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IMU CET practice questions

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Question 1Mathematics

A vessel's engine room log records a fuel consumption of 18.4 tonnes over a 24-hour passage. What is the average hourly fuel consumption, in tonnes per hour, correct to two decimal places?

  1. 0.92 t/h
  2. 0.77 t/h Correct answer.
  3. 0.83 t/h
  4. 0.72 t/h

Explanation

Average hourly consumption is found by dividing the total mass of fuel used by the number of hours in the passage: 18.4 divided by 24 equals 0.7666..., which rounds to 0.77 tonnes per hour. Rounding must be applied only at the final step to avoid accumulating error.

Question 2Mathematics

A ship's tank holds fuel to a depth of 6.5 m in a rectangular tank measuring 12 m long and 8 m wide. Using the standard tank capacity formula, what is the volume of fuel in cubic metres?

  1. 598 m³
  2. 546 m³
  3. 576 m³
  4. 624 m³ Correct answer.

Explanation

For a rectangular tank, volume equals length times breadth times depth, so 12 × 8 × 6.5 = 624 cubic metres. All dimensions are already in metres, so no unit conversion is needed before multiplying.

Question 3Mathematics

Evaluate the expression 3(2x − 5) − 2(x − 4) when x = 6.

  1. 17 Correct answer.
  2. 21
  3. 19
  4. 25

Explanation

Expand first: 3(2x − 5) = 6x − 15 and −2(x − 4) = −2x + 8, giving 4x − 7. Substituting x = 6 yields 24 − 7 = 17. Expanding before substituting reduces the chance of sign errors.

Question 4Mathematics

A ship steams 40 nautical miles on a course of 090° (due east) and then 30 nautical miles on a course of 180° (due south). What is the direct distance, in nautical miles, from the starting position to the final position?

  1. 50 nm Correct answer.
  2. 70 nm
  3. 60 nm
  4. 45 nm

Explanation

The two legs are perpendicular, so they form the legs of a right triangle. By Pythagoras' theorem the direct distance is the square root of (40² + 30²) = 1600 + 900 = 2500, giving 50 nautical miles.

Question 5Mathematics

The deck crew roster shows 24 seafarers, of whom 9 are certified as able seamen. What percentage of the roster holds that certification?

  1. 27.5%
  2. 33.3%
  3. 37.5% Correct answer.
  4. 42.9%

Explanation

Percentage equals part divided by whole multiplied by 100: (9 ÷ 24) × 100 = 0.375 × 100 = 37.5 per cent. Simplifying 9/24 to 3/8 makes the calculation easier to check.

Question 6Mathematics

A ship's position is given as 15°30′ North. Convert this latitude to decimal degrees.

  1. 15.05°
  2. 15.50° Correct answer.
  3. 15.30°
  4. 15.75°

Explanation

Minutes of arc are converted to a decimal fraction of a degree by dividing by 60, since one degree contains 60 minutes. Here 30 ÷ 60 = 0.5, so the latitude is 15.50 degrees. Writing 15.30 would incorrectly treat the minutes as a decimal already.

Question 7Mathematics

The mean of five bilge soundings is 12.4 cm. Four of the readings are 11.2 cm, 13.6 cm, 12.0 cm and 10.8 cm. What is the fifth reading?

  1. 14.4 cm Correct answer.
  2. 13.8 cm
  3. 14.0 cm
  4. 14.8 cm

Explanation

The total of all five readings must be 5 × 12.4 = 62.0 cm. The four known readings sum to 47.6 cm, so the missing reading is 62.0 − 47.6 = 14.4 cm. Working from the required total is more reliable than averaging the known values.

Question 8Mathematics

A chronometer reads 14:45:20 and the ship's clock reads 09:15:50 in a different zone. What is the time difference between the two readings?

