Sally has two IRAs. IRA #1 earns interest at 8% effective annually and IRA #2 earns interest at 10% effective annually. She has not made any contributions since January 1, 2005, when the amount in IRA #1 was twice the amount in IRA #2. The sum of the two accounts on January 1, 2013 was $73,000. Determine how much was in IRA #2 on January 1, 2005. ? < $12,750 ? $12,750 but < $13,000 ? $13,000 but < $13,250 ? $13,250 but < $13,500 ? $13,500
Sally has two IRAs. IRA #1 earns interest at 8% effective annually and IRA #2 earns interest at 10% effective annually. She has not made any contributions since January 1, 2005, when the amount in IRA #1 was twice the amount in IRA #2. The sum of the two accounts on January 1, 2013 was
12,750 ?
13,000 ?
13,250 ?
13,500 ? $13,500
The Correct Answer and Explanation is:
Let’s solve the problem step-by-step and determine how much was in IRA #2 on January 1, 2005.
Step 1: Define variables
Let:
- xx = amount in IRA #2 on Jan 1, 2005
- Then, IRA #1 on Jan 1, 2005 = 2x2x
- Time from Jan 1, 2005 to Jan 1, 2013 = 8 years
- IRA #1 grows at 8% effective annually
- IRA #2 grows at 10% effective annually
Step 2: Future values after 8 years
- IRA #1:
2x×(1.08)82x \times (1.08)^8 - IRA #2:
x×(1.10)8x \times (1.10)^8
Total value in 2013 = $73,000, so: 2x(1.08)8+x(1.10)8=73,0002x(1.08)^8 + x(1.10)^8 = 73,000
Step 3: Plug in interest values
Calculate:
- (1.08)8≈1.85093(1.08)^8 \approx 1.85093
- (1.10)8≈2.14359(1.10)^8 \approx 2.14359
Now plug in: 2x(1.85093)+x(2.14359)=73,0003.70186x+2.14359x=73,0005.84545x=73,0002x(1.85093) + x(2.14359) = 73,000 \\ 3.70186x + 2.14359x = 73,000 \\ 5.84545x = 73,000
Solve for xx: x=73,0005.84545≈12,486.50x = \frac{73,000}{5.84545} \approx 12,486.50
Step 4: Determine the correct range
- x≈12,486.50x \approx 12,486.50
- So, IRA #2 had less than $12,750 on Jan 1, 2005
✅ Correct answer: < $12,750
Explanation
This problem involves compound interest and solving a system involving exponential growth. Sally has two IRAs growing at different effective annual rates: IRA #1 at 8% and IRA #2 at 10%. She made no further contributions after January 1, 2005. At that time, the balance in IRA #1 was twice that of IRA #2. After 8 years, the combined balance of both accounts was $73,000.
To determine the original amount in IRA #2, we define the amount in IRA #2 in 2005 as xx, making the amount in IRA #1 2x2x. Over 8 years, IRA #1 grows to 2x×(1.08)82x \times (1.08)^8, and IRA #2 grows to x×(1.10)8x \times (1.10)^8. Adding these together and equating to $73,000 gives the equation: 2x(1.08)8+x(1.10)8=73,0002x(1.08)^8 + x(1.10)^8 = 73,000
Substituting the values (1.08)8≈1.85093(1.08)^8 \approx 1.85093 and (1.10)8≈2.14359(1.10)^8 \approx 2.14359, we simplify and solve for xx: 2x(1.85093)+x(2.14359)=73,000⇒5.84545x=73,000⇒x≈12,486.502x(1.85093) + x(2.14359) = 73,000 \Rightarrow 5.84545x = 73,000 \Rightarrow x \approx 12,486.50
Thus, the amount in IRA #2 in 2005 was approximately $12,486.50, which is less than $12,750. This matches the answer choice “< $12,750”.
This kind of problem is common in actuarial exams and financial planning, requiring a solid understanding of exponential growth and financial equations.
