GRE · GRE · GRE General Test

Original classroom practice within the reviewed shorter-test task types. This does not simulate an adaptive test or predict a scaled or writing score. Earlier official public forms remain exposed practice; their version and writing-source holds still apply.

• Analyse discourse and draw conclusions
• Distinguish assumptions, evidence and author perspective
• Select relevant points and understand text organisation

assumption 假设: An unstated premise needed by an argument.

discourse 语篇: Connected language with structure and meaning.

Reading comprehension includes single-answer, select-one-or-more and sentence-selection tasks. Follow each instruction exactly.

Identify the conclusion separately from observations. An assumption is an unstated bridge the argument needs.

For select-all tasks, test each option independently. One correct option does not make the others false.

An author can describe a theory before challenging it. Track whose position each sentence reports.

Sales rose after advertising increased. “Advertising caused the rise” assumes no sufficient alternative explanation, such as a price reduction. A passage that only reports the timing supports an association, not that causal conclusion.

Graduate verbal tasks reward precise scope. Avoid an answer whose always or only overstates the passage.

Original passage: [1] A town library's loans rose after it began sending reminders. [2] During the same month, a nearby library closed for repairs. [3] The increase therefore does not establish that reminders alone caused more borrowing. [4] A comparison with an otherwise similar library unaffected by closure would help assess that claim.

What is the main purpose? A. Prove reminders never work. B. Limit a causal inference and suggest a comparison. C. Show repairs reduce reading. D. Recommend closing libraries.

Model and reasoning

B. Sentence 3 limits the inference from the observed timing; sentence 4 proposes evidence. A overstates uncertainty as proof of no effect. C changes the outcome from loans to reading and is not established. D turns a reported closure into a recommendation. The author accepts the rise in loans while questioning its explanation.

For the same passage, select every supported statement: A. Loans increased. B. Reminders caused none of the increase. C. Another change may help explain the increase. Then select the sentence that proposes a way to distinguish explanations.

Check after attempting

A and C; sentence 4. A is stated in sentence 1. C follows from the competing event in sentence 2 and caution in sentence 3. B is unsupported: failure to establish an exclusive cause does not establish zero contribution. Sentence 2 provides a competing observation; sentence 4 proposes a comparison. Judge each option independently and select the complete requested sentence.

Original classroom practice within the reviewed shorter-test task types. This does not simulate an adaptive test or predict a scaled or writing score. Earlier official public forms remain exposed practice; their version and writing-source holds still apply.

• Complete one-, two- and three-blank passages using context
• Select two sentence-equivalence choices yielding coherent equivalent meanings
• Use logical cues rather than vocabulary similarity alone

arduous 艰难的: Requiring much effort.

equivalence 等义关系: Matching meaning in the completed sentence.

For text completion, predict each blank from the passage. Support for a later blank may appear earlier.

A two- or three-blank item requires all blanks correct; do not treat choices as independent mini-scores.

Sentence equivalence requires two choices that both fit and produce equivalent sentence meanings.

A pair of synonyms can both fail the sentence. Check grammar, tone and logic as well as word meaning.

Although the design looked simple, building it was arduous. Although signals a contrast between appearance and effort. For equivalence, difficult and demanding could both fit this sentence; easy and effortless would reverse the needed contrast.

Do not search only for dictionary synonyms. The completed sentence controls equivalence.

Complete the original sentence: “Although the reviewer admired the report's detail, she found its conclusion ____ because its crucial experiment had not been repeated.” A. conclusive B. premature C. exhaustive D. irrelevant E. inevitable

Model and reasoning

B, premature. The unreplicated crucial experiment makes a settled conclusion too early. Although contrasts admiration for detail with criticism of the conclusion. Conclusive and inevitable oppose the criticism; exhaustive describes completeness, not unsupported timing; irrelevant would require evidence that the conclusion concerns the wrong issue. Predict “too early” before consulting the choices.

Sentence Equivalence: select exactly two of six choices. “Far from endorsing the proposal without reservation, the committee offered only ____ support.” A. qualified B. unconditional C. cautious D. unanimous E. enthusiastic F. immediate

Check after attempting

A and C. Both produce support limited by reservations. Unconditional contradicts the explicit contrast. Unanimous concerns how many members agree; immediate concerns timing. Enthusiastic concerns intensity and does not preserve the required limited-support meaning. Contextual fit and equivalent whole-sentence meaning are both necessary; two pleasant-sounding words are insufficient.

Original classroom practice within the reviewed shorter-test task types. This does not simulate an adaptive test or predict a scaled or writing score. Earlier official public forms remain exposed practice; their version and writing-source holds still apply.

• Compare quantities across all allowed values
• Use arithmetic, ratios, divisibility and powers
• Recognise when the relationship cannot be determined

constraint 约束条件: A condition limiting allowed values.

indeterminate 不能确定的: Not fixed by the given information.

Quantitative Comparison has four fixed relationships: A greater, B greater, equal, or cannot determine.

Check all constraints. A variable can be negative, zero or fractional unless the stem restricts it.

Use strategic substitutions to disprove a fixed relation. They can show indeterminacy but a few examples do not prove universality.

Simplify both quantities without silently dividing by a value that could be zero or negative.

Given x²=9, x can be 3 or −3. Quantity A=x; B=0. A is greater for 3 and smaller for −3, so the relationship cannot be determined. If x is positive, only 3 is allowed and A is greater.

Figures are not necessarily drawn to scale. Given labels and constraints carry the evidence.

Quantitative Comparison uses these relationships: A. Quantity A is greater. B. Quantity B is greater. C. Equal. D. Cannot be determined. Given real x with $x^2=16$, compare Quantity A: x, Quantity B: 0. Repeat with the added condition $x>0$.

Model and reasoning

Without the added condition, D: x=4 makes A greater, while x=-4 makes B greater. With $x>0$, A, because only x=4 is permitted. The principal square root of 16 is four, but the equation has both roots. One chosen example cannot prove a relationship over all allowed values.

With the same four answer categories, let $a>0$ and $b<0$. Compare Quantity A: $a/b$, Quantity B: $b/a$. Give admissible examples if the relationship can vary, and identify when equality occurs.

Check after attempting

D. For a=2,b=-1, A=-2 and B=-1, so B is greater. For $a=1$,b=-2, A=-1/2 and B=-2, so A is greater. For $a=1$,b=-1 they are equal. More generally $a/b=b/a$ means $a^2=b^2$, and the stated opposite signs give a=-b. Multiplying an inequality by the negative product ab reverses direction; ignoring this sign would reverse conclusions.