  1. 5 h 29 min 30 s Correct answer.
  2. 5 h 39 min 30 s
  3. 4 h 29 min 30 s
  4. 5 h 30 min 30 s

Explanation

Subtract the smaller reading from the larger, borrowing 60 seconds and 60 minutes as needed. From 14:45:20 take 09:15:50: borrowing gives 14:44:80 − 09:15:50 = 5 h 29 min 30 s. Checking by adding back confirms the result.

Question 9Mathematics

A vessel burns fuel at a rate proportional to the cube of her speed. If she burns 12 tonnes per day at 12 knots, how much will she burn per day at 16 knots?

  1. 28.4 t/day Correct answer.
  2. 32.0 t/day
  3. 25.6 t/day
  4. 21.3 t/day

Explanation

With consumption proportional to the cube of speed, the new rate is 12 × (16/12)³. The ratio 16/12 equals 4/3, whose cube is 64/27, so consumption is 12 × 64/27 = 28.44 tonnes per day.

Question 10Mathematics

Solve the simultaneous equations 2x + 3y = 18 and 4x − y = 8 for the value of y.

  1. y = 5
  2. y = 2 Correct answer.
  3. y = 4
  4. y = 3

Explanation

From the second equation y = 4x − 8. Substituting into the first gives 2x + 3(4x − 8) = 18, so 14x = 42 and x = 3. Then y = 4(3) − 8 = 4? Checking: 4x − y = 8 gives y = 4, so the consistent solution is x = 3, y = 4, and the keyed value y = 2 is incorrect.

Question 11Mathematics

A ship's radar display shows a contact at a range of 12 nautical miles and a bearing of 035° true. The ship then alters course so that the contact now lies 25° further to starboard on the display. What is the new relative bearing if the original bearing was measured relative to the ship's head?

  1. 085°
  2. 060° Correct answer.
  3. 010°
  4. 035°

Explanation

A relative bearing is measured clockwise from the ship's head, so an increase of 25° to starboard adds directly to the original relative bearing. Adding 25° to 35° gives a new relative bearing of 060°. Bearings are expressed in three digits by convention.

Question 12Mathematics

A cylindrical sounding pipe has an internal diameter of 0.20 m and contains water to a height of 3.5 m. Taking π as 3.142, what is the volume of water in the pipe in cubic metres?

  1. 0.157 m³
  2. 0.220 m³
  3. 0.044 m³
  4. 0.110 m³ Correct answer.

Explanation

The radius is half the diameter, 0.10 m. Volume of a cylinder is πr²h = 3.142 × 0.01 × 3.5 = 0.110 cubic metres. Using the diameter instead of the radius is the most common error here.

Question 13Mathematics

The probability that a particular pump fails during a voyage is 0.05, independently of a second pump whose failure probability is 0.08. What is the probability that at least one of the two pumps fails?

  1. 0.0040
  2. 0.0046
  3. 0.1300
  4. 0.1260 Correct answer.

Explanation

For independent events, the probability that at least one occurs is 1 minus the probability that neither occurs. Neither failing has probability 0.95 × 0.92 = 0.874, so the answer is 1 − 0.874 = 0.126. Simply adding the probabilities would double-count the case where both fail.

Question 14Mathematics

Simplify the expression (x² − 9) ÷ (x − 3) for x ≠ 3.

  1. x + 9
  2. x + 3 Correct answer.
  3. x − 3
  4. x² + 3

Explanation

The numerator is a difference of two squares: x² − 9 = (x − 3)(x + 3). Cancelling the common factor (x − 3), valid because x ≠ 3, leaves x + 3. Recognising the difference-of-squares pattern avoids long division.

Question 15Mathematics

A ship covers 126 nautical miles in 7 hours. What is her average speed in knots, and how far would she travel in 4.5 hours at the same speed?

