Tires and Wheels Lesson Test (40 Questions)

Answer: C

Correct answer: The tread. The tread is the thick outer band of rubber molded with grooves and blocks, and it is the only part designed to wear against pavement. Its pattern provides grip, channels water and gives a visible wear reference. The bead, by contrast, is the steel-reinforced inner edge that locks the tire onto the rim, and the valve core and hub bearing are not part of the tire's rolling surface at all.

• A: Incorrect: The core permits air filling and retention.
• B: Incorrect: The bead seals against the wheel.
• C (correct): Correct: The tread forms the tire-road interface under normal driving.
• D: Incorrect: The bearing supports rotation inside the hub.

Answer: C

Correct answer: Radial tire construction. In a radial tire, the body plies run straight across from bead to bead at roughly 90 degrees to the direction of travel, with stiff belts laid under the tread. This lets the sidewall flex while the tread stays flat, improving wear, fuel economy and handling. In the code 215/65R15, 215 is the section width in millimeters, 65 is the aspect ratio, and 15 is the rim diameter in inches. A common mistake is reading R as radius.

• A: Incorrect: R does not identify track use.
• B: Incorrect: Direction or orientation is marked separately when applicable.
• C (correct): Correct: R describes how the carcass cords are arranged.
• D: Incorrect: The 15 is the designated rim diameter, not radius.

Answer: B

Correct answer: Vehicle Tire and Loading Information Label. Each vehicle maker chooses a cold inflation pressure based on that vehicle's weight, load capacity and handling, and prints it on a placard usually found on the driver's door jamb; the owner's manual repeats it. The number molded on the tire sidewall is different: it is the maximum the tire can safely hold, not the pressure the vehicle should run. Confusing the two often leads to overinflation, a harsher ride and reduced grip.

• A: Incorrect: Recommended values vary by vehicle and load specification.
• B (correct): Correct: The door-area placard specifies the manufacturer-recommended pressures.
• C: Incorrect: That is a tire limit, not the vehicle operating setting.
• D: Incorrect: This is not the standardized tire-pressure source.

Answer: C

Correct answer: A tire pressure gauge. A tire pressure gauge presses on the valve stem's core, letting air push against a spring, diaphragm or electronic sensor that converts the force into a reading in psi or kPa. Other shop tools sound similar but measure different things: a tread-depth gauge measures groove depth, a balancer finds uneven mass, and a torque wrench measures twisting force on fasteners. Checking pressure regularly with your own gauge is more reliable than relying on gas-station gauges.

• A: Incorrect: This measures groove depth rather than pressure.
• B: Incorrect: This detects rotational imbalance, not air pressure.
• C (correct): Correct: A suitable calibrated gauge measures the inflation pressure.
• D: Incorrect: This measures applied fastening torque.

Answer: D

Correct answer: Help evacuate water from the contact area. When a tire rolls over a wet road, water must be pushed out of the footprint so rubber can touch pavement. Grooves and sipes act as channels that carry that water to the sides and rear of the contact patch. If water builds up faster than the grooves can clear it, the tire rides up on a film of water and hydroplanes. That is why worn, shallow grooves greatly reduce wet grip, and why no tread can guarantee traction in deep water.

• A: Incorrect: Hydroplaning can occur despite tread grooves.
• B: Incorrect: They have no electrical charging role.
• C: Incorrect: The grooves do not affect bolt installation torque.
• D (correct): Correct: Groove geometry aids water displacement and wet traction.

Answer: D

Correct answer: Rotational mass imbalance. No wheel and tire is perfectly uniform; a heavy spot creates a force that swings around with each revolution and grows rapidly with speed, so the vibration often appears around highway speeds. A balancing machine spins the assembly, finds the heavy area, and tells the technician where to clip or stick small weights to offset it. Do not confuse this with alignment, which adjusts wheel angles such as toe and camber rather than mass distribution.

• A: Incorrect: Pressure must be checked with an inflation gauge.
• B: Incorrect: No balancing operation changes rubber age.
• C: Incorrect: Toe is adjusted during alignment service.
• D (correct): Correct: Balance weights reduce speed-sensitive vibration associated with imbalance.

Answer: A

Correct answer: A sidewall bulge. A bulge forms when the internal cords or belts in the sidewall are broken, often from hitting a pothole or curb, so air pressure pushes the rubber outward at the weak spot. Because the sidewall has no reinforcing plies left there, the tire can fail suddenly at speed. Deep tread does not help, since tread depth tells you only about wear, not about the structure underneath. A sidewall bulge generally means the tire should be replaced.

• A (correct): Correct: A bulge may signal serious internal structural damage.
• B: Incorrect: Cleanliness alone is not tire structural damage.
• C: Incorrect: A correctly fitted cap is expected.
• D: Incorrect: Tire identification markings are normal.

Answer: D

Correct answer: Its last four date-code digits. The tire identification number begins with DOT and ends with a four-digit date code for tires made since 2000. The first two digits are the week of manufacture and the last two are the year, so a code ending in 1522 means the fifteenth week of 2022. Knowing the date helps you judge tire age, because rubber deteriorates over time even when the tread looks fine. Speed ratings and size numbers describe capability and fit, not dates.