Original classroom practice within the reviewed shorter-test task types. This does not simulate an adaptive test or predict a scaled or writing score. Earlier official public forms remain exposed practice; their version and writing-source holds still apply.

• Solve algebraic, geometric and data-analysis problems
• Select all valid options where requested
• Enter numerical answers in the required form and use the on-screen calculator appropriately

conditional probability 条件概率: Probability within a specified condition.

numerical entry 数值填答: An answer entered as a number rather than an option.

GRE General quantitative content includes arithmetic, algebra, geometry and data analysis; calculus is not required.

Read whether a problem asks for one answer, all answers or numerical entry. Check units and form before submitting.

For data interpretation, identify the denominator and whether a percentage is conditional on a subgroup.

The on-screen calculator is available for General quantitative sections. Estimate first to catch an entry error.

A class has 12 morning and 8 afternoon students. Of them, 3 morning and 4 afternoon students cycle. P(cycles)=7/20. P(afternoon given cycles)=4/7. The second denominator includes only cyclists.

Do not assume a geometric diagram is to scale or replace a conditional denominator with the whole population.

In an original survey table, morning students number 30 and evening students 20. Twelve morning and eight evening students cycle. Enter the probability that a randomly chosen student cycles as a decimal. Enter the probability that a cyclist is an evening student as a fraction.

Model and reasoning

Total students are 50 and cyclists 20. The unconditional probability is $20/50=0.4$. Restricting to cyclists gives evening probability $8/20=2/5$. The denominator changes from 50 to 20 because the second question supplies a condition. An evening student cycling has the reversed probability $8/20$ here only by numerical coincidence: the evening group and cyclist group both happen to have size 20.

A rectangular display has area $48\ \mathrm{cm^2}$, positive integer side lengths and perimeter at most $32\ \mathrm{cm}$. Select all possible lengths of its longer side: 6, 8, 12, 16, 24. Show why your list is complete.

Check after attempting

8 and 12. Let the longer side be L and shorter side $48/L$. Positive integer factor pairs are (1,48),(2,24),(3,16),(4,12),(6,8). Their perimeters are 98,52,38,32,28 cm. Only the last two meet the bound, with longer lengths 12 and 8. Six is the shorter side of the valid 6-by-8 rectangle, so it fails the requested role. Test both the geometric restriction and the word “longer”; numerical compatibility alone is insufficient.

Original classroom practice within the reviewed shorter-test task types. This does not simulate an adaptive test or predict a scaled or writing score. Earlier official public forms remain exposed practice; their version and writing-source holds still apply.

• Develop a focused response to the specific Issue instruction
• Use reasons, examples and qualifications
• Organise and revise analytical writing for clarity

qualification 限定条件: A limit that makes a position more precise.

objection 反对意见: A reason against a position.

The standard short GRE has one 30-minute Analyze an Issue task. Its specific instruction controls what you must discuss.

Choose a position with a reasoned scope. You may qualify a general claim rather than simply agree or disagree.

Explain examples. An anecdote alone does not show why a policy produces the claimed effect.

Consider a serious objection and answer it. Revise organisation and clarity; the emphasis is analysis, not ornate vocabulary.

A qualified position might support work experience in professional courses while allowing research alternatives in theoretical programmes. Explain how experience builds practical judgement, then consider unequal access to placements and a fair alternative.

Do not practise Analyze an Argument as a current standard GRE General writing component. Older materials can contain retired tasks.

Original Issue prompt: “Universities should require every degree student to complete a work placement.” Discuss how far you agree. Explain circumstances in which the recommendation would or would not be useful, using reasons and examples. Plan a qualified position before reading the model.

Model and reasoning

I support a requirement to connect university study with work, but a single compulsory placement is too narrow for every degree and every student. Universities should normally offer supervised practical experience while allowing a demanding alternative when a placement would add little learning or create an unfair barrier. The important test is whether the experience helps students apply and examine what they have learned.

In many professional subjects, a placement can reveal a gap between understanding a rule and using it. Consider a hypothetical engineering student who can calculate an ideal load correctly. In a supervised workshop, the student may also need to explain assumptions to a technician, notice a damaged component and record a change in the design. These activities require judgement that a written calculation alone may not show. A university can strengthen this learning by asking the student to connect a workplace decision with a course principle and explain the limits of the model used.

This benefit does not mean any period in an office is educational. If a placement consists mainly of unrelated routine work, attendance becomes evidence of time spent rather than of learning. A useful requirement needs supervision, appropriate duties and reflection. Where these conditions cannot be met, a carefully assessed project with a real client may achieve more than a nominal placement. The university should judge the demonstrated learning, rather than assume that a workplace address guarantees it.

Access also matters. A student with caring responsibilities may be unable to accept a distant or unpaid role. Requiring that role could reward financial flexibility more than ability. This is a reason to provide funded local opportunities and flexible schedules, rather than a reason to abandon practical learning entirely. If adequate support remains unavailable, an alternative project should ask for comparable independence, communication and evaluation, so that the adjustment preserves academic demand.

Some theoretical programmes raise a further difficulty. A student preparing for research in pure mathematics may gain more from an extended investigation than from an available commercial placement. Research still involves planning, criticism and communication, even when it produces no immediate product. Allowing a supervised research route recognises a different form of application without suggesting that theoretical students need no experience beyond lectures.

A universal work-placement rule therefore confuses a valuable educational aim with one particular method. Universities should require substantial application and reflection, and they should make placements a strong option where suitable work and support exist. They should also accept well-designed alternatives when these better meet the learning goal. This qualified policy preserves the benefits of experience while avoiding an empty or unfair attendance requirement.

Without reusing the model's examples, write a paragraph arguing the strongest case for a universal placement requirement. Then explain one condition that weakens that case and revise your thesis accordingly. This is original classroom practice, not an official scored response.

Check after attempting

A defensible case is that a common minimum prevents departments from offering application only to students already confident enough to seek it. Develop how universal participation could broaden access and create a clear responsibility for the university. A weakening condition is lack of suitable supervised places: universality alone cannot create educational work. A revised thesis can require application for all while accepting equivalent projects where placements fail the stated educational standard. Other well-developed positions are valid. Review task fit, mechanisms, examples, counterargument and clarity; there is no uniquely correct stance or official score here.

Original classroom practice within the reviewed shorter-test task types. This does not simulate an adaptive test or predict a scaled or writing score. Earlier official public forms remain exposed practice; their version and writing-source holds still apply.