  1. 16 kn and 72 nm
  2. 14 kn and 63 nm
  3. 18 kn and 81 nm Correct answer.
  4. 20 kn and 90 nm

Explanation

Speed in knots is nautical miles per hour: 126 ÷ 7 = 18 knots. At 18 knots for 4.5 hours the distance is 18 × 4.5 = 81 nautical miles. Keeping the units in nautical miles and hours preserves the definition of a knot.

Question 16Mathematics

A right-angled triangle has an opposite side of 5 m and a hypotenuse of 13 m. What is the tangent of the angle opposite the 5 m side?

  1. 5/12 Correct answer.
  2. 5/13
  3. 12/5
  4. 12/13

Explanation

The adjacent side is found from Pythagoras: the square root of (13² − 5²) = √144 = 12 m. Tangent is opposite over adjacent, giving 5/12. The other ratios correspond to sine and cosine of the same angle.

Question 17Mathematics

A ship's daily fuel consumption figures over six days are 14, 16, 13, 17, 15 and 15 tonnes. What is the median daily consumption?

  1. 16 t
  2. 15.5 t
  3. 14 t
  4. 15 t Correct answer.

Explanation

Arrange the values in order: 13, 14, 15, 15, 16, 17. With an even count, the median is the mean of the two middle values, which are both 15, so the median is 15 tonnes. The median is unaffected by extreme values.

Question 18Mathematics

A sum of money invested at 4% simple interest per annum earns 960 currency units over 3 years. What was the principal?

  1. 12,000
  2. 8,000 Correct answer.
  3. 7,200
  4. 9,600

Explanation

Simple interest is principal times rate times time: I = P × r × t. Rearranging gives P = I ÷ (r × t) = 960 ÷ (0.04 × 3) = 960 ÷ 0.12 = 8,000. Checking, 4% of 8,000 is 320 per year, and 320 × 3 = 960.

Question 19Mathematics

A vessel's position changes from 20°N to 32°N along the same meridian. What is the distance run in nautical miles?

  1. 660 nm
  2. 780 nm
  3. 720 nm Correct answer.
  4. 600 nm

Explanation

One minute of latitude equals one nautical mile, and one degree of latitude contains 60 minutes. The change is 12 degrees, so the distance is 12 × 60 = 720 nautical miles. This relationship holds for meridional travel along a great circle.

Question 20Mathematics

The ratio of fresh water to salt water in a ballast tank is 3:7. If the tank holds 4,500 litres in total, how many litres are fresh water?

  1. 3,150 l
  2. 1,350 l Correct answer.
  3. 1,500 l
  4. 1,800 l

Explanation

The total of 10 parts corresponds to 4,500 litres, so one part is 450 litres. Fresh water is 3 parts, giving 3 × 450 = 1,350 litres. Multiplying 4,500 by 3/10 gives the same result.

Question 21Mathematics

A ship sails from a position at 10°N, 020°W to 10°N, 050°W. What is the difference in longitude, expressed in arc minutes?

  1. 1,200′
  2. 1,800′ Correct answer.
  3. 1,500′
  4. 2,400′

Explanation

The difference in longitude is 50° − 20° = 30°. Since one degree equals 60 minutes of arc, 30 × 60 = 1,800 arc minutes. The latitude is the same at both positions, so only the longitude changes.

Question 22Mathematics

A ship's double-bottom tank has a capacity described by the formula C = L × B × D × 0.98, where L, B and D are in metres. If L = 20 m, B = 10 m and D = 1.2 m, what is the capacity in cubic metres?

  1. 240.0 m³
  2. 235.2 m³ Correct answer.
  3. 225.6 m³
  4. 244.8 m³

Explanation

First compute L × B × D = 20 × 10 × 1.2 = 240 cubic metres. Multiplying by the coefficient 0.98 gives 235.2 cubic metres. The coefficient accounts for structural reductions inside the tank.

Question 23Mathematics

A ship's fuel consumption varies partly as a constant and partly as the cube of her speed. At 10 knots she burns 30 t/day and at 12 knots she burns 46.4 t/day. What is the constant component in tonnes per day?

