Transmission and Drivetrain Lesson Test

Answer: D

Correct answer: Match engine output to wheel-speed and torque needs. An engine makes useful power only within a limited rpm range, but a vehicle needs high torque to start moving and lower torque at much higher wheel speed when cruising. The transmission bridges that gap with selectable gear ratios: low gears multiply torque for starting and climbing, and high gears let the engine turn slower at highway speed. Ignition, cooling and exhaust are separate systems that do not choose ratios.

• A: An exhaust system carries gases away and cannot regenerate gasoline.
• B: The cooling radiator removes heat; it does not select gear ratios.
• C: Ignition parts initiate combustion and do not choose wheel torque ratios.
• D (correct): Correct: The transmission selects ratios to match shaft speed and torque to driving demands.

Answer: D

Correct answer: 48 N·m. Torque equals force times the length of the lever arm: 120 N × 0.4 m = 48 N·m.

• A: This comes from dividing the force by the lever-arm length instead of multiplying.
• B: This treats 0.4 m as 4 m, a decimal-place slip in the other direction.
• C: This is a decimal-place slip; 120 times 0.4 is 48, not 4.8.
• D (correct): Torque is force times lever-arm length, and 120 N times 0.4 m equals 48 N·m.

Answer: C

Correct answer: A synchronizer, which matches gear and shaft speeds before they lock. Before a sliding collar can lock a gear to its shaft, both must turn at the same speed. The synchronizer uses a cone-shaped friction ring to speed up or slow the gear until it matches, which prevents grinding.

• A: An idler only changes rotation direction, as in reverse; it does not match speeds during forward shifts.
• B: The differential works at the axle and has nothing to do with smooth gear engagement in the gearbox.
• C (correct): Its friction cone brings the gear to shaft speed so the collar can engage without clashing teeth.
• D: The clutch does disconnect the engine, but grinding in just one gear points to the part that matches speeds for that gear.

Answer: C

Correct answer: The transmission, final drive, and differential. A transaxle combines the transmission, final drive, and differential in one unit. Its output goes directly to two half-shafts, which saves space in front-wheel-drive cars.

• A: The converter couples engine and transmission, but a transaxle is defined by including the final drive and differential.
• B: A transfer case splits power front and rear in 4WD; it is not part of a transaxle.
• C (correct): A transaxle puts the gearbox, final drive gearing, and differential together and sends power straight to the half-shafts.
• D: These parts sit at the engine end of the drivetrain and are not what defines a transaxle.

Answer: B

Correct answer: Engine torque is disconnected from the transmission input shaft. With the pedal up, spring force squeezes the friction disc between the flywheel and pressure plate so the engine and gearbox turn together. Pushing the pedal down releases that force, disconnecting engine torque so gears can be changed or the car can stop without stalling.

• A: The clutch only connects or disconnects engine torque; the driver still moves the shift lever to change gears.
• B (correct): Pressing the pedal releases the spring clamping force on the friction disc, so the engine no longer drives the gearbox input.
• C: This is backward; the disc is clamped with the pedal up, and pressing the pedal releases the clamp.
• D: The differential is at the axle end of the drivetrain and is not controlled by the clutch pedal.

Answer: B

Correct answer: To change the direction the output shaft turns. Two external gears in mesh turn in opposite directions. Placing an idler between them makes the output turn the opposite way from normal forward gears, which is how a manual gearbox produces reverse. The idler's own tooth count cancels out of the overall ratio.

• A: Allowing different wheel speeds is the job of the differential.
• B (correct): Adding a third gear between two meshing gears flips the output rotation so the vehicle can back up.
• C: Matching speeds before engagement is the job of a synchronizer, not an idler.
• D: That describes an overdrive ratio, not an idler; an idler does not change the overall ratio.

Answer: A

Correct answer: Overdrive. Ratios greater than 1 are reductions, exactly 1 is direct drive, and ratios less than 1 are overdrives. Since 0.72 is less than 1, this gear is an overdrive.

• A (correct): A ratio below 1 makes the output spin faster than the input, which is overdrive.
• B: Direct drive is exactly 1:1, with input and output turning at the same speed.
• C: A reduction has a ratio greater than 1 and makes the output turn slower than the input.
• D: Low range is an extra reduction in a transfer case, with a ratio well above 1.

Answer: C

Correct answer: Direct drive. With the ratio written as input speed over output speed, 1:1 means both shafts turn at the same rate. That is direct drive; ratios above 1 are reductions and ratios below 1 are overdrives.

• A: In neutral no torque path is completed, so the output is not driven at all.
• B: Overdrive is a ratio below 1:1, where the output turns faster than the input.
• C (correct): A 1:1 ratio means input and output turn together at the same speed, which is direct drive.
• D: A reduction is a ratio above 1:1, where the output turns slower than the input.

Answer: B

Correct answer: They increase ideal wheel torque potential while reducing wheel speed at fixed engine rpm. A gear ratio trades speed for torque. If a low gear has a 3:1 reduction, the output turns one-third as fast but, ignoring friction, delivers about three times the torque, so power stays about the same. That is why first gear is used to start moving or climb steep hills. The tempting error is thinking gearing adds power or energy; it only changes the mix of speed and torque, and friction loses a little.

