Area and volume

What the reference sheet gives you, and why scaling a length by k scales volume by k³.

~30 min · prequestion, worked examples, retrieval practice

The reference sheet is a boundary, not a safety net — every volume formula this exam uses is printed on it, and not one formula for surface area is. Geometry & Trigonometry is about 15% of the Math section, and area and volume is the skill point inside it that most often arrives as a multi-step application rather than a formula lookup. Points here are almost never lost to hard algebra. They are lost hunting the sheet for something that was never printed, to a diameter dropped into a slot that wanted a radius, and to the most-tested unprinted fact in the whole domain — that doubling a length does not double an area.

Before you read on

Two or three questions on exactly what this lesson teaches. Being wrong here is fine — it's the fastest way to find out what to pay attention to next.

Question 1
Medium

Before any teaching. A right circular cylinder is enlarged so that its radius and its height are both 3 times as large as the original's. The volume of the enlarged cylinder is how many times the volume of the original?

Question 2
Medium

The Digital SAT gives you a reference sheet at the start of every Math module. Which of the following formulas is NOT printed on it, and therefore has to be known from memory?

Question 3
Hard

A shipping container has a volume of 2 cubic feet. What is its volume in cubic inches? (1 foot = 12 inches.)

What the question is actually testing

This skill point covers everything the exam asks about the size of a figure: the area of a plane shape, the volume of a solid, the surface area of a solid, and the ratios that connect two figures of the same shape at different sizes. College Board publishes the domain weight — Geometry & Trigonometry is about 15% of the Math section, which its own specifications put at 5 to 7 of the 44 questions.

What College Board does not publish is how those 5 to 7 questions split across the domain's four skill points, or how they distribute between the first module and the harder second one. The widely repeated prep-industry claim that geometry clusters in the easier module is reverse-engineered from released tests — plausible, useful for planning, and not an official figure. Do not build a strategy that depends on it.

Every Math module opens with a reference sheet. It is the only place on the entire exam where formulas are handed to you, and the single most valuable thing to know about it is where it stops. Every volume formula the exam needs is printed. Not one surface-area formula is. Neither is the trapezoid, the parallelogram, the equilateral triangle, the arc of a circle, or the rule that governs what happens when a shape is scaled up — which is the fact the hardest items in this domain are built on.

Weak path: an item describes a grain silo — a cylinder with a dome on top — and you scan the sheet for a silo, find nothing, and start improvising. Strong path: you already know the sheet contains no composite shapes and never will, so you cut the solid into printed pieces, price each piece separately, and add. Same knowledge, different first move, about ninety seconds apart.

Weak path on a scaling item: compute the original volume, compute the new dimensions, compute the new volume, divide. Strong path: notice that the question only ever asked for a ratio, and that a ratio of volumes is the cube of a ratio of lengths — so no dimension needs to be known, and neither does the formula. Four lines against one, and the one-line version cannot make an arithmetic slip.

Foundations — area, volume and surface area from zero (skip this block if 2πrh already means something to you)

If you can say what 2πrh is the area of, and why a cone carries a one-third while a cylinder does not, skip ahead. If either question is unfamiliar, this is the highest-value block on the page — every item later assumes these objects, and the exam will not teach them to you.

AREA counts unit squares. A rectangle 5 units by 3 units holds 3 rows of 5 squares, which is why A = ℓw: you are multiplying rows by columns. Area therefore always carries SQUARE units — cm², in², ft² — and that exponent is a running check on whether you computed the right kind of thing.

TRIANGLE: A = ½bh, printed on the sheet. A triangle is exactly half of a rectangle built on the same base with the same height, which is where the ½ comes from. The h is the PERPENDICULAR distance from the base to the opposite vertex — the straight-down height — not the length of a slanted side. Diagrams routinely label both, and only one of them is the height.

PARALLELOGRAM: A = bh, not printed. Cut the triangle off one end and slide it to the other and you have a rectangle of the same base and height, so a parallelogram has the same area as that rectangle. TRAPEZOID: A = ½(b₁ + b₂)h, not printed — average the two parallel sides, then multiply by the perpendicular distance between them. Both formulas are short, neither is on the sheet, and the trapezoid in particular shows up as the cross-section of a prism.

