EPFO EO/AO 2023 General Ability Test · Question 102 of 120

Correct — C, (c) 14. The water is simply moved from one container to another, so its volume does not change. Find the volume of the tank, then ask how long a cylinder of the given cross-section must be to hold that same volume. The arithmetic is cleanest in centimetres, because the pipe's radius is given in centimetres. A rectangular parallelepiped is a cuboid, so the tank measures 200 cm x 200 cm x 110 cm and its volume is 200 x 200 x 110 = 44,00,000 cubic centimetres. The pipe is a cylinder of radius 1 cm, so its cross-sectional area is pi r squared = (22/7) x 1 x 1 = 22/7 square centimetres. A cylinder's volume is cross-sectional area x length, so the length needed is volume divided by area: length = 44,00,000 / (22/7) = 44,00,000 x 7 / 22 = 2,00,000 x 7 = 14,00,000 cm. That is 14,000 m, or 14 km, which is option (c). Two details in the wording are doing work. First, the paper supplies pi = 22/7 on its own line, and the answer comes out as an exact whole number of kilometres only because the 22 in the numerator of the volume cancels the 22 in the denominator of the area. A clean cancellation of that kind is the paper telling you that the intended route has been found. Using pi = 3.14 instead gives 44,00,000 / 3.14 = 14,01,274 cm, which is 14.01 km and still rounds to the same option, so the choice of pi does not change the answer, only the tidiness. Second, 'minimum length' is asked because a longer pipe would also hold the water, with some length left empty; the minimum is the length at which the pipe is exactly full, which is the equality solved above. The metre-based route gives the same figure and is worth writing out once, because it is where unit slips happen: the volume is 2 x 2 x 1.1 = 4.4 cubic metres, the radius is 1 cm = 0.01 m, the area is (22/7) x 0.01 x 0.01 = 0.00031428 square metres, and 4.4 / 0.00031428 = 14,000 m. Note that in centimetres the radius is 1, so squaring it changes nothing and a candidate who forgets to square gets away with it; in metres, forgetting to square turns 0.0001 into 0.01 and the answer collapses to 140 m. Choosing centimetres removes that hazard.

• (a)12 — Twelve kilometres of this pipe holds 12,00,000 x 22/7 = 37,71,429 cubic centimetres, which is short of the tank's 44,00,000 by about one seventh. Working backwards, a tank that filled exactly 12 km of the pipe would have to be 94.3 cm deep on the same 2 m square base, not 110 cm. The option is there for a candidate who has divided or multiplied by 7/22 the wrong way round at the last step, or who has trimmed the awkward 1.1 down to 1 and then rounded downward as well.
• (b)13 — This is the one wrong option with a specific misreading behind it. The height is printed as 1.1 m with a raised middle dot as the decimal point, and a candidate who reads it as a plain 1 m gets a volume of 40,00,000 cubic centimetres and a length of 40,00,000 x 7 / 22 = 12,72,727 cm, which is about 12.7 km and sits temptingly between options (a) and (b). Checked directly, 13 km of pipe holds 13,00,000 x 22/7 = 40,85,714 cubic centimetres, which corresponds to a tank depth of 102.1 cm rather than the 110 cm printed. The lesson is to read the raised dot as a decimal point every time it appears in this booklet, which prints decimals that way in more than one place.
• (d)15 — Fifteen kilometres of pipe holds 15,00,000 x 22/7 = 47,14,286 cubic centimetres, which is more than the tank contains, so the pipe would not be full and 15 km cannot be the minimum length. Turned around, filling 15 km of this pipe would need a tank 117.9 cm deep on the same base, close to 1.2 m rather than the 1.1 m given. The option exists to catch a candidate who has rounded the depth upward or who has treated 'minimum length' as an invitation to leave a safety margin; the phrase asks for the exact length at which the pipe is just full, so the equality is the answer and any larger figure is wrong.

This is a volume-conservation problem, the commonest single idea in mensuration questions set for recruitment papers. When a fixed quantity of liquid is poured, drained or recast from one shape into another, the volume is the invariant: compute it once in the source shape and set it equal to the expression for the target shape. Here the source is a cuboid, whose volume is length x breadth x height, and the target is a cylinder, whose volume is pi r squared x length. Because a cylinder's cross-section is constant, its volume is simply cross-sectional area multiplied by length, so the unknown length falls out of one division. The same template solves a great many other items: a cylindrical rod melted and recast into spheres, water flowing through a pipe at a given speed filling a tank in a given time, a conical vessel emptied into a cylindrical one. The two disciplines that decide whether the arithmetic works are choosing one unit and staying in it, and using the value of pi that the paper supplies. Where a question hands you pi = 22/7 it is almost always because a factor of 7 or 22 in the data is meant to cancel, and the cancellation is the confirmation that the intended route has been found.

The EO/AO General Ability Test sets mensuration inside its quantitative blocks, usually in the applied form seen here rather than as a bare formula recall. The examiner's real test is unit handling: a radius in centimetres, a tank in metres and an answer demanded in kilometres, with three conversions between them and only one chance to get each right. The habit rewarded is to convert everything to a single unit before computing anything, and to keep the final conversion, from centimetres to kilometres in this case, as a separate written step rather than doing it in the head at the end of a long division.

• A rectangular parallelepiped is a cuboid; its volume is length x breadth x height.
• A cylinder's volume is pi r squared x length, that is, cross-sectional area multiplied by length.
• Tank volume here: 200 x 200 x 110 = 44,00,000 cubic centimetres, equal to 4.4 cubic metres.
• Pipe cross-section with radius 1 cm and pi = 22/7: area = 22/7 square centimetres.
• Required length = 44,00,000 x 7 / 22 = 14,00,000 cm = 14,000 m = 14 km.
• 1 cubic metre = 10,00,000 cubic centimetres, and 1 km = 1,00,000 cm.
• Where a paper supplies pi = 22/7 the data are usually arranged so that 7 or 22 cancels exactly.

• Mixing metres and centimetres inside one calculation; convert everything first.
• Reading the raised middle dot in 1.1 m as anything other than a decimal point.
• Forgetting to square the radius, which is harmless when the radius is 1 cm but fatal when it is 0.01 m.
• Using pi = 3.14 when the paper has supplied 22/7 and the data are built to cancel with it.
• Treating 'minimum length' as needing a margin; it is the length at which the pipe is exactly full.

Mensuration in EO/AO papers usually arrives as a conversion between two solids with the volume held constant, or as a flow problem in which a cross-sectional area and a speed together give a volume per unit time. Both reduce to the same first move, which is to write down the volume in one unit and then divide by whatever the target shape contributes per unit length or per unit time. Expect the numbers to be arranged so that the supplied value of pi cancels, and treat a whole-number answer as evidence that the route was the intended one.

No directly related past PYQ was found.

• practice — not a real PYQ

  A cylindrical vessel of radius 7 cm is filled with water to a height of 20 cm. If the water is poured into a cuboidal tank with a base 22 cm by 10 cm, what is the height of the water in the tank ? (Take pi = 22/7)

  • (a)12 cm
  • (b)14 cm
  • (c)16 cm
  • (d)20 cm

  Answer(b) 14 cm

• practice — not a real PYQ

  Water flows through a pipe of internal radius 1 cm at 7 metres per second into an empty tank. How much water, in litres, enters the tank in one minute ? (Take pi = 22/7)

  • (a)66 litres
  • (b)110 litres
  • (c)132 litres
  • (d)220 litres

  Answer(c) 132 litres