Mechanical Aptitude Test Guide: Format, Topics, Scoring

B: 30 lb. The turning effects about the pivot must balance: load × load arm = effort × effort arm.

1. Measure each arm from the pivot, not from the end of the bar. The load arm is 1 ft and the effort arm is 5 − 1 = 4 ft.
2. 120 × 1 = effort × 4, so the effort is 120 ÷ 4 = 30 lb.
3. Sense check: the effort arm is four times longer, so the force needed is four times smaller.

The trap: Using the whole 5 ft length as the effort arm, which gives 24 lb. The arm is the distance from the pivot.

D: 90 lb. The supporting rope segments share the load equally, so the pull is the load divided by the number of segments.

1. Count only the segments that hold up the moving block: 4. The free end you pull is not one of them.
2. 360 ÷ 4 = 90 lb.
3. The price is rope: to lift the block 1 ft you pull 4 ft through your hands.

The trap: Counting the free end as a fifth segment, which gives 360 ÷ 5 = 72 lb.

A: 150 lb. Work in equals work out: force × ramp length = weight × height.

1. Work to lift the crate straight up: 600 × 3 = 1,800 ft·lb.
2. Pushing along the ramp does the same work over 12 ft: force × 12 = 1,800.
3. Force = 1,800 ÷ 12 = 150 lb. The ramp is four times as long as it is high, so the force is a quarter of the weight.

The trap: Putting the length and height the wrong way round (600 × 12 ÷ 3 = 2,400 lb). A ramp always reduces the force, so an answer bigger than the weight cannot be right.

C: 150 rpm. Meshing gears pass the same number of teeth every minute, so teeth × speed is equal for both.

1. Teeth passing per minute: 40 × 90 = 3,600.
2. Gear B has fewer teeth, so it must turn faster to pass the same number: 3,600 ÷ 24 = 150 rpm.
3. Sense check: the driven gear is smaller, so it spins faster than 90 rpm.

The trap: Multiplying by the ratio the wrong way (90 × 24 ÷ 40 = 54 rpm). The smaller gear always turns faster.

D: 3. Each gear turns the opposite way to the gear it meshes with, so the directions alternate along the row.

1. A turns clockwise, so B turns anticlockwise, C clockwise, D anticlockwise and E clockwise.
2. The clockwise gears are A, C and E: 3.
3. Shortcut: the gears at odd places in the row (1st, 3rd, 5th) match the first gear.

The trap: Counting the meshes (4) instead of the gears. Gear E has turned the same way as A after an even number of meshes.

B: 60 rpm. Work through the train one pair at a time. Gears on the same shaft turn at the same speed.

1. A to B: 20 × 360 = 40 × B, so B turns at 180 rpm.
2. C is on the same shaft as B, so C also turns at 180 rpm.
3. C to D: 15 × 180 = 45 × D, so D turns at 60 rpm.
4. Check: each pair slowed the train down (20 to 40 teeth, then 15 to 45), so D must be well below 360 rpm.

The trap: Stopping after the first pair (180 rpm), or treating C as a new input at 360 rpm. C turns with B, not with A.

A: 600 lb. The fluid carries the same pressure everywhere, and pressure is force divided by area.

1. Pressure in the fluid: 40 ÷ 3 = 13.3 lb per sq in (leave it as a fraction if you prefer).
2. The large piston has 45 ÷ 3 = 15 times the area, so it exerts 15 times the force.
3. 40 × 15 = 600 lb.
4. The cost is distance: the large piston moves 15 times less far than the small one.

The trap: Dividing instead of multiplying (40 × 3 ÷ 45 = 2.7 lb). A hydraulic jack multiplies force, so the answer must be larger than 40 lb.

C: 8 atm. At a steady temperature, pressure × volume stays constant (Boyle's law), so squeezing the gas raises the pressure in proportion.

1. Pressure × volume before: 2 × 6 = 12.
2. After: pressure × 1.5 = 12, so the pressure is 12 ÷ 1.5 = 8 atm.
3. Check: the volume fell to a quarter (6 to 1.5), so the pressure rises to four times (2 to 8).

The trap: Doing the ratio the wrong way (2 × 1.5 ÷ 6 = 0.5 atm). Squeezing a gas can only raise its pressure.

D: Bar B (copper). Heat flows fastest through the material with the highest thermal conductivity, and metals conduct far better than glass.

1. With the same size and the same heat, only the material differs.
2. Roughly, from best to worst conductor: copper, aluminium, steel, glass.
3. So the far end of the copper bar warms first.

The trap: Choosing aluminium because it feels light and cool to hold. Copper conducts heat better than aluminium, and steel is much worse than both.

B: 450 lb. Take turning effects about the right support: the left support's push must balance the load's turning effect.

1. The load is 12 − 3 = 9 ft from the right support. Left support × 12 = 600 × 9.
2. Left support = 5,400 ÷ 12 = 450 lb. The right support carries the other 150 lb, because the two must add up to 600 lb.
3. Sense check: the load is nearer the left end, so the left support takes the larger share.

The trap: Swapping the supports (150 lb is the right support). The support closer to the load always carries more.

C: 3 A. Reduce the parallel pair to one resistor, add it to the series resistor, then use current = voltage ÷ resistance.

1. The pair in parallel: (6 × 3) ÷ (6 + 3) = 2 Ω. Check: it is smaller than the smaller resistor, as it must be.
2. Total resistance: 2 + 2 = 4 Ω.
3. Current = 12 ÷ 4 = 3 A.

The trap: Adding all three resistors as if they were in series (11 Ω), which gives about 1.1 A. Side-by-side resistors share the current, so they offer less resistance, not more.

A: Both A and C. A path for current still exists through C, and A is in series with that path, so both stay lit.

1. Take B out of the circuit: the current has one route left, through A and then through C.
2. Both bulbs are on that route, so both still light.
3. They change in brightness (C gets brighter and A dimmer, because the total resistance goes up), but neither goes out. If A had burned out instead, the whole loop would open and every bulb would go dark.

The trap: Treating B as if it were in series, so that its failure breaks the loop. Only a failed bulb in the series part, such as A, puts every bulb out.