CIRCLE: A = πr² and C = 2πr, both printed, both written in terms of the RADIUS. Problems state diameters far more often than radii, and every printed formula wants r. Build the habit of writing r = d/2 as your first line, before touching a formula — a diameter used where a radius belongs multiplies an area by 4 and a volume by 8, and the exam puts both of those wrong numbers in the choices on purpose.

VOLUME counts unit cubes, and carries CUBIC units. For any prism or cylinder — any solid with two identical parallel ends and straight sides — the volume is the area of one end times how far it extends: V = (area of the base) × (height). That single sentence covers both printed formulas, because ℓwh is just (ℓw) × h and πr²h is just (πr²) × h. The general form V = Bh is not printed, and it is the one worth carrying, because it handles prisms whose base is a triangle, a trapezoid or a hexagon — cases the sheet cannot help you with at all.

POINTED SOLIDS take a third. A cone is V = (1/3)πr²h and a rectangular pyramid is V = (1/3)ℓwh, both printed. The pattern is worth stating so the sheet stops being a list: if the solid tapers to a single point, its volume is one third of the prism with the same base and the same height. If it has two identical parallel ends, it does not. And the h in both is the perpendicular height from apex to base, never the slanted edge running up the side.

SPHERE: V = (4/3)πr³, printed. A HEMISPHERE is half of that, and no hemisphere formula appears anywhere on the sheet — you halve it yourself. Domes, bowls and rounded silo tops are all hemispheres wearing a costume.

SURFACE AREA is the total area of the outside of a solid, and NOTHING about it is printed. You build it by adding up faces. A closed box is 2(ℓw + ℓh + wh) — three pairs of matching rectangles. A closed cylinder is 2πr² + 2πrh: two circular ends, plus the label, which unrolls into a rectangle whose width is the circumference 2πr and whose height is h. A sphere is 4πr², the one formula here that has to be memorised outright with no derivation to fall back on. Surface area always carries SQUARE units even though it wraps a solid, which is the fastest check that you computed the thing that was asked for.

The reference sheet: the exact boundary

PRINTED, in full. Areas: circle A = πr² with circumference C = 2πr; rectangle A = ℓw; triangle A = ½bh. Right triangles: c² = a² + b², plus the two special triangles — 30-60-90 with sides x, x√3, 2x, and 45-45-90 with sides s, s, s√2. Volumes: rectangular solid ℓwh; right circular cylinder πr²h; sphere (4/3)πr³; right circular cone (1/3)πr²h; rectangular pyramid (1/3)ℓwh. Then three facts in words: there are 360 degrees of arc in a circle, 2π radians of arc in a circle, and 180 degrees in the angles of a triangle.

NOT PRINTED, and this list is the real content of this lesson. Any surface area at all — box, cylinder, cone, sphere, none of them. Trapezoid area ½(b₁ + b₂)h. Parallelogram area bh and rhombus area ½d₁d₂. Equilateral triangle area (s²√3)/4 and regular hexagon area (3√3/2)s². The general prism V = Bh. Arc length and sector area. The equation of a circle, (x − h)² + (y − k)² = r². The distance and midpoint formulas. The space diagonal of a box, √(ℓ² + w² + h²). Every trigonometric ratio. And the scaling rules in the next section — the most-tested unprinted fact in the domain.

How to actually use the sheet. Open it once at the start of the module, look at it, close it. It lives behind a button on a screen, and a student who opens it six times has spent a question's worth of time on lookups. The volume formulas are worth knowing cold regardless: the sheet is insurance against a blank, not a method for working.

The habit the sheet quietly demands: every printed circular formula is written in terms of a RADIUS, and problems overwhelmingly state a DIAMETER. Before you touch a formula, write down r explicitly. This one line prevents the highest-frequency error in the domain, and it costs two seconds.

Scaling: k, k², k³

One sentence carries most of the hard items in this domain. If every linear dimension of a figure is multiplied by k, then every length is multiplied by k, every area — including surface area — is multiplied by k², and every volume is multiplied by k³. It holds for any shape whatsoever, including shapes with no formula, because it is a fact about multiplication rather than a fact about geometry.

