This Exam FM sample reference tests Equation of Value with Quarterly Deposits. Only choice C has the correct deposit count and exponents, so it is the valid equation of value. The result agrees with the published answer key, choice C.
How to solve this Equation of Value with Quarterly Deposits question
Setup
Setup
Convert 4.2% nominal monthly to a monthly rate of 0.0035. Quarterly deposits occur every three months, and exactly forty-one occur before month 124.
jm=0.042/12=0.0035,1+jq=(1.0035)3
Model
Model
Value all deposits and the terminal target at time 0. The immediate amount X remains undiscounted, each 100 deposit is discounted 3k months, and 1.9X is discounted 124 months.
41 deposits occur before month 124
Compute
Compute
Only choice C has the correct deposit count and exponents, so it is the valid equation of value.
X+k=1∑41(1.0035)3k100=(1.0035)1241.9X
Answer
Answer
The calculation gives X +∑ 3k for equation of value with quarterly deposits, matching published choice C.
choice C(C)
Continue without hunting through PDFs
All 461 Exam FM sample solutions in syllabus order
The searchable 2210-page manual includes this complete solution, its error analysis, and one original worked variant for every active reference.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AChoice A (124) does not match the checked equation of value with quarterly deposits result (X +∑ 3k); no distinct standard one-step error is identifiable.
BChoice B (X +∑ 3( k −1)) does not match the checked equation of value with quarterly deposits result (X +∑ 3k); no distinct standard one-step error is identifiable.
DChoice D (X +∑ k −1) does not match the checked equation of value with quarterly deposits result (X +∑ 3k); no distinct standard one-step error is identifiable.
EChoice E (X +∑ k −1) does not match the checked equation of value with quarterly deposits result (X +∑ 3k); no distinct standard one-step error is identifiable.
Original practice · fully worked
Original variant: quarterly deposits valued against two later withdrawals
A fund earns 2% per quarter. Deposits of X are made at the end of quarters 1 through 8. The fund pays 2,000 at the end of quarters 10 and 12. Which equation of value at time 8 correctly determines X?
A X(1 + 1.02 + ⋯ + 1.02⁷) = 2,000(1.02² + 1.02⁴)
B X(1 + 1.02 + ⋯ + 1.02⁸) = 2,000(v² + v⁴)
C X(1 + v + ⋯ + v⁷) = 2,000(v² + v⁴)
D X(1 + 1.02 + ⋯ + 1.02⁷) = 2,000(v² + v⁴)
E X(1 + 1.02 + ⋯ + 1.02⁷) = 4,000v⁴
Variant answer in brief
The deposit side has eight terms from power 0 through 7, and the withdrawal side is 2,000 times v squared plus v to the fourth. Choice D is the only equation with both timing patterns correct.
Setup
Setup
Choose time 8. The deposit at quarter 8 has no accumulation, while the first deposit accumulates for seven quarters.
v=1/1.02
Model
Model
Discount the quarter-10 and quarter-12 withdrawals two and four quarters back to the same comparison date.
Xk=0∑71.02k=2000v2+2000v4
Compute
Compute
The deposit side has eight terms from power 0 through 7, and the withdrawal side is 2,000 times v squared plus v to the fourth.
Xk=0∑71.02k=2000(v2+v4)
Answer
Answer
Choice D is the only equation with both timing patterns correct.
The 2210-page Financial Mathematics Proof Manual reorganizes 461 verified Exam FM solutions by syllabus skill and adds formula proofs, error patterns, and original worked practice.
