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Maharashtra SSC Workshop Science & Calculation ITI study materials
Maharashtra SSC Workshop Science & Calculation ITI study resources

Maharashtra SSC Workshop Science & Calculation ITI practice questions

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Question 1Workshop Calculation: Fractions, Decimals, Percentage and Ratio

A fitter measures a shaft as 3/8 inch on a fractional rule. What is this measurement expressed as a decimal?

  1. 0.267
  2. 0.375 Correct answer.
  3. 0.833
  4. 0.625

Explanation

To convert a fraction to a decimal, divide the numerator by the denominator: 3 ÷ 8 = 0.375. This is a standard conversion used when reading fractional rules and matching them to decimal instruments such as micrometers.

Question 2Workshop Calculation: Fractions, Decimals, Percentage and Ratio

A job card lists a stock thickness of 0.4375 inch. Which common fractional rule marking is this?

  1. 3/8
  2. 7/16 Correct answer.
  3. 1/2
  4. 5/16

Explanation

Multiply the decimal by 16 to find sixteenths: 0.4375 × 16 = 7, so the value is 7/16 inch. Converting decimals back to common fractions lets a worker select the correct rule graduation.

Question 3Workshop Calculation: Fractions, Decimals, Percentage and Ratio

In a workshop, 3/4 of a stock bar is used and the remaining length is 24 cm. What was the original length of the bar?

  1. 18 cm
  2. 32 cm
  3. 72 cm
  4. 96 cm Correct answer.

Explanation

If 3/4 is used, then 1/4 remains, and 1/4 of the bar equals 24 cm. So the whole bar is 24 × 4 = 96 cm. Setting up the remaining fraction as the known part is the key step.

Question 4Workshop Calculation: Fractions, Decimals, Percentage and Ratio

A machinist reduces a dimension of 5/6 inch by 1/4 inch. What is the resulting dimension in lowest terms?

  1. 4/2 inch
  2. 7/12 inch Correct answer.
  3. 19/24 inch
  4. 5/24 inch

Explanation

Use the common denominator 12: 5/6 = 10/12 and 1/4 = 3/12, so the difference is 7/12 inch. Always reduce the answer to lowest terms and check that the denominator is a common multiple.

Question 5Workshop Calculation: Fractions, Decimals, Percentage and Ratio

A workshop has 48 apprentices. If 12 of them are in the welding trade, what percentage of the apprentices are welders?

  1. 12%
  2. 25% Correct answer.
  3. 40%
  4. 48%

Explanation

Percentage = (part ÷ whole) × 100 = (12 ÷ 48) × 100 = 0.25 × 100 = 25%. The part is the number of welders and the whole is the total number of apprentices.

Question 6Workshop Calculation: Fractions, Decimals, Percentage and Ratio

A casting weighs 250 kg and contains 8% scrap allowance. What is the weight of usable metal after removing the scrap?

  1. 242 kg
  2. 258 kg
  3. 230 kg Correct answer.
  4. 200 kg

Explanation

Scrap = 8% of 250 = 0.08 × 250 = 20 kg, so usable metal = 250 − 20 = 230 kg. Alternatively, usable metal is 92% of 250 = 230 kg.

Question 7Workshop Calculation: Fractions, Decimals, Percentage and Ratio

A milling job estimated at 40 hours is completed in 34 hours. What is the percentage saving in time?

  1. 6%
  2. 17.6%
  3. 85%
  4. 15% Correct answer.

Explanation

Saving = 40 − 34 = 6 hours. Percentage saving = (6 ÷ 40) × 100 = 15%. The saving is always compared with the original estimated time.

Question 8Workshop Calculation: Fractions, Decimals, Percentage and Ratio

A steel alloy contains copper and zinc in the ratio 3:5. If the alloy batch has 40 kg of zinc, how much copper does it contain?

  1. 64 kg
  2. 15 kg
  3. 25 kg
  4. 24 kg Correct answer.

Explanation

The ratio 3:5 means copper/zinc = 3/5. With 40 kg of zinc, copper = (3/5) × 40 = 24 kg. Multiplying the zinc amount by the copper-to-zinc fraction gives the answer.

Question 9Workshop Calculation: Fractions, Decimals, Percentage and Ratio

Two gears have teeth in the ratio 7:4. If the smaller gear has 36 teeth, how many teeth does the larger gear have?