• A: An ideal fixed reduction cannot increase both torque and rpm at once; it trades one for the other.
• B (correct): Correct: A larger fixed reduction multiplies torque by trading for reduced output rotational speed.
• C: Gearing cannot create mechanical energy or shaft power from nothing.
• D: Stopping distance primarily depends on braking and grip rather than low gear torque.

Answer: D

Correct answer: FWD. Drivetrain abbreviations name the axle that receives engine torque: FWD sends it to the front wheels, RWD to the rear, and AWD or 4WD to both. Most compact and midsize cars use FWD because mounting the engine and transaxle together up front saves space and weight. Because the abbreviations look alike, read the first letter carefully; it is easy to mix up front and rear when choosing quickly.

• A: A transmission in neutral does not normally supply propulsive wheel torque.
• B: This construction explicitly delivers propulsion only to the rear axle.
• C: RWD identifies a rear driven axle rather than normally driven front tires.
• D (correct): Correct: Front-wheel drive sends motive torque through the front axle shafts to the front tires.

Answer: D

Correct answer: 600 rpm. The ratio is driven teeth over driving teeth:

a 3:1 reduction. Output speed = 1,800 rpm ÷ 3 = 600 rpm, slower than the input as a reduction should be.

• A: This assumes meshing gears always turn at the same speed, ignoring the difference in tooth counts.
• B: This multiplies by the ratio instead of dividing, which would make a larger driven gear spin faster.
• C: This removes only one-third of the input speed instead of dividing the input speed by 3.
• D (correct): The ratio is 45/15 = 3:1, so the output speed is 1,800 divided by 3.

Answer: A

Correct answer: 540 N·m. Multiply input torque by the ratio and by the efficiency written as a decimal:

N·m. The ideal value would be 600 N·m, so losses take away 60 N·m.

• A (correct): Output torque is 250 × 2.4 × 0.90, which equals 540 N·m.
• B: This divides the torque by the ratio, but a reduction raises torque rather than lowering it.
• C: This divides by the efficiency, which wrongly makes the real output larger than the ideal output.
• D: This is the ideal output torque; it leaves out the 90 percent efficiency.

Answer: C

Correct answer: 450 rpm. Ratios in series multiply:

so the overall ratio is 6:1. Axle speed = 2,700 rpm ÷ 6 = 450 rpm.

• A: This applies only the final drive and leaves out the third-gear reduction.
• B: This adds the ratios to get 5.35:1, but ratios in series multiply.
• C (correct): The overall ratio is 1.6 × 3.75 = 6:1, and 2,700 ÷ 6 = 450 rpm.
• D: This is only the transmission output speed; it leaves out the final-drive reduction.

Answer: A

Correct answer: 156 rpm. For an open differential, the two axle speeds add up to twice the carrier speed:

The inside wheel turns

which is 24 rpm below the carrier, mirroring the outside wheel's 24 rpm above it.

• A (correct): The axle speeds add to twice the carrier speed, so the inside wheel turns 360 − 204 = 156 rpm.
• B: This adds the carrier and outside-wheel speeds instead of subtracting from twice the carrier speed.
• C: This assumes both wheels turn at the same speed, which is exactly what a differential avoids in a turn.
• D: This is only the difference between the outside wheel and the carrier, not a wheel speed.

Answer: B

Correct answer: 20.8 m/s. Each revolution moves the vehicle one circumference: 1.95 m × 640 rpm = 1,248 m per minute. Divide by 60 seconds per minute:

• A: This divides the wheel speed by the circumference instead of multiplying them.
• B (correct): Speed is 1.95 m × 640 rpm = 1,248 m per minute, and dividing by 60 gives 20.8 m/s.
• C: This is the distance covered per minute; it was not divided by 60 to convert to meters per second.
• D: This divides by 3,600 as if converting from per hour, but rpm is per minute.

Answer: C

Correct answer: Sends power to front and rear driveshafts and may add a low range. A transfer case sits behind the transmission and always drives the rear driveshaft. When 4WD is engaged it also drives a front driveshaft, and many units add a low range of about 2:1 to 2.7:1 for steep or slow driving.

• A: A fluid coupling between engine and transmission is a torque converter.
• B: That describes a transaxle, not a transfer case.
• C (correct): It mounts behind the transmission, feeds the rear and, in 4WD, the front driveshaft, and often offers a low-range reduction.
• D: Side-to-side speed difference is handled by each axle's differential.

Answer: A

Correct answer: A continuously variable transmission (CVT). In a CVT, the pulley halves move to change the effective diameter where the belt rides on each pulley. The ratio changes smoothly across a wide range, so there are no distinct shift points.

• A (correct): Moving pulley halves change the belt's working diameter, so the ratio changes smoothly without distinct gears.
• B: A dual-clutch unit uses two clutches and fixed gear pairs, not variable pulleys.
• C: A manual gearbox has fixed gear pairs selected by the driver through a clutch, not a belt drive.
• D: A conventional automatic uses planetary gearsets with fixed steps, not a belt and pulleys.

