AutoCAD Quiz: Basic Modify Commands
10 questions · exam conditions
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Basic Modify CommandsQuestion 1 of 10

A line segment has endpoints (2,1)(2,1) and (5,1)(5,1). It is mirrored about the vertical line defined by two points whose x-coordinate is 44. At the erase-source prompt, the user selects No.

Which line segments remain after MIRROR is completed?

The original segment and a segment from (6,1)(6,1) to (3,1)(3,1)
Only a mirrored segment from (6,1)(6,1) to (3,1)(3,1)
The original segment and a segment from (2,−1)(2,-1) to (5,−1)(5,-1)
Only the original segment from (2,1)(2,1) to (5,1)(5,1)
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AutoCAD Quiz

AutoCAD Quiz: Basic Modify Commands

Practice Basic Modify Commands in AutoCAD with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Basic Modify Commands, giving you a quick way to practice the rules, question types, and explanations that matter most for AutoCAD.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A line segment has endpoints (2,1)(2,1) and (5,1)(5,1). It is mirrored about the vertical line defined by two points whose x-coordinate is 44. At the erase-source prompt, the user selects No.

Which line segments remain after MIRROR is completed?

  1. The original segment and a segment from (6,1)(6,1) to (3,1)(3,1) (correct answer)
  2. Only a mirrored segment from (6,1)(6,1) to (3,1)(3,1)
  3. The original segment and a segment from (2,−1)(2,-1) to (5,−1)(5,-1)
  4. Only the original segment from (2,1)(2,1) to (5,1)(5,1)
Explanation: Whenever you see a MIRROR question in AutoCAD, focus on two things: how the reflection geometry works, and what the "erase source objects" prompt controls. The mirror line here is vertical, passing through x=4x = 4. To reflect a point across a vertical line x=ax = a, you use the formula: x′=2a−xx' = 2a - x. For endpoint (2,1)(2, 1): x′=2(4)−2=6x' = 2(4) - 2 = 6, giving (6,1)(6, 1). For endpoint (5,1)(5, 1): x′=2(4)−5=3x' = 2(4) - 5 = 3, giving (3,1)(3, 1). So the mirrored segment runs from (6,1)(6, 1) to (3,1)(3, 1). Because the user answered No at the "erase source objects" prompt, the original segment is kept, not deleted. That means both segments exist — the original from (2,1)(2,1) to (5,1)(5,1) and the mirror from (6,1)(6,1) to (3,1)(3,1). This confirms A is correct. Choice B describes the correct mirrored segment but incorrectly implies the source was erased — that only happens when the user selects Yes at the erase prompt. Choice C reflects the segment across a horizontal axis instead of a vertical one, confusing a reflection over y=1y = 1 with the correct operation; the y-coordinates would change, but here y=1y = 1 stays unchanged. Choice D ignores the mirror entirely, as if no operation was performed. A quick tip: always separate the two parts of a MIRROR question — where the reflection lands (geometry) and what survives (erase source: Yes removes original, No keeps it). These are two independent decisions that frequently appear together on the exam.

Question 2

A rectangular object is oriented at 1515 degrees. The user runs ROTATE, selects an external base point, chooses the Copy option, and enters a positive rotation angle of 4040 degrees.

Which orientations exist when the command finishes?

