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rosijanka [135]
3 years ago
6

MN is translated 4units down and 3units to the right to formPQ

Mathematics
1 answer:
yanalaym [24]3 years ago
8 0
7 that is what t should be
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Matthew invested $3,000 into two accounts. One account paid 3% interest and the other paid 8% interest. He earned 4% interest on
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<u>Answer:</u>

<em>Mathew invested</em><em> $600 and $2400</em><em> in each account.</em>

<u>Solution:</u>

From question, the total amount invested by Mathew is $3000. Let p = $3000.

Mathew has invested the total amount $3000 in two accounts. Let us consider the amount invested in first account as ‘P’

So, the amount invested in second account = 3000 – P

Step 1:

Given that Mathew has paid 3% interest in first account .Let us calculate the simple interest (I_1) earned in first account for one year,

\text {simple interest}=\frac{\text {pnr}}{100}

Where  

p = amount invested in first account

n = number of years  

r = rate of interest

hence, by using above equation we get (I_1) as,  

I_{1}=\frac{P \times 1 \times 3}{100} ----- eqn 1

Step 2:

Mathew has paid 8% interest in second account. Let us calculate the simple interest (I_2) earned in second account,

I_{2} = \frac{(3000-P) \times 1 \times 8}{100} \text { ------ eqn } 2

Step 3:

Mathew has earned 4% interest on total investment of $3000. Let us calculate the total simple interest (I)

I = \frac{3000 \times 1 \times 4}{100} ----- eqn 3

Step 4:

Total simple interest = simple interest on first account + simple interest on second account.

Hence we get,

I = I_1+ I_2 ---- eqn 4

By substituting eqn 1 , 2, 3 in eqn 4

\frac{3000 \times 1 \times 4}{100} = \frac{P \times 1 \times 3}{100} + \frac{(3000-P) \times 1 \times 8}{100}

\frac{12000}{100} = \frac{3 P}{100} + \frac{(24000-8 P)}{100}

12000=3P + 24000 - 8P

5P = 12000

P = 2400

Thus, the value of the variable ‘P’ is 2400  

Hence, the amount invested in first account = p = 2400

The amount invested in second account = 3000 – p = 3000 – 2400 = 600  

Hence, Mathew invested $600 and $2400 in each account.

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