• Predict each blank from contrast, support and qualification cues
• Use one choice from each column and check the completed whole
• Distinguish independent answer columns from interdependent passage meanings

conjecture 推测: A proposed explanation not established by the available evidence.

dogmatic 武断的: Treating a view as certain without allowing appropriate doubt.

Read the complete passage before starting a blank. Mark contrast cues such as although, support cues such as because, and a qualification that limits an apparent claim. Predict a plain-language role for each blank before looking for a difficult word. A later sentence may supply the clearest evidence for an earlier blank.

For two or three blanks, begin with the one whose context is most constrained. Use its proposed meaning to check the remaining passage, then return and verify the first choice. Answer columns do not restrict one another mechanically, but the resulting ideas must form one coherent account. Do not choose a superficially positive word merely because the subject sounds admirable.

A correct response needs one answer for every blank. The official task gives three choices per blank for multi-blank items, with no partial credit. Our additional checkbox drill labels each column explicitly: select exactly one entry from each labelled blank. It checks the full answer set locally; it is a practice adaptation, not the official column interface.

Read the completed passage using every chosen word. Check grammar, logical direction and strength. Reject a word that contradicts an explicit limitation even if it pairs well with one other blank. Keep a record of why each distractor fails: wrong polarity, wrong degree, wrong logical relation or wrong meaning in context.

Original two-blank case: “Although the proposal was initially (i) ____, the pilot’s consistent results made its critics (ii) ____.” Blank i: plausible / dubious / settled. Blank ii: reconsider / celebrate failure / lose interest. Choose dubious and reconsider: initial doubt contrasts with subsequent evidence prompting reassessment. Original three-blank case: “The curator’s account is (i) ____ about the object’s origin: it distinguishes verified records from (ii) ____, and its final conclusion is therefore (iii) ____ rather than absolute.” i: cautious / dogmatic / irrelevant; ii: certainty / conjecture / documentation; iii: qualified / categorical / unrelated. Cautious, conjecture and qualified preserve the passage’s evidential restraint.

Independent columns do not mean independent interpretations. Getting two of three blanks right is not a correct complete answer.

Original two-blank task: “The instrument's early readings were (i) ____, so the team treated the apparent trend as (ii) ____ rather than established.” Choose one per column. Blank i: erratic / consistent / definitive. Blank ii: provisional / inevitable / irrelevant.

Model and reasoning

erratic; provisional. Unstable readings justify a tentative interpretation. Consistent and definitive oppose the reason for caution. Inevitable contradicts “rather than established”; irrelevant discards the trend instead of limiting confidence in it. The “so” connects the evidence quality to the status of the inference. Both choices must work in the completed passage.

Original three-blank task: “Although the archive is (i) ____, its missing years do not make every conclusion (ii) ____: patterns repeated in the surviving independent records still justify a (iii) ____ claim.” Blank i: incomplete / comprehensive / fabricated. Blank ii: untenable / decisive / modest. Blank iii: qualified / universal / unrelated. Choose one per column and explain every rejection.

Check after attempting

incomplete; untenable; qualified. Missing years support incomplete, not comprehensive. Fabricated asserts falsification that the passage never supplies. The negation “do not make every conclusion” allows some justified inference: untenable means indefensible. Decisive would praise conclusions rather than name the criticism being rejected; modest is not the relevant failure. Repeated independent records support a limited claim, so qualified fits. Universal exceeds the incomplete evidence; unrelated breaks the stated support. The blanks are semantically dependent even though the answer columns are separate.

Original classroom practice within the reviewed shorter-test task types. This does not simulate an adaptive test or predict a scaled or writing score. Earlier official public forms remain exposed practice; their version and writing-source holds still apply.

• Select exactly two of six choices for an equivalence task
• Check both contextual fit and the meaning of the resulting sentences
• Reject synonym pairs that reverse the passage’s logic

modest 有限的: Small in amount or degree in the stated context.

evasive 回避的: Avoiding a direct response to the issue.

Predict what belongs in the blank from the whole sentence. Contrast and degree matter: “only a few errors” supports a different judgement from “errors throughout.” Avoid choosing pairs before reading the context. The pair must yield two coherent sentences that convey alike meanings, rather than merely share a dictionary association.

Group possible equivalents after establishing the needed sense. A word can have several meanings, and a candidate synonym may match the wrong one. “Reserved” can mean restrained in manner or set aside in advance. The complete sentence selects the sense, so test each candidate using a plain paraphrase.

Sentence Equivalence has six choices and requires two selections. There is no partial credit for one correct word, and selecting a third word invalidates the answer. The practice check requires exactly the correct pair; it does not award an official GRE score.

Compare the remaining plausible pair for scope and tone. Small versus negligible, uncertain versus impossible, or critical versus hostile may differ in strength. If the passage supports modest gains, do not infer that no gains occurred. Keep a vocabulary record with the contextual sense and one contrasting use rather than learning isolated translation pairs only.

Original case: “Despite the team’s ambitious targets, the pilot produced only ____ improvements.” Choices: dramatic, modest, erratic, slight, unprecedented, comprehensive. Modest and slight both fit the contrast between high ambitions and limited improvement. Dramatic and unprecedented may suggest a shared strong-result direction, but they do not fit only in this contrast. Original second case: “Far from being candid, the witness remained ____ about the missing receipt.” Evasive and noncommittal can preserve reluctance to give a direct account; eloquent concerns fluency and is not equivalent to either.

Do not choose a pair solely because the words are related. Degree, direction and local sense must all match.

Select exactly two of six options: “The curator openly acknowledged gaps in the records; her account was therefore unusually ____ about what remained unknown.” A. candid B. evasive C. ornate D. forthright E. repetitive F. hostile

Model and reasoning

A and D. Candid and forthright both express openness about uncertainty. Evasive reverses the meaning. Ornate describes style; repetitive describes recurrence; hostile adds an unsupported attitude. The sentence praises direct acknowledgement, not certainty about every detail.

Select exactly two of six options: “Despite its imposing appearance, the device proved remarkably ____ to operate; first-time users completed the procedure without assistance.” A. cumbersome B. intuitive C. unwieldy D. straightforward E. delicate F. novel

Check after attempting

B and D. Both make operation readily understandable in this context, supported by unaided new users. Cumbersome and unwieldy form a tempting negative pair but conflict with the evidence and contrast. Delicate concerns fragility or care, and novel concerns newness; neither matches ease of operation. Equivalent completed meanings matter more than choosing a dictionary synonym pair in isolation.