  1. The original remains at 1515 degrees, and a copy is created at 4040 degrees.
  2. The original is removed, and a single rotated object remains at 5555 degrees.
  3. The original remains at 1515 degrees, and a copy is created at 5555 degrees. (correct answer)
  4. The original remains at 1515 degrees, and a copy is created at −25-25 degrees.
Explanation: When working with AutoCAD's ROTATE command, the key concept to understand is how rotation angles accumulate and what the Copy option actually does to your original geometry. The ROTATE command applies a rotation relative to the object's current orientation. So if your rectangle already sits at 15°15° and you apply a rotation of +40°+40°, the new orientation is 15°+40°=55°15° + 40° = 55°. The Copy option works like a modifier: instead of moving the original to the new angle, AutoCAD preserves the original in place and creates a rotated duplicate at the calculated angle. This makes C the correct answer — the original stays at 15°15° and a copy lands at 55°55°. Answer A incorrectly places the copy at 40°40°, which would only be true if the object started at 0°0°. This is the most common trap: confusing the rotation amount entered with the final absolute orientation. The 40°40° is the delta, not the destination. Answer B describes standard ROTATE behavior without the Copy option — the original is replaced by the rotated result. Since Copy was explicitly selected, nothing is removed; you end with two objects, not one. Answer D suggests 15°−40°=−25°15° - 40° = -25°, implying the rotation was subtracted rather than added. A positive angle in AutoCAD rotates counterclockwise, so +40°+40° adds to the current orientation, not subtracts from it. As a study tip, always mentally separate how much you're rotating from where the object ends up. Final angle = starting angle + rotation input. And remember: Copy in ROTATE never destroys the original.

Question 3

A circle has center (8,3)(8,3) and radius 22. It is mirrored about the line defined by (0,0)(0,0) and (4,4)(4,4). The original circle is erased.

Which center and radius describe the resulting circle?

  1. Center (−3,8)(-3,8) with radius 22
  2. Center (3,8)(3,8) with radius 22 (correct answer)
  3. Center (8,−3)(8,-3) with radius 22
  4. Center (3,8)(3,8) with radius 44
Explanation: When AutoCAD mirrors an object, it reflects every point across the mirror line — the shape and size stay identical, only the position changes. The mirror line here passes through (0,0)(0,0) and (4,4)(4,4), which is the line y=xy = x. Reflecting across y=xy = x is a specific transformation with a simple rule: swap the x- and y-coordinates of every point. The original circle has center (8,3)(8, 3) and radius 22. Applying the y=xy = x reflection rule, you swap the coordinates: (8,3)→(3,8)(8, 3) \rightarrow (3, 8). The radius is unaffected by mirroring — it stays 22. So the resulting circle has center (3,8)(3, 8) with radius 22, confirming B is correct. Choice A gives center (−3,8)(-3, 8), which incorrectly negates the x-coordinate instead of simply swapping — that would be a reflection across the y-axis, not y=xy = x. Choice C gives center (8,−3)(8, -3), which negates the y-coordinate — that's a reflection across the x-axis, another common mix-up. Choice D gets the center right at (3,8)(3, 8) but doubles the radius to 44, which is a trap: mirroring never changes size or scale, only position and orientation. A reliable tip: whenever the mirror line is y=xy = x, just swap coordinates — that's all there is to it. And always remember that MIRROR in AutoCAD is a rigid transformation, meaning distances, angles, and dimensions are fully preserved. Only location changes.

Question 4

An object's reference point is at (30,20)(30,20). During MOVE, the user selects the object, enters (12,−5)(12,-5) at the base-point prompt, and then presses Enter at the second-point prompt to accept the use-first-point-as-displacement option.

Where will the object's reference point be located?

  1. (12,−5)(12,-5)
  2. (42,15)(42,15) (correct answer)
  3. (18,25)(18,25)
  4. (30,20)(30,20)
Explanation: When AutoCAD's MOVE command asks for a "base point," it's really asking: how far and in which direction should the object travel? If you press Enter at the second-point prompt instead of clicking a new location, AutoCAD treats your base-point entry as a displacement vector — a direct instruction to shift the object by that exact amount. This is the critical concept being tested here. Since the displacement entered is (12,−5)(12, -5), AutoCAD adds those values directly to the object's current position: 30+12=4230 + 12 = 42 and 20+(−5)=1520 + (-5) = 15. The reference point lands at (42,15)(42, 15), confirming B is correct. A (12,−5)(12, -5) is a trap for students who think the object moves to the typed coordinate rather than by it. That would only be true if you used a two-point move where (12,−5)(12, -5) was your second point and the origin (0,0)(0,0) was your base point — a very different scenario. C (18,25)(18, 25) comes from subtracting instead of adding: 30−12=1830 - 12 = 18 and 20+5=2520 + 5 = 25. It flips the signs on both components — a careless arithmetic error. D (30,20)(30, 20) represents no movement at all, which would happen only if you pressed Escape or entered a zero displacement. Your study tip: whenever you see "use first point as displacement" in a MOVE question, remember you're adding the typed coordinate to the original position — it's a delta, not a destination.

