MDCB Math questions with verified answers
__ is th eOP factor for a 6 by 12 field in a 15 by 15 applicator. The OFs
of as 6 by6 = 1.008 and 12 by 12= 1.001 Ans✓✓✓ 1.008x1.001^.5=
1.004
__ is the Air gap correction factor for a 6x12 fs shaped in at 15 by 15
applicator .980 and .991 Ans✓✓✓ (.980x.991)^.5= .985
_MU to deliver 180cgy to 90% dose contour for a 6mev beam op= .985
beam calibration is=1.03 Ans✓✓✓ 180/ .986x1.03= 177.4
% error calc Ans✓✓✓ final- initial/ initial x 100
NOT USED for change in depths or dose error calc
any time distance use ISF
40Gy in 20Fx- On 10th day of tx wedge was left out for field 2. What is
the total dose given that day?
field #1 weight= 100
field #2 weight= 80 Wedge factor .682
field #3 weight= 80 wedge factor .682 Ans✓✓✓ 1x+ .8x+.8x=200
2.6x=200
,200/2.6= 77
1(77)+ .8(77)+ .8(77)= 200
field 1 = 77Cgy
field 2= 62
field 3= 62 =62/.682= 91
229cgy
6mv beam 100sad machine. 12 by 15 cm fs, sep is 20, tray factor 1.
__mus needed to give 200cgy to midline through ap pa ports weighted
3:2 if ssd of 100cm used for both. PDD 1.5cm 13.5 x 13.15 100 ssd=
68.3% OF is 1.022 Dcal is 1.01cgy/mu Ans✓✓✓ 200/ .683x1.022x1.01=
283.6
3x+2x= 283.6
5x=283.6=
56.72(3)+ 56.72(2)
AP = 170mu PA is 114MU
a 20cm seperation patient is txed to midplane using a 100cm SAD
teqnique. If the same patient is then treated using a 100cm SSD
technqiue the MUs would increase by? Ans✓✓✓ ssd+ depth 10cm=
110
,anytime you have 2 distances changing*** ISF and need a new MU
110/100^2=1.21x100-100=21%
A 5.5cm thick cerrobend block was used to shield a field on a patient
who is treated on a 18mv linear accelerator. If 40Gy in 20fxs was
delivered to dmax in the open field, what is the dmax dose to center of
the shielded area under the block? there is 9% transmission at 5.5cm
Ans✓✓✓ 4000 x.09= 360cGy
A 6mv tx designed to be given at 100cm is given at 90cm. What is the
error in the dose delivered? Ans✓✓✓ ISF
100/90^2x 100= 123-100= 23% overdose
A collimated fs is 10 x10 cm2. 30% of field is blocked. Effective field size
is approx how many cm2? Ans✓✓✓ 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
, A dose of 30gy in 10 fx was given to a depth of 7cm from a 6mv single
posterior field using an isocentric technique at 100 cm SAD. The patient
was incorrectly treated 5 times at 100SSD. The approximate dose at
7cm after 5 fxs is? Ans✓✓✓ To find dose at depth
(TD/ TMR @ISO) x (SAD/SSD=D)^2 x tmr at new depth
original/actual ^2
(100/107)^2 = .872 x 300cgy a day= 262cgy x 5 fx= 13.1gy
A Lesion extending to 3.5cm is treated w a 12mev e beam at 100 SSD/
1cm of bolus is used. If 200cgy is delivered what is the skin dose and
the maximum tissue dose, given dmax depth=3cm.
