Question

Two investments totaling $50,000 produce an annual income of $4795. One investment yields 8% per year, while the other yields 7% per year. How much is invested at each rate?

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

• Two investments totaling $ 50,000 produce an annual income of $ 4795.

• One investment yields 8% per year, while the other yields 7% per year.

Let assume that

• Investment in first case is $ x

• Investment in second case is $ y

So, According to statement

Two investments totaling to $50,000

$\rm \implies\:x + y = 50000$

$\rm \implies\: \boxed{\tt{ y = 50000 - x}} - - - - (1)$

According to second condition

Two investments totaling $ 50,000 produce an annual income of $ 4795. One investment yields 8% per year, while the other yields 7% per year.

$\rm :\longmapsto\:\dfrac{8}{100}x + \dfrac{7}{100}y = 4795$

$\rm :\longmapsto\:8x + 7y = 479500$

$\rm :\longmapsto\:8x + 7(50000 - x) = 479500$

$\rm :\longmapsto\:8x + 350000 - 7x = 479500$

$\rm :\longmapsto\:x = 479500 - 350000$

$\bf\implies \:x = 29500$

On substituting the value of x in equation (1), we get

$\bf\implies \:y = 20500$

Hence,

• The investment in first case is $ 29500

• The investment in second case is $ 20500

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Basic Concept Used :-

To solve systems using substitution, follow this procedure:

• Select one equation and solve it to get one variable in terms of second variables.

• In the second equation, substitute the value of variable evaluated in Step 1 to reduce the equation to one variable.

• Solve the new equation to get the value of one variable.

• Substitute the value found in to any one of two equations involving both variables and solve for the other variable.