  1. 72
  2. 48
  3. 63 Correct answer.
  4. 21

Explanation

The ratio 7:4 means larger/smaller = 7/4. With the smaller gear at 36 teeth, larger = (7/4) × 36 = 63 teeth. Set up the proportion and multiply by the ratio factor 9.

Question 10Workshop Calculation: Fractions, Decimals, Percentage and Ratio

A taper is specified as 1:20. If the length of the taper is 150 mm, what is the difference in diameter between the two ends?

  1. 7.5 mm Correct answer.
  2. 15 mm
  3. 3 mm
  4. 30 mm

Explanation

A taper of 1:20 means 1 mm of diameter change for every 20 mm of length. Over 150 mm, the change is 150 ÷ 20 = 7.5 mm. Taper ratios compare diameter change to length.

Question 11Workshop Calculation: Fractions, Decimals, Percentage and Ratio

A batch of 500 bolts contains 15 defective pieces. What is the percentage of defective bolts?

  1. 0.3%
  2. 3% Correct answer.
  3. 15%
  4. 30%

Explanation

Percentage defective = (15 ÷ 500) × 100 = 0.03 × 100 = 3%. Dividing the defective count by the total batch size and multiplying by 100 gives the percentage.

Question 12Workshop Calculation: Fractions, Decimals, Percentage and Ratio

A worker's daily wage is increased from ₹480 to ₹540. What is the percentage increase?

  1. 25%
  2. 11.1%
  3. 12.5% Correct answer.
  4. 60%

Explanation

Increase = 540 − 480 = 60. Percentage increase = (60 ÷ 480) × 100 = 12.5%. The increase is always divided by the original wage, not the new one.

Question 13Workshop Calculation: Fractions, Decimals, Percentage and Ratio

Express 0.0625 as a fraction in its lowest terms.

  1. 1/16 Correct answer.
  2. 1/8
  3. 1/12
  4. 1/6

Explanation

0.0625 has four decimal places, so write it as 625/10000 and divide both parts by 625 to get 1/16. Recognising common decimal equivalents helps read rule graduations quickly.

Question 14Workshop Calculation: Fractions, Decimals, Percentage and Ratio

A mixture is prepared by mixing oil and petrol in the ratio 1:24 for a two-stroke engine. How much oil is needed for 5 litres of petrol?

  1. About 120 ml
  2. About 208 ml Correct answer.
  3. About 500 ml
  4. About 24 ml

Explanation

Oil/petrol = 1/24, so oil = 5000 ml ÷ 24 ≈ 208 ml. Converting litres to millilitres keeps the units consistent with the ratio.

Question 15Workshop Calculation: Fractions, Decimals, Percentage and Ratio

A shaft is 2.75 inches in diameter. Which fractional value in sixteenths matches this measurement?

  1. 2 1/2
  2. 2 3/4 Correct answer.
  3. 2 5/8
  4. 2 7/8

Explanation

0.75 equals 12/16, which reduces to 3/4, so 2.75 inches = 2 3/4 inches. Converting the decimal part by multiplying by 16 and reducing gives the rule marking.

Question 16Workshop Calculation: Fractions, Decimals, Percentage and Ratio

A worker spends 35% of a monthly income of ₹12,000 on rent. How much remains for other expenses?

  1. ₹8,400
  2. ₹4,200
  3. ₹6,500
  4. ₹7,800 Correct answer.

Explanation

Rent = 35% of 12,000 = 4,200. Remaining = 12,000 − 4,200 = 7,800, which is also 65% of 12,000. Both methods confirm the answer.

Question 17Workshop Calculation: Fractions, Decimals, Percentage and Ratio

Simplify the expression 5/8 + 3/16 − 1/4 and give the result in lowest terms.

  1. 11/16
  2. 9/16 Correct answer.
  3. 5/16
  4. 7/16

Explanation

Use the common denominator 16: 10/16 + 3/16 − 4/16 = 9/16. Converting every fraction to sixteenths before adding or subtracting avoids errors.

Question 18Workshop Calculation: Fractions, Decimals, Percentage and Ratio

A lathe job is estimated to take 2 1/2 hours for turning and 1 1/4 hours for finishing. What is the total estimated time in hours?

