Auto Information Practice Test 2: Symptom Diagnosis

Answer: B

Correct answer: The pads are close to the end of their life. Many brake pads carry a small metal tab that touches the rotor and squeals when the pad is nearly worn out. It is a warning to inspect and replace the pads soon, before metal-to-metal grinding damages the rotor.

• A: ABS action is felt as pedal pulsing during hard or slippery stops, not as a squeal in light braking.
• B (correct): Wear tabs are designed to squeal against the rotor when the pad material is nearly used up.
• C: A slipping drive belt squeals with engine speed or accessory load, not specifically when the brakes are applied.
• D: Tire pressure does not make a squeal from the brake area each time the brakes are applied.

Answer: D

Correct answer: A small round gear on the end of the steering shaft. The steering shaft ends in a small round gear, the pinion. When the driver turns the wheel, the pinion rotates and its teeth drive the toothed rack left or right, which moves the tie rods.

• A: That describes the rack, which the pinion drives.
• B: That is the outer tie rod end, which links the tie rod to the steering arm.
• C: That is the steering arm, the last link before the wheel pivots.
• D (correct): The pinion is the small gear turned by the steering shaft; its teeth push the rack sideways.

Answer: A

Correct answer: The meter supplies its own test current, which outside voltage or other paths would upset. An ohmmeter supplies its own small test current and calculates resistance from it. Any live voltage adds to that and gives false readings or damages the meter, and other parallel paths can let current bypass the part. Unpower and disconnect the part to get a true reading.

• A (correct): Live voltage adds to the meter's own test current, giving false readings or damage, and other paths can make a bad part look good.
• B: Resistance does not simply double with power applied; the problem is outside voltage interfering with the meter.
• C: Isolation, not removal from the vehicle, is what matters; unplugging the part is usually enough without unbolting it.
• D: Cooling lowers a winding's resistance only slightly; it never falls to zero.

Answer: B

Correct answer: DOT 5. DOT 3, DOT 4 and DOT 5.1 are all glycol-based fluids with different boiling points. DOT 5 is silicone-based and should not be mixed with them, so always use the type named on the reservoir cap or in the owner's manual.

• A: Despite the similar number, DOT 5.1 is glycol-based like DOT 3 and DOT 4.
• B (correct): DOT 5 is silicone-based, a different chemistry from the glycol-based DOT 3, 4 and 5.1.
• C: DOT 4 is a glycol-based fluid with a higher boiling point than DOT 3.
• D: DOT 3 is a common glycol-based fluid, not silicone.

Answer: B

Correct answer: An oil pressure sensor, or sender. A pressure switch only changes state when pressure drops below a limit, which suits a warning light. A gauge that tracks pressure continuously needs a sensor (sender) whose output varies with pressure.

• A: A switch is only on or off at a set threshold, so it can light a warning lamp but cannot drive a smoothly moving gauge.
• B (correct): A sensor produces a signal that changes in proportion to pressure, which is what a moving gauge needle needs.
• C: Level and pressure are different measurements; a level sensor does not report how hard oil is being pushed.
• D: This sensor reports coolant temperature, not pressure in the lubrication circuit.

Answer: D

Correct answer: Clean and tighten the battery terminals. Corrosion on the posts acts like extra resistance in series with the starter, so less current reaches it and the engine cranks slowly even though the battery is charged. Cleaning and tightening the connections is the least intrusive first step; test further only if the slow crank remains.

• A: A slow crank with heavy corrosion on the posts has not yet been traced to the starter, so this skips the cheap evidence.
• B: The battery shows full charge; the visible problem is in the connections, not its size.
• C: A jump start may get the truck going but leaves the high-resistance connection in place.
• D (correct): Corrosion adds resistance that limits starter current, and cleaning and tightening is a quick, low-cost check that may fix it.

Answer: A

Correct answer: 9:1. Compression ratio compares total volume at BDC with clearance volume at TDC. Here the BDC volume of 540 cc already includes the clearance, so the ratio is simply the quotient below, written 9:1.

• A (correct): The ratio is the volume at BDC divided by the volume at TDC: 540 divided by 60 is 9.
• B: This divides the swept volume alone (540 minus 60, or 480 cc) by 60; the top of the ratio must be the full BDC volume.
• C: This adds the 60 cc clearance to 540 cc a second time; 540 cc already includes the clearance space.
• D: The ratio is written with the larger BDC volume first; 1:9 turns the comparison upside down.

Answer: B

Correct answer: 14.25:1, slightly rich. Divide air mass by fuel mass.

Since 14.25 is lower than 14.7, there is more fuel for the air than stoichiometric, so the mixture is slightly rich.

• A: The ratio is calculated correctly, but a value below 14.7 means more fuel than stoichiometric, which is rich.
• B (correct): 456 divided by 32 is 14.25, which is below 14.7, so the mixture has extra fuel and is slightly rich.
• C: This divides fuel by air; AFR always puts the air mass on top.
• D: This adds the two masses instead of dividing them; an AFR in the hundreds is a sign of that mistake.

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: 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.