It runs backwards, and backwards is where the harder items live. Areas in the ratio 9 : 1 means lengths in the ratio 3 : 1. Volumes in the ratio 8 : 27 means lengths in the ratio 2 : 3. Coming back from an area, take the square root; coming back from a volume, take the cube root. A cube of volume 3,375 cubic centimetres has an edge of ∛3,375 = 15 cm, and no formula beyond that is needed.

The word that licences all of this is SIMILAR. Similar figures have every dimension scaled by the same k. When only one dimension changes, the rule does not apply and you go back to the formula: triple only the radius of a cylinder and the volume is multiplied by 9; triple only the height and it is multiplied by 3; triple both and you get 27. The exam writes all three of those numbers into the same answer set.

Percentages are the most-missed form of this. Increasing every edge of a cube by 20% multiplies the volume by 1.2³ = 1.728 — a 72.8% increase, not 20% and not 60%. Percentage changes compound across dimensions; they do not add. The same fact in reverse: shrinking every dimension by 10% leaves 0.9³ = 0.729 of the volume, a 27.1% loss, not a 30% one.

Unit conversion is scaling wearing a different hat, and it is the most reliable place in the domain to lose a point. 1 foot = 12 inches, so 1 square foot = 144 square inches and 1 cubic foot = 1,728 cubic inches. 1 yard = 3 feet, so 1 cubic yard = 27 cubic feet. 1 metre = 100 centimetres, so 1 cubic metre = 1,000,000 cubic centimetres. Convert the LENGTHS first and compute once, or compute first and convert with the cubed factor — but never apply a linear factor to a volume, and never trust a conversion whose exponent you did not consciously choose.

Mechanism

Why the exponents, and why the one-third

An area is a product of two lengths and a volume is a product of three. That is the entire reason for the exponents, and it is worth holding as an operation rather than a rule: multiply every length by k, and a product of two lengths picks up the factor k twice, while a product of three picks it up three times. Nothing about the shape enters the argument — it works identically for a circle, a trapezoid, a grain silo, or a blob with no formula at all, which is precisely why a scaling item can be answered without knowing what the figure is or what its dimensions are. The units say the same thing out loud: cm² has a 2 in it and cm³ has a 3, and if you can read the exponent off the unit you can read the exponent off the scale factor. This is also the whole reason one cubic foot is 1,728 cubic inches rather than 12 — a unit conversion is a scaling with k = 12, applied to three dimensions at once.

The one-third on cones and pyramids looks arbitrary and is not. A cube can be dissected into exactly three congruent square pyramids: pick one vertex of the cube as the shared apex, and each of the three faces that does not touch that vertex becomes the base of one pyramid. The three fill the cube exactly, with nothing left over, so each holds one third of it — and each has base s² and height s, giving V = ⅓ × base × height. The same relationship survives changing the base shape and the height, which is why the reference sheet carries a third on the cone and a third on the pyramid and no third anywhere else. If a solid tapers to a point, it holds a third of the prism it fits inside. If it has two identical parallel ends, it holds all of it.

And the reason the box and the cylinder are the same formula in two costumes: a prism is a stack of identical copies of its base, and a cylinder is a stack of identical circles. The volume is the area of one layer times how tall the stack is, V = Bh — with ℓwh and πr²h being that sentence with a rectangle and a circle substituted in. The sheet prints the two special cases and leaves the general one to you, which is exactly the gap that a prism with a trapezoidal cross-section is designed to find.

