1 - Quia



A radiograph is made using 90kVp at 300 mA for 0.035 seconds. The quality control technologist asks that you repeat this film using 40% less mAs. What new time would be needed if you are to continue to use the 300 mA station? |1. 300 x .035=10.5

10.5-40%=6.3

6.3/300=.02

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|Correct answer: .02 | |

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|You radiograph the lumbar spine of an elderly woman and note that she has advanced|15% rule |

|osteroporosis with substantial calcium loss. The film fails to demonstrate the |90-15%=76.5kv |

|needed diagnostic information because of the lack of contrast. What change in the |150 x 2=300mAs |

|technique would you make if you originally used 90 kVp at 150 mAs? | |

|Correct answer: 76.5 kV @ 300mAs | |

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|You made a cross-table lateral radiograph of a broken femur employing a 6:1 |Grid conversion formula |

|stationary grid and using 74 kVp at 300 mA for 0.05 sec. The orthopedic physician |8:1 3 x 15=22.5 |

|applies a long leg splint to the patient and requests that you repeat the film. |6:1 2 |

|Another technologist has taken the 6:1 grid on a portable and you must use an 8:1 | |

|grid for your follow up film. What new mAs would be needed to provide the same | |

|radiographic density? | |

|Correct answer: 22.5 | |

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|You made a portable chest radiograph of a recumbent patient and are now asked to |mAs1 = (D1)2 |

|repeat the examination with the patient in the erect position. Your original |mAs2 (D2)2 |

|technique was 86 kVp and 1.6 mAs at a 44-in distance. When you repeat the study in| |

|the erect position you will be using a 68-in distance. What new mAs would be |1.6 = (44)2 = 1936 |

|needed to obtain equal density? |x (68)2 4624 |

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| |3.82 |

|Correct answer: 3.82 | |

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|You make an AP abdominal radiograph of a patient who is unable to hold his breath.|The goal here is to lower the exposure time. I chose to half it. |

|Because of excessive motion you must repeat the film. Your original technique was |300 x .15 = 45 |

|80 kVp at 300 mA for 0.15 sec. How would you change this technique to try to |.15/2 = .08 sec |

|eliminate the motion while maintaining the same density? |300 x 2 = 600mA |

|Correct answer: 600 mA | |

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|You made a PA chest radiograph of a patient using 120 kVp at 1.125 mAs. The |Three steps of the 15% rule |

|radiologist requests a repeat PA chest film with the kVp not to exceed 80. What |120 – 15% = 102 – 15% = 86.7 – 15% = 73.7kv |

|new technique would provide the radiologist with a film of the same density using |1.125 x 2 x 2 x 2 = 9mAs |

|no more than 80 kVp? | |

|Correct answer: 73.7 kV @ 9 mAs | |

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|You are told to increase the density of a radiograph by 50%. Your original |600 x .05 = 30 |

|technique was 90 kVp at 600 mA for 0.05 sec. What new technique would be used if |30 + 50% = 45 |

|you were to change only the time, leaving the mA and kVp the same? |45/600 = .08 |

|Correct answer: .08 | |

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|A radiograph is produced with an 8:1 grid using 90 kVp at 400 mA for 0.08 sec. You|. 12:1 = 4 x 32 = 42.6 |

|are asked to repeat the examination using a 12:1 grid. What new mAs would be |8:1 3 |

|needed? | |

|Correct answer: 42.6 | |

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|A radiograph is produced with a 16:1 grid using 100 kVp at 800 mA for 0.02 sec. |4 x 16 = 12.8 |

|You are asked to repeat the examination using a 12:1 grid. What new mAs would be |5 |

|needed? | |

|Correct answer: 12.8 | |

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|A radiograph is produced with a 5:1 grid using 70 kVp at 800 mA for 0.02 sec. You |2 NONE |

|are asked to repeat the examination using a 6:1 grid. What changes in mAs would be|2 |

|needed? | |

|Correct answer: none | |

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|A radiograph is produced with a 16:1 grid using 120 kVp at 600 mA for 1/10 sec. |120 – 15% = 102 |

|You are asked to repeat the examination using only 100 kVp. What new mAs would be |60 x 2 = 120mAs |

|needed? | |

|Correct answer: 120 mAs | |

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|A radiograph is produced with a 12:1 grid using 100 kVp at 600 mA for 1/30 sec. |100 – 15% = 85 – 15% = 72 |