Answer: C

Correct answer: 11.2:1. The final drive follows the transmission, so the overall reduction is their product.

• A: This adds the two ratios, but each stage scales the speed left by the one before, so they must be multiplied.
• B: This uses only the transmission ratio and leaves out the final drive.
• C (correct): Ratios in series multiply: 3.2 × 3.5 = 11.2.
• D: This divides 3.5 by 3.2 instead of multiplying the two ratios.

Answer: B

Correct answer: The stator. A torque converter has an impeller driven by the engine, a turbine connected to the transmission input, and a stator between them. The stator redirects the fluid coming back from the turbine, which multiplies torque when the turbine is turning much slower than the impeller.

• A: The turbine is driven by the fluid and connects to the transmission input; it receives the flow rather than redirecting it back.
• B (correct): The stator sits between turbine and impeller and redirects returning fluid, which multiplies torque at low turbine speed.
• C: A pressure plate belongs to a manual friction clutch, not to a fluid torque converter.
• D: The impeller, or pump, turns with the engine and throws fluid outward; it does not redirect the returning flow.

Answer: D

Correct answer: 265 rpm. For an open differential, the carrier turns at the average of the two axle speeds.

• A: This is the difference between the wheel speeds, not the carrier speed.
• B: This is the sum of the wheel speeds; it must be divided by 2 to get the carrier speed.
• C: This is half the difference, which is how far each wheel differs from the carrier, not the carrier speed itself.
• D (correct): Carrier speed is the average of the two axle speeds: (285 + 245) ÷ 2 = 265 rpm.

Answer: B

Correct answer: CV joint. A constant-velocity (CV) joint lets the axle shaft transmit torque smoothly while the wheel moves up and down and turns for steering. Its balls or rollers need grease, so a rubber or plastic accordion-style boot seals it. If the boot tears, grease escapes and grit enters, and the joint soon wears; a clicking noise in tight turns is a classic sign of a worn outer CV joint. The other parts listed are not half-shaft joints.

• A: The cooling fan moves airflow and is not a grease-filled half-shaft joint.
• B (correct): Correct: Flexible joint boots retain grease around the CV joint while axle angles change.
• C: Spark plugs are ignition components rather than axle torque connectors.
• D: The brake pedal controls deceleration request and has no rotating axle boot.

Answer: B

Correct answer: Use a lower gear so engine braking holds speed back. With the vehicle in gear and the throttle closed, the wheels drive the engine, and its internal friction and pumping resistance slow the vehicle. A lower gear makes the engine turn faster at the same road speed, so engine braking is stronger and the service brakes stay cooler, as long as the engine stays below its maximum safe rpm.

• A: A high gear gives the weakest engine braking, so the brakes must absorb more of the energy and may fade.
• B (correct): A lower gear spins the engine faster at the same road speed, giving stronger engine braking and sparing the brakes.
• C: Neutral breaks the torque path, removing engine braking and putting all the work on the service brakes.
• D: Overdrive lowers engine rpm, which weakens engine braking; it suits level cruising, not long descents.

Answer: A

Correct answer: A slipping clutch with worn facings. When engine speed rises but road speed does not keep up, torque is being lost somewhere in the path, and in a manual drivetrain a slipping clutch with worn friction facings is a common cause. The symptom points toward slip but does not prove it, so a qualified technician should confirm the diagnosis.

• A (correct): When worn facings cannot hold engine torque, engine speed rises without matching acceleration.
• B: A worn U-joint tends to cause clunking or vibration, but it still transmits torque without slipping.
• C: A worn synchronizer usually causes grinding when shifting into a particular gear, not a rising engine speed under load.
• D: A bad wheel bearing typically makes a growl or hum that rises with road speed, not engine flare.

Answer: A

Correct answer: 400 W. Convert 1.5 kN to 1,500 N. Work is W = F

and power is work divided by time:

• A (correct): Work = 1,500 N × 12 m = 18,000 J, and power = 18,000 J ÷ 45 s = 400 W.
• B: This divides the force by the distance instead of multiplying them to find work.
• C: This leaves the force as 1.5 instead of converting 1.5 kN to 1,500 N.
• D: This is the work in joules; it still has to be divided by the 45 s to get power.

Answer: C

Correct answer: 3,200 rpm. With the ratio written as input over output, output speed equals input speed divided by the ratio.

• A: This adds 750 rpm instead of dividing by the ratio.
• B: This treats the gear as direct drive, but 0.75:1 is not 1:1.
• C (correct): Output speed = input ÷ ratio = 2,400 ÷ 0.75 = 3,200 rpm, faster than the input as an overdrive should be.
• D: This multiplies by 0.75; an overdrive must make the output faster, not slower.

Answer: D

Correct answer: The torque converter transmits power through fluid, which allows slip. A torque converter couples engine and transmission through moving fluid rather than a friction clutch. The fluid lets the engine-driven impeller spin while the turbine is held still, so the engine idles in gear with no pedal input.

