Question

Juana invests $1,500 in an account accumulating 3% interest according to the equation , where v represents the value of the account after y years. Marquez and Calvin invest the same amount of money at the same rate. Marquez invests three years before Juana, and Calvin invests two years after Juana. By what factor would Calvin’s investment need to be increased to equal Marquez’s investment at any time after Calvin’s investment is made?

Answer 1

The answer will be D.

Answer 2

Answer:

By a factor of 1.16 Calvin’s investment need to be increased to equal Marquez’s investment at any time after Calvin’s investment is made.

Step-by-step explanation:

Given Juana invests $1,500 in an account accumulating 3% interest.

We know $A=P(1+\frac{r}{100})^n$

Where, A is amount

P is invested amount

r is rate of interest

n is time

Given  v represents the value of the account after y years.

P = $ 1,500 , r = 3%

Thus, substituting in above , we get,

$V=1500(1+\frac{3}{100})^y$

$\Rightarrow V=1500(1+0.03)^y$

$\Rightarrow V=1500(1.03)^y$  ...........(1)

This equation (1) represents the amount in Juana's account.

Also, Marquez and Calvin invest the same amount of money at the same rate.

That is P = $ 1,500  r = 3%

Marquez invests three years before Juana thus time becomes  (y +3) years

Thus, the amount in Marquez's account is given by,

$\Rightarrow V=1500(1.03)^{(y+3)}$   ......(2)

Calvin invests two years after Juana thus time becomes  (y - 2) years.

Thus, the amount in Calvin's account is given by,

$\Rightarrow V=1500(1.03)^{(y-2)}$   ......(3)

Thus, the factor by which Calvin’s investment need to be increased to equal Marquez’s investment at any time after Calvin’s investment is made is given by

let the factor be x

Then

$x=\dfrac{\text{amount in Marquez's account}}{\text{amount in Calvin's account}}$

substitute, we get,

$x=\frac{1500(1.03)^{(y+3)}}{1500(1.03)^{(y-2)}}$

Simplify , we get,

$x=\frac{(1.03)^y(1.03)^3}{(1.03)^y(1.03)^{-2}}}\\\\\ x=\frac{(1.03)^3}{(1.03)^{-2}}}$

Simplify , we get,

Using rule of exponent $a^{-n}=\frac{1}{a^n}$

$x=(1.03)^{3+2}\\\\ x=(1.03)^5=1.159$(appox)

Thus, by a factor of 1.16 Calvin’s investment need to be increased to equal Marquez’s investment at any time after Calvin’s investment is made.