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Fuseina has ¢40000 to invest.She decides to put part of the money in an account paying 4% interest and the reminder into stocks paying 6% interest . Her financial advisor tells her that the total annual income from these investments should be¢2040. How much should she invest at each rate?

Respuesta :

Answer:

The amounts invested in the account = ¢ 18000

and the amount invested in the stock = ¢40000-¢18000= ¢22000.

Step-by-step explanation:

Given that the account paying 4% interest and the stocks paying 6% interest.

Assuming she put ¢ x in the account, so

Income (interest) from the account = 4% of x = ¢ 0.04x...(i)

She has a total of ¢ 40000 to invest, so the remaining amount = ¢(40000-x).

As she put the remaining amount into stock, so

Income (interest) from the stock=6% of (40000-x)=¢(2400-0.06x)...(ii)

As total income from the investment is ¢ 2040, so on adding both the incomes from equations (i) and (ii), we have

[tex]2040=0.04x + (2400-0.06x) \\\\\Rightarrow 2040=2400 - 0.02x \\\\\Rightarrow 0.02x = 2400-2040=360 \\\\\Rightarrow x = 360/0.02=18000.[/tex]

So, the amount invested in the account = ¢ 18000

and the amount invested in the stock = ¢40000-¢18000= ¢22000.

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