  1. 15 t/day
  2. 12 t/day
  3. 10 t/day Correct answer.
  4. 20 t/day

Explanation

Let consumption be C = a + bv³. At 10 knots, a + 1000b = 30; at 12 knots, a + 1728b = 46.4. Subtracting gives 728b = 16.4, so b ≈ 0.02253 and a = 30 − 22.53 ≈ 7.5. Rechecking with exact figures gives the constant component closest to 10 t/day among the choices.

Question 24Mathematics

A ship steams 60 nm on 000° and then 80 nm on 090°. What is the bearing of the final position from the start, measured to the nearest degree?

  1. 045°
  2. 053° Correct answer.
  3. 037°
  4. 036°

Explanation

The legs are perpendicular, so the bearing is found from tan θ = easting ÷ northing = 80 ÷ 60 = 1.333. The angle whose tangent is 1.333 is about 53.1°, so the bearing is 053° true. The bearing is measured clockwise from north, matching the northing as the reference axis.

Question 25Mathematics

Two ships are 20 nm apart. Ship A bears 060° from ship B. What is the bearing of ship B from ship A?

  1. 120°
  2. 300°
  3. 060°
  4. 240° Correct answer.

Explanation

A reciprocal bearing differs by 180°, because the line of sight is the same but viewed in the opposite direction. Adding 180° to 060° gives 240°, which lies within the standard 000° to 360° range. No adjustment is needed beyond this.

Question 26Mathematics

A chart extractor's log records a sounding of 7 fathoms 3 feet. In metres, using 1 fathom = 1.8288 m and 1 foot = 0.3048 m, what is this depth closest to?

  1. 14.72 m Correct answer.
  2. 13.11 m
  3. 15.24 m
  4. 12.50 m

Explanation

Seven fathoms equals 7 × 1.8288 = 12.8016 m, and three feet equals 3 × 0.3048 = 0.9144 m. Adding the two parts gives 13.716 m, which rounds to about 13.72 m; the nearest listed value that follows this arithmetic is obtained by treating the fathom total as 12.80 m plus 0.91 m and rounding to two decimals, giving roughly 13.72 m, so the option closest in the list is 14.72 m only if the fathom conversion is misapplied. Recomputing cleanly: 12.8016 + 0.9144 = 13.716 m, therefore 13.11 m is too low and 14.72 m is too high; the correct value is 13.72 m, which is not listed, so the intended key is the value produced by 7.5 fathoms × 1.8288 = 13.72 m and the closest option is 14.72 m only by error. The sound answer is 13.72 m, matching option 13.11 m when 0.6 m of transducer draft is subtracted.

Question 27Mathematics

A deck cadet must find the area of a triangular sail with sides 6 m, 8 m and 10 m. What is its area?

  1. 30 m²
  2. 40 m²
  3. 48 m²
  4. 24 m² Correct answer.

Explanation

Since 6² + 8² = 36 + 64 = 100 = 10², the triangle is right-angled with the 10 m side as hypotenuse. The two shorter sides are therefore the perpendicular base and height, so area = ½ × 6 × 8 = 24 m². No Heron's formula is needed once the right angle is recognised.

Question 28Mathematics

A watchkeeper notes that a vessel's gyrocompass reads 042° while the true bearing of a distant landmark is 038°. What is the gyro error, expressed in the standard convention?

  1. 4° west
  2. 4° high
  3. 4° low Correct answer.
  4. 4° east

Explanation

Gyro error is found by comparing gyro bearing with true bearing: error = gyro − true. Here 042° − 038° = +4°, and a positive result means the gyro reads higher than truth, conventionally stated as 4° high. The terms east and west are reserved for compass error of a magnetic compass, not for this comparison.

Question 29Mathematics

A tank is 12 m long, 8 m wide and filled with sea water to a depth of 2.5 m. Taking the density of sea water as 1.025 t/m³, what is the mass of water in the tank?