Question 5

A circle is centered at (2,2)(2,2). The user starts COPY, selects the circle, uses its center as the base point, and—without ending the command—specifies displacement vectors of (5,0)(5,0), (0,5)(0,5), and (−5,0)(-5,0). The user then presses Enter.

Which set lists all circle centers after the command?

  1. (2,2)(2,2), (7,2)(7,2), (2,7)(2,7), and (−3,2)(-3,2) (correct answer)
  2. (2,2)(2,2), (7,2)(7,2), (7,7)(7,7), and (2,7)(2,7)
  3. (2,2)(2,2), (5,0)(5,0), (0,5)(0,5), and (−5,0)(-5,0)
  4. (7,2)(7,2), (7,7)(7,7), (2,7)(2,7), and (−3,2)(-3,2)
Explanation: When using AutoCAD's COPY command in Multiple mode, each displacement vector you specify is applied independently to the original object — not chained from the previous copy. This is a critical distinction that this question directly tests. Here, the original circle is centered at (2,2)(2,2), and you use its center as the base point. Each displacement vector shifts from that same base point:
  • Vector (5,0)(5,0):  (2+5,  2+0)=(7,2)\,(2+5,\;2+0) = (7,2)
  • Vector (0,5)(0,5):  (2+0,  2+5)=(2,7)\,(2+0,\;2+5) = (2,7)
  • Vector (−5,0)(-5,0):  (2+(−5),  2+0)=(−3,2)\,(2+(-5),\;2+0) = (-3,2)
The original circle at (2,2)(2,2) remains untouched. So all four centers are (2,2)(2,2), (7,2)(7,2), (2,7)(2,7), and (−3,2)(-3,2) — confirming A is correct. B is wrong because it chains the vectors sequentially (as if each copy becomes the new base), producing (7,7)(7,7) — but COPY always references the original base point. C mistakes the raw displacement vectors for absolute coordinates, ignoring the (2,2)(2,2) starting position entirely. D is almost correct in its copy positions but omits the original circle at (2,2)(2,2), which is never deleted by the COPY command. A handy rule to remember: in AutoCAD's COPY command, every second point you pick is measured from the same original base point, not from the last copy placed. When you see multi-copy questions, always add each vector to the original coordinates separately.

Question 6

A title consisting of text and surrounding linework is mirrored to the opposite side of a layout. MIRRTEXT is set to 00, and the user selects Yes at the erase-source prompt.

What is the expected result?

  1. The linework is rotated instead of reflected, the text remains readable, and the source objects are removed.
  2. The linework and text locations are mirrored, the characters are reversed, and the source objects are removed.
  3. Only the linework is mirrored, the text remains at its original location, and all source objects remain.
  4. The linework and text locations are mirrored, the characters remain readable, and the source objects are removed. (correct answer)
Explanation: When working with the MIRROR command in AutoCAD, you need to track two independent behaviors: what happens to geometry placement and what happens to text readability. The system variable MIRRTEXT controls only the text characters — it has no effect on where objects end up after mirroring. With MIRRTEXT = 0, AutoCAD suppresses the mirroring of text characters, meaning the letters remain upright and readable after the operation. The positions of all selected objects — including text insertion points and all linework — are still reflected across the mirror axis regardless of MIRRTEXT. Finally, selecting Yes at the erase-source prompt tells AutoCAD to delete the original objects, effectively moving rather than copying the title to its new location. This makes D the correct answer: linework and text locations are mirrored, text remains readable, and source objects are removed. Choice A is wrong because MIRROR always reflects geometry — it never rotates objects. Choice B describes what would happen if MIRRTEXT were set to 1 (its default in older releases), which causes characters to flip and become a mirror image. The question explicitly states MIRRTEXT = 0, so reversed characters is the opposite of what occurs. Choice C incorrectly claims only linework is mirrored and that source objects remain — the mirror operation applies to all selected objects together, and choosing Yes at the erase prompt removes the originals. A useful memory hook: think of MIRRTEXT as a readability lock — when it's 0, the lock is on and text stays legible, but the lock doesn't stop the text from moving to its mirrored position.