• A: Incorrect: A speed symbol encodes speed capability, not a calendar.
• B: Incorrect: TPMS does not display a production date.
• C: Incorrect: Rim diameter does not reveal manufacturing week.
• D (correct): Correct: Modern date codes use two digits for week and two for year.

Answer: B

Correct answer: Indirect TPMS. An underinflated tire has a slightly smaller rolling radius, so it must spin a bit faster than the others to cover the same distance. Indirect TPMS uses the anti-lock brake wheel-speed sensors to spot that difference and warns the driver. Direct TPMS instead places a pressure sensor and transmitter inside each wheel. A weakness of the indirect method is that if all four tires lose pressure evenly, the speed difference may be too small to notice.

• A: Incorrect: It controls coolant flow based on temperature.
• B (correct): Correct: Indirect systems infer changes using wheel speeds and other signals.
• C: Incorrect: It adjusts charging output, not tire behavior.
• D: Incorrect: Direct TPMS measures inflation pressure with pressure sensors.

Answer: D

Correct answer: Tire rotation. Front and rear tires wear differently; on most front-wheel-drive cars, the front tires handle steering, most braking and all driving force, so they wear faster. Rotating the tires on a schedule, often every 5,000 to 8,000 miles, moves each assembly to a new position so wear evens out and the set lasts longer. Alignment and balancing are separate services: one sets wheel angles, the other corrects mass distribution, and neither changes which corner a tire occupies.

• A: Incorrect: This removes air from hydraulic brake circuits.
• B: Incorrect: Alignment adjusts wheel angles rather than mounting position.
• C: Incorrect: Balancing adjusts mass distribution on an assembly.
• D (correct): Correct: Rotation relocates assemblies according to an allowed pattern.

Answer: A

Correct answer: 6 psi. Subtract measured from specified cold pressure:

• A (correct): Correct: Subtract measured from specified cold pressure: 34 − 28 = 6 psi.
• B: Incorrect: The stated values are six apart.
• C: Incorrect: That is the target, not the shortfall.
• D: Incorrect: Adding both readings does not give the correction.

Answer: A

Correct answer: 135 mm. The aspect ratio 60 means the sidewall is 60 percent of the section width.

• A (correct): Sidewall height is 60 percent of the 225 mm width: 225 times 0.60 equals 135 mm.
• B: This divides the width by 0.60 instead of multiplying, giving a sidewall taller than the tire is wide.
• C: This subtracts the aspect ratio from the width, which mixes a percentage with a length.
• D: This reads the aspect ratio as a length, but it is a percentage of the tire's width.

Answer: B

Correct answer: 248 kPa. A kilopascal is a much smaller unit than a psi, so converting from psi to kPa must give a larger number. Multiplying by 6.895 does that. A quick estimate checks the result: 36 × 7 is 252, slightly above 248, which fits because 6.895 is a little less than 7. If an answer comes out smaller than the starting value, or wildly larger, the conversion factor was divided, added or misplaced rather than multiplied.

• A: This divides by the conversion factor instead of multiplying; a kilopascal is smaller than a psi, so the number should grow.
• B (correct): Multiplying 36 psi by 6.895 kPa per psi gives about 248 kPa.
• C: This adds the conversion factor to the pressure instead of multiplying by it.
• D: This misplaces the decimal point, giving a value ten times too large.

Answer: B

Correct answer: At 30 psi or lower. A 25 percent drop from 40 psi is

leaving

Pressures between 30 and 40 psi are underinflated but may not trigger the lamp.

• A: This is 25 percent of 40 psi, which is the amount of the drop, not the pressure remaining.
• B (correct): A 25 percent drop from 40 psi is 10 psi, so the warning is required once pressure reaches 30 psi.
• C: This subtracts 15 psi, a 37.5 percent drop, which is well past the threshold.
• D: A 5 psi drop is only 12.5 percent, half of the required 25 percent threshold.

Answer: D

Correct answer: The reading is warm and may exceed the cold value. Driving flexes the tire and heats the air inside, and by the gas laws warmer air at nearly constant volume exerts more pressure, often a few psi more. The vehicle placard value is a cold specification, meaning the tire has sat for a few hours or been driven only briefly. Letting air out of a hot tire to reach the cold number leaves it underinflated once it cools, which is the most tempting mistake here.

• A: Incorrect: The sidewall marking is not vehicle-specific recommendation.
• B: Incorrect: That risks underinflation when the tire cools.
• C: Incorrect: A single warm measurement is not leak evidence.
• D (correct): Correct: Heat from operation commonly increases inflation pressure.

Answer: A

Correct answer: TPMS can miss lesser underinflation and does not detect every defect. TPMS is a warning system, not an inspection tool. U.S. rules require the lamp to light when a tire falls about 25 percent below the recommended cold pressure, so a tire can be noticeably low with the lamp still off. TPMS also cannot see cuts, bulges, embedded nails or worn tread. That is why a monthly check with a gauge and a visual walk-around remains necessary, even on vehicles with working TPMS.