Worked examples

Fully worked — a composite solid, cut into printed pieces

  1. 01Problem: "A grain silo consists of a right circular cylinder 12 feet tall topped by a hemispherical dome. The radius of both the cylinder and the dome is 5 feet. What is the total volume of the silo, in cubic feet, in terms of π?"
  2. 02First move: there is no silo on the reference sheet and there never will be. Cut it into shapes that ARE printed — a cylinder plus half a sphere — price each one, and add.
  3. 03Cylinder: V = πr²h = π(5²)(12) = π(25)(12) = 300π cubic feet. Note that the 12 is the cylinder's own height; the dome sits on top of it and adds nothing to it.
  4. 04Dome: a hemisphere is half a sphere, and no hemisphere formula is printed. Halve the printed one yourself: ½ · (4/3)πr³ = (2/3)π(5³) = (2/3)π(125) = 250π/3 cubic feet.
  5. 05Add, over a common denominator: 300π + 250π/3 = 900π/3 + 250π/3 = 1,150π/3 cubic feet.
  6. 06Sanity-check the size before committing. 1,150/3 ≈ 383.3, so the total is about 383.3π ≈ 1,204 cubic feet. Separately: the cylinder alone is about 942 and the dome about 262, and 942 + 262 ≈ 1,204. ✓ The two routes agree, so the halving and the common denominator both survived.
  7. 07Last check, and the one that decides the point: the question asked for a volume in terms of π, so 1,150π/3 is the answer as written. A choice reading 1,150/3 has silently dropped the π, and a choice reading 1,204 has silently evaluated it — both appear in real answer sets.

One step hidden — similarity does the work, no formula needed

  1. 01Problem: "A conical paper cup holds 96 cubic centimetres when full. A second cup, similar in shape to the first, has a height 1.5 times as great. What is the volume of the second cup, in cubic centimetres?"
  2. 02The word SIMILAR is the licence: every linear dimension of the second cup is 1.5 times the first, not just the height. Without that word you would need the radius separately and the problem would be unsolvable as stated.
  3. 03Volume scales by the cube of the linear factor: k³ = 1.5³ = 3.375. Notice what is not needed — the radius, the height, and the cone formula itself. The ratio is the whole computation.

Two steps hidden — scaling run backwards, from an area to a volume

  1. 01Problem: "A closed rectangular box measures 4 cm by 5 cm by 10 cm. A second box is similar to the first and has a surface area 4 times as great. What is the volume of the second box, in cubic centimetres?"
  2. 02Surface area is an area, so it scales by k². A surface-area ratio of 4 gives k² = 4, so k = 2 — the second box is twice the first in every dimension. No surface-area formula was needed to get there, which is convenient, because none is printed.
  3. 03Volume of the first box: V = ℓwh = 4 × 5 × 10 = 200 cubic centimetres.

Solve alone

  1. 01Problem: "A sphere has the same radius as the base of a right circular cylinder, and the cylinder's height is equal to the sphere's diameter. What fraction of the cylinder's volume does the sphere occupy?"

In your own words

In one sentence: why does multiplying every linear dimension of a solid by k multiply its surface area by k² but its volume by k³ — and why does that same argument explain why one cubic foot is 1,728 cubic inches rather than 12?

Named traps

Diameter dropped into a radius slot
Every circular formula on the reference sheet — area, circumference, cylinder, cone, sphere — is written in terms of r, and problems state diameters far more often than radii. Using d where r belongs multiplies an area by 4 and a volume by 8, and both of those wrong numbers are reliably present in the choices, because they are the two most predictable errors on the whole skill. The fix is mechanical: write r = d/2 as a separate line before any formula gets touched. The mirror image is just as costly — solving correctly for r and reporting it when the question asked for the diameter.
Slant used as height
The h in (1/3)πr²h and (1/3)ℓwh is the PERPENDICULAR height from the apex straight down to the base, not the slanted edge running up the outside. The same error appears one dimension down, in ½bh, where the height must be perpendicular to the base and not the length of a leaning side. Diagrams supply both numbers precisely because only one of them is correct, and the slant is always the larger of the two — so an answer built on it is always too big.
Hunting for a formula that was never printed
Spending thirty seconds scanning the sheet for a trapezoid, a hexagon, a surface area, a hemisphere, or a composite solid, all of which are absent. Nothing about surface area is printed anywhere on the sheet, and there is no general prism formula on it either. Knowing the boundary converts that lost time into an immediate decision: build the formula from faces, or cut the solid into printed pieces.
Linear factor applied to an area or a volume
Doubling a length and doubling the area, or converting cubic feet to cubic inches by multiplying by 12. Areas take k² and volumes take k³, always, and the unit itself is announcing the exponent: ft² takes 12², ft³ takes 12³. This is the same error whether it arrives dressed as similarity, as a unit conversion, or as a percentage — and as a percentage it is worst, because +20% on each edge is +72.8% on the volume, not +20% and not +60%.
Cube law applied to a partial scaling
k³ requires SIMILAR figures — every dimension multiplied by the same k. When a problem changes only the radius, or only the height, the rule does not apply and you go back to the formula. Tripling only the radius of a cylinder multiplies its volume by 9; tripling only the height multiplies it by 3; tripling both multiplies it by 27. The exam puts all three in the same answer set and lets the word "similar" — present or absent — decide which is right.
The wrong quantity answered
The geometry is finished and every line of it is correct, and the number reported is r when the question asked for d, or r² when it asked for r, or the surface area when it asked for the volume, or the total volume when it asked for the volume added. This is the characteristic error at the top of the score range, and the unit is the cheapest detector available: an answer in square centimetres cannot be a volume, and an answer in cubic centimetres cannot be a surface area. Before selecting, read the last clause of the question again and name the unit it wants.

