(Originally published on 2026-01-23; last reviewed on 2026-01-23; last updated on 2026-01-23)

This article contains twenty-four rules questions featuring cards from Magic‘s newest set Lorwyn Eclipsed.

The first couple of questions focus on the set’s new and returning mechanics, similar to the update quizzes of several judge organizations; the rest focus on individual cards.

I usually recommend reading up on the Comprehensive Rules changes for a set before taking the quiz, but that’s not needed this time; a simplified overview like WotC’s mechanics article is more than sufficient.

Quiz style

This quiz follows the style most judge organizations use for rules questions on their update and certification tests.

• Each question lists multiple possible answers, usually five.
• For most questions, only one of those answers is correct.
• If more than one answer can be correct, the question will end with the phrase “Select all that apply.”
  • The number of correct answers can vary from one to all of them.
  • For the final score, such a question is all-or-nothing: selecting only some of the correct answers, but not all of them, or selecting an incorrect answer in addition to the correct ones, means that the question is failed.
• The name of the active player always starts with an “A” and the name of the nonactive player always starts with an “N.” In scenarios describing Two-Headed Giant games, the names of the two players on the active team start with an “A” and the names of the two players on the nonactive team start with an “N.”
• Each player has all the resources needed to perform the described actions, such as mana, cards in hand, or permanents to sacrifice. Usually, those resources are not described in detail unless they matter for the scenario.
• The scenario description contains all the relevant information about the game state. If something is not mentioned, then it’s not part of the game state or it doesn’t influence the correct answer.
• All game actions taken by the players are legal. The only exception are questions where one or more of the potential answers claim the opposite; usually, those answers start with “This can’t happen as described.” In such cases, a game action may have been illegal. Determining whether that’s the case is key to that question.
• Each question provides Oracle text for all cards mentioned in the scenario. You can click on the text to show or hide card images. Italicized card names in the scenario description should also have a tooltip displaying the card’s rules text for quick reference.

Judge organizations differ in what resources they allow for different kinds of tests. Update quizzes are generally open-internet, while certification tests are open-book or closed-book. Pick whatever works best for you.

The questions

Question 1

Alex controls a 1/1 black and red Goblin creature token, a 2/2 black and green Elf creature token, and Loading Zone. Sourbread Auntie enters the battlefield under Alex’s control and its ability triggers. Which answer best describes what happens if Alex chooses to blight while Sourbread Auntie’s ability resolves?

1. Alex can choose either the 2/2 Elf creature token or Sourbread Auntie. Alex puts four ‑1/‑1 counters on that creature and creates two 1/1 black and red Goblin creature tokens.
2. Alex cannot choose any of their creatures, so they cannot choose to blight in this scenario. Nothing happens.
3. Alex can choose any one of their three creatures. Alex puts four ‑1/‑1 counters on that creature and creates two 1/1 black and red Goblin creature tokens.
4. Alex can choose any one of their three creatures. Alex puts four ‑1/‑1 counters on that creature and creates four 1/1 black and red Goblin creature tokens.
5. Alex can choose any one of their three creatures. Alex puts four ‑1/‑1 counters on that creature but does not create any tokens.

Click to reveal the correct answer.

Answer C is correct.

If a player is instructed to blight, they may choose any creature they control, even if its toughness is less than or equal to the number of the blight instruction [CR 701.68a]. Alex may blight any one of their three creatures.

Sourbread Auntie’s ability reads, in part, “you may blight 2. If you do, create […].” This phrasing signifies that blight 2 is an optional cost [CR 118.12]. Alex may choose to pay that cost, even if the replacement effect of Loading Zone’s ability will modify the outcome so that it does not match the original cost [CR 118.11]. Likewise, the part after “If you do” depends only on Alex’s decision to pay the optional cost, and not on what happens as they actually pay that cost; it occurs exactly once if Alex chooses to pay the cost. As a result, Alex puts four ‑1/‑1 counters on the chosen creature and creates two Goblin creature tokens.

Question 2

Alice controls a 1/1 blue and black Faerie creature token. Napoleon has Disruptor of Currents in his hand. Alice casts Dream Seizer. Which answer best describes what happens when Dream Seizer enters the battlefield and its ability triggers?

1. Alice chooses both a creature she controls and whether to blight it as she puts the ability onto the stack. When the ability resolves, if Alice did choose to blight the creature, she puts a ‑1/‑1 counter on it and Napoleon discards a card. If the creature is not on the battlefield anymore but Alice did choose to blight it, Napoleon still discards a card.
2. Alice chooses a creature she controls as she puts the ability onto the stack, but she does not choose whether to blight that creature at that time. When the ability resolves, Alice decides whether to blight the previously chosen creature. If the chosen creature is not on the battlefield anymore, she cannot choose to blight it and Dream Seizer’s ability has no effect.
3. Alice puts the ability onto the stack. When the ability resolves, she chooses whether to blight a creature, and if so, which one. If Napoleon responds with Disruptor of Currents and returns a creature to Alice’s hand before the ability resolves, she can only choose to blight the remaining creature or to not blight at all.
4. Alice chooses both a creature she controls and whether to blight it as she puts the ability onto the stack. When the ability resolves, if Alice did choose to blight the creature, she puts a ‑1/‑1 counter on it if it’s still on the battlefield. Napoleon discards a card only if a ‑1/‑1 counter is actually put on the chosen creature.
5. Alice puts the ability onto the stack. When the ability resolves, she chooses whether to blight a creature, and if so, which one. If Alice blights a creature, Napoleon has an opportunity to cast Disruptor of Currents before he has to discard a card.

Click to reveal the correct answer.

Answer C is correct.

When a player puts a triggered ability onto the stack, they choose the ability’s mode, its targets, and how to distribute counters, damage, etc. among those targets [CR 603.3c–d]. All other decisions are made only when the ability resolves [CR 603.5][CR 608.2d]. The blight keyword action does not target the creature, so in this scenario, Alice does not decide whether and what she blights until Dream Seizer’s ability resolves [CR 701.68a].

As the ability resolves, Alice decides whether to blight; if she does, she then chooses a creature she controls and puts a ‑1/‑1 counter on it, and Napoleon has to discard a card. If Napoleon wants to cast Disruptor of Currents, he needs to do so before Dream Seizer’s ability resolves and therefore before Alice makes any decisions; while the ability resolves, Napoleon may not cast spells [CR 117.2e][CR 117.1a].

Question 3

Alex controls Inversion Behemoth and The Sibsig Ceremony. They cast Kithkeeper. How many 1/1 green and white Kithkin creature tokens does Alex create when Kithkeeper’s enters-the-battlefield ability resolves?