• A (correct): Correct: NHTSA recommends monthly checks regardless of TPMS status.
• B: Incorrect: Neither direct nor indirect TPMS provides tread-depth inspection.
• C: Incorrect: TPMS does not certify structural tire condition.
• D: Incorrect: TPMS warns only past a set threshold, so an off lamp does not confirm exact placard pressure.

Answer: B

Correct answer: A TPMS system malfunction. U.S. TPMS rules call for a distinct malfunction signal: the lamp typically flashes for about a minute at startup and then stays lit. This tells the driver the system itself cannot monitor pressures, often because of a dead sensor battery, a sensor not programmed after a tire change, or a module fault. A steady lamp without flashing usually means low pressure instead. Because monitoring is not working, check all tires manually with a gauge until the system is repaired.

• A: Incorrect: A single malfunction signal cannot prove that.
• B (correct): Correct: Flashing followed by steady illumination commonly indicates a monitoring fault.
• C: Incorrect: This lamp specifically relates to TPMS function.
• D: Incorrect: A malfunction indication does not diagnose overinflation.

Answer: D

Correct answer: Inflation history is worth investigating, but wear alone is not conclusive. Classic textbook patterns say overinflation wears the center because the tread crowns outward, while underinflation wears both shoulders. However, real wear also depends on load, tire design, wheel width, driving style and how often the tire spent time at different pressures. A good technician treats a wear pattern as a clue that points to likely causes, then checks pressure records, alignment and suspension before concluding. Assuming one certain cause from a single pattern is the error to avoid.

• A: Incorrect: Battery health does not directly produce that tread pattern.
• B: Incorrect: A wear pattern cannot certify alignment geometry.
• C: Incorrect: Low inflation is classically linked to both-shoulder wear, and wear alone cannot prove any single cause.
• D (correct): Correct: Center wear can relate to inflation and other variables; inspect systematically.

Answer: D

Correct answer: Inspect inflation, suspension and alignment rather than guessing one failure. Wear concentrated on one edge often points to wheel angles, such as excessive camber or incorrect toe, but worn bushings, ball joints, bent parts or pressure problems can contribute too. A sound diagnosis checks inflation first, then inspects suspension components for looseness, and finally measures alignment. Replacing or adjusting only one thing without measuring risks leaving the real cause in place, so the new tire wears the same way.

• A: Incorrect: The sidewall maximum is not the vehicle's inflation target, and pressure alone does not correct geometry.
• B: Incorrect: Engine lubricant does not set wheel angles.
• C: Incorrect: Continued uneven wear can compromise tire safety, and the cause would remain.
• D (correct): Correct: One-edge wear may involve geometry or other mechanical influences.

Answer: A

Correct answer: Alignment sets wheel angles; balancing corrects rotating mass distribution. Alignment adjusts how the wheels point relative to the vehicle and the road, mainly camber, caster and toe, which affects pulling, steering feel and edge wear. Balancing spins each wheel-and-tire assembly and adds small weights to even out heavy spots, which addresses vibration that grows with speed. A useful memory hook: alignment deals with angles, balancing deals with weight. Mixing them up leads to the wrong fix for a symptom.

• A (correct): Correct: The procedures address different physical problems.
• B: Incorrect: Neither simply sets inflation pressure.
• C: Incorrect: This reverses their basic functions.
• D: Incorrect: Neither service is defined by those geometric changes.

Answer: C

Correct answer: Its specified rolling direction must be preserved. A directional tire has a tread pattern, often V-shaped, designed to pump water out efficiently only when rolling one way; an arrow on the sidewall shows that direction. A crossed rotation moves a tire to the opposite side of the car, which would make it spin backward. Directional tires are therefore usually rotated straight front to back on the same side, unless they are dismounted and remounted on their wheels to reverse them.

• A: Incorrect: Direction marking does not imply dry-only use.
• B: Incorrect: Speed capability is not lost simply by relocation.
• C (correct): Correct: Moving across vehicle sides may reverse rolling direction unless remounted appropriately.
• D: Incorrect: Directional tires still require inflation.

Answer: C

Correct answer: Week 24 of 2023. For tires made since 2000, the date code at the end of the DOT number has four digits: the first two give the week of manufacture (01 to 53) and the last two give the year. So 2423 reads as week 24 of 2023, roughly mid-June. The trap is reading the digits as a calendar date or a size, but the code never includes a day of the month, and tire size appears in a separate marking.

• A: Incorrect: The code does not encode a calendar day.
• B: Incorrect: The code is not a tire-size string.
• C (correct): Correct: The first two digits indicate week and the final two indicate year.
• D: Incorrect: The four digits are not a pressure-temperature pair.

Answer: B

Correct answer: Approximately 1.6 mm. 2/32 in = 0.0625 in ≈ 1.5875 mm.

• A: Incorrect: That is much shallower than 2/32 inch.
• B (correct): Correct: 2/32 in = 0.0625 in ≈ 1.5875 mm.
• C: Incorrect: That equals roughly 1/4 inch.
• D: Incorrect: The fraction is in inches, not millimeters.

