juana invests $1,500 in an account accumulating 3% interest according to the equation $v = 1500(1.03)^y$…

juana invests $1,500 in an account accumulating 3% interest according to the equation $v = 1500(1.03)^y$, 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 calvins investment need to be increased to equal marquezs investment at any time after calvins investment is made?\n\n86.3%\n97.1%\n115.0%\n115.9%

juana invests $1,500 in an account accumulating 3% interest according to the equation $v = 1500(1.03)^y$, 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 calvins investment need to be increased to equal marquezs investment at any time after calvins investment is made?\n\n86.3%\n97.1%\n115.0%\n115.9%

Answer

Explanation:

Step1: Find Marquez's investment formula

Marquez invests 3 years before Juana. If we use $y$ as the number of years since Juana's investment, then the number of years for Marquez's investment is $y + 3$. So the value of Marquez's investment $M=1500(1.03)^{y + 3}$.

Step2: Find Calvin's investment formula

Calvin invests 2 years after Juana. So the number of years for Calvin's investment is $y-2$. The value of Calvin's investment $C = 1500(1.03)^{y - 2}$.

Step3: Find the factor

Let the factor be $x$. We want $xC = M$. So $x\times1500(1.03)^{y - 2}=1500(1.03)^{y + 3}$. Divide both sides by $1500(1.03)^{y - 2}$, we get $x=\frac{(1.03)^{y + 3}}{(1.03)^{y - 2}}$. Using the exponent - rule $\frac{a^m}{a^n}=a^{m - n}$, we have $x=(1.03)^{(y + 3)-(y - 2)}=(1.03)^{5}$. Calculate $(1.03)^{5}=1.159274$. As a percentage, it is approximately $115.9%$.

Answer:

115.9%