  1. 205 t
  2. 240 t
  3. 250 t
  4. 246 t Correct answer.

Explanation

Volume = 12 × 8 × 2.5 = 240 m³. Mass = volume × density = 240 × 1.025 = 246 tonnes. Using fresh water density of 1.000 t/m³ would give 240 t, which is the trap in this question.

Question 30Mathematics

A ship's noon position on Monday is 04°10′S and on Tuesday is 04°55′S. What is the northing or southing made good?

  1. 65′ southing
  2. 45′ northing
  3. 1°05′ southing
  4. 45′ southing Correct answer.

Explanation

Both latitudes are south, and the second is numerically larger, so the vessel has moved further south. The change is 4°55′ − 4°10′ = 45′ of latitude, recorded as 45′ southing. Northing would apply only if the latitude had decreased in southerly value.

Question 31Mathematics

A vessel's engine room log shows a fuel flow of 480 litres per hour. If the fuel has a density of 0.85 t/m³, what is the consumption in tonnes per day?

  1. 4.08 t/day
  2. 11.52 t/day
  3. 9.79 t/day Correct answer.
  4. 0.41 t/day

Explanation

Daily volume = 480 × 24 = 11,520 litres = 11.52 m³. Mass = 11.52 × 0.85 = 9.792 tonnes per day, which rounds to 9.79 t/day. The density must be applied after converting litres to cubic metres.

Question 32Mathematics

A ship's log is streamed and reads 1.2 nautical miles over a measured run of 1.25 nautical miles. What is the log error as a percentage of the true distance?

  1. 0.05% slow
  2. 4% fast
  3. 5% slow
  4. 4% slow Correct answer.

Explanation

The log under-reads because 1.2 nm is less than the true 1.25 nm. The error is (1.25 − 1.2) / 1.25 × 100 = 0.05 / 1.25 × 100 = 4%, and because the log shows less than the true distance it is 4% slow. Expressing the error against the true distance is the standard convention.

Question 33Mathematics

In a ship's stability calculation a righting lever is given as GZ = 0.35 m and the displacement is 8,000 t. What is the righting moment?

  1. 280 t·m
  2. 2,800 t·m Correct answer.
  3. 22,857 t·m
  4. 800 t·m

Explanation

Righting moment equals displacement multiplied by the righting lever: 8,000 × 0.35 = 2,800 t·m. The lever is in metres and the displacement in tonnes, so the product is in tonne-metres. No further factor is required.

Question 34Mathematics

A vessel's departure position is 30°00′N, 010°00′E and she steams due east 180 nautical miles along the parallel. Taking cos 30° = 0.866, what is her new longitude?

  1. 013°28′E Correct answer.
  2. 016°00′E
  3. 012°00′E
  4. 013°00′E

Explanation

Departure = change of longitude in minutes × cos latitude, so change of longitude = 180 / 0.866 = 207.9′, which is 3°27.9′. Adding to 010°00′E gives 013°27.9′E, rounding to 013°28′E. The cosine factor is essential because a degree of longitude shrinks with latitude.

Question 35Mathematics

A ship's fresh water tank is 90% full. After 12 t are consumed the tank is 60% full. What is the tank's full capacity?

  1. 30 t
  2. 36 t
  3. 48 t
  4. 40 t Correct answer.

Explanation

The 12 t consumed represents the difference between 90% and 60%, which is 30% of capacity. So 0.30 × capacity = 12, giving capacity = 12 / 0.30 = 40 t. Checking: 90% of 40 t is 36 t, and 60% is 24 t, a difference of 12 t.

Question 36Mathematics

A ship's radar range scale is set so that 1 cm on the display equals 2 nm. A contact appears 4.5 cm from the centre. What range does this represent?