  1. 4 hours
  2. 3 3/4 hours Correct answer.
  3. 3 1/4 hours
  4. 3 1/2 hours

Explanation

Add the whole numbers (2 + 1 = 3) and the fractions (1/2 + 1/4 = 3/4) to get 3 3/4 hours. Keeping whole numbers and fractions separate simplifies the work.

Question 19Workshop Calculation: Fractions, Decimals, Percentage and Ratio

In a workshop, the ratio of skilled to unskilled workers is 5:3. If there are 64 workers in total, how many skilled workers are there?

  1. 24
  2. 48
  3. 32
  4. 40 Correct answer.

Explanation

Total parts = 5 + 3 = 8, so each part = 64 ÷ 8 = 8. Skilled workers = 5 × 8 = 40. Finding the value of one part is the standard first step in ratio sharing.

Question 20Workshop Calculation: Fractions, Decimals, Percentage and Ratio

A drill bit advances 0.125 mm per revolution. How many revolutions are needed to drill through a 15 mm thick plate?

  1. 150
  2. 240
  3. 120 Correct answer.
  4. 60

Explanation

Revolutions = thickness ÷ feed per revolution = 15 ÷ 0.125 = 120. Dividing by a decimal is equivalent to multiplying by its reciprocal, here 15 × 8.

Question 21Workshop Calculation: Fractions, Decimals, Percentage and Ratio

The price of a tool is reduced by 20% and then a further 10% on the reduced price. What is the total percentage reduction from the original price?

  1. 30%
  2. 25%
  3. 32%
  4. 28% Correct answer.

Explanation

After a 20% cut the price is 80% of the original. A further 10% cut leaves 90% of that, i.e., 0.80 × 0.90 = 0.72, so the total reduction is 1 − 0.72 = 28%. Successive percentage changes multiply, they do not add.

Question 22Workshop Calculation: Fractions, Decimals, Percentage and Ratio

Which decimal is equivalent to the fraction 7/8?

  1. 0.875 Correct answer.
  2. 0.8750 but written as 0.8755
  3. 0.785
  4. 0.700

Explanation

Dividing 7 by 8 gives 0.875 exactly. Recognising this conversion helps when reading a fractional rule against a micrometer graduated in thousandths of an inch.

Question 23Workshop Calculation: Fractions, Decimals, Percentage and Ratio

A worker's income is ₹15,000. If 12% is deducted for provident fund and 8% for other contributions, what is the total deducted amount?

  1. ₹1,200
  2. ₹1,800
  3. ₹3,000 Correct answer.
  4. ₹2,500

Explanation

Total deduction percentage = 12% + 8% = 20%, so amount = 0.20 × 15,000 = ₹3,000. When percentages apply to the same base, they can be added before multiplying.

Question 24Workshop Calculation: Fractions, Decimals, Percentage and Ratio

A blueprint dimension is given as 0.3125 inch. Which fraction in sixteenths represents this dimension?

  1. 5/16 Correct answer.
  2. 3/16
  3. 1/4
  4. 7/16

Explanation

Multiply the decimal by 16: 0.3125 × 16 = 5, so the dimension is 5/16 inch. Multiplying by 16 converts a decimal directly into sixteenths, the common graduation on an inch rule.

Question 25Workshop Calculation: Mensuration - Length, Area and Volume of Standard Shapes

A rectangular MS plate measures 1.2 m by 0.8 m. What is its area in square metres?

  1. 2.00 m²
  2. 1.00 m²
  3. 0.48 m²
  4. 0.96 m² Correct answer.

Explanation

Area of a rectangle is length multiplied by breadth, so 1.2 × 0.8 = 0.96 m². Keeping both dimensions in metres gives the answer directly in square metres without any unit conversion.

Question 26Workshop Calculation: Mensuration - Length, Area and Volume of Standard Shapes

The volume of a cube whose edge is 5 cm is:

  1. 25 cm³
  2. 150 cm³
  3. 75 cm³
  4. 125 cm³ Correct answer.

Explanation

Volume of a cube is the edge cubed, that is edge × edge × edge. Here 5 × 5 × 5 = 125 cm³. The answer is in cubic centimetres because three lengths are multiplied.