1. Alex creates one or two Kithkin creature tokens, depending on their choices.
2. Alex creates one, two, or three Kithkin creature tokens, depending on their choices.
3. Alex creates two or three Kithkin creature tokens, depending on their choices.
4. Alex creates two Kithkin creature tokens, regardless of any choices they may get to make.
5. Alex creates one Kithkin creature token, regardless of any choices they may get to make.

Click to reveal the correct answer.

Answer A is correct.

When Kithkeeper enters the battlefield, its own ability and that of The Sibsig Ceremony trigger. If two abilities trigger at the same time and they are controlled by the same player, that player decides the order in which they put those abilities onto the stack [CR 603.3b]. The ability put onto the stack last resolves first [CR 405.5].

Kithkeeper’s ability determines the value of X only as it resolves, and only once [CR 608.2h]. It also has one instruction to create X Kithkin creature tokens rather than X instructions to create one Kithkin creature token, so all tokens enter the battlefield at the same time and do not count as a green permanent toward the value of X [CR 608.2c].

If they decide that Kithkeeper’s ability resolves first, Alex controls a colorless permanent, a black one, and a white one as that ability resolves. Colorless is not a color, so Alex creates two Kithkin creature tokens [CR 105.4]. If they instead decide that The Sibsig Ceremony’s ability resolves first, Alex controls a colorless permanent and two black permanents as Kithkeeper’s ability resolves. Colorless is still not a color and each color is only counted once, even if Alex controls multiple permanents of that color, so Alex creates one Kithkin creature token.

Question 4

Andrew controls Stiltzkin, Moogle Merchant and has Flock Impostor in his graveyard. He casts Emptiness using its evoke ability and spends {W}{W} to pay for its cost. Assuming that Emptiness resolves and enters the battlefield, which of the following statements are correct? Select all that apply.

1. Andrew can choose the order in which Emptiness’s first ability and the evoke triggered ability resolve.
2. If Andrew puts Flock Impostor onto the battlefield with Emptiness’s first ability and targets Emptiness with Flock Impostor’s triggered ability, he returns Emptiness to his hand before he has to sacrifice it.
3. Andrew has to put the evoke triggered ability onto the stack first, followed by Emptiness’s first ability.
4. Andrew has to sacrifice Emptiness before he can respond with Stiltzkin’s ability and before Emptiness’s first ability resolves.
5. If Andrew responds to Emptiness’s abilities with Stiltzkin’s ability and has Nina gain control of Emptiness, Emptiness will still be on the battlefield after everything has resolved and the stack is once again empty.

Click to reveal the correct answer.

Answers A and B are correct.

Sacrificing a permanent that was cast for its evoke cost is the effect of an enters-the-battlefield triggered ability [CR 702.74a]. That ability triggers at the same time as the permanent’s other enters-the-battlefield abilities. The permanent’s controller decides the order in which they put all those abilities onto the stack, effectively determining the order in which the abilities resolve [CR 603.3b][CR 405.5]. In this scenario, Andrew decides whether Flock Impostor returns to the battlefield before or after Emptiness is sacrificed.

If Emptiness’s first ability resolves first and Flock Impostor returns to the battlefield, its enters-the-battlefield ability triggers and Andrew puts that ability onto the stack, on top of the evoke triggered ability [CR 405.2]. Since Emptiness is still on the battlefield, it is a legal target for Flock Impostor’s ability. As the topmost object on the stack, it resolves before the evoke triggered ability and Andrew returns Emptiness to his hand [CR 405.5]. When the evoke triggered ability resolves, Emptiness is not on the battlefield, so nothing happens.

Regardless of the order of abilities, Emptiness is only sacrificed when the evoke triggered ability resolves, so Andrew can respond to that ability and activate Stiltzkin’s ability.

When the evoke triggered ability resolves, it instructs the current controller of Emptiness to sacrifice it rather than the controller of the ability itself [CR 702.74a]. Thus, if Emptiness is on the battlefield when that ability resolves, Emptiness is sacrificed by whoever controls it—Andrew or Nina.

Question 5

Which of the following statements regarding Changeling Wayfinder are correct? Select all that apply.

1. If Alex casts Osseous Exhale, they may reveal Changeling Wayfinder from their hand to pay the spell’s additional cost.
2. Alex may sacrifice Changeling Wayfinder to activate Jolene, Plundering Pugilist’s ability.
3. If Alex exiles Changeling Wayfinder with Ardyn, the Usurper’s triggered ability, the resulting token is all creature types.
4. If Alex activates Secret Tunnel’s last ability, they cannot choose Changeling Wayfinder and a creature manifested with Manifest Dread as the targets.
5. While Changeling Wayfinder is enchanted by Sugar Coat, it has no creature types.

Click to reveal the correct answer.

Answers A, D, and E are correct.

Changeling is a characteristic-defining ability and thus functions in all game zones and even outside the game [CR 702.73a]. In Alex’s hand, Changeling Wayfinder is all creature types and Alex may reveal it when instructed to behold a Dragon.

Changeling grants the object it’s on all creature types; it does not grant artifact types, enchantment types, or any other set of subtypes. Treasure is an artifact type, not a creature type, so Changeling Wayfinder is not a Treasure and may not be sacrificed to activate Jolene, Plundering Pugilist’s ability [CR 205.3g].

An object can only have subtypes that match one of its current card types [CR 205.1a][CR 205.3d]. Sugar Coat replaces all of the enchanted permanent’s card types with the artifact card type. While Changeling Wayfinder is enchanted by Sugar Coat, it is an artifact, not a creature, and thus has no creature types.

Changeling Wayfinder is all creature types and shares a creature type with any creature that has at least one creature type. A manifested creature has no creature types, so it cannot share a creature type with anything, including another manifested creature [CR 701.40a].

If a copy effect modifies the values it copies and those values include a characteristic-defining ability for the affected characteristic, then that characteristic-defining ability is not copied [CR 707.9d]. If Alex creates a token that’s a copy of Changeling Wayfinder with Ardyn, the Usurper’s triggered ability, the token is just a Demon and does not have changeling.

Question 6

Amelia controls Dawn’s Light Archer and a 2/2 black and green Elf creature token. She casts Celestial Reunion and chooses 5 as the value of X. For the additional cost, she chooses Elf as the creature type and beholds Dawn’s Light Archer and the Elf creature token. Nolan responds with Noggle the Mind, targeting Dawn’s Light Archer. Which answer best describes what happens as Celestial Reunion resolves?

1. If Amelia finds an Elf creature card, she puts it onto the battlefield.
2. Celestial Reunion does not resolve. Nothing happens.
3. If Amelia finds a creature card that is an Elf and/or a Noggle, she puts it onto the battlefield.
4. Amelia does not put the found creature card onto the battlefield, regardless of its creature types.
5. If Amelia finds a creature card that is both an Elf and a Noggle, she puts it onto the battlefield.

