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Veronika [31]
3 years ago
6

Need help on this one...

Mathematics
1 answer:
lara31 [8.8K]3 years ago
5 0

Answer: 1,865,311 people in 11 years

800,000(1+.08)^11

1,865,311 population in 11 yeas

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$6,000 dollars is invested in two different accounts earning 3% and 5% interest. at the end of one year, the two accounts earned
Anettt [7]

<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%

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