RD Practice Questions (2026) — Free Sample Questions with Answers

Answer: A

Reject the null hypothesis; the difference is statistically significant: Correct. Because p (0.03) is below alpha (0.05), the null hypothesis of no difference is rejected and the result is statistically significant.

Fail to reject the null; the difference is not statistically significant: Wrong. A p-value below alpha leads to rejecting the null hypothesis.

There is a 97% probability that the new diet caused the LDL change: Wrong. A p-value is not the probability that the hypothesis is true. It is the probability of results this extreme if the null were true.

The result proves the difference is large enough to be clinically important: Wrong. Statistical significance says nothing about the size or clinical importance of the effect. Look at the effect size and confidence interval.

If p < alpha, reject H0 (statistically significant). Statistical significance is not the same as clinical significance. A Type I error is rejecting a true H0, and a Type II error is failing to reject a false H0.

Answer: D

Recommended Dietary Allowance: Wrong. The RDA meets the needs of 97–98% of individuals. The proportion below it overestimates the prevalence of inadequacy in a group.

Adequate Intake: Wrong. An AI is set when an EAR cannot be determined. Intakes below the AI cannot be read as a prevalence of inadequacy.

Tolerable Upper Intake Level: Wrong. The UL is used to assess the risk of excessive intake, not inadequacy.

Estimated Average Requirement: Correct. Using the EAR cut-point method, the share of the group with usual intakes below the EAR estimates the prevalence of inadequate intake.

For groups: use the EAR to estimate the prevalence of inadequacy and the UL for the risk of excess. For individuals: plan with the RDA or AI. Never use the RDA as a cut-point for groups.

Answer: C

Stroke related to hypertension as evidenced by MRI findings: Wrong. Stroke is a medical diagnosis. The 'P' must be a nutrition problem that the RDN can resolve or improve.

Inadequate oral intake related to poor intake as evidenced by low intake: Wrong. The etiology and the signs restate the problem. The statement is circular and gives no root cause or measurable evidence.

Inadequate oral intake related to dysphagia as evidenced by <50% of needs eaten for 7 days: Correct. It names a nutrition problem, a specific cause the RDN can target, and measurable signs and symptoms.

Inadequate oral intake related to dysphagia as evidenced by need for diet education: Wrong. 'Need for education' is an intervention, not objective or subjective evidence of the problem.

PES: Problem (nutrition diagnosis) related to Etiology (root cause) as evidenced by Signs and Symptoms (measurable data). The intervention targets the etiology, and monitoring tracks the signs and symptoms.

Answer: B

1,320 kcal and 90 g protein: Wrong. 1,320 is the daily volume in mL (55 × 24). It was not multiplied by the 1.5 kcal/mL density.

1,980 kcal and 90 g protein: Correct. Volume = 55 mL/h × 24 h = 1,320 mL. Energy = 1,320 × 1.5 = 1,980 kcal. Protein = 1.32 L × 68 g/L = 89.8 g, about 90 g.

1,980 kcal and 68 g protein: Wrong. 68 g is the protein in one liter. The patient receives 1.32 L, so about 90 g.

2,970 kcal and 135 g protein: Wrong. These figures apply the 1.5 density a second time (1,980 × 1.5 and 90 × 1.5).

Tube feeding math: daily volume = rate × hours. kcal = volume × kcal/mL. Protein = liters × g/L. Free water = volume × the formula's water fraction.

Answer: C

About 12.5 lb: Wrong. 12.5 lb is the edible portion only (200 oz ÷ 16) and ignores cooking and trim losses.

About 16.3 lb: Wrong. This adds 30% to the edible amount (12.5 × 1.3) instead of dividing by the 70% yield.

About 17.9 lb: Correct. EP needed = 50 × 4 oz = 200 oz. AP = 200 ÷ 0.70 = 285.7 oz ÷ 16 = 17.9 lb.

About 8.8 lb: Wrong. This multiplies by the yield (12.5 × 0.70) instead of dividing, which would leave too little meat.

As-purchased quantity = edible portion needed ÷ yield percentage (as a decimal). Dividing by yield always gives a larger AP amount.

Answer: C

From 135°F to 41°F within a total of 4 hours: Wrong. The Food Code uses a two-stage process totaling 6 hours, with a strict first stage of 2 hours.

From 140°F to 70°F in 4 hours, then to 41°F in 2 more hours: Wrong. The stage times are reversed. The first, fastest stage (to 70°F) must be done within 2 hours.

From 135°F to 70°F in 2 hours, then to 41°F in 4 more hours: Correct. Cooked TCS food must cool from 135°F to 70°F within 2 hours and to 41°F or less within a total of 6 hours.

From 135°F to 70°F in 4 hours, then to 41°F in 2 more hours: Wrong. Allowing 4 hours for the first stage leaves food too long in the range where bacteria grow fastest.

Two-stage cooling: 135°F → 70°F in 2 hours, then to 41°F or below within 6 hours total. Use shallow pans, ice baths or ice wands, and blast chillers.