Original classroom practice within the reviewed shorter-test task types. This does not simulate an adaptive test or predict a scaled or writing score. Earlier official public forms remain exposed practice; their version and writing-source holds still apply.

• Evaluate each of three select-all options against the passage
• Locate the sentence fulfilling the exact requested rhetorical role
• Distinguish a supported inference from an attractive unsupported claim

concession 让步: Acknowledgement of a point while limiting what follows from it.

scope 范围: The persons, conditions or degree of certainty covered by a claim.

Read a passage for its claim and structure as well as local details. Identify what the author accepts, reports or questions. For select-one-or-more items, classify each option as supported or unsupported using the passage. An unsupported statement need not be disproved; absence of evidence is enough to reject it as a passage-based conclusion.

For inference, retain the text’s conditions and degree of confidence. If a study examined volunteers at one site, do not infer the result holds for every person or site. A statement consistent with the passage can still be unsupported. Select-all reading tasks use three choices, and require all correct choices with no extra selections.

Select-in-Passage asks for a sentence performing a specified role within the allowed text. Name the requested function: evidence, concession, explanation, challenge or conclusion. Read complete sentences rather than choosing the one containing a keyword. A sentence repeating the topic may do a different job. The local lesson uses numbered sentences and option labels; official delivery selects a sentence in the passage itself.

Create a short evidence note for every selected answer. For a rejected option, identify the extra assumption or scope change. This makes review productive and separates vocabulary difficulty from reasoning errors. These original short passages practise question types; they do not replicate adaptive routing or create an official verbal score.

Original passage: [1] A workshop extended its opening by one hour during a four-week trial. [2] Evening attendance increased, but the same month a nearby workshop closed. [3] The director therefore cannot attribute the entire increase to longer hours. [4] A second trial, at a site without a nearby closure, would help separate the explanations. Supported statements: attendance increased; another event provides an alternative explanation. Unsupported: longer hours had no effect. For “select the sentence stating a limitation on the causal conclusion,” choose sentence 3, not sentence 2: sentence 2 reports the confounding observation, while sentence 3 explicitly states the limit.

“Not proved to cause the increase” does not mean “proved to have no effect.” Match the requested sentence function exactly.

Original passage: [1] A laboratory replaced handwritten booking with an online system. [2] Booking errors fell during a two-month trial. [3] The laboratory also reduced the number of users during that period. [4] Although the new system may have helped, the trial cannot establish its effect under normal demand. [5] Testing it with the usual number of users would address this limitation.

Select every supported statement: A. Errors fell during the trial. B. Lower demand may affect the interpretation. C. Online booking cannot reduce errors. Select the sentence that explicitly limits the generalisation.

Model and reasoning

A and B; sentence 4. A restates sentence 2. B follows from the changed user count and stated limitation. C turns lack of decisive evidence into evidence of impossibility. Sentence 3 reports the confounding change; sentence 4 states why the normal-demand conclusion remains limited. Select the whole sentence, not just its caution phrase.

For that passage, identify the role of sentence 5 and explain whether a successful normal-demand trial would prove that every laboratory should adopt the system. Name one additional fact needed for that wider recommendation.

Check after attempting

Sentence 5 proposes a test addressing the specified limitation. A successful trial could support an effect in this laboratory under usual demand, but it would not establish a universal policy. Other laboratories may have different booking tasks, user access or costs. For example, evidence that the system remains usable for laboratories with limited internet access would inform transfer. The answer must connect the new fact to the proposed scope, not merely ask for unspecified “more research”.

評価対象の短いテスト形式における教室での実践。これは適応型テストをシミュレートするものではなく、スケール化された点数や作文スコアを予測するものでもない。過去の公式公開用フォームは依然として練習用として提供されており、そのバージョンおよび作文出典の適用は引き続き有効である。

• 個別の観察、仮定、結論を分離すること
• 特定の議論を強化または弱める証拠を選択すること
• 仮定の否定を用いて、議論がそれに依存しているかテストすること

自己選択: 関連するベースラインの違いを生む可能性のあるグループへの個人の選好。

必須仮定: その失敗が与えられた推論を損なう前提条件。

短い議論を証拠、結論、および欠落している橋渡し（ブリージ）に翻訳せよ。仮定とは、推論が必要とするものの、まだ確立されていない要素である。単に合理的に聞こえる、あるいは道徳的に望ましいように見える文を探すのは避けるべきである。目標は、これらの前提からこの結論への推論である。

因果に関する主張については、代替的な説明、逆方向の因果関係、そして関連するベースラインの違いをテストせよ。ある結果が変化に続いて発生したという証拠は順序性を示すだけであり、変化そのものの効果を隔離するものではない。重要な条件が十分に類似している、または考慮されている場合のみ、比較群は有効である。

強化と弱化は絶対的ではなく、方向性を持つ。新たな関連事実は、結論をより plausible にしたり、不 plausible にしたりするが、それを証明したり反証したりするわけではない。その事実が link をどのように変化させるかを問う：欠落的なメカニズムを裏付けたり、競合する要因を取り除いたり、選択バイアスを露出させたり、反対の条件下での結果を示したりするかどうかである。

必須仮定のタスクについては、候補となる仮定を慎重に否定し、提示された議論がその支持を失うかどうかを問う。この手法は依存性の同定であり、普遍的な真実の同定ではない。ある選択肢は結論に対して十分であっても、必須よりも強い場合がある。主張の範囲を正確に保つ：平均的な改善は、すべての参加者の改善とは異なる。

原文の議論：「任意の復習ワークショップを選択した学生の方が、選ばなかった学生よりも点数が高かった。したがって、ワークショップがより高い点数を引き起こした。」欠落している橋渡し：自己選択が差を十分に説明していなかった。ワークショップ参加者が開始前にすでに高い点数を得ていた場合、結論は弱まる。同等の学生が無作為に割り当てられ、出席の有無だけが異なっていた場合、因果の推論は強まる。ワークショップの人気や鮮やかな教材だけでは、効果の隔离是不可能的である。「すべての参加者が改善した」場合、集団全体の平均向上だけでは不十分である。なぜなら、個人の成果は異なるからである。

事実がトピックに関連していても、推論に影響を与えないことがあります。それが支持を変えるメカニズムを特定してください。

元の議論：「任意の計画セッションを選択した参加者は、プロジェクトをより早く完了した。したがって、このセッションが完了の迅速化を引き起こした。」この推論を最も弱めるものはどれか？ A. 参加者は非参加者よりも過去のプロジェクトを既に早く完了していた。 B. 部屋は最近塗装された。 C. セッションでは印刷された資料を使用した。 D. 多くの参加者がセッションを好んだ。