Click to reveal the correct answer.

Answer A is correct.

The earliest moment where Nolan may cast Noggle the Mind is when he receives priority after Amelia has finished casting Celestial Reunion [CR 117.1a][CR 117.3c–d]. At that point, Amelia has paid the total cost of Celestial Reunion and has chosen the creature type Elf for it.

The effect of Celestial Reunion does not check again whether the beheld objects are still the chosen type. Neither does the found card have to share a creature type with the beheld objects in order to be put onto the battlefield. The only requirement is that it has the chosen creature type. If Amelia finds an Elf creature card, she puts it onto the battlefield. A creature card that is a Noggle but not an Elf is put into her hand.

Question 7

Alex controls Eirdu, Carrier of Dawn, two 1/1 blue and black Faerie creature tokens, Satoru, the Infiltrator, and no other permanents. Alex has no mana in their mana pool. Which answer best describes what happens if Alex casts Rottenmouth Viper?

1. Alex may cast Rottenmouth Viper by tapping Eirdu and the two Faerie creature tokens and sacrificing those three permanents. When Rottenmouth Viper enters the battlefield, Satoru’s ability does not trigger.
2. Alex may not cast Rottenmouth Viper in this scenario.
3. Alex may not cast Rottenmouth Viper by tapping Eirdu and the two Faerie creature tokens and sacrificing those three permanents. They may cast Rottenmouth Viper by tapping all four creatures and sacrificing the two tokens. When Rottenmouth Viper enters the battlefield, Satoru’s ability triggers.
4. Alex may not cast Rottenmouth Viper by tapping Eirdu and the two Faerie creature tokens and sacrificing those three permanents. They may cast Rottenmouth Viper by tapping all four creatures and sacrificing the two tokens. When Rottenmouth Viper enters the battlefield, Satoru’s ability does not trigger.
5. Alex may cast Rottenmouth Viper by tapping Eirdu and the two Faerie creature tokens and sacrificing those three permanents. When Rottenmouth Viper enters the battlefield, Satoru’s ability triggers.

Click to reveal the correct answer.

Answer E is correct.

While casting Rottenmouth Viper, Alex first declares how many permanents they intend to sacrifice for the additional cost, but they do not sacrifice any permanents yet [CR 601.2b]. Next, Alex calculates Rottenmouth Viper’s total cost, using the previously declared number; that cost is locked in and does not change afterward [CR 601.2f]. If Alex declares that they sacrifice three permanents, Rottenmouth Viper’s total cost is {2}{B} and “sacrifice three nonland permanents.”

Then, Alex pays those costs in any order, meaning that they can either pay the mana component first and then sacrifice three nonland permanents, or sacrifice the permanents first and then pay {2}{B} [CR 601.2h]. If Alex pays the mana component first, Eirdu is still on the battlefield, so Rottenmouth Viper has convoke and Alex may tap three creatures to pay for {2}{B} [CR 702.51a][CR 611.3b]. Once the mana component of the total cost is paid, it does not matter if Rottenmouth Viper loses convoke because Eirdu gets sacrificed afterward to pay the other cost.

Alex pays the mana component, but they do not spend any mana to do so; they only tap creatures. Therefore, Satoru’s ability triggers when Rottenmouth Viper enters the battlefield.

Question 8

Koh, the Face Stealer enters the battlefield under Alex’s control and Alex exiles Ashling, Rimebound (a double-faced card whose front face is Ashling, Rekindled) with Koh’s enters-the-battlefield ability. Then, Alex activates Koh’s ability and chooses the exiled card. Which answer best describes the triggered abilities Koh has and what happens during their resolution?

1. Koh has both triggered abilities of Ashling, Rekindled. If Alex pays {U} while the second of those abilities resolves, Koh transforms. The card representing it is turned face down and no other ability triggers.
2. Koh has both triggered abilities of Ashling, Rekindled. If Alex pays {U} while the second of those abilities resolves, Koh transforms. The card representing it is not turned over, but the other ability does trigger.
3. Koh has both triggered abilities of Ashling, Rekindled. If Alex pays {U} while the second of those abilities resolves, nothing happens. The other ability does not trigger.
4. Koh has both triggered abilities of Ashling, Rekindled. If Alex pays {U} while the second of those abilities resolves, Koh transforms. The card representing it is not turned over and the other ability does not trigger.
5. Koh has both triggered abilities of Ashling, Rimebound. If Alex pays {R} while the second of those abilities resolves, nothing happens. The other ability does not trigger.

Click to reveal the correct answer.

Answer C is correct.

Outside the battlefield and the stack, a double-faced card has only the characteristics of its front face [CR 712.8a]. In this scenario, Koh has the abilities of Ashling, Rekindled and not those of Ashling, Rimebound, regardless of the face that was up when that card left the battlefield.

A permanent can only transform if it is represented by a modal or nonmodal double-faced card or by a double-faced token [CR 701.27a][CR 712.4c]. Koh is not a double-faced card, so it cannot transform and if the triggered ability instructs Alex to transform Koh, nothing happens. The card representing that permanent is not turned over and the other ability does not trigger.

Question 9

Arnold controls Bristlebane Outrider and Nadia controls a 2/2 black and green Elf creature token. Arnold casts Brigid’s Command, choosing the first mode with Bristlebane Outrider as the target and the fourth mode with Bristlebane Outrider and Nadia’s Elf creature token as the targets. Nadia responds and casts Nameless Inversion, targeting Bristlebane Outrider. Which answer best describes what happens when Brigid’s Command resolves?

1. Arnold creates a token that’s a copy of Bristlebane Outrider. Then Bristlebane Outrider fights the Elf creature token.
2. The Elf creature token deals 2 damage to Bristlebane Outrider. Bristlebane Outrider deals no damage.
3. Nothing happens. Brigid’s Command has no effect.
4. Bristlebane Outrider fights the Elf creature token.
5. Bristlebane Outrider deals 6 damage to the Elf creature token. The Elf creature token deals no damage.

Click to reveal the correct answer.

Answer D is correct.

Nameless Inversion resolves before Brigid’s Command [CR 117.7]. When Brigid’s Command resolves, Bristlebane Outrider is a 6/2 creature with no creature types. It is not a Kithkin, so it is not a legal target for the spell’s first mode. It is still a legal target for the fourth mode, along with Nadia’s Elf creature token. If at least one target of a spell is legal, the spell resolves and does as much as possible [CR 608.2b].

If a resolving spell targets an object multiple times and the object is an illegal choice for one target, but a legal choice for another, the parts for which it is a legal target still affect it [CR 608.2b]. In this scenario, Bristlebane Outrider is an illegal target for the first mode, so Arnold does not create a token. It is a legal target for the fourth mode, so Bristlebane Outrider fights Nadia’s Elf creature token.

