ASE practice questions, with worked answers

C: worn or stuck piston rings in that cylinder. Oil temporarily seals the rings. If the reading rises, the rings are leaking. If it stays low, look at the valves or head gasket.

The trap: Choosing a valve. Oil seals rings, not valves, so a reading that rises with oil points to the rings.

B: B only. Technician A is wrong: A rich mixture makes black smoke. Technician B is right: Blue-gray smoke is the sign of burning oil.

A: 0.0027 in; over the limit, so the cylinder needs machining or the block replacing. Taper is the top-of-travel diameter minus the bottom: 3.6248 − 3.6221 = 0.0027 in. That is more than the 0.0020 in limit, so the bore must be machined oversize (or sleeved) before new rings will seal.

1. Taper = top diameter − bottom diameter.
2. 3.6248 − 3.6221 = 0.0027 in.
3. 0.0027 in is over the 0.0020 in limit.

The trap: Halving the difference (0.0014 in). Halving is for out-of-round on some charts, never for taper.

A: A only. Where the air comes out tells you where it leaks. Air at the throttle body or intake means an intake valve; at the tailpipe, an exhaust valve; at the oil filler or PCV, the rings; bubbles in the coolant, a head gasket or crack.

1. Air is heard at the throttle body.
2. That path runs past an intake valve.
3. Rings would vent into the crankcase, so B is wrong.

The trap: Treating a high percentage as proof of ring wear. The percentage says how bad; the sound says where.

A: A pressure and flow test with the gauge between pump and gear, briefly closing the valve. The tester shows pump pressure with the valve closed (only for a few seconds) and system pressure while steering.

The trap: Holding the valve closed for long. It is closed only for a few seconds, or the pump overheats.

A: To the left, the side with less positive caster. The wheel with more positive caster pushes the vehicle toward the other side, so it drifts toward the side with less positive caster: the left (+2.50°).

B: To the right. A vehicle drifts toward the side with more positive camber. The right side is more positive (+0.5° against −0.4°), a cross-camber of 0.9°, so it drifts right.

1. Find the more positive side: +0.5° on the right.
2. The vehicle drifts toward it.
3. So it drifts to the right.

The trap: Using the caster rule. Caster pulls toward the less positive side; camber pulls toward the more positive side.

A: 135 mm. The aspect ratio (60) is the sidewall height as a percentage of the section width (225 mm): 225 × 0.60 = 135 mm.

1. Width: 225 mm.
2. Aspect ratio: 60%.
3. 225 × 0.60 = 135 mm.

The trap: Reading the 60 as millimetres. It is a percentage of the width.

A: A only. A pull that moves to the other side when the tires are swapped follows the tire, so the tire is the cause (often a belt shifted inside it). An alignment change would not move with the tire, so B is wrong.

1. The pull changed direction when the tires moved.
2. So the cause travels with the tire.
3. A is right; adjusting camber would not fix it.

The trap: Going straight to alignment. Rule out tires first; it is quicker and cheaper.

A: They are hygroscopic and absorb moisture from the air, lowering their boiling point. Water in glycol-based fluid lowers the boiling point and causes corrosion. Keep containers closed and use fresh fluid.

The trap: Thinking brake fluid evaporates. The risk is the water it absorbs from the air.

B: Machine it; it will be 9.98 in, within the machining limit. After machining: 9.97 + 0.01 = 9.98 in. The limit for machining is 10.03 in, so it can be machined.

C: Both A and B. Technician A is right: Many systems re-learn piston position after pads are changed. Technician B is right: Low voltage can prevent the motors from applying or releasing.

A: 2,000 pounds. Pressure = force ÷ area = 400 ÷ 0.8 = 500 psi, the same everywhere in the closed system. Caliper force = pressure × area = 500 × 4 = 2,000 pounds.

1. Pressure: 400 ÷ 0.8 = 500 psi.
2. Pressure is equal throughout the system.
3. Force: 500 × 4 = 2,000 pounds.

The trap: Stopping at 500. That is the pressure in psi, not the force at the caliper.

A: Replace the rotor; it would be 1.021 in, under the minimum. After machining the rotor would be 1.039 − 0.018 = 1.021 in, which is under the 1.024 in minimum, so it must be replaced. A rotor may only be machined if it stays at or above the minimum after the cut.