The 800-level margin

At this point every printed formula is automatic and k³ is second nature. Six things are still moving points: hollowed solids, cross-sections, scaling run backwards, inscribed figures, the off-sheet formulas nobody warns you about, and a short list of execution errors that survive knowing all of the above.

HOLLOWED SOLIDS AND SUBTRACTION. A composite is not always an addition. A pipe is a cylinder minus a cylinder; a washer is a disc minus a disc; a box with a hole drilled through it is a box minus a cylinder. The efficient form for anything with a uniform cross-section is V = (cross-sectional area) × (length), so a pipe of outer radius R and inner radius r and length L is π(R² − r²)L in one line. The error the item is built to catch is subtracting the radii before squaring them: (R − r)² is not R² − r², and for R = 4, r = 3 the two differ by a factor of seven. Square first, subtract second — always.

CROSS-SECTIONS AND NETS. Two presentations recur. A NET is a solid unfolded flat: its area IS the surface area of the solid, so a net question is an addition problem, not a spatial one — count the faces on the page and add them. A CROSS-SECTION is a slice: the cross-section of a cylinder parallel to its base is a circle of radius r, and a slice through the axis is a rectangle 2r by h. Any solid with a constant cross-section is (that area) × (length), which is how you handle a prism whose base is a trapezoid, a hexagon, or an L-shape the sheet has never heard of.

SCALING BACKWARDS, AND NON-INTEGER FACTORS. Given a volume ratio, take the cube root to recover the linear ratio; given an area ratio, take the square root. Volumes of 54 and 128 are in the ratio 27 : 64, so the lengths are in the ratio 3 : 4. The same machinery answers questions with no ratio in sight: a cube of volume 3,375 cm³ has edge ∛3,375 = 15 cm and surface area 6(15²) = 1,350 cm². And the percentage form once more, because it is where 1500-level students actually lose this: +20% per edge is ×1.728 on volume, −10% per edge is ×0.729, and "the volume increased by 33.1%" means each edge grew by 10%, since 1.1³ = 1.331.

INSCRIBED FIGURES AND THE SPACE DIAGONAL. A sphere inscribed in a cube touches all six faces, so the sphere's DIAMETER equals the cube's edge — and the sphere fills πs³/6 ÷ s³ ≈ 52% of the cube, a useful sanity bound. A cube inscribed in a sphere touches at its corners, so the cube's SPACE DIAGONAL equals the sphere's diameter, and the space diagonal of a box is √(ℓ² + w² + h²) — Pythagoras applied twice, once across the base and once up. For a cube that is s√3. None of this is on the sheet, and "inscribed" is the word that tells you which of the two relationships is in play.

THE OFF-SHEET FORMULAS WORTH CARRYING. Trapezoid ½(b₁ + b₂)h. Parallelogram bh. Rhombus or kite ½d₁d₂. Equilateral triangle (s²√3)/4, which falls straight out of a 30-60-90 split: the height of an equilateral triangle of side s is (s√3)/2, so the area is ½ · s · (s√3)/2. Regular hexagon (3√3/2)s², because a regular hexagon is six equilateral triangles meeting at the centre. Closed box surface area 2(ℓw + ℓh + wh). Closed cylinder 2πr² + 2πrh. Sphere surface area 4πr². Sector area (θ/360)πr² and arc length (θ/360)(2πr) in degrees, or ½r²θ and rθ in radians. Nine formulas, none printed, and between them they cover every figure the exam uses that the sheet omits.