Question 10

Alex controls Meanders Guide, Wanderbrine Preacher, and Sword of the Paruns. Both creatures are untapped. Wanderbrine Preacher has two ‑1/‑1 counters on it and is equipped by Sword of the Paruns. Alex attacks with Meanders Guide. Which answer best describes what happens after Alex taps Wanderbrine Preacher as Meanders Guide’s ability resolves?

1. Alex can choose Wanderbrine Preacher as the target of the reflexive triggered ability and return it to the battlefield.
2. Alex can choose Wanderbrine Preacher as the target of the reflexive triggered ability and return it to the battlefield. Alex gains 2 life.
3. Alex cannot choose Wanderbrine Preacher as the target of the reflexive triggered ability. Alex gains 2 life.
4. Alex can choose Wanderbrine Preacher as the target of the reflexive triggered ability, but it does not return to the battlefield. Alex gains 2 life.
5. Alex cannot choose Wanderbrine Preacher as the target of the reflexive triggered ability.

Click to reveal the correct answer.

Answer B is correct.

When Alex taps Wanderbrine Preacher during the resolution of Meanders Guide’s ability, both Wanderbrine Preacher’s ability and the reflexive triggered ability created by Meanders Guide’s ability trigger [CR 603.2]. Alex puts both abilities onto the stack the next time a player receives priority [CR 603.3]. Nothing else happens while Meanders Guide’s ability resolves.

After that ability has resolved, state-based actions are performed [CR 117.3b][CR 117.5]. Wanderbrine Preacher has toughness 0 and is put into Alex’s graveyard [CR 704.5f]. Then, Alex puts the triggered abilities onto the stack and selects targets as necessary [CR 603.3d]. At that time, Wanderbrine Preacher is in Alex’s graveyard, so it is a legal target for the reflexive triggered ability. When that ability resolves, Alex returns Wanderbrine Preacher to the battlefield.

The Wanderbrine Preacher that is the source of the other triggered ability is not on the battlefield anymore, but that has no bearing on the ability [CR 113.7a]. It resolves and Alex gains 2 life.

Question 11

Anna controls Bre of Clan Stoutarm and Feisty Changeling. She activates Bre’s ability and has Feisty Changeling gain flying and lifelink, then attacks Nigel with Feisty Changeling. After combat damage has been dealt, Nigel casts Boros Charm, targeting Anna with the first mode. Which answer best describes what happens during Anna’s end step if the top card of her library is Gwen Stacy (a modal double-faced card whose back face is Ghost-Spider)?

1. Bre’s ability does not trigger.
2. Bre’s ability triggers and when it resolves, Anna exiles Gwen Stacy. She may cast Ghost-Spider but not Gwen Stacy.
3. Bre’s ability triggers and when it resolves, Anna exiles Gwen Stacy. She may cast either Gwen Stacy or Ghost-Spider.
4. Bre’s ability triggers and when it resolves, Anna exiles Gwen Stacy. She may not cast the card.
5. Bre’s ability triggers and when it resolves, Anna exiles Gwen Stacy. She may cast Gwen Stacy but not Ghost-Spider.

Click to reveal the correct answer.

Answer E is correct.

The intervening “if” clause of Bre’s ability simply checks whether Anna gained life this turn; it does not check whether her current life total is greater than it was at the start of the turn. Therefore, the ability triggers at the beginning of Anna’s end step.

The effect of Bre’s ability checks the resulting spell’s mana value, not the exiled card’s mana value. Outside the battlefield and the stack, a double-faced card has only the characteristics of its front face, whereas on the stack, a modal double-faced card has the characteristics of the face that is up [CR 712.8a][CR 712.8f]. In this scenario, casting Gwen Stacy results in a spell with mana value 2, while casting Ghost-Spider results in a spell with mana value 5. Anna gained only 2 life this turn, so she may cast only Gwen Stacy this way.

Question 12

Alex controls Moonshadow and no other creatures and has two Bear Cubs in their hand. Which of the following statements are correct? Select all that apply.

1. If Moonshadow has a ‑1/‑1 counter on it and Alex activates Reverberating Summons’s ability, Moonshadow’s ability triggers once.
2. If Moonshadow has a ‑1/‑1 counter on it and Alex activates Reverberating Summons’s ability, Moonshadow’s ability triggers twice.
3. If Moonshadow has a ‑1/‑1 counter on it and Alex activates Reverberating Summons’s ability, Moonshadow’s ability triggers three times.
4. If Moonshadow has no counters on it and Alex activates Spiral into Solitude’s ability, Moonshadow’s ability does not trigger, regardless of any choices Alex may have.
5. If Moonshadow has no counters on it and Alex activates Spiral into Solitude’s ability, Moonshadow’s ability may trigger, depending on Alex’s choices.

Click to reveal the correct answer.

Answers B and E are correct.

While paying the total cost to cast a spell or activate an ability, its controller may pay the individual costs in any order [CR 602.2b][CR 601.2h]. In the case of Spiral into Solitude’s ability, the total cost consists of three costs: a mana payment, the blight action, and sacrificing Spiral into Solitude. Alex can first blight 1 and put a ‑1/‑1 counter on Moonshadow, and then sacrifice Spiral into Solitude; since Moonshadow has a ‑1/‑1 counter on it when Spiral into Solitude is put into Alex’s graveyard, its ability triggers. Alternatively, Alex can sacrifice Spiral into Solitude before performing the blight action; in that case, Moonshadow has no ‑1/‑1 counters on it (yet) when Spiral into Solitude is put into Alex’s graveyard, and its ability does not trigger.

The total cost to activate Reverberating Summons’s ability consists of three costs: a mana payment, discarding all cards in hand, and sacrificing Reverberating Summons. Alex can pay those costs in any order, and each cost paid is a separate event for Moonshadow’s triggered ability. The instruction to discard their hand, however, is a single instruction, so Alex discards all cards simultaneously; since the discarded cards are put into the graveyard all at the same time, Moonshadow’s ability triggers exactly once for that cost payment. In total, Moonshadow’s ability triggers twice, regardless of the order in which Alex pays the costs.

Question 13

Alex controls Goliath Daydreamer and casts Flush Out from their hand. Which answer best describes what happens as Flush Out resolves?

1. Alex exiles Flush Out with a dream counter on it. If they cast the card while Goliath Daydreamer’s second ability resolves, Alex may cast either Stormshriek Feral or Flush Out.
2. Alex exiles Flush Out with a dream counter on it and shuffles their library.
3. Alex exiles Flush Out with a dream counter on it. If they cast the card while Goliath Daydreamer’s second ability resolves, Alex may cast Stormshriek Feral, but not Flush Out.
4. Alex shuffles Flush Out into their library.
5. Alex chooses between exiling Flush Out with a dream counter on it and shuffling it into their library.