|You are asked to repeat the examination using 80 kVp. What new mAs would be |80mAs |

|needed? | |

|Correct answer: 80 mAs | |

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|A radiograph is produced with an 8:1 grid using 80 kVp at 300 mA for 1.5 sec. You |80 + 15% = 92 + 15% = 105.8 |

|are asked to repeat the examination using 100 kVp. What new mAs would be needed? |112.5mAs |

|Correct answer: 112.5 mAs | |

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|A radiograph is produced with a 6:1 grid using 70 kVp at 100 mA for 1/15 sec. You |70 + 15% = 80.5kv |

|are asked to repeat the examination using 80 kVp. What new mAs would be needed? |3.35mAs |

|Correct answer: 3.35 mAs | |

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|A radiograph is made using 10 mAs and 50-speed (detail) screens. What new mAs |Screen conversion formula |

|would be needed if 100-speed screens were substituted? |To 100 (flip) 50 x 10 = 5mAs |

| |From 50 100 |

|Correct answer: 5 mAs | |

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|A radiograph is made using 200-speed screens, and the radiologist requests that |Screen conversion formula |

|you repeat the study using 50-speed (detail) screens. If your original technique |To 50 (Flip) 200 x 3.5 |

|called for 3.5 mAs, what new mAs would be needed? |From 200 50 |

| |14mAs |

|Correct answer: 14 mAs | |

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|A radiograph made with 200-speed screens must be repeated using 100-speed screens.|Screen conversion formula |

|If your original technique called for 5 mAs, what new mAs would be needed? |100 (Flip) 200 x 5 = 10mAs |

| |200 100 |

|Correct answer: 10 mAs | |

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|Which technique will give you the greatest density? | |

|70 kVp 500 mA 1/10 sec 40"SID 6:1 grid | |

|75 kVp 300 mA 1/15 sec 44"SID 5:1 grid | |

|80 kVp 100 mA 1/20 sec 36"SID 8:1 grid | |

|65 kVp 200 mA 1/10 sec 36"SID no grid | |

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|Correct answer: 65 kVp 200 mA 1/10 sec 36"SID no grid | |

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|Which technique will give you the greatest density? | |

|110 kVp 400 mA 0.10 sec 44"SID 12:1 grid | |

|120 kVp 600 mA 0.07 sec 40"SID 16:1 grid | |

|90 kvp 500 mA 0.15 sec 36"SID 12:1 grid | |

|120 kVp 800 mA 0.2 sec 40"SID 16:1 grid | |

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|Correct answer: 120 kVp 800 mA 0.2 sec 40"SID 16:1 grid | |

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|Which technique will give you the greatest density? | |

|100 kVp 300 mA 0.15 sec 48"SID 12:1 grid | |

|120 kVp 200 mA 0.07 sec 44"SID 16:1 grid | |

|95 kVp 400 mA 0.2 sec 36"SID 12:1 grid | |

|100 kVp 400 mA 0.2 sec 30"SID 16:1 grid | |

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|Correct answer: 100 kVp 400 mA 0.2 sec 30"SID 16:1 grid | |

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|Which technique will give you the lowest density? | |

|70 kVp 500 mA 0.035 sec 40"SID 6:1 grid | |

|75 kVp 300 mA 0.025 sec 44"SID 5:1 grid | |

|80 kVp 100 mA 0.03 sec 36"SID 8:1 grid | |

|65 kVp 200 mA 0.02 sec 36"SID no grid | |

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|Correct answer: 80 kVp 100 mA 0.03 sec 36"SID 8:1 grid | |

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|Determine the size of the projected image of an object that measures 1-in if using|40 = x = 1.05 |

|an SID of 40-in and an SOD of 38-in. |38 1 |

|Correct answer: 1.05 | |

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|Determine the projected image size of an object that measures 14-cm when using an |x = 108 = 15 |

|SID of 108-in and an SOD of 100-in. |14 100 |

|Correct answer: 15 | |

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|What will be the projected image size of a 2-in object if the OID is 0.5-in and |x = 60.5 = 2.01 |

|the SOD is 60-in? |2 60 |

|Correct answer: 2.01 | |

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|What will be the projected image size of a 4-in object if the OID is 1.5-in and |x = 39.5 = 4.15 |