Answer: D

Correct answer: More casing flex creates rolling energy losses. A tire is constantly bending as it rolls: the part entering the contact patch flattens and then springs back. Rubber is not perfectly elastic, so each flex cycle turns some energy into heat, a loss called rolling resistance. A soft, underinflated tire deflects more, wasting more energy, and the engine must burn extra fuel to make up for it. The added flexing and heat also speed wear and can damage the tire.

• A: Incorrect: Inflation does not directly set charging voltage.
• B: Incorrect: A soft tire usually flattens into a larger footprint, and patch size is not the basic mechanism.
• C: Incorrect: Tire pressure cannot eliminate air drag.
• D (correct): Correct: Deformation and heat dissipate additional energy.

Answer: C

Correct answer: The code is a reference to a standardized load-capacity table. The load index is a code number, not a weight. Each number corresponds to a load listed in a standard table, and the scale is not linear; for example, index 91 means 615 kg (1,356 lb) per tire, while 100 means 800 kg. You look the number up rather than calculate it. The index appears after the size, often paired with a speed rating letter, as in 91V.

• A: Incorrect: Bolt count is a wheel fitment specification.
• B: Incorrect: Load capacity and pressure are distinct markings.
• C (correct): Correct: A numbered index must be interpreted using its rating scale.
• D: Incorrect: No simple linear conversion like that applies; the index must be looked up in the table.

Answer: C

Correct answer: Improper clamping even if threads appear to fit. Lug nuts and bolts are held tight mainly by the contact between their seat and the matching seat in the wheel, such as conical, ball or flat types. If the shapes do not match, the fastener touches only a small ring or edge, so clamping force is uneven and can drop as the metal deforms. Wheels can then loosen, studs can break, and a wheel can come off, even though the threads turned on normally.

• A: Incorrect: Fastener mismatch cannot provide balance.
• B: Incorrect: Vehicle cold pressure does not follow seat shape.
• C (correct): Correct: Fastener seat geometry must match the wheel specification.
• D: Incorrect: Mounting hardware does not alter tread rubber depth.

Answer: B

Correct answer: Use a pattern specifically permitted by the vehicle and tire manufacturers. A staggered setup puts wider tires on one axle, usually the rear, so those tires cannot simply move to the other axle. In many cases rotation is limited to swapping side to side on the same axle, and only if the tires are not directional; some combinations cannot be rotated at all. The vehicle and tire makers specify the allowed pattern. Assuming a standard pattern always works is the mistake that leads to incorrect fitment.

• A: Incorrect: Crossing can violate size and directional restrictions.
• B (correct): Correct: Staggered sizing may prevent ordinary front-to-rear rotation.
• C: Incorrect: Physical position is not permission to mount an incompatible size.
• D: Incorrect: Inspection remains necessary on all configurations.

Answer: B

Correct answer: Wheel imbalance is possible, but other rotating or suspension faults must be checked. Vibration that grows with road speed comes from something rotating at wheel speed, and imbalance is the most common cause. However, bent wheels, out-of-round or separated tires, worn tie-rod ends, loose wheel bearings and other suspension faults can feel similar. Good diagnosis starts with the likely cause but verifies by inspection, for example checking balance and rim runout and looking for tread separation. Jumping to a single certain cause often wastes money and misses safety issues.

• A: Incorrect: A battery replacement is unrelated to the identified symptom.
• B (correct): Correct: Speed-dependent vibration has multiple possible sources.
• C: Incorrect: Toe is not the sole cause of vibration, and imbalance is a more common one.
• D: Incorrect: Pressure alone cannot exclude structural damage or imbalance.

Answer: A

Correct answer: It indicates when tread reaches a built-in wear reference. Treadwear indicators are small raised bars molded across the bottom of the main grooves at 2/32 inch, the common legal minimum in many states. As tread wears, the surrounding rubber drops toward the bars until they sit flush with the tread surface, signaling the tire needs replacement. Because they are simply rubber ridges, they show depth only, not pressure, age or alignment, though uneven bar exposure can hint at irregular wear.

• A (correct): Correct: A raised bar becomes level with surrounding worn tread near the limit.
• B: Incorrect: It is a molded rubber bar with no pressure sensor.
• C: Incorrect: Manufacturing date is shown by an identification code.
• D: Incorrect: It provides no camber or toe reading.

Answer: B

Correct answer: Aging, pressure, condition and manufacturer guidance still matter. Rubber ages through oxidation and heat even when a tire is never driven, and spares often sit in hot trunks or under the vehicle exposed to the elements. Deep tread shows only that the rubber is not worn, not that it is still sound. Check the date code, look for cracking, verify pressure, since spares often lose air unnoticed, and follow the vehicle and tire makers' age guidance. Many makers suggest replacing tires after about six to ten years regardless of tread.

• A: Incorrect: Balance does not assess aging or deterioration.
• B (correct): Correct: Unused tires age even when tread is not worn away.
• C: Incorrect: Heat and time can affect compounds and structure.
• D: Incorrect: The two markings serve different purposes.