モデル解答と推理

A. 既存の完了時の違いは、自己選択による説明を提供する。B と C は結果との明示的な関連がない；D は満足度を測定しており、因果効果ではない。A はセッションに何ら影響がなかったことを証明するものではない。しかし、観測されたグループ間の違いが声称される原因を確立するという推論を弱める。

組織者は次のように主張する。「ガイドは、すべての申請者がそれを読めるため、登録誤りをすべて防止する。」アクセス性に関する必須仮定を述べ、その否定をテストし、実際にすべての人がガイドを読むというより強い主張と区別せよ。「すべて」を証明せずとも、推奨事項を強化する证据を提案せよ。

解答後の確認

必須のアクセス性の橋渡しは、申請者のいずれもガイドの関連する指示を理解できない者がいないことである。これを否定すると、少なくとも一人の申請者が理解できないことが判明し、普遍的な防止に関する主張の支持は失われる。アクセス性は実際の読書よりも弱く、どちら单独的にもすべての行動が正しく行われることを保証しない。Comparable な申請者間でガイドを使用した方が誤りが少ないという比較は、有効性主張を強化するが、依然としてすべての申請者に誤りゼロであることを証明するものではない。他の必須仮定、例えば、指示が考えられる誤りを網羅していることは、その否定が正確な普遍的な主張に結びつく場合に有効である。

Original classroom practice within the reviewed shorter-test task types. This does not simulate an adaptive test or predict a scaled or writing score. Earlier official public forms remain exposed practice; their version and writing-source holds still apply.

• Select every allowed answer without assuming a fixed number
• State numerical answers in the requested representation and unit
• Use estimation and substitution to verify a computed response

candidate solution 候选解: A value produced by a solving step that still needs checking.

rounding 舍入: Reporting a value at a stated precision after calculation.

Read the response instruction before calculation. Select-one asks for one option; select-one-or-more can require several; Numerical Entry has no options to cue an approximate answer. Identify restrictions such as integer, positive or rounded to a stated precision. In select-all tasks, test every option against those restrictions.

Solving a transformed equation yields candidates. A denominator restriction or a sign requirement in an unsquared radical can remove one. Substituting into the original is more reliable than recognising a familiar-looking list of roots. For inequality options, test the strict or inclusive boundary as written.

A requested fraction and a requested decimal require different entry forms. Equivalent fractions are accepted in official Numeric Entry; simplify if needed to fit the boxes. Apply rounding only at the end when instructed. Check units: a percentage answer 25 differs from a probability answer 0.25. The free local numerical drill below requests an exact terminating decimal; it is a teaching check, not a replacement for the official two-box fraction interface.

Estimate before using the on-screen General-test calculator and verify after entry. An implausible order of magnitude often reveals misplaced parentheses or a percent converted incorrectly. The GRE General mathematical level does not require calculus; sophisticated reasoning can still arise from elementary constraints and careful instructions.

Original select-all: x is real and (x−2)(x+3)=0. Choices −3, −2, 0, 2, 3. Select −3 and 2, and verify each factorisation root in the original. If the stem additionally requires $x>0$, only 2 remains. Original numerical entry: three of eight equally likely counters are marked; enter the probability as a decimal. P(marked)=3/8=0.375. Enter 0.375, not 37.5 (a percent value) or 3 (the count).

The number of correct select-all choices comes from the constraints, not from a guessed pattern. Numeric form is part of the task.

Select all real solutions of $\sqrt{x+6}=x$ from -3, -2, 0, 2, 3. Then enter the probability of drawing one marked counter from a bag with seven marked and thirteen unmarked counters, as a decimal.

Model and reasoning

3 only; 0.35. The original radical requires x nonnegative. Squaring gives $x^2-x-6=(x-3)(x+2)=0$. The candidate -2 fails both the sign condition and substitution. At 3 both sides equal three. For the probability, the denominator is twenty, so $P=7/20=0.35$. Enter a probability, not the percent value 35.

Select every listed integer satisfying $-2\le x<4$ and $x^2>1$: -3, -2, -1, 0, 1, 2, 3, 4. Then enter $7/12$ as a decimal rounded to the nearest hundredth.

Check after attempting

-2, 2 and 3; 0.58. The first restriction keeps integers -2 through 3. The strict squared inequality excludes -1,0,1. Four fails the strict upper endpoint; -3 fails the lower bound. The decimal $7/12=0.58333\ldots$ rounds to 0.58. The rounding instruction here is explicit; an exact fraction response would require the fraction representation instead. Check each listed value and retain every valid one.

Original classroom practice within the reviewed shorter-test task types. This does not simulate an adaptive test or predict a scaled or writing score. Earlier official public forms remain exposed practice; their version and writing-source holds still apply.

• Use prime factorisation for divisibility, greatest common factors and least common multiples
• Interpret remainders with the original divisor
• Combine ratios and percentage changes with the correct reference total

least common multiple 最小公倍数: The smallest positive integer divisible by each given positive integer.

remainder 余数: The nonnegative amount left after division by a positive integer, smaller than that divisor.

An integer divisible by d can be written dk; a remainder r gives dk+r with 0≤r<d for a positive divisor. Prime factorisation identifies divisibility and common multiples. The greatest common factor takes shared prime powers at their smaller exponents; the least common multiple takes every needed prime power at its larger exponent.

Keep odd/even and positive/negative properties separate. A product of two odd integers is odd; a product with an even integer is even. A statement about integers need not hold for real numbers. For quantitative comparison, exploit a remainder form to expose possible cases rather than assume the smallest positive example is the only value.

A ratio a:b allocates a whole into a+b equal parts when the groups exhaust it. A percentage change multiplies the original value by 1+r or 1−r with r expressed as a decimal. Successive changes multiply factors. A part-of-a-part calculation changes the reference group at each stage; label the denominator before multiplying.

Estimate magnitude and check integer restrictions after solving. Fractions, roots and negative exponents follow their ordinary domain restrictions; even roots represent the nonnegative principal value. Do not cancel a term in a sum as if it were a factor. Exact arithmetic can be faster and safer than decimal calculator entries.

Known: $18=2\cdot3^2$ and $24=2^3\cdot3$. Shared smaller powers give GCF six; needed larger powers give LCM seventy-two. If $n=5k+2$, then $2n=5(2k)+4$, so the remainder on division by five is four.