Click to reveal the correct answer.

Answer E is correct.

After resolving Flush Out’s effect, Alex gets instructed to put the spell into their graveyard [CR 608.2n]. Two replacement effects want to modify that event: the replacement effect created by Goliath Daydreamer’s ability, and the replacement effect generated by rule 720.3d. As the controller of the affected object, Alex chooses one of the two replacement effects and applies it to the upcoming event [CR 616.1].

If Alex chooses the replacement effect created by Goliath Daydreamer’s ability, the upcoming event is modified to “Alex exiles the spell with a dream counter on it.” The spell is not scheduled to be put into their graveyard anymore, so the other replacement effect does not apply [CR 616.1f]. Alex exiles the card and does not shuffle their library. Neither Goliath Daydreamer’s second ability nor the Comprehensive Rules mandate that an omen card cast from exile must be cast as an Omen or as a non-Omen, so Alex can choose between casting Stormshriek Feral and Flush Out when Goliath Daydreamer’s second ability resolves.

If Alex chooses the replacement effect generated by rule 720.3d, the upcoming event is modified to “Alex shuffles the spell into their library.” The spell is not scheduled to be put into Alex’s graveyard anymore, so the other replacement effect does not apply [CR 616.1f]. Alex shuffles the card into their library and nothing is exiled.

Question 14

Alan controls Kaervek, the Punisher and has Emptiness and Moonstone Harbinger in his graveyard. He casts Duress, targeting Nicole. Kaervek’s ability triggers and Alan chooses Emptiness as its target. Which answer best describes what happens if Alan chooses to cast the copy of Emptiness as Kaervek’s ability resolves?

1. Alan may cast the copy using its evoke ability. The copy’s first two abilities don’t trigger when it enters the battlefield, regardless of what mana Alan spends to cast the copy.
2. Alan may not cast the copy using its evoke ability. If he spends {W}{W} to cast the copy, its first ability triggers when it enters the battlefield and returns Moonstone Harbinger to the battlefield. Moonstone Harbinger’s ability does not trigger from the life loss effect of Kaervek’s ability.
3. Alan may cast the copy using its evoke ability. If he spends {W}{W} to cast the copy, its first ability triggers when it enters the battlefield and returns Moonstone Harbinger to the battlefield. Moonstone Harbinger’s ability triggers from the life loss effect of Kaervek’s ability.
4. Alan may cast the copy using its evoke ability. If he spends {W}{W} to cast the copy, its first ability triggers when it enters the battlefield and returns Moonstone Harbinger to the battlefield. Moonstone Harbinger’s ability does not trigger from the life loss effect of Kaervek’s ability.
5. Alan may cast the copy without spending any mana. Alternatively, he may cast the copy using its evoke ability. If Alan spends {W}{W} to cast the copy, its first ability triggers when it enters the battlefield and returns Moonstone Harbinger to the battlefield. Moonstone Harbinger’s ability does not trigger from the life loss effect of Kaervek’s ability.

Click to reveal the correct answer.

Answer D is correct.

Kaervek’s ability does not instruct Alan to cast the copy without paying its mana cost or for any other specific cost. Therefore, Alan has to pay the mana cost or an alternative cost. The evoke static ability functions in any zone from which the object it’s on may be cast, so he may choose to cast the copy using its evoke ability [CR 702.74a].

Once Alan has finished casting the copy, the rest of Kaervek’s ability resolves and he loses 2 life. The copy of Emptiness has not resolved yet, much less its enters-the-battlefield abilities, so Moonstone Harbinger is not on the battlefield and the life loss does not cause its ability to trigger.

When the copy of Emptiness resolves, it enters the battlefield. Depending on the mana Alan spent to cast it, its first and/or second ability triggers.

Question 15

Alex controls Wartime Protestors and Lasting Tarfire. During their precombat main phase, Alex casts and resolves Firdoch Core. Which answer best describes the remainder of their turn?

1. Wartime Protestors’s ability triggers. When it resolves, Alex puts a +1/+1 counter on Firdoch Core and it gains haste until end of turn. If Firdoch Core becomes a creature after that point but before the end step begins, Lasting Tarfire’s ability does not trigger during Alex’s end step.
2. Wartime Protestors’s ability triggers. When it resolves, Alex puts a +1/+1 counter on Firdoch Core and it gains haste until end of turn. If Firdoch Core becomes a creature after that point but before the end step begins, Lasting Tarfire’s ability triggers during Alex’s end step.
3. Wartime Protestors’s ability triggers. If Firdoch Core is not a creature when the ability resolves, Alex does not put a +1/+1 counter on Firdoch Core and it does not gain haste until end of turn.
4. Wartime Protestors’s ability does not trigger.
5. Wartime Protestors’s ability triggers. If Firdoch Core is not a creature when the ability resolves, Alex does not put a +1/+1 counter on Firdoch Core but it does gain haste until end of turn.

Click to reveal the correct answer.

Answer A is correct.

The trigger event of Wartime Protestors’s ability specifies an “Ally,” not an “Ally creature.” Therefore, that ability triggers when a noncreature Ally such as Firdoch Core enters the battlefield.

The ability’s effect refers to “that creature”; however, this simply means the permanent that caused the ability to trigger, regardless of whether it actually is a creature [CR 700.7]. When Wartime Protestors’s ability resolves, Alex puts a +1/+1 counter on Firdoch Core and it gains haste, even if it isn’t a creature at that moment.

As Alex’s end step begins, Lasting Tarfire’s ability checks whether they put a counter on a creature. This check looks at the characteristics of the permanent at the time the counter was put on it [CR 608.2i]. If Firdoch Core becomes a creature only after Wartime Protestors’s ability has resolved, Lasting Tarfire’s ability does not trigger during Alex’s end step.

Question 16

Abigail controls Morselhoarder, which has no counters on it. Ned controls Gathering Stone and Nightmare Sower. Abigail casts Pyrrhic Strike. Which answer best describes the casting process?