Answer: B

Correct answer: 634 mm. Sidewall height is

and the rim is

• A: This adds 17 directly to the sidewalls without converting the rim from inches to millimeters.
• B (correct): Each sidewall is 225 times 0.45, or 101.25 mm; two sidewalls plus the 431.8 mm rim give about 634 mm.
• C: This treats the aspect ratio 45 as a 45 mm sidewall instead of a percentage of width.
• D: This adds only one sidewall to the rim, but the diameter spans a sidewall above and below the rim.

Answer: A

Correct answer: 125 N. Force depends on speed squared.

• A (correct): The speed ratio is 75/30 = 2.5, and squaring it gives 6.25, so the force becomes 20 times 6.25 = 125 N.
• B: This scales the force in direct proportion to speed, but imbalance force grows with the square of speed.
• C: This scales by the square root of the speed ratio, which makes the force grow far too slowly.
• D: This cubes the speed ratio instead of squaring it.

Answer: A

Correct answer: 2,880 N. The grip limit is

Braking and cornering must satisfy

• A (correct): The limit is 0.9 times 4,000 N, or 3,600 N; the square root of 3,600 squared minus 2,160 squared is 2,880 N.
• B: This adds the squares instead of subtracting, giving more than the tire's total grip.
• C: This is the full grip limit and ignores the braking force that is already using part of the budget.
• D: This subtracts the braking force directly from the limit, but the forces combine at right angles, not by simple subtraction.

Answer: C

Correct answer: 24.5 m/s. Find the circumference, then multiply by revolutions per second.

That is about 88 km/h.

• A: This divides the circumference by the rotation rate instead of multiplying.
• B: This multiplies the diameter by the rotation rate and forgets the factor of $\pi$ needed for circumference.
• C (correct): Circumference is $\pi$ times 0.65 m, about 2.04 m, and 2.04 m times 12 rev/s is about 24.5 m/s.
• D: This uses the radius in place of the diameter when finding circumference, halving the distance per revolution.

Answer: C

Correct answer: 744 N·m. Use

• A: This uses the full diameter of 0.62 m as the lever arm instead of the radius.
• B: This halves the radius, as if the force acted midway between the center and the tread.
• C (correct): Torque is force times radius: 2,400 N times 0.31 m equals 744 N$\cdot$m.
• D: This divides the force by the radius; torque multiplies force by the lever arm.

Answer: D

Correct answer: Inflate the right front to 33 psi, then road-test to see whether the drift remains. Start with the simplest cause. The low right front tire can make the car pull toward that side, so inflate it to the label value and road-test. If the drift remains, then consider alignment, a dragging brake or a tire pull.

• A: Matching at the low value leaves both tires underinflated; the label value is the target.
• B: A pull has several possible causes, and nothing here shows the tires are damaged.
• C: Alignment is one possible cause, but simple causes such as uneven pressure should be corrected and ruled out first.
• D (correct): Uneven side-to-side pressure is a simple, likely cause of pull, so correct it before moving on to alignment or brake checks.

Answer: D

Correct answer: 260 kPa. Use

• A: This misplaces the decimal point, giving a value ten times too small.
• B: This misplaces the decimal point, giving a value ten times too large.
• C: This multiplies force by area instead of dividing.
• D (correct): Pressure is force divided by area: 3,900 N divided by 0.015 square meters is 260,000 Pa, or 260 kPa.

Answer: C

Correct answer: A 3/16-inch nail hole in the central tread area. Repairs are generally limited to holes about 1/4 inch (6 mm) or smaller in the central tread. Shoulder and sidewall damage is usually not repairable, and a proper repair is made from inside the tire after it is removed from the wheel.

• A: Sidewall damage is usually not repairable, even when the hole is small.
• B: Shoulder damage is generally not repairable because that area flexes heavily in turns.
• C (correct): This is under the roughly 1/4-inch limit and in the central tread, the area where repairs are generally allowed.
• D: At about 12 mm, this is larger than the roughly 1/4-inch (6 mm) repair limit.

Answer: A

Correct answer: 370 kg. The stated equal-distribution assumption gives

per wheel.

• A (correct): Correct: The stated equal-distribution assumption gives 1480/4 = 370 kg per wheel.
• B: Incorrect: This assigns the entire mass to every wheel.
• C: Incorrect: This divides by five instead of four.
• D: Incorrect: This multiplies by four rather than divides.

Answer: A

Correct answer: Recheck when cold rather than automatically venting to 33 psi warm. Driving warms the air inside a tire, and warmer air at nearly constant volume reads higher, so a 4 psi rise after a drive is common. The placard's 33 psi applies when the tire is cold, typically after sitting for three hours or more. If you vent the warm tire down to 33, it will read low once it cools. Recheck when cold and adjust then; only an unusually large or rising difference suggests another problem.

• A (correct): Correct: A warm tire often measures higher; the placard target applies to cold conditions.
• B: Incorrect: That may leave the tire underinflated after cooling.
• C: Incorrect: One warm gauge measurement cannot establish TPMS failure.
• D: Incorrect: The maximum marking is not the vehicle cold target.