DENSITY, RATE AND CAPACITY, which is how hard area-and-volume items usually disguise themselves. Mass = density × volume, so a solid's weight is a geometry problem with one multiplication attached. A tank filling at a constant rate is volume ÷ rate = time, and the quantity that matters is almost always the volume ADDED, not the total volume — a level rising from 12 inches to 20 inches is eight inches of new water, and an answer computed from twenty inches is arithmetically flawless and wrong. Read whether the question is about a state or about a change before computing anything.

EXECUTION ERRORS, which is where the last few points genuinely sit. Reporting r² because the equation ended there. Forgetting to halve the sphere for a hemisphere. Adding the dome's radius to the cylinder's height when the diagram never said to. Leaving π symbolic when the question asked for a decimal, or evaluating it when the question said "in terms of π". Mixing units inside a single composite — a length in feet multiplied by a radius in inches. Answering in square units what was asked in cubic. And a format rule that is College Board's own: π cannot be typed into a student-produced response field, so a grid-in on this skill is always designed so that π cancels or so that a decimal is wanted — and if you are entering a decimal, the field must be filled, since a truncated repeating decimal is scored as wrong.

Retrieval — with feedback on every choice

Question 1
Medium

A right circular cylinder has a volume of 245π cubic inches and a height of 5 inches. What is the diameter of the cylinder's base, in inches?

Reference — not a study method, a lookup
AREA AND VOLUME — reference card
ON THE SHEET: circle A = πr² and C = 2πr; rectangle A = lw; triangle A = ½bh.
ON THE SHEET: c² = a² + b²; 30-60-90 is x, x√3, 2x; 45-45-90 is s, s, s√2.
ON THE SHEET: box lwh; cylinder πr²h; sphere (4/3)πr³; cone (1/3)πr²h; pyramid (1/3)lwh.
NOT ON THE SHEET: any surface area at all. Build it by adding faces.
NOT ON THE SHEET: trapezoid ½(b1 + b2)h; parallelogram bh; rhombus ½d1d2.
NOT ON THE SHEET: equilateral triangle (s²√3)/4; regular hexagon (3√3/2)s².
NOT ON THE SHEET: V = Bh for a general prism; sector and arc; circle equation; space diagonal.
Worth memorising: closed box SA = 2(lw + lh + wh); closed cylinder SA = 2πr² + 2πrh; sphere SA = 4πr².
Prism or cylinder: V = (base area) × height. Tapers to a point: multiply that by 1/3.
Hemisphere = half a sphere. No hemisphere formula is printed — halve it yourself.
Every printed circular formula wants r. Write r = d/2 on its own line before anything else.
The h in a cone, pyramid or triangle is PERPENDICULAR height, never the slant.
SCALING: multiply every length by k -> lengths ×k, areas and surface areas ×k², volumes ×k³.
Backwards: area ratio -> square root; volume ratio -> cube root. Volumes 8:27 means lengths 2:3.
Only for SIMILAR figures. One dimension tripled on a cylinder is ×9 (radius) or ×3 (height), not ×27.
Percents compound: +20% per edge is ×1.728 (+72.8%); -10% per edge is ×0.729 (-27.1%).
Units are scaling: 1 ft³ = 1,728 in³; 1 yd³ = 27 ft³; 1 m³ = 1,000,000 cm³.
Hollow solids: square first, subtract second. π(R² - r²)L, never π(R - r)²L.
Uniform cross-section: V = (cross-sectional area) × length. A net's area IS the surface area.
Rate items: volume ADDED ÷ rate = time. Ask whether the question is about a state or a change.
Before answering: is it a length, an area or a volume? The unit's exponent is the check.

Every item on this page is Meridian-original, written to match the Digital SAT's format and difficulty — it is not a real SAT question. The only source that matches the live test exactly is College Board's own Bluebook and Question Bank.

Question 1Medium

A right circular cylinder has a volume of 245π cubic inches and a height of 5 inches. What is the diameter of the cylinder's base, in inches?