1. Abigail decides whether to pay the additional cost. If she does, she chooses a creature, but does not put any counters on it at that moment. Next, Abigail chooses the appropriate number of modes and selects targets for them. After that, she may activate mana abilities. Once she has finished activating mana abilities, Abigail pays the spell’s total cost by paying {2}{W} and, if necessary, putting two ‑1/‑1 counters on the chosen creature.
2. Abigail decides whether she wants to pay the additional cost. Depending on her choice, she then chooses the appropriate number of modes and selects targets for them. Next, Abigail pays the spell’s total cost of {2}{W} and, if necessary, blight 2. She may pay the two costs in any order and may activate mana abilities in between those two costs.
3. Abigail decides whether to pay the additional cost. If she does, she then blights 2. Next, Abigail chooses the appropriate number of modes and selects targets for them. After that, she may activate mana abilities, including Morselhoarder’s ability if she blighted it. Once she has finished activating mana abilities, Abigail pays the spell’s (remaining) cost by paying {2}{W}.
4. Abigail decides whether she wants to pay the additional cost. Depending on her choice, Abigail then chooses the appropriate number of modes and selects targets for them. Next, she may activate mana abilities. Once she has finished activating mana abilities, Abigail pays the spell’s total cost by paying {2}{W} and, if necessary, blighting a creature.
5. Abigail decides whether to pay the additional cost. If she does, she then blights 2. Next, Abigail chooses the appropriate number of modes and selects targets for them. After that, she may activate mana abilities; Abigail may not activate Morselhoarder’s ability, even if she blighted it. Once she has finished activating mana abilities, Abigail pays the spell’s (remaining) cost by paying {2}{W}.

Click to reveal the correct answer.

Answer D is correct.

While casting Pyrrhic Strike, Abigail first declares whether she intends to pay the additional cost, but does not do anything else about it at that point [CR 601.2b]. Next, she chooses the appropriate number of modes [CR 601.2b]. Abigail then selects targets [CR 601.2c]. The next step is activating mana abilities [CR 601.2g]. At this point, Abigail has not paid any of the costs yet and Morselhoarder has no ‑1/‑1 counters on it, so she cannot activate its mana ability. Once she has finished activating mana abilities, Abigail pays the two costs of {2}{W} and blight 2 in any order [CR 601.2h]. She does not have an opportunity to activate mana abilities again while paying those costs [CR 601.2g].

Question 17

Alex has Mirror Image and Waxen Shapethief in their hand. They attack with Kinscaer Sentry, Mistmeadow Council, and a 1/1 green and white Kithkin creature token. Which answer best describes Alex’s options while Kinscaer Sentry’s triggered ability resolves?

1. Alex may put Mirror Image onto the battlefield. Alex may have it enter the battlefield as a copy of either Kinscaer Sentry or the Kithkin creature token. If Mirror Image enters the battlefield as a copy of Kinscaer Sentry, its ability does not trigger.
2. Alex may put Mirror Image onto the battlefield. Alex may have it enter the battlefield as a copy of any of their three creatures. If Mirror Image enters the battlefield as a copy of Kinscaer Sentry, its ability triggers; when that ability resolves, Alex may put Waxen Shapethief onto the battlefield.
3. Alex may put Mirror Image onto the battlefield. Alex may have it enter the battlefield as a copy of any of their three creatures. If Mirror Image enters the battlefield as a copy of Kinscaer Sentry, its ability does not trigger.
4. Alex may put Mirror Image onto the battlefield. Alex may have it enter the battlefield as a copy of either Kinscaer Sentry or the Kithkin creature token. If Mirror Image enters the battlefield as a copy of Kinscaer Sentry, its ability triggers; when that ability resolves, Alex may put Waxen Shapethief onto the battlefield. If they do, it has to enter the battlefield as a copy of either Kinscaer Sentry, Mirror Image, or the Kithkin creature token.
5. Alex may put Mirror Image or Waxen Shapethief onto the battlefield. If they choose Mirror Image, Alex may have it enter the battlefield as a copy of either Kinscaer Sentry or the Kithkin creature token; if Alex chooses Waxen Shapethief, it has to enter the battlefield as a copy of either Kinscaer Sentry or the Kithkin creature token. If the chosen creature card enters the battlefield as a copy of Kinscaer Sentry, its ability does not trigger.

Click to reveal the correct answer.

Answer C is correct.

The effect of Kinscaer Sentry’s ability specifies “a creature card with mana value X or less,” so it refers to the card in Alex’s hand rather than the creature that ends up on the battlefield [CR 109.2a]. Alex controls three attacking creatures, so they may put Mirror Image onto the battlefield but not Waxen Shapethief; Alex may have Mirror Image enter the battlefield as a copy of any of their creatures, even if that creature has a mana value of 4 or greater.

An ability that triggers when “[a creature] attacks” triggers if a creature is declared as an attacker during the turn-based action at the start of the declare attackers step [CR 508.3a]. If Mirror Image enters the battlefield as a copy of Kinscaer Sentry, it is an attacking creature, but Alex did not declare it as an attacker at the start of the step, so its ability does not trigger.

Question 18

Ambrose controls Dockworker Drone, which has a +1/+1 counter on it. Nora controls a tapped Figure of Fable. On a previous turn, she resolved Figure of Fable’s last ability while it was a Soldier. Ambrose casts Mutinous Massacre and chooses even when it resolves. Which answer best describes Figure of Fable after Mutinous Massacre and Dockworker Drone’s triggered ability have both resolved (or otherwise been dealt with)?

1. Figure of Fable is untapped, has haste, and has a +1/+1 counter on it. Ambrose controls it.
2. Figure of Fable is untapped and has haste. Nora controls it.
3. Figure of Fable is untapped and has haste. Ambrose controls it.
4. Figure of Fable is tapped. Nora controls it.
5. Figure of Fable is in Nora’s graveyard.

Click to reveal the correct answer.

Answer C is correct.

A permanent with protection from a specific player cannot be the target of spells and abilities controlled by that player [CR 702.16k]. Mutinous Massacre has no targets, so all of its effects affect Figure of Fable regardless of any protection abilities [CR 115.10a].

An object’s mana value is determined by its mana cost [CR 202.3]. Figure of Fable’s mana cost is {G/W} and its activated abilities do not change that characteristic, so its mana value is 1. If Ambrose chooses even, Dockworker Drone gets destroyed but not Figure of Fable. Then Ambrose gains control of Figure of Fable and untaps it, and it gains haste.

Dockworker Drone’s ability triggers when it’s destroyed, but Ambrose only puts that ability onto the stack and selects a target for it after Mutinous Massacre has resolved [CR 603.2][CR 603.3]. At that point, Ambrose controls Figure of Fable. However, Figure of Fable’s protection abilities are granted by the effect of a resolving ability, rather than the effect of a static ability. As a result, the set of players Figure of Fable has protection from is determined and locked in at the time its last ability resolves [CR 611.2d]. If the controller of Figure of Fable changes, the players it has protection from remain the same; in this scenario, Figure of Fable still has protection from Ambrose and is not a legal target for the ability of his Dockworker Drone.