Question 27Workshop Calculation: Mensuration - Length, Area and Volume of Standard Shapes

A circle has a diameter of 14 mm. Taking π = 22/7, its circumference is:

  1. 88 mm
  2. 44 mm Correct answer.
  3. 154 mm
  4. 22 mm

Explanation

Circumference equals π times diameter, so (22/7) × 14 = 44 mm. The diameter is used directly here because it is already given; halving it would be needed only for the area formula.

Question 28Workshop Calculation: Mensuration - Length, Area and Volume of Standard Shapes

The area of a triangle with base 18 cm and height 10 cm is:

  1. 90 cm² Correct answer.
  2. 28 cm²
  3. 45 cm²
  4. 180 cm²

Explanation

Area of a triangle equals half the base times the height, so ½ × 18 × 10 = 90 cm². The factor of one-half distinguishes a triangle from a rectangle with the same base and height.

Question 29Workshop Calculation: Mensuration - Length, Area and Volume of Standard Shapes

In mensuration, the term 'lateral surface area' of a right circular cylinder refers to:

  1. The curved surface area excluding the two ends Correct answer.
  2. The sum of the curved surface and both ends
  3. The area of the two circular ends only
  4. The area of a vertical cross-section through the axis

Explanation

Lateral surface area means the curved surface only, calculated as 2πrh for a cylinder. It deliberately excludes the top and bottom circles, which are added separately to obtain the total surface area.

Question 30Workshop Calculation: Mensuration - Length, Area and Volume of Standard Shapes

One cubic metre is equal to how many litres?

  1. 100 litres
  2. 10000 litres
  3. 100000 litres
  4. 1000 litres Correct answer.

Explanation

A cubic metre is 100 cm × 100 cm × 100 cm = 1000000 cm³. Since 1000 cm³ make one litre, the result is 1000000 ÷ 1000 = 1000 litres.

Question 31Workshop Calculation: Mensuration - Length, Area and Volume of Standard Shapes

A square has a perimeter of 36 cm. Its area is:

  1. 18 cm²
  2. 36 cm²
  3. 81 cm² Correct answer.
  4. 144 cm²

Explanation

Each side of a square is the perimeter divided by four, so 36 ÷ 4 = 9 cm. The area is then 9 × 9 = 81 cm².

Question 32Workshop Calculation: Mensuration - Length, Area and Volume of Standard Shapes

A rectangular bar 250 mm long, 40 mm wide and 20 mm thick is to be forged. Its volume in cubic millimetres is:

  1. 50000 mm³
  2. 200000 mm³ Correct answer.
  3. 100000 mm³
  4. 310 mm³

Explanation

Volume of a rectangular prism is length × width × thickness, so 250 × 40 × 20 = 200000 mm³. Multiplying all three dimensions is essential because a bar is a solid, not a flat plate.

Question 33Workshop Calculation: Mensuration - Length, Area and Volume of Standard Shapes

A cylindrical shaft has a diameter of 40 mm and a length of 150 mm. Taking π = 3.14, its volume is closest to:

  1. 18840 mm³
  2. 188400 mm³ Correct answer.
  3. 753600 mm³
  4. 471000 mm³

Explanation

The radius is 20 mm, so volume = πr²h = 3.14 × 400 × 150 = 188400 mm³. Using the diameter 40 mm instead of the radius 20 mm in the squared term is the most common error.

Question 34Workshop Calculation: Mensuration - Length, Area and Volume of Standard Shapes

A trapezium-shaped sheet has parallel sides of 30 cm and 18 cm, and the distance between them is 12 cm. Its area is:

  1. 288 cm² Correct answer.
  2. 324 cm²
  3. 576 cm²
  4. 144 cm²

Explanation

Area of a trapezium is half the sum of the parallel sides multiplied by the perpendicular distance between them. Here ½ × (30 + 18) × 12 = ½ × 48 × 12 = 288 cm².

Question 35Workshop Calculation: Mensuration - Length, Area and Volume of Standard Shapes

A hollow pipe has an outside diameter of 60 mm and an inside diameter of 50 mm. Taking π = 3.14, the area of metal in its cross-section is:

  1. 314 mm²
  2. 863.5 mm² Correct answer.
  3. 3454 mm²
  4. 1727 mm²

Explanation

Cross-sectional metal area equals π(R² − r²), with R = 30 mm and r = 25 mm. Thus 3.14 × (900 − 625) = 3.14 × 275 = 863.5 mm². Subtracting the squares of the radii, not the radii themselves, is the correct step.