  • A7

    7 is the radius, and the question asked for the diameter. Every step before this one is correct; the error is in the last line. The reference sheet's formulas are all written in r, so an answer that comes out of one of them is a radius until you deliberately convert it.

  • 14

    Correct. V = πr²h gives 245π = πr²(5), so r² = 245/5 = 49 and r = 7. The question asked for the diameter, which is 2r = 14. Check by substituting back: π(7²)(5) = π(49)(5) = 245π ✓.

  • C49

    49 is r², reported before the square root was taken. The equation ended at r² = 49 and the working stopped one line early — a pure execution error, and the reason it is worth writing the target quantity down before starting.

  • D98

    This doubles r² instead of doubling r: 2 × 49 = 98. The doubling that converts a radius to a diameter has to happen after the square root, not before it, because squaring and doubling do not commute — 2(7²) = 98 while (2 × 7)² = 196, and neither is the diameter.

Traps tested: Radius reported for diameter · Unsimplified intermediate reported · Doubling applied to squared value

Question 2Hard

A cube-shaped shipping crate is redesigned so that each edge is 20% longer than the corresponding edge of the original crate. The volume of the redesigned crate is what percent greater than the volume of the original crate?

  • A20%

    This applies the percentage as though volume were a length. Each of the three dimensions grew by 20%, and volume is the product of all three, so the factor 1.2 is applied three times — the volume cannot possibly grow by the same percentage as a single edge.

  • B44%

    44% is the increase in AREA, not volume: 1.2² = 1.44. This is the correct answer to a different question — what happens to one face of the crate, or to its total surface area. Volume needs one more dimension, and therefore one more factor of 1.2.

  • C60%

    This multiplies 20% by the three dimensions and adds the results. Percentage changes across dimensions compound rather than add: the second 20% is charged on a length that the first 20% already grew. The same error turns three successive 10% raises into 30% when the true figure is 33.1%.

  • 72.8%

    Correct. Every linear dimension is multiplied by k = 1.2, so the volume is multiplied by k³ = 1.2³ = 1.728 — an increase of 0.728, or 72.8%. Check with numbers: an edge of 10 gives a volume of 1,000; an edge of 12 gives 1,728; and 728/1,000 = 72.8% ✓.

Traps tested: Linear scaling applied to volume · Area factor used for volume · Percent multiplied not compounded

Question 3Hard

A closed cylindrical can has a radius of 3 centimeters and a height of 8 centimeters. What is the total surface area of the can, in square centimeters?

  • A48π

    48π is the lateral surface alone, 2πrh — the label wrapped around the side, with both circular ends left off. The word "closed" in the problem is what rules this out: a closed can has a top and a bottom, and each contributes πr² = 9π.

  • B57π

    This counts the side and exactly one circular end: 48π + 9π = 57π. A closed cylinder has two ends, so the circular term carries a factor of 2. The error is easy to make precisely because no surface-area formula is printed on the reference sheet — you are assembling it from faces, and one face went missing.

  • 66π

    Correct. No surface-area formula is on the reference sheet, so build it: two circular ends give 2πr² = 2π(9) = 18π, and the side unrolls into a rectangle of width 2πr and height h, giving 2πrh = 2π(3)(8) = 48π. Total: 18π + 48π = 66π square centimeters. ✓ Sanity check on units — the answer is an area, so square centimeters is right.

  • D72π

    72π is the VOLUME, πr²h = π(9)(8), computed with the one formula that is printed on the sheet. It is the reflexive answer for exactly that reason. The unit is the detector: a volume comes out in cubic centimeters and the question asked for square centimeters.

Traps tested: Closed solid treated as open · Faces undercounted · Volume computed for surface area

Question 4Hard

A rectangular aquarium with a base measuring 30 inches by 18 inches contains water to a depth of 12 inches. Water is then added at a constant rate of 90 cubic inches per minute. How many minutes does it take for the depth of the water to reach 20 inches?

  • 48

    Correct. The question is about a CHANGE, so the depth that matters is 20 − 12 = 8 inches. Volume added = 30 × 18 × 8 = 540 × 8 = 4,320 cubic inches, and 4,320 ÷ 90 = 48 minutes. Check: after 48 minutes the tank holds 6,480 + 4,320 = 10,800 cubic inches, and 10,800 ÷ 540 = 20 inches of depth ✓.