Question 19

Astrid controls Valley Flamecaller and Nicolas controls Ghalta, Primal Hunger and Loading Zone. Astrid attacks with Valley Flamecaller and Nicolas blocks it with Ghalta. Before combat damage, Astrid casts Barbed Bloodletter and attaches it to Valley Flamecaller with its enters-the-battlefield ability. Which answer best describes what happens during the combat damage step?

1. Either ten ‑1/‑1 counters or nine ‑1/‑1 counters are put on Ghalta, depending on Astrid’s choices.
2. Five ‑1/‑1 counters are put on Ghalta.
3. Either ten ‑1/‑1 counters are put on Ghalta, or eight ‑1/‑1 counters are put on Ghalta and 1 damage is marked on it. Nicolas’s choices determine which alternative happens.
4. Ten ‑1/‑1 counters are put on Ghalta.
5. Either ten ‑1/‑1 counters or nine ‑1/‑1 counters are put on Ghalta, depending on Nicolas’s choices.

Click to reveal the correct answer.

Answer D is correct.

As the combat damage step begins, the creatures assign combat damage [CR 510.1]. Valley Flamecaller is 4/5, so it assigns 4 combat damage to Ghalta. Likewise, Ghalta assigns 12 combat damage to Valley Flamecaller. Then, the assigned combat damage is converted into a damage-dealing event [CR 510.2][CR 120.4].

First, Astrid and Nicolas apply any replacement or prevention effects that modify damage [CR 120.4b]. There is only one such effect, that of Valley Flamecaller’s ability, and it increases the damage dealt to Ghalta by 1. Once that’s done, the amount of damage being dealt is locked in.

Next, the results of the damage are determined [CR 120.4c]. Valley Flamecaller has wither, so the 5 damage dealt to Ghalta results in Astrid putting five ‑1/‑1 counters on it [CR 120.3d]. Ghalta has neither wither nor infect, so the 12 damage dealt to Valley Flamecaller causes 12 damage to be marked on it [CR 120.3e]. Then, Astrid and Nicolas apply any replacement effects that modify those results. There is only one such effect, that of Loading Zone’s ability, and it doubles the number of ‑1/‑1 counters Astrid puts on Ghalta.

The two replacement effects are applied in different steps of this process, so there is no choice involved for either player.

Finally, the damage event occurs [CR 120.4d]. Valley Flamecaller deals 5 damage to Ghalta, Astrid puts ten ‑1/‑1 counters on Ghalta, Ghalta deals 12 damage to Valley Flamecaller, and Valley Flamecaller has 12 damage marked on it.

Question 20

Alvin controls Figure of Fable, which is a 4/5 Kithkin Soldier. During his precombat main phase, he activates Figure of Fable’s last ability. Nancy responds and casts End-Blaze Epiphany with X = 5, targeting Figure of Fable. Alvin responds to End-Blaze Epiphany with Pillar Launch, targeting Figure of Fable. After everything has resolved, he attacks Nancy with Figure of Fable. Nancy casts Blessed Alliance and targets Alvin with the third mode. Which answer best describes what happens when Alvin sacrifices Figure of Fable?

1. The delayed triggered ability created by End-Blaze Epiphany triggers. When it resolves, Nancy does not exile any cards from the top of her library.
2. The delayed triggered ability created by End-Blaze Epiphany triggers. When it resolves, Nancy exiles the top seven cards of her library.
3. The delayed triggered ability created by End-Blaze Epiphany triggers. When it resolves, Nancy exiles the top six cards of her library.
4. The delayed triggered ability created by End-Blaze Epiphany does not trigger.
5. The delayed triggered ability created by End-Blaze Epiphany triggers. When it resolves, Nancy exiles the top nine cards of her library.

Click to reveal the correct answer.

Answer E is correct.

When End-Blaze Epiphany resolves, Figure of Fable does not have protection from it, so End-Blaze Epiphany deals 5 damage to it and creates the delayed triggered ability [CR 608.2b]. That ability triggers when Figure of Fable dies for any reason, including being sacrificed for Blessed Alliance.

A permanent with protection from a specific player cannot be the target of spells and abilities controlled by that player [CR 702.16k]. Blessed Alliance does not target the attacking creature, so Alvin still sacrifices Figure of Fable when Blessed Alliance resolves [CR 115.10a]. The delayed triggered ability does not target Figure of Fable either, so it can determine information about it, such as its power.

Exiling cards from the top of Nancy’s library is part of the effect of the delayed triggered ability, and the number of cards is only determined when that ability resolves [CR 608.2h]. At that point, Figure of Fable is not on the battlefield anymore, so its last known information is used [CR 608.2h]. The last time Figure of Fable was on the battlefield, it was affected by the effect of its last ability and by the effect of Pillar Launch, so its power was 9. Nancy exiles nine cards from the top of her library.

Question 21

Alex controls Curator Beastie and Glaring Fleshraker, and has Painter’s Servant and Friendly Teddy in their graveyard. Alex casts Raise the Past. Which answer best describes what happens as Raise the Past resolves?

1. Painter’s Servant and Friendly Teddy enter the battlefield with two +1/+1 counters on each of them. Glaring Fleshraker’s last ability does not trigger.
2. Painter’s Servant enters the battlefield with no +1/+1 counters or with two +1/+1 counters on it, depending on Alex’s choices. Friendly Teddy enters the battlefield with two +1/+1 counters on it. Glaring Fleshraker’s last ability does not trigger.
3. Painter’s Servant and Friendly Teddy enter the battlefield with two +1/+1 counters on each of them. Glaring Fleshraker’s last ability triggers twice.
4. Painter’s Servant enters the battlefield with no +1/+1 counters on it. Friendly Teddy enters the battlefield with two +1/+1 counters on it. Glaring Fleshraker’s last ability triggers once.
5. Painter’s Servant and Friendly Teddy enter the battlefield with no +1/+1 counters on them. Glaring Fleshraker’s last ability does not trigger.

Click to reveal the correct answer.

Answer B is correct.

When Raise the Past resolves, Alex puts Painter’s Servant and Friendly Teddy onto the battlefield at the same time [CR 608.2f]. To determine how the two creatures enter the battlefield, the process described in rules 614.12 and 616 is used.

For the event of Painter’s Servant entering the battlefield, there are two replacement effects that could modify it: that of Curator Beastie’s second ability and that of Painter’s Servant’s first ability. Since Alex has not yet chosen a color, the effect of Painter’s Servant’s second ability does not do anything and Painter’s Servant is determined to be a colorless creature [CR 607.5a]. Therefore, both replacement effects are applicable and as the would-be controller of Painter’s Servant, Alex chooses either one and applies it to the upcoming event [CR 616.1].