Answer: D

Correct answer: It rated in the top class in straight-line braking tests on wet pavement. UTQG traction grades, from highest to lowest AA, A, B and C, come from straight-line braking tests on wet pavement. An AA grade is useful information, but it does not describe cornering, hydroplaning resistance, snow grip or tread life.

• A: Snow and ice performance is not part of the UTQG traction test at all.
• B: The test does not measure cornering or hydroplaning, so the grade says nothing about them.
• C: Tread life is shown by the separate treadwear number, not by the traction letter.
• D (correct): The traction grade comes only from wet, straight-line braking tests, and AA is the highest class.

Answer: A

Correct answer: About 7 mm larger. Old tire: h = 205 x 0.60 = 123 mm, rim = 16 x 25.4 = 406.4 mm, so

New tire: h = 225 x 0.45 = 101.25 mm, rim = 18 x 25.4 = 457.2 mm, so

The new tire is about

or about 7 mm, larger.

• A (correct): Old: 2(123) + 406.4 = 652.4 mm; new: 2(101.25) + 457.2 = 659.7 mm, a difference of about 7 mm.
• B: This comes from treating the aspect ratios 60 and 45 as sidewall heights in millimeters instead of percentages of width.
• C: This counts only one sidewall per tire; the diameter crosses two sidewalls plus the rim.
• D: This compares only the two sidewalls and forgets that the rim grew from 16 to 18 inches.

Answer: B

Correct answer: About 27 N. First convert frequency to angular speed:

rad/s.

• A: This plugs 12 rev/s in as $\omega$ without converting to radians per second.
• B (correct): Convert first: $\omega =2\pi (12)\approx 75.4$ rad/s, then $F=0.025\times 0.19\times 75.4^2\approx 27$ N.
• C: This squares the 12 but multiplies by $2\pi$ only once, so the conversion factor is not squared.
• D: This uses $\omega$ correctly but forgets to square it.

Answer: B

Correct answer: The highest speed the tire is rated for under proper conditions. In a code such as 94H, the number is the load index and the letter is the speed symbol. H corresponds to a maximum rated speed of 130 mph under proper conditions; like the load index, it is a limit, not a goal.

• A: A speed symbol is a limit, not a recommended operating speed.
• B (correct): The speed symbol is a maximum rating, not a target speed to reach.
• C: The letter after the load index is the speed symbol; it says nothing about rubber hardness.
• D: H is not a road-use category; it is the speed symbol for a 130 mph maximum rating.

Answer: C

Correct answer: The combined demand of 3,000 N exceeds the 2,700 N limit, so the tire loses grip. The grip limit is

Because 3,000 N is greater than 2,700 N, the tire cannot supply both forces and will lose grip and skid.

• A: Braking and cornering draw on one shared budget; they must be combined, not checked one at a time.
• B: The forces act at right angles, so they combine as a square root of squares, not by simple addition.
• C (correct): Limit: 0.75 x 3,600 = 2,700 N. Combined demand: $\sqrt{2400^2+1800^2}=3000$ N, which is more than the limit.
• D: Subtracting the two forces has no physical meaning here; they must be combined vectorially.

Answer: B

Correct answer: Week 11 of 2021. In the four-digit DOT date code, the first two digits are the production week and the last two are the year. So 1121 means week 11 of 2021, which is in mid-March. The code tells when the tire was made, not how well it was stored.

• A: The first two digits are a week number, not a month.
• B (correct): The first two digits give the week and the last two give the year.
• C: This reverses the order; the week comes first and the year last.
• D: The code is week then year; it does not contain a day of the month.

Answer: C

Correct answer: About 35 psi. Since kPa = psi x 6.895, reverse it to get psi = kPa / 6.895.

• A: This multiplies by 6.895, which is the direction for converting psi to kPa.
• B: This simply drops a zero instead of dividing by the conversion factor 6.895.
• C (correct): Going from kPa to psi means dividing: 240 / 6.895 is about 34.8, or about 35 psi.
• D: This subtracts 6.895 from 240 instead of dividing by it.

Answer: B

Correct answer: The bead. The beads are loops of steel wire at the tire's inner edges. Air pressure pushes them against the rim, sealing the air in and locking the tire in place.

• A: The shoulder is the rounded edge where tread meets sidewall, not the part that seals on the rim.
• B (correct): The beads are wire loops at the tire's inner edges that seal against and grip the rim.
• C: Steel belts lie under the tread to give it strength; they do not touch the rim.
• D: The sidewall is the flexible side between shoulder and rim; the bead is at its inner edge.

Answer: D

Correct answer: Whether the tire has been run underinflated or overloaded. An underinflated or overloaded tire sags, so its edges carry more of the load and both shoulders wear faster. Center wear suggests overinflation and one-edge wear suggests alignment, so both-shoulder wear points to checking pressure and load first.

• A: Overinflation tends to wear the center of the tread, not both shoulders.
• B: Imbalance is linked to vibration and cupping, not even wear on both shoulders.
• C: Camber or toe problems usually wear only one edge, not both edges equally.
• D (correct): Wear on both shoulders points toward low pressure or too much load, though it is a clue, not proof.