  • B72

    This is the time to add the water already in the tank: 30 × 18 × 12 = 6,480, and 6,480 ÷ 90 = 72. It computes the correct kind of quantity for the wrong interval — the twelve inches that were there before the tap was opened.

  • C120

    This uses the full final volume, 30 × 18 × 20 = 10,800, divided by 90. It answers how long the tank would take to fill to 20 inches from empty, but the tank started with 12 inches already in it. Deciding whether a question is about a state or about a change, before computing, is what separates these two answers.

  • D4,320

    4,320 is the volume of water added, in cubic inches — the correct intermediate quantity, reported instead of the time. The division by the rate is the step that converts cubic inches into minutes, and the unit named in the question is the check that catches it.

Traps tested: Wrong interval computed · Total used instead of change · Wrong quantity answered

Question 5Hardest on the test

A straight section of metal pipe is 10 feet long. Its outer radius is 4 inches and the pipe wall has a uniform thickness of 1 inch. What is the volume of metal in this section of pipe, in cubic inches? (1 foot = 12 inches.)

  • A70π

    The geometry here is right and the units are not: this uses a length of 10 rather than converting 10 feet into 120 inches. Every other measurement in the problem is in inches, and the answer was asked for in cubic inches, so the length has to be converted before it is multiplied by anything.

  • B120π

    This subtracts the radii before squaring them: π(4 − 3)²(120) = 120π. But (R − r)² is not R² − r². Here (4 − 3)² = 1 while 4² − 3² = 7 — a factor of seven apart. The rule is square first, subtract second, and this distractor exists on every hollowed-solid item ever written.

  • 840π

    Correct. The metal is a cylinder with a cylinder removed. Outer radius 4, wall thickness 1, so inner radius 3. Length 10 ft = 120 in. Volume = π(4²)(120) − π(3²)(120) = π(16 − 9)(120) = π(7)(120) = 840π cubic inches. Check by the cross-section route: the annular face has area π(16 − 9) = 7π square inches, and 7π × 120 = 840π ✓.

  • D1,920π

    This is the solid cylinder, π(4²)(120), with nothing removed — the hole the pipe is defined by was never subtracted. It is the largest of the four choices, and a pipe must contain less metal than a solid rod of the same outside dimensions, which rules it out before any arithmetic.

Traps tested: Unit mismatch in composite · Difference of radii squared · Hollow region not subtracted

Question 6Hardest on the test

A concrete drainage channel is a solid right prism 9 centimeters long. Its cross section is a trapezoid whose parallel sides measure 5 centimeters and 11 centimeters and whose height — the perpendicular distance between those parallel sides — is 4 centimeters. What is the volume of the prism, in cubic centimeters?

  • A96

    This divides the correct volume by 3, applying the one-third that belongs to cones and pyramids. The third is earned by tapering to a point: a solid with two identical parallel ends holds the full base-times-height, and a prism has two identical parallel ends by definition.

  • 288

    Correct. Neither the trapezoid's area nor the general prism formula is on the reference sheet, so both come from you. Trapezoid area = ½(b₁ + b₂)h = ½(5 + 11)(4) = ½(16)(4) = 32 square centimeters. Volume of any prism = (base area) × (length) = 32 × 9 = 288 cubic centimeters. Check by decomposition: the trapezoid splits into a 5-by-4 rectangle (20) plus a triangle with base 6 and height 4 (12), and 20 + 12 = 32 ✓.

  • C396

    This treats the cross section as an 11-by-4 rectangle: 44 × 9 = 396. Using the longer parallel side alone overcounts, since the trapezoid narrows to 5 at the other edge. The formula averages the two parallel sides for exactly this reason — ½(5 + 11) = 8 is the width of the rectangle with the same area.

  • D576

    This drops the ½ from the trapezoid formula: (5 + 11)(4)(9) = 576, exactly twice the correct answer. The factor is the whole point of the formula — you are averaging the two parallel sides, not adding them — and it is the most common casualty of a formula that has to be recalled rather than read off the sheet.

Traps tested: One third applied to a prism · Single base used · Trapezoid half omitted

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