On Nov 1 2006 Abby invests 1000 in an account earning 10 sim

On Nov 1, 2006, Abby invests $1000 in an account earning 10% simple annual interest. On the same day, Ben invests $1000 at a nominal annual interest rate of X, convertible monthly. On September 1, 2010; the accumulated value s of Abby\'s and Ben\'s investments are equal. Calculate X. A. 8.50% B. 8.54% C. 8.56% D. 8.52% E. 8.48%

Solution

Abby invests $1000 earning 10 % simple interest on November 1, 2006 and Ben invests the same day, the same amount i.e. $ 1000 at a nominal rate of x convertible monthly. The accumulated values of both the investments are equal on September 1, 2010, i.e. after 3 years, 9 months = 3.75 years. The simple interest on $1000 for 3.75 years @ 10 % is $ 1000*3.75*10/100 =$ 375. Thus, the accumulated value of Abby’s investment on September1, 2010 is $1375.

Then, the accumulated value of Ben’s investment on September1, 2010 is also $1375. Therefore,        1000[ 1+x/1200]^12*3.75 = 1375 or, (1+x/1200)⁴⁵ = 1375/1000 = 1.375. On taking logarithm of both the sides, we have 45 log(1+x/1200)= log 1.375 = 0.138302698. Then log(1+x/1200) = 0.138302698/45 = 0.003073393293. Therefore, (1+x/1200) = 10^{0.003073393293} = 1.007101849. Then x/1200 =0. 007101849 so that x = 1200*0. 007101849 = 8.52 (approximately). The answer D is correct.