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MDCB Math Problems – Questions With Right Solutions

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MDCB Math Problems – Questions With Right Solutions

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MDCB Math Problems – Questions With Right
Solutions
The surface dose rate of a Strontium-90 (Sr-90) eye applicator was
4.5 cGy/second on May 15. If a patient is to be treated 2 years later
on May 15, what is the treatment time if 1000 cGy is prescribed?
(The half-life of Sr-90 is 28 years.) Correct Answer - 234 secs

A=Ao(.5)^n n= t/t1/2

a=1000/4.5=222.2

222.2= x (.5)^2/28
222.2= x (.5)^.071
222.2=x.9519
222.2/.9519= 233.4

Iodine 125 has an exposure rate constant of 1.1 r/hr/mci and a half
life of 60.2 days. The milligram hour radium equivalency of a
permanent implant of a 1 mci iodine seed is therefore? Correct
Answer - 277

1.44x60.2x 24(hrs)x1.1/8.25)= 277

formula- find t mean= 1.44 x t1/2
ER radium= 8.25rcm3/hr mci

If the exposure rate constant is .87 Rcm2 Hr-1 mci-1 and the average
patient transmission factor is .20 the exposure rate at 12.5cm for a
patient who has been injected with 20mci of tc99m is? Correct
Answer - 22 mr formula = Er= Activity x exposure rate constx
1/d^2

,ER= 20mCi x .87 x 1/12.5^2 x .2
.02272 x1000=22.7
or 22r/hr

The exposure rate emitting from a brachytherapy safe is 25mR/hr at
1 m from the source. The minimum thickness of shielding necessary
to reduce the exposure to an adjacent secretary to an acceptable
level would be? How many TVL? Correct Answer - 2 TVL

A simulation for an isocentric treatment on a linear accelerator SAD
is 100cm with a 12x15 fs. The tx is given on a co 60 unit SAD 80cm.
The fs to be used is>=? Correct Answer - 12X 15 > 100
NORMINAL sad LINAC
80> NOMINAL SAD CO60
fs defined at nominal SAD

A simulation film is taken at 130 SFD for making blocks to be used
on a linac with an SAD of 100cm and a source to tray distance of
65cm. The tx is rescheduled to a linac with an SAD of 80cm and
source to try dx of 52cm. The block may be accurately produced is
the source to film distance is set to? Correct Answer - 104cm



80/100 x 130

THe HVL for Co60 is 1.1 cm of lead. What thickness of lead would
reduce the transmitted dose to 2%? Correct Answer - 6 HVL=
6x 1.1= 6.6cm

1 HVL= 50%
2 HVL= 25%
3 HVL= 12.5%
4HVL= 6.25%

,5 HVL= 3.125%
6 HVL= 1.56%

in the triangle above the value of X (slant on triangle) is? Given
height is 4cm and angle is 30 degrees? Correct Answer -
tang(30)= 4/y

.577y=4

4/.577= 6.9cm

Calculate the given dose of a 100cm SSD tx with a tumor dose of
60cgy prescribed to a depth of 11cm
PDD @1.5cm=100
PPD @ 5.5cm=81.1
PDD @ 11cm= 67.1
PDD@ 12.5cm= 62.5 Correct Answer - 60cgy/ .671= 89cGy

A patient is treated for lung cancer using POP AP/PA beams 180Cgy
a day equally weighted fields. An anterior 30 Degree wedge is used
daily with a factor of .617. This factor was accidently left out of the
calc for the first 4 days of tx but the wedge was used. What total
dose did the patient receive after 4 days of tx? Correct Answer -
we know patient got 90Cgy a day from posterior
dose with wedge in = 90x .617= 55.53cgy
dose with wedge out= 90cGy
90+55.53=145.53x 4 (fx)= 582Cgy

or .617=90/x= 145.86 x 4= 582

Calculate the dose to the cord at 6cm depth from 100cm SSD PA tx
that has a tumor dose of 90cgy defined at a depth of 12cm

PDD 1.5cm= 100

, PDD 6cm= 82.1
PDD 12cm= 64.1 Correct Answer - find given dose x new pdd

90/ .641= 140x .821= 115

A collimated fs is 10 x10 cm2. 30% of field is blocked. Effective field
size is approx how many cm2? Correct Answer - find open
area= 10x 10= 100cm
30% blocked so 30% of 100= 30cm
100(open field)- 30cm (blocked field)= 70cm

then take the square root of 70= 8.4cm

The mass attenuation coefficient of lead at 140kev is 2.6cm2/g and
the density of lead is 11.3 g/cm3. What is the percentage
transmission through a .5mm thick lead apron with an effective
energy of 140 KEV? Correct Answer - um (mass atten
coefficient)=u (linear atten coefficient/ P (lead)

2.6= u/ 11.3
u=29.38
thickness of .5mm= .05cm
e^ -ux
e^-29.38x.05

.23 x 100= 23%

A patient is treated with a 6MV photon beam 12.5 x12.5 cm2 at 110
cm SAD to a 9cm depth. the machine is calibrated to deliver a 1.0 cgy
per mu at dmax for a 10x 10 cm2 field at 100 SSD. The number of
MUS required to deliver 150cGy at the iso is?

TMR for 6MV photons
depth 10x10 12 x 12 14x 14

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Subido en
24 de junio de 2025
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