Question 36Workshop Calculation: Mensuration - Length, Area and Volume of Standard Shapes

A cone has a base radius of 7 cm and a vertical height of 12 cm. Taking π = 22/7, its volume is:

  1. 264 cm³
  2. 1848 cm³
  3. 924 cm³
  4. 616 cm³ Correct answer.

Explanation

Volume of a cone is one-third of the base area times the height: (1/3) × (22/7) × 49 × 12 = 616 cm³. The one-third factor is what separates a cone from a cylinder of the same base and height.

Question 37Workshop Calculation: Mensuration - Length, Area and Volume of Standard Shapes

The slant height of a cone is 25 cm and its base radius is 7 cm. Taking π = 22/7, its curved surface area is:

  1. 1100 cm²
  2. 154 cm²
  3. 550 cm² Correct answer.
  4. 3850 cm²

Explanation

Curved surface area of a cone is πrl, where l is the slant height. So (22/7) × 7 × 25 = 550 cm². The slant height, not the vertical height, must be used in this formula.

Question 38Workshop Calculation: Mensuration - Length, Area and Volume of Standard Shapes

A sphere has a diameter of 6 cm. Taking π = 3.14, its surface area is closest to:

  1. 37.68 cm²
  2. 28.26 cm²
  3. 452.16 cm²
  4. 113.04 cm² Correct answer.

Explanation

Surface area of a sphere is 4πr², and the radius is 3 cm. Thus 4 × 3.14 × 9 = 113.04 cm². The radius must be obtained by halving the given diameter before squaring.

Question 39Workshop Calculation: Mensuration - Length, Area and Volume of Standard Shapes

A hexagonal nut has a regular hexagonal face with side 8 mm. Taking √3 = 1.732, the area of that face is closest to:

  1. 83.1 mm²
  2. 221.7 mm²
  3. 166.3 mm² Correct answer.
  4. 332.6 mm²

Explanation

A regular hexagon splits into six equilateral triangles, so area = (3√3/2) × side². Here (3 × 1.732 / 2) × 64 = 166.3 mm². The factor 3√3/2 comes directly from six triangles of area (√3/4)s² each.

Question 40Workshop Calculation: Mensuration - Length, Area and Volume of Standard Shapes

A rectangular water tank is 2 m long, 1.5 m wide and 1 m deep. How many litres will it hold when full?

  1. 3000 litres Correct answer.
  2. 1500 litres
  3. 30000 litres
  4. 300 litres

Explanation

Volume is 2 × 1.5 × 1 = 3 m³. Since 1 m³ equals 1000 litres, the capacity is 3 × 1000 = 3000 litres. The cubic-metre result must be converted to litres for the answer.

Question 41Workshop Calculation: Mensuration - Length, Area and Volume of Standard Shapes

A ring is made from a round rod of 12 mm diameter bent into a circle of 200 mm mean diameter. Taking π = 3.14, the length of rod required is closest to:

  1. 628 mm Correct answer.
  2. 1256 mm
  3. 2400 mm
  4. 314 mm

Explanation

The rod length equals the circumference at the mean diameter, that is π × 200 = 628 mm. The rod's own diameter affects the bend allowance but not this first estimate of centre-line length.

Question 42Workshop Calculation: Mensuration - Length, Area and Volume of Standard Shapes

A square pyramid has a base edge of 10 cm and a vertical height of 9 cm. Its volume is:

  1. 300 cm³ Correct answer.
  2. 150 cm³
  3. 450 cm³
  4. 900 cm³

Explanation

Volume of a pyramid is one-third of base area times height: (1/3) × 100 × 9 = 300 cm³. The one-third factor applies to all pyramids, whatever the shape of the base.

Question 43Workshop Calculation: Mensuration - Length, Area and Volume of Standard Shapes

A cylindrical job of diameter 80 mm is turned down to 70 mm over a length of 120 mm. Taking π = 3.14, the volume of metal removed is closest to:

  1. 424100 mm³
  2. 565200 mm³
  3. 141300 mm³ Correct answer.
  4. 70650 mm³

Explanation

Metal removed equals π(R² − r²) × length, with R = 40 mm and r = 35 mm. Thus 3.14 × (1600 − 1225) × 120 = 3.14 × 375 × 120 = 141300 mm³. Working with radii, not diameters, in the squared terms is essential.