Answer: D

Correct answer: The hub. The hub is the vehicle's rotating mount, carried by a wheel bearing and fitted with studs or bolt holes. The wheel bolts to the hub, and the tire is mounted on the wheel.

• A: The rim is part of the wheel that holds the tire; it bolts to the hub rather than being the mount.
• B: The bead is the wire edge of the tire that seals on the rim.
• C: The valve stem is the fitting used to add or check air, not a mounting point.
• D (correct): The hub is the rotating mount, carried by a bearing, that the wheel bolts onto.

Answer: A

Correct answer: Only to reach a repair shop, staying under the speed limit printed on it. A compact temporary spare is smaller and lighter than a regular tire and usually has its own pressure and a printed speed limit, commonly 50 mph. It is meant only to reach a place where the regular tire can be repaired or replaced.

• A (correct): Compact spares are short-term tires, often limited to about 50 mph, meant only to get to a shop.
• B: A temporary spare is not built for regular use and should be replaced as soon as possible.
• C: The printed speed limit applies no matter how much tread the spare has.
• D: Compact spares often need a different, higher pressure printed on the spare or label.

Answer: A

Correct answer: A flat run-flat tire looks nearly normal, so the driver may not notice the loss. Run-flat tires have stiff sidewalls that can carry the car for a limited distance after losing air. Because a flat run-flat looks almost normal, TPMS is needed to alert the driver so the maker's speed and distance limits can be followed.

• A (correct): Stiff sidewalls hold the car up after air loss, hiding the problem unless TPMS warns the driver.
• B: Run-flats are not leakier; the issue is that a loss is hard to see.
• C: Run-flats still have maker limits on distance and speed after losing air, and TPMS does not change them.
• D: TPMS only monitors and warns; it does not add air.

Answer: B

Correct answer: About 3.96 MJ. Force per tire:

For four tires: 4 x 39.6 = 158.4 N.

Convert distance: 25 km = 25,000 m.

• A: This is the loss for only one tire; the question asks for all four.
• B (correct): Per tire 0.011 x 3,600 = 39.6 N; four tires give 158.4 N; 158.4 N x 25,000 m = 3,960,000 J.
• C: This uses 25 m instead of converting 25 km to 25,000 m.
• D: This misreads the coefficient as 0.11 instead of 0.011, making the answer ten times too large.

Answer: C

Correct answer: About 746 rpm. Each revolution moves the car one circumference:

From v = Cn,

rev/s.

Convert to rpm: 12.4 x 60 = about 746 rpm.

• A: This is the rate in revolutions per second; it still needs to be multiplied by 60.
• B: This divides speed by the diameter instead of the circumference, leaving out $\pi$.
• C (correct): C = $\pi$(0.64) = 2.01 m, so n = 25 / 2.01 = 12.4 rev/s, and 12.4 x 60 = about 746 rpm.
• D: This uses $2\pi$ times the diameter, which gives twice the true circumference.

Answer: A

Correct answer: Camber. Camber is the tilt of the wheel's top inward or outward when seen from the front. Toe is seen from above and caster from the side, so only camber matches this description.

• A (correct): Camber is the inward or outward tilt of the top of the wheel seen from the front.
• B: Toe compares the front edges of a pair of wheels seen from above, pointing in or out.
• C: Caster is the forward or backward tilt of the steering axis seen from the side.
• D: Offset is the wheel's mounting-face position relative to its centerline, not an alignment tilt.

Answer: D

Correct answer: By the 2/32-inch reading, which means the tire has reached the replacement point. Tread is measured at several spots across the width because wear is rarely even, and the lowest reading drives the decision. The inner groove is at 2/32 inch, the point where tread is flush with the wear bars and the tire should be replaced. The uneven wear is also a clue to check alignment.

• A: Using the deepest groove hides the worn area where grip is actually lowest.
• B: Averaging masks uneven wear; the lowest reading is the one that counts.
• C: Convenience does not decide it; several spots are checked and the lowest is used.
• D (correct): Decisions use the lowest reading, and 2/32 inch is where the tread is flush with the wear bars.

Answer: A

Correct answer: 140 N·m. Torque equals force times the perpendicular distance from the center.

• A (correct): Torque is force times lever arm: 400 x 0.35 = 140 N$\cdot$m.
• B: This divides the force by the lever arm instead of multiplying.
• C: This treats 35 cm as 0.035 m, a decimal slip in converting to meters.
• D: This ignores the lever arm; torque depends on the distance from the center as well as the force.

Answer: C

Correct answer: About 251 kPa. Mass per tire:

so N = 460 x 9.81 = 4,512.6 N.

• A: This puts the whole van's weight on one patch instead of splitting it four ways.
• B: This divides the mass in kilograms by the area without converting to newtons.
• C (correct): 1,840 / 4 = 460 kg; 460 x 9.81 = 4,512.6 N; 4,512.6 / 0.018 = about 250,700 Pa, or about 251 kPa.
• D: This misplaces the decimal in the area, using 0.0018 m${}^2$ instead of 0.018 m${}^2$.

