrac{5x}{100}+\dfrac{3(6000-x)}{100}=220

\Rightarrow \dfrac{5x+3(6000-x)}{100}=220

\Rightarrow 5x+3(6000-x)=22000

\Rightarrow 5x+18000-3x=22000

\Rightarrow 5x-3x=22000-18000

\Rightarrow 2x=4000

\Rightarrow x=2000

Therefore, $2000 were invested at 5%

Scilla [17]3 years ago
8 0
The 2 values invested are x and 6000-x

I=PRT
interest=principal times rate times time

time is 1 year so t=1
r=0.03 ad 0.05
we want to find 5% so x is wha we solve for

the interests are
0.03(6000-x) and 0.05x
they add to 220
220=0.03(6000-x)+0.05x
220=180-0.3x+0.05x
220=180+0.02x
minus 180 both sides
40=0.02x
divide both sides by 0.02
2000=x
$2000 was invested at 5%
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(0.95+1.02+ 1.01+ 0.98 ) /4 =0.99

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n=4

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The t value for 95% confidence with df= 3 is t=3.182

Substituting the values in equation (i)

0.99 ± 3.182 * 0.0274/√4

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0.99 ± 3.182 * 0.0137

0.99 ± 0.0436

(0.9464,1.034)

8 0
2 years ago
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