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A brother and a sister invested part of their $600 of allowance money at 4% and the remainder as 7%. Their annual income from these two investments was equivalent to an income of 6% on the entire sum. How much was invested at each rate?

Respuesta :

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Define x and y :
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Let the amount they invested in 4% be x
Let the amount they invested in 7% be y

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Construct equations :
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Total amount invested:
x + y = 600

The two investment is equivalent to 6% of the entire sum:
0.04x + 0.07y = 0.06 x 600
0.04x + 0.07y = 36

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Solve for x and y :
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x + y = 600    ------------------ (1)
0.04x + 0.07y = 36 ---------- (2)

From equation (1):
x + y = 600
x = 600 - y ---------- Sub into (2)
0.04(600 - y) + 0.07y = 36
24 - 0.04y + 0.07y = 36
0.03y = 12
y = 400 -------- Sub into (1)
x + 400 = 600
x = 200

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Find amount invested :
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4% investment = x = $200
7% investment = y = $400

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Answer: They invested $200 in the 4% investment
                               and $400 in the 7% investment.
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Answer:

$200 invested in 4% and $400 invested in 7%

Step-by-step explanation:

let x be the amount invested in 4%

Here is the primary equation:

0.04x + 0.07(600-x) = 0.06(600)

0.04x represents the return on the 4%

0.07(600-x) is equal to the return on the 7%

0.06(600) is equal to the annual income

You then solve it:

0.04x + 42 - 0.07x = 0.06(600)

-0.03x + 42 = 36

-0.03x = -6

x = 200

$200 invested in 4%

600 - x = 600 - 200 = $400 invested in 7%

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