For a model price of 100 CNY decreased then increased by 20%, multiply the two changes:

The price is not restored because the second percentage uses a different base.

Dividing 250 in the ratio 2:3 gives 100 and 150: each of five parts is fifty.

Equal percentage increases and decreases use different bases. A remainder must be reported relative to the named divisor.

Find the greatest common factor and least common multiple of 36 and 60. A price rises 25% then falls 20%. Find its overall percentage change.

Model and reasoning

$36=2^2\cdot3^2$, $60=2^2\cdot3\cdot5$. Shared smaller powers give GCF=12; larger required powers give LCM=180. Their product $12\cdot180=36\cdot60$ checks the result. The price multiplier is $1.25\cdot0.80=1$, so the overall change is zero percent. The two changes use different reference prices; adding +25 and -20 would give a wrong five-percent increase.

An integer n has remainder 5 on division by 8. Find the remainder of $3n+7$ on division by 8, including for negative n. A value rises 10% in one year and 20% the next. What single percentage reduction restores its original value?

Check after attempting

Write $n=8k+5$ for integer k. Then $3n+7=24k+22=8(3k+2)+6$, so the remainder is six for all integer k. The final value is 1.32 times the original. Restoring it multiplies by $1/1.32=25/33$, a reduction of $1-25/33=8/33$, or about 24.24%. A 30% reduction does not reverse compounded growth. The remainder convention uses $0\le r<8$ even when n is negative.

Original classroom practice within the reviewed shorter-test task types. This does not simulate an adaptive test or predict a scaled or writing score. Earlier official public forms remain exposed practice; their version and writing-source holds still apply.

• Solve equations while preserving denominator and radical restrictions
• Analyse inequalities and simultaneous conditions without losing endpoints
• Evaluate functions and translate verbal relationships into algebra

domain restriction 定义域限制: A condition identifying inputs on which the original expression is defined.

simultaneous solution 联立解: Values satisfying every relation in a system at the same time.

Write restrictions before manipulating an equation. A denominator must be nonzero; a real even root needs a nonnegative argument. Cancelling factors preserves the expression only on its original domain. Squaring can introduce candidates because a square loses sign information; verify every candidate in the unsquared relation.

Solving a system requires the same ordered pair to satisfy all relations. Substitution or elimination can expose one, no or infinitely many linear solutions. For an inequality, multiplication or division by a negative reverses direction. For a factored quadratic inequality, order roots and test interval signs, including endpoints only when equality is allowed.

A function assigns one output to each permitted input. Evaluate the inner function first in a composition, preserving domain restrictions. Read a slope as change in output per input unit and an intercept as the output at input zero only when that input is meaningful. A verbal model should name variables and keep quantities in compatible units.

For Quantitative Comparison, preserve all allowed values. A variable with x²=4 could be −2 or 2 unless constrained. One counterexample can reject a universal relationship; several favourable substitutions cannot prove it for every real value. Algebraic structure can establish a relationship without exhaustive numerical sampling.

For √(x+2)=x, x≥0. Squaring gives x²−x−2=0, candidates 2 and −1. Only 2 satisfies the original. For (x−1)(x−4)≤0, an upward quadratic is nonpositive on 1≤x≤4, including roots. In the system x+y=10, x−y=2, adding gives 2x=12; x=6 and y=4. For f(t)=2t+1 and g(x)=x², f(g(3))=2·9+1=19, applying g first.

A transformed equation supplies candidates, not automatic original solutions. Negative inequality multipliers and strict endpoints need explicit checks.

Solve $(x^2-4)/(x-2)=5$ on its original real domain. Solve $(x+1)(x-3)\le0$ and find $f(g(2))$ for $f(t)=t^2$, $g(x)=x-3$.

Model and reasoning

The rational equation excludes x=2. For allowed x it reduces to x+2=5, giving x=3, which substitutes correctly. The quadratic is nonpositive between its roots, including them: $-1\le x\le3$. The composition applies g first: $g(2)=-1$ and $f(g(2))=1$. Cancelling a factor does not restore excluded inputs.

For real a, classify the system $x+y=4$, $2x+ay=8$ as having one, none or infinitely many solutions. Compare with the second equation $2x+ay=9$. Give solutions in every consistent case.

Check after attempting

Subtract twice the first equation. In the first system $(a-2)y=0$: for a not equal to two, y=0,x=4, uniquely. For a=2, both equations are the same relation, giving every pair $(4-t,t)$. There is no inconsistent case in this first system. With right side nine, $(a-2)y=1$. For a not equal to two, $y=1/(a-2)$ and $x=4-1/(a-2)$, uniquely. For a=2 it demands 0=1, so no solution. A zero coefficient row means inconsistency only when the right side is nonzero.

Original classroom practice within the reviewed shorter-test task types. This does not simulate an adaptive test or predict a scaled or writing score. Earlier official public forms remain exposed practice; their version and writing-source holds still apply.

• Use given relationships rather than a diagram’s apparent scale
• Connect coordinate distances, slopes and geometric properties
• Distinguish linear, area and volume scale factors

sector 扇形: A circle region bounded by two radii and their included arc.

scale factor 缩放比例: The multiplier relating corresponding lengths of similar figures.

Extract the facts a diagram supplies: lengths, right-angle marks, parallel lines, circle radii and stated shape properties. Avoid measuring the screen. Angle sums and similarity give relationships when their conditions hold. Pythagorean reasoning needs a right triangle; it does not apply to every drawn triangle.

For two coordinates, use midpoint ((x₁+x₂)/2,(y₁+y₂)/2) and distance √((x₂−x₁)²+(y₂−y₁)²). Slope is Δy/Δx when Δx≠0. A vertical line has no finite slope. Combine coordinate and shape facts where useful; perpendicular nonvertical lines have slopes whose product is −1.

A circle has circumference 2πr and area πr². An arc or sector uses its fraction of a full turn. Rectangular solids use volume lwh; cylinders use πr²h. Surface area counts exposed faces or surfaces, not the enclosed volume. Attach units so an area answer is not confused with a length or volume.

For similar figures with linear factor k, corresponding areas multiply by k² and volumes by k³. If a question provides only an area ratio, take a square root to recover the length ratio; do not copy the area ratio directly. General-test geometry uses elementary relationships, with no trigonometry or calculus required for this lesson’s tasks.