|the SOD is 38-in? |4 38 |

|Correct answer: 4.15 | |

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|Determine the percentage of magnification if the OID is 2-in and the SOD is 38-in.|SID = 38 +2 = |

| |40-38 x 100 = 5.26% |

| |38 |

|Correct answer: 5.26% | |

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|Determine the percentage of magnification if the OID is 4-in and the SOD is 72-in.|SID = 72 + 4 = 76 |

| |76 – 72 x 100 = 5.5% |

| |72 |

|Correct answer: 5.5% | |

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|Determine the percentage of magnification if the OID is 1-in and the SOD is 40-in.|SID = 40 + 1 = |

| |41 – 40 x 100 = 2.5% |

| |40 |

|Correct answer: 2.5% | |

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|Determine the magnification factor if the SID is 40-in and the SOD is 36-in. |40 – 36 x 100 = 11% |

| |36 |

|Correct answer: 11% | |

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|Determine the amount of geometric unsharpness (GU) in the following technique. |GU = (FSS x OID) / SOD |

|FSS=2.0 OID=1.5in SOD=38.5in GU=? |Gu =.0779 |

|Correct answer: .0779 | |

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|Determine the amount of geometric unsharpness (GU) in the following technique. |GU = .12 |

|FSS=0.6 SOD=30 in SID=36 in GU=? | |

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|Correct answer: .12 | |

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|Determine the amount of geometric unsharpness (GU) in the following technique. |53. GU = .0857 |

|FSS=3 SOD=70 in SID=72 in GU=? | |

|Correct answer .0857 | |

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|Determine the amount of geometric unsharpness (GU) in the following technique. |54. GU = .0285 |

|FSS=1 OID=2 in SOD=70 in GU=? | |

|Correct answer .0285 | |

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|To radiograph an extremity in a dry plaster cast requires a 100% increase in the | 2+ 100% =4 mAs |

|original mAs of 2. To radiograph the hand of a patient in a plaster cast that is 1| |

|week old you would use ___ mAs. | |

|Correct answer: 4 | |

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|Heat units for a single-phase x-ray machine are determined by multiplying kVp x mA|120 x 400 x .035 =1680 |

|x sec. The technique you are to use is 120 kVp at 400 mA for 0.035 sec. The heat | |

|units produced by this exposure are ___? | |

|Correct answer: 1680 | |

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|If 200 mA for 0.15 sec at a 40-in SID produces 4 R, an 80-in SID will produce ____|4 = 802 = 6400 = 1 R |

|R. |x 402 1600 |

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|Correct answer: 1 | |

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|To increase by 40% your original technique of 30 mAs requires a new mAs of _____. |30+ 40% =42 |

|Correct answer: 42 | |

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|The technique chart calls for 1/15 sec but your equipment requires decimal time | 1 /15 = .06 |

|settings. The decimal equivalent of 1/15 is ____. | |

|Correct answer: .06 | |

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|10 Gy is equivalent to _____ rad. |1R x ( 2.58 x 10 – 4) = 1 c/kg |

| |1rad x .01 = 1 gray |

| |1rem x .01 = 1sv |

| |1000 .01x = 10 x = 1000 |

| |.01 .01 |

|Correct answer: 1000 | |

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|Sv is equivalent to ____ rem. | 10 .01x = .1 x = 10 |

| |.01 .01 |

|Correct answer: 10 | |

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|10 rad is equivalent to ____ Gy. |1 10 x .01 = .1 |

|Correct answer: .1 | |

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|4 rem is equivalent to _____ Sv. | .04 4 x .01 = .04 |

|Correct answer: .04 | |

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|Determine the percentage of magnification if the OID is 1.5 in and the SOD is 35 |SID = 35 + 1.5 = 36.5 |

|in. |36.5 - 35 x 100 = 4.28% |

| |35 |

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|Correct answer: 4.28% | |

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|Determine the magnification factor if the SID is 72 in and the SOD is 68 in. | 72 = 1.05 |

| |68 |

|Correct answer: 1.05 | |

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|How many heat units would be produced on a single-phase radiographic unit using a |mA x kv x sec |

|technique of 90 kVp at 300 mA for 0.035 sec? |3 phase |

| |6p x 1.35 |

| |12p x 1.41 |

| |1 phase = x 1 |

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| |90 x 300 x .035 =945 |

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|Correct answer: 945 | |

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|How many heat units would be produced on a three-phase (6-pulse) radiographic unit|120 x 600 x .005 x 1.35 = 486 |

