ARRT Nuclear Medicine Boards (Math and
Stats)
1/2 + 1/3 =
a. 2/5
b. 2/6
c. 4/6
d. 5/6 - ANS-D. 5/6
or 1/3
10 mCi is equal to how many becquerels?
A. 370 Bq
B. 370 kBq
C. 370 MBq
D. 370 GBq - ANS-C. 370 MBq
1 curie = 3.7 1010 disintegrations per second [dps], and in SI units, 1 Bq = 1 dps;
therefore, 1 Ci = 37 GBq and 1 mCi = 37 MBq.
In this problem, activity in
Bq = 10mCi * (10^-3 Ci/mCi) * (3.7 *10^10 Bq/Ci)
20 mCi is equal to how many becquerels?
a. 0.74 GBq
b. 0.37 GBq
c. 0.54 GBq
d. 0.20 GBq - ANS-A. 0.74 GBq
1 curie = 3.7 1010 disintegrations per second [dps], and in SI units 1 Bq = 1 dps;
therefore, 1 Ci = 37 GBq and 1 mCi = 37 MBq.
In this problem, activity in
Bq = 20 mCi * ( 10^-3 Ci/mCi) * (3.7*10^10 Bq/Ci)
30% of 7 ml =
a. 0.23 ml.
b. 2.10 ml.
c. 2.30 ml.
d. 4.90 ml. - ANS-b. 2.10 ml.
, Percentages are values expressed as a fraction of some whole, entire value: 75% of
some number is the same as 0.75 multiplied by that number.
A 5-ml sample of a standard diluted 1:10,000 produces 27,200 counts in the well
counter. A 5-ml sample of patient plasma, counted for the same time as the diluted
standard sample, produces 99,100 counts. What is the plasma volume?
a. 10.8 liter
b. 2.74 liter
c. 1.73 liter
d. 0.91 liter - ANS-b. 2.74 liter
The dilution principle can also be used to measure an unknown volume. Using the
dilution principle, it is possible to estimate a patient's plasma volume, C1V1 = C2V2.
Because the standard may be too concentrated, a portion of the standard is diluted and
counted while the full-strength undiluted standard can be injected into the patient. The
standard can be diluted by adding 1 ml of it to a flask, which is then filled to 500 ml with
distilled water. It could then be expressed as a 1:500 diluted standard. The standard
counts have to be multiplied by the dilution factor of 500 to obtain the true standard
concentration that was injected into the patient. Taking this into consideration, the new
equation to be used to calculate the plasma volume would be C1V1 = (DF)C2V2.
A 99Mo99mTc generator is calibrated for 1.00 Ci of 99Mo at 7:00 AM Monday. The
generator is eluted daily Monday through Friday at 8:00 AM, but the workload is
especially heavy on Friday so the generator is eluted again at 3:00 PM to obtain more
99mTc. Assuming 100% elution efficiency, how many mCi of 99mTc will be eluted at
3:00 PM Friday?
a. 930 mCi
b. 710 mCi
c. 160 mCi
d. 80 mCi - ANS-c. 160 mCi
The 99mTc radioactivity expected to be eluted from the generator can be calculated by
knowing three values: (1) the activity of 99Mo in the generator, (2) the time since the
last elution of the generator, and (3) the ratio of 99mTc to 99Mo in the generator. Using
the equation, A = Ci * e -0.693 (t/65.9 hr), enter in the known values and solve for the
unknown.
A 99Mo99mTc generator is eluted Monday at 7:00 AM, producing 1.8 Ci of 99mTc in the
eluate vial, in 20 ml saline. What volume of eluate should be withdrawn from the eluate
Stats)
1/2 + 1/3 =
a. 2/5
b. 2/6
c. 4/6
d. 5/6 - ANS-D. 5/6
or 1/3
10 mCi is equal to how many becquerels?
A. 370 Bq
B. 370 kBq
C. 370 MBq
D. 370 GBq - ANS-C. 370 MBq
1 curie = 3.7 1010 disintegrations per second [dps], and in SI units, 1 Bq = 1 dps;
therefore, 1 Ci = 37 GBq and 1 mCi = 37 MBq.
In this problem, activity in
Bq = 10mCi * (10^-3 Ci/mCi) * (3.7 *10^10 Bq/Ci)
20 mCi is equal to how many becquerels?
a. 0.74 GBq
b. 0.37 GBq
c. 0.54 GBq
d. 0.20 GBq - ANS-A. 0.74 GBq
1 curie = 3.7 1010 disintegrations per second [dps], and in SI units 1 Bq = 1 dps;
therefore, 1 Ci = 37 GBq and 1 mCi = 37 MBq.
In this problem, activity in
Bq = 20 mCi * ( 10^-3 Ci/mCi) * (3.7*10^10 Bq/Ci)
30% of 7 ml =
a. 0.23 ml.
b. 2.10 ml.
c. 2.30 ml.
d. 4.90 ml. - ANS-b. 2.10 ml.
, Percentages are values expressed as a fraction of some whole, entire value: 75% of
some number is the same as 0.75 multiplied by that number.
A 5-ml sample of a standard diluted 1:10,000 produces 27,200 counts in the well
counter. A 5-ml sample of patient plasma, counted for the same time as the diluted
standard sample, produces 99,100 counts. What is the plasma volume?
a. 10.8 liter
b. 2.74 liter
c. 1.73 liter
d. 0.91 liter - ANS-b. 2.74 liter
The dilution principle can also be used to measure an unknown volume. Using the
dilution principle, it is possible to estimate a patient's plasma volume, C1V1 = C2V2.
Because the standard may be too concentrated, a portion of the standard is diluted and
counted while the full-strength undiluted standard can be injected into the patient. The
standard can be diluted by adding 1 ml of it to a flask, which is then filled to 500 ml with
distilled water. It could then be expressed as a 1:500 diluted standard. The standard
counts have to be multiplied by the dilution factor of 500 to obtain the true standard
concentration that was injected into the patient. Taking this into consideration, the new
equation to be used to calculate the plasma volume would be C1V1 = (DF)C2V2.
A 99Mo99mTc generator is calibrated for 1.00 Ci of 99Mo at 7:00 AM Monday. The
generator is eluted daily Monday through Friday at 8:00 AM, but the workload is
especially heavy on Friday so the generator is eluted again at 3:00 PM to obtain more
99mTc. Assuming 100% elution efficiency, how many mCi of 99mTc will be eluted at
3:00 PM Friday?
a. 930 mCi
b. 710 mCi
c. 160 mCi
d. 80 mCi - ANS-c. 160 mCi
The 99mTc radioactivity expected to be eluted from the generator can be calculated by
knowing three values: (1) the activity of 99Mo in the generator, (2) the time since the
last elution of the generator, and (3) the ratio of 99mTc to 99Mo in the generator. Using
the equation, A = Ci * e -0.693 (t/65.9 hr), enter in the known values and solve for the
unknown.
A 99Mo99mTc generator is eluted Monday at 7:00 AM, producing 1.8 Ci of 99mTc in the
eluate vial, in 20 ml saline. What volume of eluate should be withdrawn from the eluate