Points A(−1,2) and B(5,10) have Δx=6, Δy=8. Distance d=√(6²+8²)=10 and midpoint M=((−1+5)/2,(2+10)/2)=(2,6). For a circle of radius 6, a 60° sector is 1/6 of a full circle: area=(60/360)π6²=6π; arc length=(60/360)2π6=2π. Enlarging every length by 3 multiplies area by 9 and volume by 27, not 3.

Use labelled facts, include required π or units, and distinguish a perimeter from an area or a sector from its arc.

Points A=(-1,2) and B=(5,10) are given. Find their distance and midpoint. A right circular cylinder has radius 3 cm and height 4 cm. Find its volume and total surface area, including both ends.

Model and reasoning

Coordinate differences are six and eight. Distance is $\sqrt{6^2+8^2}=10$ and midpoint $(({-1}+5)/2,(2+10)/2)=(2,6)$. Cylinder volume is $V=\pi r^2h=\pi(3\ \mathrm{cm})^2(4\ \mathrm{cm})=36\pi\ \mathrm{cm^3}$. Total area is $S=2\pi r^2+2\pi rh=2\pi(3\ \mathrm{cm})^2+2\pi(3\ \mathrm{cm})(4\ \mathrm{cm})=42\pi\ \mathrm{cm^2}$. Volume and surface area measure different quantities.

A rectangle has diagonal 10 cm and one side 6 cm. A similar rectangle has area nine times as large. Find its perimeter. Explain which labelled facts permit Pythagoras, and whether the same calculation would hold for an arbitrary quadrilateral.

Check after attempting

A rectangle's adjacent sides are perpendicular. Its diagonal forms a right triangle, so the other side is $b=\sqrt{d^2-a^2}=\sqrt{(10\ \mathrm{cm})^2-(6\ \mathrm{cm})^2}=8\ \mathrm{cm}$. Area scale nine means length scale three. The new sides are 18 and 24 cm, giving perimeter $P=2(a+b)=2(18+24)\ \mathrm{cm}=84\ \mathrm{cm}$. An arbitrary quadrilateral need not supply a right angle, so its drawing alone would not justify the equation.

Original classroom practice within the reviewed shorter-test task types. This does not simulate an adaptive test or predict a scaled or writing score. Earlier official public forms remain exposed practice; their version and writing-source holds still apply.

• Distinguish mean, median, spread and a weighted average
• Interpret distributions and overlapping or conditional event counts
• Use the correct reference total in table-based inference

weighted average 加权平均: An average that accounts for each group’s contribution or size.

mutually exclusive 互斥的: Events that cannot occur together in the same outcome.

Mean is sum divided by count; median is the central ordered value or the mean of the middle two. Range is maximum minus minimum. An extreme value can alter mean and range without changing the median. Standard deviation measures spread about the mean; adding a constant changes the centre but not spread, while multiplying all values by c multiplies standard deviation by |c|.

Combine groups with their counts, not an unweighted mean of group means unless sizes are equal. A group of 10 with mean 6 contributes total 60; a group of 30 with mean 10 contributes total 300. The combined mean is (60+300)/40=9. A percentage of a subgroup uses that subgroup’s denominator.

For equally likely discrete outcomes, probability is favourable count over total count. Conditional probability restricts the denominator to the given condition. Without replacement changes later counts. Overlapping events need subtraction of the overlap when counting a union. Independence must be given or justified; it is not the same as mutually exclusive.

Read displays for their axes, units and sample description before making inferences. A correlation does not isolate a cause. A convenience sample may not represent the target population. Basic counting can use ordered multiplication when roles differ, or divide duplicated arrangements when order does not matter. Keep the events and outcomes explicitly defined.

A bag holds four red counters, three marked, and six blue counters, one marked. There are four marked counters in total. P(red)=4/10, P(marked)=4/10, P(red given marked)=3/4. P(two red without replacement)=(4/10)(3/9)=2/15, not (4/10)². For data 1,2,3,4,20, mean=30/5=6 and median=3. Adding 5 to each shifts mean and median by 5 but leaves their pairwise deviations and standard deviation unchanged.

A condition changes the reference population. Group means need weights, and no-replacement draws need changing denominators.

Group A has 12 observations with mean 5; group B has eight with mean 10. Find the combined mean. For the data 1,2,3,4,20 find mean, median and range. Describe how adding 7 to every value changes the standard deviation.

Model and reasoning

The combined total is $12(5)+8(10)=140$ across twenty observations, so mean is seven. An unweighted average of five and ten would incorrectly ignore group sizes. In the second data set, sum is thirty, mean six, median three and range nineteen. Adding seven shifts all values and their mean together; every deviation from the mean is unchanged, so standard deviation is unchanged.

Of 40 students, 24 study art, 18 study music and 10 study both. Find the probability of music given art and the probability of at least one subject. Are the events independent? Then explain what happens to standard deviation if all measurements are multiplied by -3.

Check after attempting

Conditioning on art gives $P(M\mid A)=10/24=5/12$. The union count is $24+18-10=32$, so union probability is 4/5. Independence would require $P(A\cap M)=P(A)P(M)$, but $10/40=1/4$ differs from $(24/40)(18/40)=27/100$. The events are not independent. Scaling every deviation by -3 scales its square by nine and the square root by three, so standard deviation triples. It cannot become negative.

評価対象の短いテスト形式における教室での実践。これは適応型テストをシミュレートするものではなく、スケール化された点数や作文スコアを予測するものでもない。過去の公式公開用フォームは依然として練習用として提供されており、そのバージョンおよび作文出典の適用は引き続き有効である。

• 固定されたテンプレートではなく、特定の指示に従って「イシュー」への回答を作成する
• 指示がある場合、理由、対立する意見、または政策の影響について触れる
• 30分以内に、具体例を伴う限定された立場を発展させる

政策の影響: 提案された規則の妥当性を判断する上で関連する、意図的または非意図的な結果。

限定された主張: 根拠のない絶対的な声明をするのではなく、その限界や条件を示す立場。

現在のGREライティング評価には、1つの30分間の「イシュー」課題がある。この課題と指示 together で回答の方向性が決まる。指示によって「ある声明が成立するのはどの場合か」「ある推奨が有効なのはどの場合か」「自分の立場に対する最大の反論は何か」などが問われる。具体例を生成する前に、求めている推論のアクションを特定する。一般的な賛成・反対テンプレートでは、中心的なタスクを省略してしまうことがある。

2つの視点に関する指示では、選んだ片方の側面だけでなく、両方の視点に関わるべきである。「主張と理由」に関する指示では、主張と提示された根拠を別々に評価する必要がある。結論は合理的でも根拠は弱い、あるいはその逆あり得る。 thesis（主張）と展開の中で、その関係を明確にし、読者に推論を委ねてはいけない。