|using a technique of 120 kVp at 600 mA for 0.005 sec? | |

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|Correct answer: 486 | |

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|How many heat units would be produced on a three-phase (12-pulse) radiographic |120 x 600 x .005 x 1.41 = 507.6 |

|unit using a thechique of 120 kVp at 600 mA for 0.005 sec? | |

|Correct answer: 507.6 | |

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|A series of six x-ray exposures is made on a single-phase unit with a technique of| |

|75 kVp at 400 mA for 0.4 sec. What is the total number of heat units produced? |75 x 400 x .04 x 6 =72,000 |

|Correct answer: 72,000 | |

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|If a patient receives a radiation dose of 6 mR when radiographed at an SID of 40 |72. 6 = (60)2 = 3600 = 2.6 |

|in, what dose in mR would be received if an SID of 60 in were used? |x (40) 1600 |

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|Correct answer: 2.6 | |

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|If a patient receives a radiation dose of 2 mR when radiographed at an SID of 60 |73. 2 = (40)2 = 4.5 |

|in, what dose in mR would be received if an SID of 40 in were used? |x (60) |

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|Correct answer: 4.5 | |

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|If a patient receives a radiation dose of 0.02 mR when radiographed at an SID of |74. .02 (44)2 1936 = .01 |

|36 in, what dose in mR would be received if an SID of 44 in were used? |x (36) 1296 |

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|Correct answer: .01 | |

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|If a patient receives a radiation dose of 0.5 mR when radiographed at an SID of 48|75. .5 = (40)2 1600 = .72 |

|in, what dose in mR would be received if an SID of 40 in were used? |x (48) 2304 |

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|Correct answer: .72 | |

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|The original technique called for 75 kVp at 45 mAs. Without changing the density |93. 8:1 3 x 45 = 67.5 |

|or scale of contrast, it is necessary to change from a 6:1 grid technique to an |6:1 2 |

|8:1 grid and also to change the distance from 40 in to 30 in. |67.5 = (40)2 1600 |

| |x (30) 900 |

| |x = 37.9 |

|Correct answer: 37.9 | |

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|The original technique called for 75 kVp at 45 mAs, decrease the contrast, use a |75 @ 45 mAs @ 6:1 grid (original technique) |

|12:1 grid, and increase the density by 100%. |75 + 15% = 86 kv |

| |45/2 = 23 mAs |

| |4 x 23 = 45 mAs |

| |2 |

| |45 mAs + 100% = 90 mAs |

|Correct answer: 86kV @ 90 mAs | |

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|It is necessary to repeat the x-ray of a patient for whom you originally used 100 |m1 = (D1)2 4 = (44)2 1936 x = 2.7 mAs 3 x 2.7 = 2 mAs |

|kVp at 200 mA for 0.02 sec, with a 12:1 grid at an SID of 44 in. You have to do |m2 (D2) x (36) 1296 4 |

|ths patient "portable" and must use an 8:1 grid at a 36 in distance. | |

|Correct answer: 2 mAs | |

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|The original technique calls for 90 kVp at 300 mA for 0.06 sec at a 40 in SID. |104 @ 18 |

|Your radiograph needs a longer scale of contrast, twice the density, and you must |18 (40)2 = 28MAS |

|use a 50 in SID. What time would be needed if you were required to use the 600 mA |x (50) |

|station? |28/600 =.05 |

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|Correct answer: .05 | |

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|The original technique calls for 90 kVp at 300 mA for 0.06 sec at a 40 in SID. |77kv @ 18 |

|Your radiograph needs a shorter scale of contrast, half the density, and you must |18 = (40)2 = 10 mAs |

|use a 30 in SID. What time would be needed if you were required to use the 400 mA |x (30) |

|station? |10 /400 =.025 |

|Correct answer: .025 | |

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|The original technique calls for 75 kVp at 500 mA for 3/20 sec at a 40 in SID. |75kv @ 75 mAs |

|Your radiograph needs less contrast, 50% more density, and you must use a 50 in |+15% and half mas |

|SID. What time would be needed if you were required to use the 600 mA station? |86kV @37.5 |

| |+50% density = 56mAs |

| |56 = (40)2 = 1600 = 87.5 |

| |x (50) 2500 |

| |87.5 /600 = .145 seconds |

|Correct answer: .145 sec | |

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