If Alex chooses the effect of Curator Beastie’s ability, the event is modified to “Painter’s Servant enters the battlefield with two +1/+1 counters on it.” The effect of Painter’s Servant’s ability is still applicable to that event, so Alex applies it next and chooses a color [CR 616.1f]. With that effect applied, the effect of Painter’s Servant’s second ability now applies while determining Painter’s Servant’s would-be appearance on the battlefield and Painter’s Servant does not appear as a colorless creature anymore. Once applied, a replacement effect is not reverted because of later replacement effects, however, so Painter’s Servant still enters the battlefield with two +1/+1 counters on it.

If Alex instead chooses the effect of Painter’s Servant’s first ability over the effect of Curator Beastie’s ability, the modified event has Painter’s Servant appear as a creature of the chosen color. Since it does not appear as a colorless creature, the effect of Curator Beastie’s ability is not applicable to the modified event and it does not get applied. Painter’s Servant enters the battlefield with no +1/+1 counters on it.

For the simultaneous event of Friendly Teddy entering the battlefield, there is only one replacement effect: that of Curator Beastie’s second ability. The effect of Painter Servant’s second ability is not considered, since it enters the battlefield at the same time as Friendly Teddy. As a result, Friendly Teddy always enters the battlefield with two +1/+1 counters.

After the two creatures have entered the battlefield, Glaring Fleshraker’s last ability checks whether that event matches its trigger event [CR 603.10]. With Painter’s Servant now on the battlefield, all permanents are the chosen color, so neither of the newly entered creatures is colorless [CR 105.2c]. Glaring Fleshraker’s last ability does not trigger for either creature.

Question 22

Nelson controls Zhao, the Moon Slayer, which has a conqueror counter on it. Amanda plays Eclipsed Realms. A few turns later, Zhao leaves the battlefield. Which answer best describes the event of Eclipsed Realms entering the battlefield and Amanda’s options after Zhao has left the battlefield?

1. Amanda does not choose a creature type as Eclipsed Realms enters the battlefield. After Zhao has left the battlefield, Amanda can tap Eclipsed Realms for colorless mana, but not for mana of any color.
2. Amanda does not choose a creature type as Eclipsed Realms enters the battlefield. After Zhao has left the battlefield, Amanda can tap Eclipsed Realms for mana of any color and can spend that mana on anything.
3. Amanda does not choose a creature type as Eclipsed Realms enters the battlefield. After Zhao has left the battlefield, Amanda immediately chooses a creature type. She can tap Eclipsed Realms for mana of any color and can spend that mana on spells of the chosen type and abilities of sources of the chosen type.
4. Amanda does not choose a creature type as Eclipsed Realms enters the battlefield. After Zhao has left the battlefield, Amanda can tap Eclipsed Realms for mana of any color, but she cannot spend that mana on anything.
5. Amanda chooses a creature type as Eclipsed Realms enters the battlefield. After Zhao has left the battlefield, Amanda can tap Eclipsed Realms for mana of any color and can spend that mana on spells of the chosen type and abilities of sources of the chosen type.

Click to reveal the correct answer.

Answer D is correct.

As Eclipsed Realms enters the battlefield, its would-be appearance on the battlefield is used to determine which replacement effects might apply to this event [CR 614.12]. The effect of Zhao’s last ability is taken into account, so Eclipsed Realms’s appearance does not include its first ability. Without that ability, there is no replacement effect and Amanda does not choose a creature type.

Once Zhao has left the battlefield, Eclipsed Realms has its usual abilities, including its last ability. If Amanda activates that ability, it produces one mana of any color. Without a chosen creature type, the mana can only be spent to cast spells of an undetermined creature type or activate abilities of sources of an undetermined creature type. This is impossible, so Amanda cannot spend that mana on anything.

Question 23

Alex controls Oko, Shadowmoor Scion, Tanufel Rimespeaker, Mischievous Sneakling, and Dawn’s Light Archer. Several turns ago, they resolved Oko’s ‑6 ability and chose Elemental as the creature type. Alex activates Oko’s ‑6 ability a second time and when it resolves, they choose Elf. What are the power and toughness of Alex’s creatures after Alex has created the second emblem?

1. Tanufel Rimespeaker is 5/7, Mischievous Sneakling is 8/8, and Dawn’s Light Archer is 7/5.
2. Tanufel Rimespeaker is 8/10, Mischievous Sneakling is 8/8, and Dawn’s Light Archer is 10/8.
3. Tanufel Rimespeaker is 2/4, Mischievous Sneakling is 8/8, and Dawn’s Light Archer is 10/8.
4. Tanufel Rimespeaker is 2/4, Mischievous Sneakling is 5/5, and Dawn’s Light Archer is 7/5.
5. Tanufel Rimespeaker is 2/4, Mischievous Sneakling is 8/8, and Dawn’s Light Archer is 4/2.

Click to reveal the correct answer.

Answer B is correct.

The emblem’s ability is linked to Oko’s ‑6 ability and refers to the creature type chosen for that ability [CR 607.1d][CR 607.2d]. If there are multiple creature types because the ‑6 ability resolved multiple times, each emblem’s ability refers to all of those creature types. As a result, each emblem’s ability affects all of Alex’s creatures that are Elementals and/or Elves.

Question 24

Adrian controls Mistmeadow Council and a 1/1 green and white Kithkin creature token. He casts Spry and Mighty and Natalie responds with Fresh Start, targeting Mistmeadow Council, and with Shock, targeting the Kithkin creature token. Which answer best describes what happens as Spry and Mighty resolves?

1. Mistmeadow Council gains trample until end of turn.
2. Adrian draws a card. Mistmeadow Council gets +1/+1 and gains trample until end of turn.
3. Nothing happens.
4. Mistmeadow Council gets ‑1/‑1 and gains trample until end of turn.
5. Adrian draws two cards. Mistmeadow Council gets +2/+2 and gains trample until end of turn.

Click to reveal the correct answer.

Answer B is correct.

Fresh Start and Shock resolve before Spry and Mighty; when the latter resolves, Adrian controls only one creature [CR 117.7]. Spry and Mighty does not target the creatures, so Adrian chooses the two creatures he controls during the spell’s resolution [CR 608.2d]. He controls only Mistmeadow Council, so he chooses it as one of the creatures; the other creature is undetermined [CR 609.3].

The game then determines the value of X. To do so, it needs to determine the powers of the two creatures and subtract the smaller one from the greater one. In this scenario, the power of one creature is -1; the other creature and thus the other power is undetermined [CR 107.1b]. If a calculation would use an undetermined number, it uses 0 instead [CR 107.2]. In this scenario, the game uses ‑1 and 0 as the two power values [CR 107.1b]. Subtracting the smaller number from the larger one yields 0 – (‑1) = 1.

The result of the calculation is X = 1, so Adrian draws one card and Mistmeadow Council gets +1/+1 and gains trample until end of turn.