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PST 1 Math review
problem solving
Question | Answer |
---|---|
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 |