• A: The differential lets wheels turn at different speeds; it does not disconnect the engine.
• B: Synchronizers match speeds during manual shifts; they do not let an automatic idle in gear.
• C: Low range adds reduction for crawling; it does not let the engine idle while stopped.
• D (correct): Because the coupling is fluid, the impeller can turn with the engine while the turbine stays nearly still.

Answer: A

Correct answer: 240 rpm. The two axle speeds of an open differential add up to twice the carrier speed:

With the left wheel at 0 rpm, the right wheel on the ice spins at 240 rpm, which is why the car may not move at all.

• A (correct): The axle speeds must add to 2 × 120 = 240 rpm, and the left wheel contributes 0 rpm.
• B: This would be true only if both wheels turned at 120 rpm; with one wheel stopped, the other must make up the difference.
• C: An open differential still sends drive to the low-grip wheel, which spins freely on the ice.
• D: This halves the carrier speed, but the average rule requires the spinning wheel to turn at twice the carrier speed.

Answer: D

Correct answer: 92.5 percent. Efficiency is output power divided by input power: 74 kW ÷ 80 kW = 0.925, which is 92.5 percent. The remaining 6 kW becomes heat in the gears, bearings and fluid.

• A: This is the fraction of power lost as heat, not the fraction delivered.
• B: This divides input by output; efficiency is output over input, and no gearbox can exceed 100 percent.
• C: This subtracts the 6 kW loss directly from 100 percent without dividing it by the 80 kW input.
• D (correct): Efficiency is output power over input power: 74 ÷ 80 = 0.925, or 92.5 percent.

Answer: B

Correct answer: A wheel bearing. Wheel bearings support the vehicle weight and cornering loads while the hub turns. When one fails, it often makes a growl that rises with road speed and changes during turns, though a technician should confirm the cause.

• A: A worn synchronizer causes grinding while shifting into a gear, not a growl that follows road speed.
• B (correct): A failing wheel bearing typically growls or hums louder with speed and changes as cornering loads shift.
• C: A worn clutch disc causes slipping, where engine speed rises without matching acceleration, not a wheel-area growl.
• D: Stator problems affect torque multiplication and acceleration, not a speed-dependent noise at one wheel.

Answer: C

Correct answer: The next gear is preselected on the other shaft, and one clutch releases as the other engages. A dual-clutch transmission works like two manual gearboxes in one case, one clutch for the odd-numbered gears and one for the even-numbered gears. While one gear drives, the computer preselects the next gear on the other shaft, so the shift happens by releasing one clutch as the other engages.

• A: That describes a continuously variable transmission, which has no fixed gears to shift between.
• B: That describes an automated manual, which still has one clutch and cannot preselect the next gear on a second shaft.
• C (correct): One clutch serves odd gears and the other even gears, so a shift is just a hand-off between clutches.
• D: Dual-clutch transmissions do not use a torque converter; they use two friction clutches.

Answer: A

Correct answer: 4.0:1. Axle torque = engine torque × transmission ratio × final drive × efficiency. So the ratio =

meaning the transmission must provide at least 4.0:1.

• A (correct): Solving 140 × ratio × 4.5 × 0.90 = 2,268 gives a ratio of 2,268 ÷ 567 = 4.0.
• B: This inverts the result, and an overdrive ratio would lower torque rather than raise it.
• C: This leaves out the final-drive reduction, so it assigns all of the needed multiplication to the transmission.
• D: This ignores the 90 percent efficiency; with losses, a 3.6:1 gear would fall short of the needed torque.

Answer: D

Correct answer: 50 rpm. Multiply all reductions in the path:

an overall ratio of 36.4:1. Axle speed = 1,820 rpm ÷ 36.4 = 50 rpm.

• A: This uses only first gear and the final drive, leaving out the low-range reduction.
• B: This uses only low range and the final drive, leaving out the first-gear reduction.
• C: This adds the three ratios to get 10.1:1, but ratios in series multiply.
• D (correct): The overall ratio is 4.0 × 2.6 × 3.5 = 36.4:1, and 1,820 ÷ 36.4 = 50 rpm.

Answer: D

Correct answer: 2,106 rpm. Wheel speed = 27 m/s ÷ 2.0 m = 13.5 revolutions per second, or 810 rpm. The overall ratio is

so engine speed =

• A: This divides wheel speed by the overall ratio; engine speed is wheel speed times the ratio.
• B: This adds the ratios to get 4.05:1 instead of multiplying them.
• C: This uses only the final drive and ignores the overdrive gear, which lowers engine speed.
• D (correct): The wheels turn 27 ÷ 2.0 × 60 = 810 rpm, and 810 × (0.8 × 3.25) = 810 × 2.6 = 2,106 rpm.

Answer: B

Correct answer: 38 kW. Ideal gears raise torque by the same factor that they lower speed, so power stays the same. Only losses reduce it: 40 kW × 0.95 = 38 kW.

• A: This is the ideal torque scaling applied to power, and it also ignores the efficiency.
• B (correct): Output power is input power times efficiency: 40 × 0.95 = 38 kW; the ratio does not change power.
• C: This divides power by the ratio, treating power like speed; the drop in speed is offset by the rise in torque.
• D: This multiplies power by the ratio, but gears trade torque for speed and cannot create power.