Answer: D

Correct answer: Hydroplaning, where water lifts the tires off the pavement. Hydroplaning occurs when the tread cannot push water away fast enough and a wedge of water lifts the tire off the road. Steering and braking grip nearly disappear. Higher speed, deeper water, worn tread and low pressure all raise the risk.

• A: Cupping is a wear pattern felt as noise or vibration, not a sudden loss of steering on water.
• B: Overloading causes flexing and heat, not a sudden lift-off on a wet road.
• C: Imbalance causes speed-related vibration, not loss of contact with the road.
• D (correct): When tires cannot clear water fast enough, a water wedge lifts them and grip nearly vanishes.

Answer: A

Correct answer: A cracked rubber valve stem. Slow leaks often come from somewhere other than a puncture: a cracked valve stem, a leaking valve core, a missing cap that lets dirt in, or corrosion where the bead meets the rim. A tire that keeps losing air should be professionally inspected.

• A (correct): Cracked valve stems, leaky valve cores and bead-to-rim leaks often cause slow air loss without a puncture.
• B: Shallow tread reduces wet grip but does not by itself let air escape.
• C: Imbalance causes vibration, not air loss.
• D: Moving a tire to another position does not create a leak.

Answer: A

Correct answer: A low tire has a smaller rolling radius, so it turns slightly faster than the others. A tire that is low sags, so its effective radius gets smaller. To cover the same distance it must turn a little faster. Indirect TPMS uses the antilock brake wheel-speed sensors to spot that faster-turning wheel.

• A (correct): The system compares antilock brake wheel-speed signals and flags the wheel spinning faster.
• B: A smaller rolling radius makes the wheel spin faster, not slower, at the same road speed.
• C: Wheel-speed sensors measure rotation, not temperature.
• D: Radio-reporting sensors are part of direct TPMS, not indirect TPMS.

Answer: C

Correct answer: About 431. Sidewall: h = 235 x 0.60 = 141 mm. Rim: 18 x 25.4 = 457.2 mm. Diameter:

= 0.7392 m.

Circumference:

Revolutions per km: 1,000 / 2.32 = about 431. Real counts differ a little because a loaded tire flattens and tread wears.

• A: This divides 1,000 m by the diameter and leaves out $\pi$.
• B: This uses only one sidewall in the diameter instead of two.
• C (correct): D = 2(141) + 457.2 = 739.2 mm; C = $\pi$(0.7392) = 2.32 m; 1,000 / 2.32 = about 431 revolutions.
• D: This uses half the circumference, as if $C=\pi D/2$.

Answer: C

Correct answer: 1,260 N less. Water lowers the friction coefficient, which lowers the grip limit.

Dry:

Wet:

The drop is

• A: 2,100 N is the wet-road limit itself, not the amount lost.
• B: 3,360 N is the dry-road limit, not the reduction.
• C (correct): Dry: 0.8 x 4,200 = 3,360 N. Wet: 0.5 x 4,200 = 2,100 N. Difference: 1,260 N.
• D: This adds the dry and wet limits instead of subtracting them.

Answer: D

Correct answer: A smaller contact patch, a harsher ride and faster wear in the tread center. Overinflation makes the tire stiff and rounded, so less rubber touches the road. The ride gets harsher, wet grip and braking can suffer, the tread center wears faster and impacts are more likely to damage the tire. The door-label cold pressure balances these tradeoffs.

• A: These are signs of underinflation, the opposite condition.
• B: A smaller contact patch usually reduces wet grip and braking.
• C: A stiff, overinflated tire is more likely to be damaged by impacts.
• D (correct): Too much pressure rounds the tire, shrinking the patch, stiffening the ride and wearing the center.

Answer: B

Correct answer: Shock absorbers, struts and other suspension parts for wear. Scooped dips around the tread are cupping, or scalloping, often felt as rhythmic noise or vibration. It usually suggests worn shocks or struts, loose suspension parts or long-running imbalance, so new tires alone may soon cup again unless the cause is fixed.

• A: Age matters for safety, but it does not explain a cupping pattern.
• B (correct): Cupping commonly points to worn shocks or struts, loose suspension parts or long-running imbalance.
• C: The traction grade is a test result and has nothing to do with this wear pattern.
• D: The spare is worth checking routinely, but it does not cause cupping on the road tires.

Answer: D

Driving warms the air in a tire, so warm readings normally run a few psi above the cold value, as the 37 psi tires show. The left rear reads 30 psi even while warm, so its cold pressure would be lower still. The TPMS lamp only warns at about a 25 percent drop, so it can stay off. Recheck the tire cold, inflate it to 34 psi and have it inspected for a leak.

• A: The 4 psi comparison ignores heating; a warm reading below the cold target means the cold pressure is lower still.
• B: Pressure rises with heat, so letting air out of hot tires would leave them low when cold.
• C: Heating raises pressure, not lowers it, so this reasoning is backward.
• D (correct): Warm tires normally read above the cold value, as the three 37 psi tires do, so a warm 30 psi points to a tire that is truly low.