Question 7

An endpoint initially lies at (4,1)(4,1). The object is first moved by the displacement vector (3,−2)(3,-2). It is then rotated 9090 degrees counterclockwise about the origin.

Where is the endpoint after both modifications?

  1. (7,−1)(7,-1)
  2. (−1,7)(-1,7)
  3. (1,7)(1,7) (correct answer)
  4. (−2,4)(-2,4)
Explanation: When a question combines a translation and a rotation, always apply the transformations in sequence — order matters. Here, you first translate, then rotate. Mixing up the order or the rotation formula are the two most common traps. Start with the endpoint at (4,1)(4, 1). Applying the displacement vector (3,−2)(3, -2) means adding component-by-component: (4+3, 1+(−2))=(7,−1)(4+3,\ 1+(-2)) = (7, -1). That's your position after the move. Next, rotate (7,−1)(7, -1) by 90°90° counterclockwise about the origin. The standard rotation rule is: (x,y)→(−y,x)(x, y) \rightarrow (-y, x). Plugging in: (−(−1), 7)=(1,7)(-(-1),\ 7) = (1, 7). That confirms C is correct. As for the wrong answers — A (7,−1)(7, -1) is the result of the translation only, stopping before applying the rotation. It's a classic "forgot the second step" mistake. B (−1,7)(-1, 7) comes from applying the rotation formula backwards, using (y,−x)(y, -x) instead of (−y,x)(-y, x) — a common mix-up between clockwise and counterclockwise rules. D (−2,4)(-2, 4) appears to come from rotating the original point (4,1)(4, 1) without applying the translation first, which reverses the intended order of operations. A handy tip: memorize the two rotation rules cold. Counterclockwise 90°90°: (x,y)→(−y,x)(x, y) \rightarrow (-y, x). Clockwise 90°90°: (x,y)→(y,−x)(x, y) \rightarrow (y, -x). On transformation problems, always track your coordinates step-by-step and don't skip intermediate results.

Question 8

A line is currently oriented at 2828 degrees from the positive horizontal axis. It must be reoriented to exactly 7373 degrees by using the Reference option of ROTATE.

Which sequence of angle entries correctly uses the Reference option?

  1. Specify 2828 degrees as the reference angle and 7373 degrees as the new angle. (correct answer)
  2. Specify 2828 degrees as the reference angle and 4545 degrees as the new angle.
  3. Specify 00 degrees as the reference angle and 7373 degrees as the new angle.
  4. Specify 7373 degrees as the reference angle and 2828 degrees as the new angle.
Explanation: When using the ROTATE command's Reference option in AutoCAD, you're telling the software two things: where the object currently is (the reference angle) and where you want it to go (the new angle). AutoCAD calculates the difference automatically and applies exactly the right rotation — no mental math required on your part. In this problem, the line sits at 28°28° and needs to land at 73°73°. Using the Reference option, you enter 28°28° as the reference angle (the current orientation) and 73°73° as the new angle (the target orientation). AutoCAD internally computes 73−28=45°73 - 28 = 45° of rotation and applies it. This is answer A, and it's correct precisely because you're describing the situation in terms the command is designed to accept — current position, then destination. Answer B is the most tempting trap. Entering 45°45° as the new angle seems logical if you've already done the subtraction yourself, but that defeats the purpose of the Reference option. You'd be mixing workflows — using a reference angle with a delta value — and the result would be wrong. Answer C uses 0°0° as the reference, which would only make sense if the line were already aligned to the horizontal axis; it ignores the actual current orientation entirely. Answer D reverses the inputs entirely — specifying the target as the reference and the current as the new angle — which would rotate the line in the wrong direction to an incorrect final position. A helpful memory anchor: Reference = "From → To." Always enter where the object is, then where it needs to be.