Answer: A

Correct answer: 500 rpm, clockwise. An idler does not change the overall ratio, so only the end gears count:

1, giving

The driving gear turns the idler counterclockwise, and the idler turns the output gear clockwise, the same direction as the input.

• A (correct): The overall ratio is 60/20 = 3:1, so 1,500 ÷ 3 = 500 rpm, and the idler makes the output turn the same way as the input.
• B: The speed is right, but with an idler between them the input and output gears turn the same way.
• C: This multiplies by the ratio, but a larger output gear must turn slower than the smaller driving gear.
• D: This counts the idler as a separate reduction; its tooth count cancels out of the overall ratio.

Answer: B

Correct answer: The locked front and rear driveshafts cannot turn at the different average speeds the axles need. In a turn, every wheel follows a different path, so the front and rear axles need slightly different average speeds. A part-time transfer case locks the front and rear driveshafts together, so on dry pavement the driveline winds up and the tires must scrub to relieve it. That is why part-time 4WD is meant for loose or slippery surfaces.

• A: Each axle's differential still allows left-right speed differences; the problem is between the front and rear axles.
• B (correct): In a turn the front and rear axles need slightly different average speeds, and a part-time system locks them together, so the driveline winds up and the tires scrub.
• C: Low fluid can cause wear or noise, but it does not explain binding that appears only in tight turns on high-grip pavement.
• D: High range is about 1:1, so engaging 4WD does not add torque multiplication; it only adds a front drive path.

Answer: A

Correct answer: The high road load from the grade and trailer is limiting acceleration. Separate a failing torque path from a heavy road load. Slip in the path shows up as engine speed rising without matching road speed. Here the two stay in step, so the drivetrain is delivering torque, and the slow acceleration comes from the steep grade and heavy trailer.

• A (correct): Engine and road speed staying in step shows the drivetrain is delivering torque; slow acceleration then reflects the heavy load.
• B: A converter that could not couple the engine would show engine speed running away from road speed, not staying in step.
• C: Slip would let engine speed flare up without a matching increase in road speed, which is not what is described.
• D: With a broken axle the truck would not be driven up the grade at all, and engine speed would not track road speed.

Answer: C

Correct answer: Start in a higher gear with gentle throttle. Wheel torque can push the car only as hard as the tire's grip allows. On ice, first gear delivers more torque than the tires can use, so they spin. Starting in a higher gear with a light throttle keeps wheel torque below the traction limit, which moves the car better than more torque would.

• A: The wheels already receive more torque than the ice can support, so adding throttle just spins them faster.
• B: More torque multiplication is the opposite of what is needed when the tires already lack grip.
• C (correct): A higher gear delivers less wheel torque, keeping it within the little grip the ice provides so the tires can push instead of spin.
• D: Neutral delivers no torque at all, and dropping into gear at high rpm would send a burst of torque that breaks traction again.

Answer: C

Correct answer: 5,100 N·m. Multiply the reductions:

an overall ratio of 20:1. Then apply efficiency to the torque:

N·m.

• A: This is the ideal value; it ignores the 85 percent efficiency.
• B: This divides by the efficiency, which wrongly makes the real torque larger than the ideal torque.
• C (correct): The overall ratio is 2.0 × 2.5 × 4.0 = 20:1, and 300 × 20 × 0.85 = 5,100 N·m.
• D: This leaves out the low-range reduction, using only 2.0 × 4.0 = 8:1.

Answer: D

Correct answer: It drops by 765 rpm. Engine rpm equals wheel rpm times the overall ratio. In fourth:

In fifth:

The engine slows by

• A: 2,295 rpm is the engine speed in fifth gear, not the amount it changes.
• B: This applies the gear change to wheel speed and leaves out the 3.4:1 final drive.
• C: This divides by the overdrive ratio, as if overdrive raised engine speed; at a fixed road speed it lowers engine speed.
• D (correct): Engine speed falls from 900 × 3.4 = 3,060 rpm to 900 × 3.4 × 0.75 = 2,295 rpm, a drop of 765 rpm.

Answer: B

Correct answer: The propeller shaft (driveshaft). Place the part in the power path: engine, clutch or torque converter, transmission, driveshaft, final drive and differential, axle shafts, wheels. In a rear-drive layout the driveshaft carries power from the transmission back to the rear axle.

• A: The flywheel is bolted to the crankshaft and sits ahead of the transmission, not after it.
• B (correct): The driveshaft runs from the back of the transmission to the rear axle, so it is the next link after the gearbox.
• C: A torque converter sits between the engine and the transmission input, before the gearbox.
• D: The hubs and bearings are at the very end of the chain, after the differential and axle shafts.

Answer: A

Correct answer: Neutral releases the clutches and bands so no drive path exists; use park and the parking brake. In neutral the transmission does not connect its input shaft to its output shaft, so engine torque stops inside the gearbox. Because nothing is locked, the vehicle is free to roll, so a parked vehicle should rely on park and the parking brake.

