Math Review Exam
88%
Q&A
2024
,1. A nurse needs to administer 0.5 mg of a drug to a patient. The drug is
available in a concentration of 2 mg/mL. How many mL of the drug
should the nurse give? (Round your answer to the nearest tenth)
A) 0.2 mL
B) 0.25 mL
C) 0.4 mL
D) 0.5 mL
Answer: B) 0.25 mL
Rationale: To find the volume of the drug, use the formula V = D/C,
where V is the volume, D is the dose, and C is the concentration. Plug in
the given values and solve for V: V = 0.5/2 = 0.25 mL.
2. A patient has a body mass index (BMI) of 27.5 kg/m^2. What is the
patient's weight in pounds if the patient is 5 feet 6 inches tall? (Round
your answer to the nearest whole number)
A) 150 lbs
B) 165 lbs
C) 171 lbs
D) 180 lbs
Answer: C) 171 lbs
Rationale: To find the weight in pounds, use the formula W = BMI x H^2
x 703, where W is the weight, BMI is the body mass index, H is the height
in inches, and 703 is a conversion factor. Convert the height to inches: 5
feet 6 inches = 66 inches. Plug in the given values and solve for W: W =
27.5 x 66^2 x 703 / (10000) = 170.8 lbs. Round to the nearest whole
number: 171 lbs.
3. A patient has a blood pressure of 150/90 mmHg. What is the patient's
pulse pressure?
A) 60 mmHg
B) 75 mmHg
C) 90 mmHg
D) 120 mmHg
Answer: A) 60 mmHg
Rationale: To find the pulse pressure, subtract the diastolic pressure from
the systolic pressure: 150 - 90 = 60 mmHg.
, 4. A nurse needs to prepare a solution of dextrose 5% in water (D5W) and
normal saline (NS) for a patient. The nurse has a bag of D5W that
contains 1000 mL and a bag of NS that contains 500 mL. How many mL
of each solution should the nurse mix to obtain a final volume of 750 mL
of D5W/NS?
A) 375 mL of D5W and 375 mL of NS
B) 500 mL of D5W and 250 mL of NS
C) 600 mL of D5W and 150 mL of NS
D) None of the above
Answer: B) 500 mL of D5W and 250 mL of NS
Rationale: To find the amount of each solution, use the formula M1V1 +
M2V2 = M3V3, where M is the concentration and V is the volume of each
solution. The concentration of D5W is 5%, or 0.05, and the concentration
of NS is 0%, or 0. Plug in the given values and solve for V1 and V2:
0.05V1 + 0V2 = (0.05)(750)
V1 = (0.05)(750)/0.05 = 750/0.05 = 500 mL
V2 = V3 - V1 = 750 - 500 = 250 mL
5. A patient has a serum sodium level of 140 mEq/L and a serum
potassium level of 4 mEq/L. What is the patient's anion gap?
A) -4 mEq/L
B) -2 mEq/L
C) +2 mEq/L
D) +4 mEq/L
Answer: C) +2 mEq/L
Rationale: To find the anion gap, subtract the sum of serum sodium and
potassium from the sum of serum chloride and bicarbonate: AG = (Na +
K) - (Cl + HCO3). Assume that the normal values for chloride and
bicarbonate are 100 mEq/L and
24 mEq/L, respectively. Plug in the given values and solve for AG: AG =
(140 +
4) - (100 +
24) =
144 -
124 =
20 mEq/L.