PDD values
1cm= 91.3
3.5cm=97.5cm
4.5cm= 73.8 Ans✓✓✓ Skin Dose= 200x 91.3/73.8= 247.2 cgy
Max tissue dose= 200x 100 (pdd at 3cm)/ 73.8= 271cgy
A machine calibrated for SSD in a 15mv beam dcal = 1.005 cgy/mu =
3cm for 10 x 10 field 100cm SSD. Now using isocentric teq new
calibration factor is? Ans✓✓✓ (100/103)^2 = .9417
1.005/.9417= 1.066cgy/mu
__ is th eOP factor for a 6 by 12 field in a 15 by 15 applicator. The OFs
of as 6 by6 = 1.008 and 12 by 12= 1.001 Ans✓✓✓ 1.008x1.001^.5=
1.004
__ is the Air gap correction factor for a 6x12 fs shaped in at 15 by 15
applicator .980 and .991 Ans✓✓✓ (.980x.991)^.5= .985
_MU to deliver 180cgy to 90% dose contour for a 6mev beam op= .985
beam calibration is=1.03 Ans✓✓✓ 180/ .986x1.03= 177.4
% error calc Ans✓✓✓ final- initial/ initial x 100
NOT USED for change in depths or dose error calc
any time distance use ISF
40Gy in 20Fx- On 10th day of tx wedge was left out for field 2. What is
the total dose given that day?
field #1 weight= 100
field #2 weight= 80 Wedge factor .682
field #3 weight= 80 wedge factor .682 Ans✓✓✓ 1x+ .8x+.8x=200
2.6x=200
,200/2.6= 77
1(77)+ .8(77)+ .8(77)= 200
field 1 = 77Cgy
field 2= 62
field 3= 62 =62/.682= 91
229cgy
6mv beam 100sad machine. 12 by 15 cm fs, sep is 20, tray factor 1.
__mus needed to give 200cgy to midline through ap pa ports weighted
3:2 if ssd of 100cm used for both. PDD 1.5cm 13.5 x 13.15 100 ssd=
68.3% OF is 1.022 Dcal is 1.01cgy/mu Ans✓✓✓ 200/ .683x1.022x1.01=
283.6
3x+2x= 283.6
5x=283.6=
56.72(3)+ 56.72(2)
AP = 170mu PA is 114MU
a 20cm seperation patient is txed to midplane using a 100cm SAD
teqnique. If the same patient is then treated using a 100cm SSD
technqiue the MUs would increase by? Ans✓✓✓ ssd+ depth 10cm=
110
,anytime you have 2 distances changing*** ISF and need a new MU
110/100^2=1.21x100-100=21%
A 5.5cm thick cerrobend block was used to shield a field on a patient
who is treated on a 18mv linear accelerator. If 40Gy in 20fxs was
delivered to dmax in the open field, what is the dmax dose to center of
the shielded area under the block? there is 9% transmission at 5.5cm
Ans✓✓✓ 4000 x.09= 360cGy
A 6mv tx designed to be given at 100cm is given at 90cm. What is the
error in the dose delivered? Ans✓✓✓ ISF
100/90^2x 100= 123-100= 23% overdose
A collimated fs is 10 x10 cm2. 30% of field is blocked. Effective field size
is approx how many cm2? Ans✓✓✓ 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
, A dose of 30gy in 10 fx was given to a depth of 7cm from a 6mv single
posterior field using an isocentric technique at 100 cm SAD. The patient
was incorrectly treated 5 times at 100SSD. The approximate dose at
7cm after 5 fxs is? Ans✓✓✓ To find dose at depth
(TD/ TMR @ISO) x (SAD/SSD=D)^2 x tmr at new depth
original/actual ^2
(100/107)^2 = .872 x 300cgy a day= 262cgy x 5 fx= 13.1gy
A Lesion extending to 3.5cm is treated w a 12mev e beam at 100 SSD/
1cm of bolus is used. If 200cgy is delivered what is the skin dose and
the maximum tissue dose, given dmax depth=3cm.
PDD values
1cm= 91.3
3.5cm=97.5cm
4.5cm= 73.8 Ans✓✓✓ Skin Dose= 200x 91.3/73.8= 247.2 cgy
Max tissue dose= 200x 100 (pdd at 3cm)/ 73.8= 271cgy
A machine calibrated for SSD in a 15mv beam dcal = 1.005 cgy/mu =
3cm for 10 x 10 field 100cm SSD. Now using isocentric teq new
calibration factor is? Ans✓✓✓ (100/103)^2 = .9417
1.005/.9417= 1.066cgy/mu