• A (correct): In neutral the gear-selecting clutches and bands are released, breaking the torque path, so the truck can roll unless park and the parking brake hold it.
• B: Neutral does not lock anything; it only breaks the path, which is why a vehicle in neutral can roll.
• C: The engine still turns the converter impeller in neutral; neutral alone does not hold a vehicle in place.
• D: The differential is not what breaks the path in neutral, and the foot brakes are not a parking device.

Answer: C

Correct answer: 300 N·m. Ratio = driven teeth ÷ driving teeth =

a 2.5:1 reduction.

Ideal output torque =

N·m. Check: a larger driven gear turns slower, so its torque must be higher.

• A: This flips the tooth ratio to 16/40 = 0.4, which would make a reduction lower the torque.
• B: This multiplies the torque by the 40 teeth instead of by the gear ratio.
• C (correct): The ratio is 40/16 = 2.5:1, and ideal output torque is 120 × 2.5 = 300 N·m.
• D: This adds the 40 teeth to the torque instead of multiplying by the ratio.

Answer: B

Correct answer: 19.0 m/s. Overall ratio =

so wheel speed =

Distance per minute = 1.9

m/min, and

• A: This adds the ratios (1.2 + 3.5 = 4.7) instead of multiplying them.
• B (correct): Overall ratio 4.2 gives 600 wheel rpm, and 1.9 × 600 = 1,140 m/min, or 19.0 m/s.
• C: This uses only the 1.2:1 gearbox ratio and leaves out the final-drive reduction.
• D: This is actually 1,140 meters per minute; it was not divided by 60 to get meters per second.

Answer: C

Correct answer: 36 kW. Power is torque times angular speed:

Divide by 1,000 to convert: 36,000 W = 36 kW.

• A: This misplaces the decimal when converting 36,000 W to kilowatts.
• B: This divides torque by angular speed instead of multiplying them.
• C (correct): P = Tω = 240 × 150 = 36,000 W, which is 36 kW.
• D: This shifts the decimal the other way; 36,000 W is 36 kW, not 360 kW.

Answer: B

Correct answer: 24 kW of power, with torque raised to 400 N·m. Gears trade speed for torque but cannot add power, so an ideal gearbox puts out the same 24 kW it takes in.

Output torque =

N·m, while output speed drops by the same factor of 2.5.

• A: This multiplies power by the ratio too, but gears cannot create power.
• B (correct): Ideal gears keep power the same, and torque is multiplied by the ratio: 160 × 2.5 = 400 N·m.
• C: This divides power by the ratio; in an ideal gearbox power is unchanged.
• D: This divides torque by the ratio; a reduction raises torque while lowering speed.

Answer: D

Correct answer: The flywheel. The flywheel is a heavy disc bolted to the crankshaft. Its mass smooths engine pulses, its outer ring gear is what the starter engages, and its face is one of the two surfaces that grip the clutch disc.

• A: The pressure plate bolts to the flywheel and clamps the disc, but it does not carry the starter ring gear.
• B: The release bearing only pushes on the diaphragm spring when the pedal is pressed.
• C: The friction disc is splined to the transmission input shaft, not bolted to the crankshaft.
• D (correct): The heavy flywheel on the crankshaft smooths pulses, carries the starter ring gear and provides a clutch friction face.

Answer: B

Correct answer: 3.6:1. Read ratios by their number: the larger the number, the greater the reduction and torque multiplication. That is why first gear, with the largest ratio (here 3.6:1), is called low gear.

• A: A ratio below 1 is an overdrive, the highest gear with the least torque.
• B (correct): The numerically largest ratio has the most reduction, so it is the lowest gear.
• C: A 1:1 ratio is direct drive, which gives no torque multiplication.
• D: This is a reduction, but a milder one than 3.6:1.

Answer: C

Correct answer: When the two drive wheels turn at different speeds, as in a turn. In a turn the outside wheel travels a longer arc than the inside wheel and must turn faster. The spider gears spin on their shaft to let one side gear speed up while the other slows down. Going straight, everything turns together as a unit.

• A: Going straight, both axles turn at carrier speed and the spider gears ride along without spinning on their shaft.
• B: Reverse changes the direction of the whole carrier; it does not by itself make the spider gears spin on their shaft.
• C (correct): Spider gears rotate on their shaft only when one side gear must turn faster than the other.
• D: With the vehicle stopped, the carrier and axles are not turning, so the spider gears are not working.

Answer: A

Correct answer: The turbine. Follow the fluid: the impeller turns with the engine and flings fluid outward, the fluid strikes the turbine, and the turbine turns the transmission input shaft. The stator redirects the fluid on its way back.

• A (correct): The turbine is struck by the fluid and drives the transmission input.
• B: The impeller, or pump, turns with the engine and throws the fluid; it is the input element.
• C: The stator sits between the other two and redirects returning fluid; it does not drive the input shaft.
• D: The lock-up clutch only joins engine and transmission mechanically at cruise; it is not the fluid-driven element.

Answer: D

Correct answer: 496.8 N·m at 900 rpm. Output torque = input torque × ratio × efficiency =

N·m.

Output speed =

Efficiency applies only to torque or power, never to speed.