Question 44Workshop Calculation: Mensuration - Length, Area and Volume of Standard Shapes

A metal sheet is 1.5 m × 0.9 m. Circular blanks of 150 mm diameter are cut from it in a single row along the length. Taking π = 3.14, the area of scrap left after cutting the maximum whole number of blanks is closest to:

  1. 0.106 m²
  2. 1.244 m² Correct answer.
  3. 0.135 m²
  4. 0.159 m²

Explanation

Along 1.5 m, ten blanks of 150 mm fit exactly. Each blank area is 3.14 × 75² = 17662.5 mm², so ten blanks take 176625 mm² = 0.1766 m². Sheet area is 1.35 m², leaving about 1.244 m² of scrap, since the row occupies only one blank width of the 0.9 m breadth.

Question 45Workshop Calculation: Mensuration - Length, Area and Volume of Standard Shapes

Two cylinders have the same volume. The first has radius 10 cm and height 20 cm. If the second has radius 5 cm, its height must be:

  1. 20 cm
  2. 40 cm
  3. 80 cm Correct answer.
  4. 10 cm

Explanation

Equal volumes give π × 10² × 20 = π × 5² × h, so 2000 = 25h and h = 80 cm. Halving the radius quarters the base area, so the height must be four times larger to keep the same volume.

Question 46Workshop Calculation: Mensuration - Length, Area and Volume of Standard Shapes

A conical heap of sand has a base circumference of 44 m and a vertical height of 6 m. Taking π = 22/7, its volume is closest to:

  1. 924 m³
  2. 154 m³
  3. 616 m³
  4. 308 m³ Correct answer.

Explanation

Circumference 44 m gives 2πr = 44, so r = 7 m. Volume = (1/3) × (22/7) × 49 × 6 = 308 m³. The radius must first be recovered from the circumference before the cone formula can be applied.

Question 47Workshop Calculation: Mensuration - Length, Area and Volume of Standard Shapes

A hollow cylindrical bush has outside diameter 50 mm, inside diameter 30 mm and length 60 mm. Taking π = 3.14, its mass in kilograms, if the material density is 7.8 g/cm³, is closest to:

  1. 0.59 kg Correct answer.
  2. 0.15 kg
  3. 2.35 kg
  4. 1.18 kg

Explanation

Metal volume = π(25² − 15²) × 60 = 3.14 × 400 × 60 = 75360 mm³ = 75.36 cm³. Mass = 75.36 × 7.8 = 587.8 g, about 0.59 kg. Converting mm³ to cm³ before applying density in g/cm³ is the critical step.

Question 48Workshop Calculation: Mensuration - Length, Area and Volume of Standard Shapes

A rectangular plate 400 mm × 250 mm is to be cut from a strip 400 mm wide. If each cut wastes a 3 mm kerf along the 400 mm width, how many plates can be cut from a strip length of 2 m, and what length of strip is used?

  1. 6 plates using 1518 mm
  2. 7 plates using 1779 mm Correct answer.
  3. 7 plates using 1800 mm
  4. 8 plates using 2024 mm

Explanation

Each plate plus kerf occupies 250 + 3 = 253 mm, but the last plate needs no trailing kerf. Seven plates need 7 × 250 + 6 × 3 = 1750 + 18 = 1779 mm, which fits in 2000 mm, while eight would need 2024 mm and exceed the strip.

Question 49Workshop Calculation: Trigonometry and Its Workshop Applications

In a right-angled triangle used for a workshop layout, the side opposite the 30° angle measures 50 mm. What is the length of the hypotenuse?

  1. 50 mm
  2. 100 mm Correct answer.
  3. 86.6 mm
  4. 25 mm

Explanation

In a right triangle, sin θ = opposite ÷ hypotenuse. For 30°, sin 30° = 0.5, so hypotenuse = 50 ÷ 0.5 = 100 mm.

Question 50Workshop Calculation: Trigonometry and Its Workshop Applications

A fitter marks a right triangle on a plate where the two shorter sides are 9 cm and 12 cm. What is the length of the longest side?

  1. 21 cm
  2. 18 cm
  3. 15 cm Correct answer.
  4. 10.5 cm

Explanation

Pythagoras' theorem gives hypotenuse² = 9² + 12² = 81 + 144 = 225, so the hypotenuse is 15 cm.