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PST 1 Math review

problem solving

QuestionAnswer
mA 1200 Time 1/4 sec. How much mAs? 300 mAs
Time .15 sec mAs 150. How much mA? 1000 mA
mA 800 mAs 160. How much time? .2 or 1/5 sec
mA 300 mAs 120. How much time? .4 or 2/5 sec
Time 1/5 sec mAs 90. How much mA? 450 mA
In order to maintain receptor exposure, what mAs will be required to compensate for the changes in SID? 32" and 40mAs changed to 48" and _____ mAs 90 mAs
In order to maintain receptor exposure, what mAs will be required to compensate for the changes in SID? 60" and 45mAs changed to 40" and _____ mAs 20mAs
In order to maintain receptor exposure, what mAs will be required to compensate for the changes in SID? 40" and 150mAs changed to 72" and _____ mAs 486 mAs
In order to maintain receptor exposure, what mAs will be required to compensate for the changes in SID? 40" and 200mAs changed to 36" and _____ mAs 162mAs
In order to maintain receptor exposure, what mAs will be required to compensate for the changes in SID? 36" and 20mAs changed to 72" and _____ mAs 80mAs
State the percentage of increase and decrease from Exposure A to Exposure B. 70kVp to 84kVp, increase or decrease by _____% INCREASE; 20%
State the percentage of increase and decrease from Exposure A to Exposure B. 75kVp to 100kVp, increase or decrease by _____% INCREASE; 33.33%
State the percentage of increase and decrease from Exposure A to Exposure B. 20mAs to 15mAs, increase or decrease by _____% DECREASE; 25%
State the percentage of increase and decrease from Exposure A to Exposure B. 30mAs to 21mAs, increase or decrease by _____% DECREASE; 30%
State the percentage of increase and decrease from Exposure A to Exposure B. 40mAs to 64mAs, increase or decrease by _____% INCREASE; 60%
68mAs was used to produce an image. If mAs is increased to 92, what is the percent of increase? 35.3% (35.29%)
2mAs is used to produce an image. If you are instructed to increase the mAs by 40%, what new mAs will you use? Round to one decimal place. 2.8mAs
An image is made using 0.06 sec and 600 mA. If time is changed to 0.03 sec, what mA will be required to maintain receptor exposure? 1200mA
State required mAs to maintain receptor exposure when the indicated change in SID is made. All other factors remain same. 60" and 10 mAs to 30" and _____ mAs 2.5mAs
State required mAs to maintain receptor exposure when the indicated change in SID is made. All other factors remain same. 40" and 200 mAs to 28" and _____ mAs 98mAs
State required mAs to maintain receptor exposure when the indicated change in SID is made. All other factors remain same. 36" and 40 mAs to 72" and _____ mAs 160mAs
State required mAs to maintain receptor exposure when the indicated change in SID is made. All other factors remain same. 48" and 80 mAs to 72" and _____ mAs 180mAs
State required mAs to maintain receptor exposure when the indicated change in SID is made. All other factors remain same. 56" and 128 mAs to 21" and _____ mAs 18mAs
An image is made using 58kVp and 3 mAs. If mAs is decreased to 1.5mAs, what kVp must be used to maintain receptor exposure (density) within 1 kVp accuracy? 67 kVp
An image is made using 48mAs and 79kVp. If kVp remains at 79, what mAs must you use in order to decrease receptor exposure (density) by 50% 79 kVp and 24 mAs
State the formula for mAs, mA or Time when 2 of the 3 are known. mA X Time = mAs
How do you find percentage difference in mAs or kVp? Highest - Lowest / original x 100 1. find the difference 2. Divide the difference by the original 3. Move decimal 2 places to right (multiply by 100)
Application of kVp 15% rule: double receptor exposure halve receptor exposure increase contrast decrease contrast original kVp x 1.15 original kVp x 0.85 decrease kVp by 15% and 2x mAs increase kVp by 15% and 1/2x mAs
How do you maintain receptor exposure with mAs when SID changes? mAs Distance Compensation formula; Distance Square Law mAs 1 / mAs 2 = (SID1) squared / (SID2) squared
Maintain exposure rate with changes in SID: Inverse Square Law; I1 / I2 = (SID2) squared / (SID1) squared
Created by: Larobbins