• A: The torque is right, but efficiency was also applied to speed; gear teeth do not slip, so speed is 900 rpm.
• B: This is the ideal torque; it leaves out the 92 percent efficiency.
• C: This divides torque and multiplies speed, the opposite of what a reduction does.
• D (correct): Torque: 180 × 3 × 0.92 = 496.8 N·m. Speed follows the ratio alone: 2,700 ÷ 3 = 900 rpm.

Answer: B

Correct answer: It resists large wheel-speed differences so more torque reaches the wheel with traction. An open differential sends equal torque to both axles, limited by the wheel with less grip, so a wheel on ice can leave the other wheel with almost nothing. A limited-slip differential resists that speed difference so more torque goes to the wheel that has traction.

• A: That describes a locker that is always engaged, which would scrub tires in turns on dry pavement.
• B (correct): Clutch packs, special gearing or viscous fluid limit slip so the gripping wheel can still push.
• C: A limited-slip unit does not add a reduction; the final drive sets the ratio.
• D: That is the job of a torque converter, not a differential.

Answer: A

Correct answer: Universal joints for the angle and a slip yoke for the length. The propeller shaft runs from the transmission to the rear axle, which moves with the suspension. Universal joints let the shaft angle change, and a slip yoke lets the shaft get slightly longer or shorter.

• A (correct): U-joints at the shaft ends flex with the angle, and the slip yoke slides to let the length change.
• B: A center bearing supports a two-piece shaft; it does not let the length change.
• C: Spider and side gears are inside the differential and let the wheels turn at different speeds.
• D: A wheel bearing supports the hub and a pressure plate is a clutch part; neither is part of the driveshaft.

Answer: B

Correct answer: A sliding collar moved by a shift fork. In a constant-mesh gearbox all gear pairs stay meshed, and the gears on the output shaft spin freely on bearings. Moving the shift lever moves a fork, which slides a collar to lock the chosen gear to the shaft, so only that pair sets the ratio.

• A: Hydraulic bands select ratios in automatic planetary transmissions, not in a manual gearbox.
• B (correct): The shift lever moves a fork that slides a collar to lock the free-spinning gear to its shaft.
• C: In a constant-mesh gearbox the gear pairs are always in mesh; the collar does the locking.
• D: The clutch disc connects the engine to the input shaft; it does not select individual gears.

Answer: A

Correct answer: 7.6:1. Low range is a separate reduction in the transfer case that multiplies whatever the transmission is doing. Combined reduction =

1, before the final drive is applied.

• A (correct): Ratios in series multiply: 3.8 × 2.0 = 7.6.
• B: This adds the ratios, but each stage scales the speed left by the one before it.
• C: This divides the ratios instead of multiplying them.
• D: This ignores the low range, which adds its own reduction on top of first gear.

Answer: D

Correct answer: It carries hydraulic pressure that applies clutches and bands. Automatic transmission fluid transmits power in the torque converter, carries hydraulic pressure to apply clutches and bands, lubricates and cools. Because shifts depend on that pressure, low or worn fluid can cause slipping or harsh shifts.

• A: Engine coolant, not transmission fluid, cools the cylinder heads and block.
• B: Brake calipers are applied by brake fluid from the master cylinder, a separate system.
• C: Ignition timing is controlled by the ignition system, not by transmission fluid.
• D (correct): Shifting in an automatic depends on fluid pressure applying the clutches and bands.

Answer: B

Correct answer: By holding one member still, driving another and taking output from the third. A planetary set has a sun gear, planet gears on a carrier and an outer ring gear. Hydraulically applied clutches and bands decide which member is held, which is driven and which delivers output, and each combination gives a different ratio.

• A: Sliding gears into mesh describes an old manual design, not a planetary set.
• B (correct): Choosing which of the sun, carrier and ring is held, driven or output sets the ratio.
• C: Variable pulleys belong to a CVT.
• D: Two clutches for odd and even gears on separate input shafts describe a dual-clutch transmission.

Answer: C

Correct answer: Electric motors make strong torque from zero rpm across a very wide speed range. A gasoline engine has a narrow useful speed band, so it needs several ratios. An electric motor produces strong torque from zero rpm and runs over a very wide speed range, so a single reduction, plus a differential, is enough.

• A: They still use a fixed reduction, often around 8:1 to 10:1.
• B: Motor speed varies widely with road speed; that is why one ratio is enough.
• C (correct): Because the motor works well from standstill to high speed, one ratio can cover all driving.
• D: They still need a reduction gear, a differential and drive shafts to the wheels.

Answer: A

Correct answer: An automated manual transmission. The clues are one friction clutch, actuators doing the clutch and shift work, and a manual-style gearbox with no torque converter. That combination is an automated manual.

• A (correct): An automated manual uses one clutch and actuators to shift a manual-type gearbox.
• B: A dual-clutch unit uses two clutches, one for odd gears and one for even gears.
• C: A CVT uses a belt or chain on variable pulleys and has no gear steps.
• D: A conventional automatic uses a torque converter and planetary gearsets.