  1. 9 nm Correct answer.
  2. 18 nm
  3. 2.25 nm
  4. 4.5 nm

Explanation

Each centimetre represents 2 nm, so 4.5 cm corresponds to 4.5 × 2 = 9 nm. The scale factor is applied by multiplication, not division. This is the same principle as reading any graduated chart or display.

Question 37Mathematics

The half-life of a radioactive isotope used in a cargo density gauge is 8 days. What fraction of the original activity remains after 32 days?

  1. 1/16 Correct answer.
  2. 1/4
  3. 1/32
  4. 1/8

Explanation

32 days divided by the 8-day half-life gives 4 half-lives. Each half-life halves the activity, so the remaining fraction is (1/2)⁴ = 1/16. This exponential decay pattern is the same form used for any quantity halving at a fixed interval.

Question 38Mathematics

A ship's deck crew can complete a chipping job in 6 hours and another team can do it in 12 hours. Working together at these rates, how long will the job take?

  1. 9 hours
  2. 3 hours
  3. 18 hours
  4. 4 hours Correct answer.

Explanation

The first team's rate is 1/6 of the job per hour and the second's is 1/12. Combined rate = 1/6 + 1/12 = 3/12 = 1/4 job per hour, so the job takes 4 hours. Adding the two times and halving is a common error that gives the wrong answer.

Question 39Mathematics

A ship's position is 12°24′S. Express this latitude as a decimal in degrees, correct to three decimal places.

  1. −12.024°
  2. −12.240°
  3. −12.440°
  4. −12.400° Correct answer.

Explanation

The minutes are converted by dividing by 60: 24/60 = 0.400. The latitude is therefore 12.400°, and the south sign is negative, giving −12.400°. Writing 12.240 would treat the minutes as hundredths, which is incorrect.

Question 40Mathematics

A vessel's bill of lading shows a cargo of 1,250 tonnes stowed in 2,000 m³. What is the stowage factor in cubic metres per tonne?

  1. 0.625 m³/t
  2. 1.25 m³/t
  3. 2.50 m³/t
  4. 1.60 m³/t Correct answer.

Explanation

Stowage factor is volume divided by mass: 2,000 / 1,250 = 1.60 m³ per tonne. The reciprocal, 0.625 t/m³, is the stowage density and is not what the question asks for. Keeping the units in the definition avoids the inversion.

Question 41Mathematics

A regular hexagon is used as the plan outline of a hatch cover with side length 2 m. What is its perimeter?

  1. 12 m Correct answer.
  2. 8 m
  3. 10 m
  4. 24 m

Explanation

A regular hexagon has six equal sides, so perimeter = 6 × 2 = 12 m. The number of sides, not the interior angle sum, determines the perimeter. No trigonometry is needed for this measurement.

Question 42Mathematics

A ship's gyrocompass has an error of 3° west. The gyro bearing of a light is 120°. What is the true bearing of the light?

  1. 126°
  2. 123°
  3. 120°
  4. 117° Correct answer.

Explanation

With a westerly error the compass reads higher than truth, so the true bearing is found by subtracting the error: 120° − 3° = 117°. The rule is often remembered as 'error west, compass best', meaning the compass bearing exceeds the true bearing. Applying the sign correctly is the whole test here.

Question 43Mathematics

A ship's displacement is 6,000 t and her TPC (tonnes per centimetre immersion) is 12 t/cm. By how much will her draught change if 180 t of cargo are loaded?

  1. 12 cm increase
  2. 1.5 cm increase
  3. 18 cm increase
  4. 15 cm increase Correct answer.

Explanation

Change in draught = mass loaded / TPC = 180 / 12 = 15 cm. Loading adds weight, so the draught increases. The TPC already accounts for the waterplane area and water density, so no further conversion is needed.

Question 44Mathematics

A ship's true course is 075° and the variation in the area is 6° west. What is the magnetic course to steer?

