Free CPA TCP (Tax Compliance & Planning) Entity Tax Planning Practice Questions

Entity tax planning on the CPA TCP exam covers entity selection and conversion strategies, compensation planning (reasonable compensation, deferred compensation), retirement plan optimization, and tax-efficient business structuring.

179 Questions
65 Easy
63 Medium
51 Hard
2026 Syllabus

Sample Questions

Question 1 Easy
What is the maximum annual contribution a self-employed individual can make to a SEP-IRA for 2025?
Solution
C is correct.

A self-employed individual may contribute to a SEP-IRA up to 25% of net self-employment income (compensation), after adjusting for the deduction for one-half of self-employment tax and the SEP contribution itself. The effective rate is approximately 20% of net self-employment earnings before the SEP deduction. The dollar cap for 2025 is 70,000 (indexed for inflation).
Question 2 Medium
An employer establishes a Section 125 cafeteria plan. Which of the following benefits can be offered through the cafeteria plan on a pre-tax basis?
Solution
A is correct.

Under IRC Section 125, a cafeteria plan allows employees to choose between cash (taxable salary) and qualified benefits on a pre-tax basis. Qualified benefits that can be offered include health insurance premiums, dental and vision coverage, dependent care assistance (up to 5,000 per year under Section 129), health FSA contributions (up to 3,300 for 2025), adoption assistance, and certain other benefits.
Question 3 Hard
An actuary fits an AR(2) model to a stationary time series: Yt=4+0.5Yt−1−0.2Yt−2+ϵtY_t = 4 + 0.5 Y_{t-1} - 0.2 Y_{t-2} + \epsilon_t, where ϵt∼N(0,σ2)\epsilon_t \sim N(0, \sigma^2). What is the unconditional mean μ=E[Yt]\mu = E[Y_t] of this process?
Solution
C is correct.

For a stationary AR(2) process Yt=c+ϕ1Yt−1+ϕ2Yt−2+ϵtY_t = c + \phi_1 Y_{t-1} + \phi_2 Y_{t-2} + \epsilon_t, taking expectations of both sides and using E[Yt]=E[Yt−1]=E[Yt−2]=μE[Y_t] = E[Y_{t-1}] = E[Y_{t-2}] = \mu and E[ϵt]=0E[\epsilon_t] = 0:
μ=c+ϕ1μ+ϕ2μ\mu = c + \phi_1 \mu + \phi_2 \mu
μ(1−ϕ1−ϕ2)=c\mu(1 - \phi_1 - \phi_2) = c
μ=c1−ϕ1−ϕ2=41−0.5−(−0.2)=41−0.5+0.2=40.7=5.714\mu = \frac{c}{1 - \phi_1 - \phi_2} = \frac{4}{1 - 0.5 - (-0.2)} = \frac{4}{1 - 0.5 + 0.2} = \frac{4}{0.7} = 5.714
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