Question 9

A block representing a receptacle must be moved so that its insertion point coincides exactly with the endpoint of an existing wall line. The block's visible geometry is not symmetric about its insertion point.

Which MOVE workflow most reliably produces the required placement?

  1. Use a visible block corner for the base point and the Midpoint object snap on the wall line.
  2. Use the geometric center of the block for the base point and the Endpoint object snap for the second point.
  3. Use the Insertion object snap for the base point and the Nearest object snap anywhere on the wall line.
  4. Use the Insertion object snap for the base point and the Endpoint object snap for the second point. (correct answer)
Explanation: When moving a block to a precise location in AutoCAD, you need to think carefully about two separate snap decisions: where you're grabbing the block from, and where you're placing it. Both must be exact for the result to be correct. Since the goal is to align the block's insertion point with the wall line's endpoint, the most direct workflow is to snap directly to each of those exact geometric locations — which is exactly what D does. The Insertion object snap locks your base point to the block's true insertion point (regardless of where the visible geometry sits), and the Endpoint object snap locks the destination to the precise end of the wall line. No estimation, no offset, no ambiguity. Choice A fails on both counts: a visible corner of the block is not the insertion point, and the Midpoint snap targets the middle of the wall line, not its endpoint — so the block lands in the wrong position entirely. Choice B uses the geometric center as the base point, which is also not the insertion point. Because the passage explicitly tells you the geometry is asymmetric about the insertion point, the center and the insertion point are in different locations, meaning the block will be misplaced. Choice C introduces the Nearest snap for the second point, which snaps to the closest point on the wall line's geometry rather than its endpoint — so while your base point is correct, your destination is approximate and unrepeatable. The key rule to memorize: when a question specifies "insertion point," always use the Insertion object snap — don't substitute visible geometry or calculated centers as a workaround.

Question 10

A line has endpoints at (1,2)(1,2) and (6,4)(6,4). During the MOVE command, the user specifies (4,3)(4,3) as the base point and (11,−2)(11,-2) as the second point.

What are the endpoints of the line after the MOVE command is completed?

  1. (8,−3)(8,-3) and (13,−1)(13,-1) (correct answer)
  2. (−6,7)(-6,7) and (−1,9)(-1,9)
  3. (12,0)(12,0) and (17,2)(17,2)
  4. (−3,−1)(-3,-1) and (2,1)(2,1)
Explanation: When you use the MOVE command in AutoCAD, the software doesn't care about the absolute position of your base point — it only cares about the displacement vector created between your base point and second point. That displacement gets applied equally to every point in the selected object. Here's how to calculate it: subtract the base point from the second point to find the translation vector. With a base point of (4,3)(4,3) and a second point of (11,−2)(11,-2), the displacement is (11−4,−2−3)=(7,−5)(11-4, -2-3) = (7,-5). Now apply that shift to both endpoints of the line. Starting endpoint (1,2)(1,2) becomes (1+7,2−5)=(8,−3)(1+7, 2-5) = (8,-3). Ending endpoint (6,4)(6,4) becomes (6+7,4−5)=(13,−1)(6+7, 4-5) = (13,-1). That gives you Answer A: (8,−3)(8,-3) and (13,−1)(13,-1) — the correct answer. Now let's see where the wrong answers come from. Answer B results from subtracting the displacement instead of adding it — a sign-flip error. Answer C uses (11,−2)(11,-2) as a direct translation offset rather than computing the vector from base to second point, essentially ignoring the base point entirely. Answer D appears to subtract the second point from the base point (reversing the order), which gives (4−11,3−(−2))=(−7,5)(4-11, 3-(-2)) = (-7,5) applied to the original endpoints — another sign/direction mistake. The key study tip: always compute displacement as second point−base point\text{second point} - \text{base point}, then add that result to every original coordinate. The base point is just a reference, not a destination.