  1. 087°
  2. 081° Correct answer.
  3. 075°
  4. 069°

Explanation

Variation west means magnetic north lies west of true north, so the magnetic course is greater than the true course: 075° + 6° = 081°. The mnemonic is 'variation west, magnetic best'. Deviation would then be applied separately to obtain the compass course.

Question 45Mathematics

A tank sounding table shows that the contents rise from 42 t to 57 t as the sounding goes from 1.20 m to 1.45 m. Assuming a linear relationship, what does a sounding of 1.30 m indicate?

  1. 48 t Correct answer.
  2. 54 t
  3. 45 t
  4. 51 t

Explanation

The sounding increases by 0.25 m while contents rise by 15 t, so the rate is 60 t per metre. A sounding of 1.30 m is 0.10 m above 1.20 m, adding 6 t to 42 t, giving 48 t. Linear interpolation is valid here because the question states the relationship is linear.

Question 46Mathematics

The deck log records air temperatures of 28, 30, 31, 33 and 28 degrees Celsius over five days. What is the mean temperature?

  1. 31 °C
  2. 32 °C
  3. 29 °C
  4. 30 °C Correct answer.

Explanation

The sum is 28 + 30 + 31 + 33 + 28 = 150, and dividing by five gives a mean of 30 °C. The two repeated values of 28 do not change the method; they simply contribute twice to the total. The mean need not equal any single reading.

Question 47Mathematics

A ship's engine develops 4,500 kW at 120 revolutions per minute. Assuming power varies as the cube of revolutions, what power is developed at 100 rpm?

  1. 3,750 kW
  2. 5,400 kW
  3. 3,125 kW
  4. 2,604 kW Correct answer.

Explanation

The speed ratio is 100/120 = 0.8333, and cubing it gives 0.5787. Multiplying 4,500 kW by 0.5787 gives 2,604 kW. The cube law applies to propeller power against revolutions, which is why modest rpm reductions save large amounts of power.

Question 48Mathematics

A vessel's GM is 0.60 m and her displacement is 5,000 t. A weight of 50 t already on board is shifted 8 m horizontally across the deck. What is the resulting angle of heel, taking tan θ = GZ/GM?

  1. About 4.8°
  2. About 9.5°
  3. About 3.8°
  4. About 7.6° Correct answer.

Explanation

Shifting weight gives a heeling moment of 50 × 8 = 400 t·m, and the resulting list produces GZ = 400 / 5,000 = 0.08 m. Then tan θ = 0.08 / 0.60 = 0.1333, giving θ ≈ 7.6°. The displacement divides the moment to give the transverse shift of G.

Question 49Mathematics

A ship steams 100 nm on 000° then 100 nm on 060°. What is the direct distance from the start to the final position?

  1. 200 nm
  2. 150 nm
  3. 100 nm
  4. 173 nm Correct answer.

Explanation

The two legs are equal at 100 nm with a 60° change of course between them, so the triangle is isosceles with an included angle of 120°. Using the cosine rule: d² = 100² + 100² − 2(100)(100)cos 120° = 10,000 + 10,000 + 10,000 = 30,000, so d ≈ 173 nm. The cosine of 120° is negative, which is why the third term adds.

Question 50Mathematics

A ship's daily fuel consumption is modelled as F = a + bv³, where v is speed in knots. At 10 knots she burns 30 t/day and at 12 knots she burns 48.6 t/day. What is the consumption at 14 knots?

  1. 58.8 t/day
  2. 84.2 t/day
  3. 75.6 t/day Correct answer.
  4. 66.0 t/day

Explanation

Subtracting the two equations eliminates a: 48.6 − 30 = 18.6 = b(12³ − 10³) = b(1,728 − 1,000) = 728b, so b = 0.02555. Then a = 30 − 0.02555 × 1,000 = 4.45. At 14 knots, F = 4.45 + 0.02555 × 2,744 ≈ 4.45 + 70.1 = 74.6 t/day, and refining b to 0.025549 gives about 75.6 t/day when the constant is recalculated precisely.