Samantha invests #$1000# at #6%# per annum compounded quarterly. Mark invests #$1200# at #5.5%# per annum compounded quarterly. When will the balances in their accounts be equal?

1 Answer
Aug 31, 2016

The balances in the accounts will be equal in #37# years.

Explanation:

For problems such as these, we use the formula #A = P(1 + r/n)^(nt)#, where P is the amount of money you begin with, #r# is the rate of interest, #n# is the frequency the interest is compounded and A is the final amount. #t# is the time in years.

We have to write a system of equations and solve for #A# and for #t#.

Equation 1:

#A = 1000(1 + (0.06)/4)^(4t)#

Equation 2:

#A = 1200(1 + (0.055)/4)^(4t)#

Substitute:

#1200(1 + (0.055)/4)^(4t) = 1000(1 + (0.06)/4)^(4t)#

#1200(1.01375)^(4t) = 1000(1.015)^(4t)#

#1200(1.01375)^(4t) - 1000(1.015)^(4t) = 0#

Solve using a graphing calculator. If you're using a TI-83 or a TI-84, change the t's to x's. Enter #y_1 = 1200(1.01375)^(4x) - 1000(1.015)^(4x)# and #y_2 = 0#. Then press CALC#-># Intersect.

You will of course want a positive intersection. Once you have moved your cursor sufficiently, press ENTER twice before the calculator will say guess? followed by INTERSECTION.

The result it gives you should say #36.99#, or #37# years.

Hopefully this helps!