1. Thickness after machining: 1.039 − 0.018 = 1.021 in.
2. Compare with the minimum: 1.021 < 1.024.
3. Under the minimum, so replace.

The trap: Checking only today's thickness. The rule is about the thickness after machining.

A: A only. Air compresses, so a spongy, low pedal points to air in the lines or calipers; bleeding fixes it. A booster fault makes the pedal hard, not spongy, so B is wrong.

1. Spongy means something is compressing.
2. Fluid does not compress; air does.
3. A booster fault gives a hard pedal, so only A is right.

The trap: Blaming the booster for any pedal complaint. A booster changes effort, not sponginess.

D: In series, so the current flows through the meter. An ammeter must be part of the current path. Connected in parallel, it can blow its fuse or damage the meter.

The trap: Connecting the ammeter across the load, as you would a voltmeter. An ammeter goes in series.

A: 20 A. Current = 3 × 55 ÷ 12 = 13.75 A. With 25% margin: 17.19 A. The smallest standard fuse at or above that is 20 A.

A: 2.4 A. The two 6 Ω resistors in parallel give (6 × 6) ÷ (6 + 6) = 3 Ω. Add the 2 Ω in series: 5 Ω. I = E ÷ R = 12 ÷ 5 = 2.4 A.

1. Parallel pair: 36 ÷ 12 = 3 Ω.
2. Total: 3 + 2 = 5 Ω.
3. Current: 12 ÷ 5 = 2.4 A.

The trap: Adding all three resistors (14 Ω, 0.86 A). Only resistors in series add.

A: A only. A ground-side voltage drop under load should be small (typically 0.2 to 0.3 V at most; follow the maker's limit). 0.9 V means resistance in the ground cable or its connections. A voltage drop reading on a cable says nothing on its own about how much current the starter draws; that needs a current test.

1. The meter spans the ground path under load.
2. 0.9 V is well over a typical limit: resistance in that path.
3. Starter draw is measured with an ammeter, so B is wrong.

The trap: Reading a high drop as a starter fault. The drop is across the cable, so the fault is in the cable or its ends.

A: Clearing erases freeze frame data and resets monitors, which can hide the fault and delay an inspection. Record codes and freeze frame first. After the repair, clear them and confirm the fix with the monitor.

The trap: Clearing codes to "see if it comes back". That erases the freeze frame data you need.

A: A only. Technician A is right: Switching across 0.45 volts shows closed-loop fuel control. Technician B is wrong: A steady downstream signal is normal behind a working converter.

A: +18%; the PCM is adding fuel because the engine is running lean. Total trim is short-term plus long-term: 0% and +18% give +18%. Positive trim adds fuel to correct a lean condition.

A: A vacuum leak. Total trim at idle is +22%: the PCM is adding a lot of fuel for a lean condition. A vacuum leak lets in a fixed amount of unmetered air, which is a big share of the airflow at idle and a small share at 2,500 rpm, so the trims come back toward zero as rpm rises. A weak pump gets worse with load.

1. Idle total: +4 + 18 = +22%, lean.
2. The lean condition fades at higher rpm.
3. Unmetered air at idle: a vacuum leak.

The trap: Picking the weak fuel pump. A fuel delivery problem is worst under load, not at idle.

A: 10%. A sequential injector fires once every two crankshaft turns. Two turns at 2,400 rpm take 120,000 ÷ 2,400 = 50 ms. Duty cycle = 5 ÷ 50 = 10%.

1. Time for two turns: 120,000 ÷ 2,400 = 50 ms.
2. Duty cycle = on time ÷ cycle time.
3. 5 ÷ 50 = 10%.

The trap: Using one turn (25 ms) and getting 20%. A sequential injector fires once per engine cycle, which is two turns.

A: A only. A working catalyst stores oxygen, so the downstream sensor reads fairly steady while the upstream one switches. When the downstream copies the upstream, the catalyst is not doing its job, which is what P0420 reports. A switching upstream sensor is working, so replacing it is not the fix.

1. A good catalyst steadies the downstream signal.
2. Here the downstream copies the upstream.
3. So the catalyst is weak: A is right, B is not.

The trap: Replacing the upstream sensor because a code mentions oxygen. A sensor that switches normally is working.