Answer: D

Correct answer: 1,110 rpm. Carrier speed is the average of the two axle speeds:

The driveshaft is the input to a 3.7:1 final drive, so it turns 3.7 times faster than the carrier:

• A: This multiplies the outside wheel speed by 3.7 instead of using the carrier's average speed.
• B: This adds the two wheel speeds without dividing by 2 before applying the final drive.
• C: This divides the carrier speed by the ratio, but the driveshaft is the input side and turns faster.
• D (correct): Carrier speed is (336 + 264) ÷ 2 = 300 rpm, and the driveshaft turns 3.7 times faster: 300 × 3.7 = 1,110 rpm.

Answer: A

Correct answer: CV joints keep output speed steady even at the large angles of steered wheels. A simple universal joint at an angle makes its output speed rise and fall slightly during each turn, which is acceptable only at small angles. Front wheels steer through large angles while driving, so CV joints with balls or rollers in grooved races keep the speed smooth.

• A (correct): A simple U-joint speeds up and slows down each turn at an angle; CV joints avoid this at large angles.
• B: CV joints pass rotation through at the same ratio; the final drive provides reduction.
• C: Allowing left and right wheels to turn at different speeds is the differential's job.
• D: Disconnecting engine torque is the clutch's job.

Answer: A

Correct answer: About 82.1 kW. Each stage passes on only its efficiency fraction of what it receives. After the transmission:

After the final drive:

about 82.1 kW.

• A (correct): Efficiencies in series multiply: 90 × 0.95 × 0.96 = 82.08 kW.
• B: This subtracts 5 percent and 4 percent from 90 kW together, but the second loss applies to the reduced power.
• C: This includes only the transmission loss and leaves out the final-drive loss.
• D: This assumes no losses, but real gears and bearings turn some power into heat.

Answer: D

Correct answer: The 4.10:1 truck's engine turns faster, giving more pulling torque but more revolutions per mile. At a fixed road speed both trucks' wheels turn at the same rpm, and engine rpm = wheel rpm × overall ratio. The numerically higher 4.10:1 final drive therefore makes the engine turn faster, giving more torque at the wheels at the cost of more engine revolutions per mile.

• A: Equal road speed means equal wheel rpm, but different ratios give different engine rpm.
• B: A numerically lower ratio is a taller gearing that lowers engine rpm at a given speed.
• C: The higher ratio raises cruising rpm; the 3.08:1 truck would be the economical one.
• D (correct): Engine rpm equals wheel rpm times overall ratio, so the numerically higher ratio means higher engine speed.

Answer: D

Correct answer: Second gear. Engine rpm = wheel rpm × final drive × gear ratio, and

per unit of gear ratio.

First: 8,820 rpm (too high). Second: 5,292 rpm. Third: 3,528 rpm. Fourth: 2,520 rpm. A lower gear brakes harder, so the lowest gear that stays under 6,000 rpm, second, is best.

• A: First would spin the engine at 700 × 3.6 × 3.5 = 8,820 rpm, far over the 6,000 rpm limit.
• B: Third is safe at 3,528 rpm, but second gives stronger braking while still under the limit.
• C: Fourth turns the engine only 2,520 rpm, giving the weakest engine braking of the safe choices.
• D (correct): Engine speed in second is 700 × 3.6 × 2.1 = 5,292 rpm, the highest value still under the limit.

Answer: C

Correct answer: The release (throwout) bearing. With the pedal up, the diaphragm spring in the pressure plate clamps the friction disc against the flywheel. Pressing the pedal moves the release, or throwout, bearing against that spring, which removes the clamping force and disconnects engine torque.

• A: The starter drive gear engages the flywheel's ring gear only during cranking.
• B: A synchronizer matches gear speeds inside the gearbox; it does not touch the clutch spring.
• C (correct): The release bearing presses the diaphragm spring so the pressure plate stops squeezing the disc.
• D: The disc facings are what get clamped; they do not push on the spring.

Answer: C

Correct answer: 400 rpm and 375 N·m. Each stage ratio is driven teeth ÷ driving teeth:

Stages in series multiply:

1.

Output speed =

and ideal output torque =

N·m. Speed went down and torque went up, as a reduction should.

• A: This adds the stage ratios (2.5 + 3 = 5.5) instead of multiplying them.
• B: This uses only the first stage's 2.5:1 ratio and leaves out the second stage.
• C (correct): Stage ratios 30/12 = 2.5 and 45/15 = 3 multiply to 7.5, giving 3,000 ÷ 7.5 = 400 rpm and 50 × 7.5 = 375 N·m.
• D: The speed is right, but the torque was divided by the ratio; a reduction raises torque.

Answer: D

Correct answer: About 1,284 N·m. Overall ratio =

Total torque = 220 × 13.26 × 0.88 ≈ 2,567 N·m.

An open differential sends equal torque to each axle, so each gets about 2,567 ÷ 2 ≈ 1,284 N·m.

• A: This is the total torque at the differential; it is not split between the two axles.
• B: This splits the torque correctly but leaves out the 88 percent efficiency.
• C: This adds the ratios (3.4 + 3.9 = 7.3) instead of multiplying them.
• D (correct): Total axle torque is 220 × 3.4 × 3.9 × 0.88 ≈ 2,567 N·m, and an open differential splits it equally, about 1,284 N·m per axle.