政策に関する指示の場合、影響を説明し、それがあなたの立場にどう関わるかを述べる。意図的な利益と plausible な非意図的なコストを分離する。具体的なメカニズムや状況を提示する：誰が受益するか、必要な資源は何か、推奨案がどこで失敗するか、どのような制限が改善するか。仮定的な具体例は推理を補うために有用だが、根拠のない捏造された統計を測定された証拠として提示してはならない。

必須の段落数に拘束されることなく、柔軟な計画を使う。推奨される練習のタイムスケジュールは、解釈と構成に4分、執筆に23分、校正に3分である。タスクとの適合性、展開、論理的な組織、明瞭な言語をチェックする。教室でのオリジナルの回答は教員によってレビューされ、局所的な形成評価が公式なGREライティングスコアを生むわけではない。現在の「イシュー」の練習は、廃止された「アーグメンテ」課題と混同してラベル付けしてはならない。

元の主張と理由のケース：主張：大学は、すべての学生が実験科目を独立した読書に置き換えることを認めるべきである。理由：意味のある学習はすべて本から得られる。限定された学説は、全般的な置き換えを拒否しつつ、読書を準備として受け入れることができる。この理由は、直接測定、手順上のエラー、そして協同による学習を見落としている。モデル開発：「仮想的な導入測量コースにおいて、学生は読書後に不確かさを正しく説明できても、使用時に機器の調整を誤る可能性がある。指導付きの実験では、ルールを知るのと実際に適用するとの違いが明らかになる。したがって、読書は常に実験を代替するのではなく、実験を補完するものである。検証済みで同等の実践経験を持つ学生には、適切な代替課題が与えられることができるため、実験科目を維持することは、個人の事情を無視することにはならない。」これは推奨事項とその理由の両方を評価し、メカニズムを提供し、説明付きの例外を導入している。

単に例が印象的であるとラベル付けするのではなく、その例が主張、必要な指示、および結論の限界とどのように関連しているかを説明せよ。

元の「イシュー」プロンプト。主張：「公立学習センターは書面ガイドを通じてのみ支援を提供すべきである」。理由：「すべての来場者に同一の情報を与えることは最も公正な政策である」。主張とその理由にどれだけ同意するか議論せよ。自分の立場に影響を与える状況と影響を説明せよ。

モデル解答と推理

書面ガイドは公共サービスを明確かつ一貫性のあるものにできるが、学習センターにおける唯一の支援源であってはならない。また、その政策のための理由にも疑問を抱く。公平性は信頼できる情報を含むことを要求するが、すべての人に同一の文章を与えることが、すべての人がそれを使用できることを保証するわけではない。より良い政策は、共有された書面ガイドと適切な個人的支援を組み合わせるものである。

この議論には実質的な強みがあります。訪問者は、担当スタッフによって回答が異なるため、矛盾する指示を受け取るべきではありません。明確なガイドは、営業時間、予約ルール、利用可能なサービスなどを、訪問者が後から参照できる形式で記載できます。かつてスタッフ間でコース登録について異なる回答をしていた仮想的な施設では、共通の文書化された手順により混乱を軽減できます。また、 staff が定型の説明を繰り返す必要がなくなり、難しい質問に時間を割けるようになります。これらはガイドへの投資における実用的な理由です。

しかし、情報の一貫性は、アクセスの平等とは異なります。読解力が限られた訪問者はガイドを追随できない可能性があります。別の訪問者は各文節を理解しても、不規則なケースにどのルールが適用されるか分からないかもしれません。両者が自信のある常連客と同じ文書だけを与えられれば、施設は同等に扱ったように見えますが、学習機会には差が生じます。個人的な説明はこの障壁を取り除き、根本的なルールを変えずに済みます。したがって、公平性は、一律のコミュニケーション方法ではなく、一貫した意思決定と適切な支援を支持します。

提案された制限は、その根拠が無視しているコストを生む可能性もあります。例えば、訪問者が予約の指示を誤解して不完全なフォームを繰り返し提出した場合、短い対話で即座に問題を特定できるでしょう。その対話を拒否することは、訪問者の遅延を増やし、staff に追加の修正作業を生じさせます。施設は、最初の問い合わせで節約された時間だけでなく、プロセス全体を比較すべきです。この例は仮想的ですが、一見効率的な政策がリソースを無駄にするメカニズムを説明しています。

限定的なサービスにおいて、文書のみによるサポートが合理的である場合もあります。経験者向けの単純なセルフサービスルームでは、明確な案内とアクセシブルなガイドだけで十分かもしれません。それでも、施設は例外的な問題の報告方法を提供し、実際にユーザーが指示を理解しているか確認する必要があります。限定的なケースは、すべてのサービスおよびすべての訪問者に制限を拡張する正当性はありません。

したがって、私は文書ガイドを一般的な参照として維持し、staff が説明・適用することを認めます。staff は反復される困難を記録し、ガイドを改善すべきであり、不規則な判断は共通のルールに基づいて説明責任を負うべきです。このアプローチは、同一の核心情報としての価値を認めつつ、同一のコミュニケーションが常に公平であるという主張を否定します。これは、訪問者が文書を解釈する能力が支援を受けるための隠れた条件となることを防ぎながら、一貫性を維持します。

この課題を用いて3つの指示ファミリーを比較してください：(a) 推奨事項が機能するタイミングを評価する、(b) 主張とその明記された理由の双方を議論する、(c) 自身の立場に対する最大の挑戦に対処する。それぞれに対して、異なる計画文を作成してください。固定された賛成/反対テンプレートがタスクを見落とす可能性がある理由を説明してください。

解答後の確認

(a)「文書のみによるサポートは、経験者向けの単純な定型サービスでは機能するが、アクセシビリティや不規則なケースが説明を必要とする場面では失敗する。」(b)「私は限定された文書サポートを否定し、理由が提唱する一貫性への妥当な懸念と、同じ言葉が公平性と等価であるという unsupported な結論を区別する。」(c)「最大の挑戦は、个人的サポートが不一致の助言をもたらす可能性があることだ。共通のルール、記録、staff の研修により、そのリスクを低減できる。」他の計画も、要求された行動を果たす限り有効です。条件付きのない一般的な立場、理由に対する判断、または答えた挑戦は、文章が流暢であっても指示を省略する可能性があります。推論と言語については人間のレビューを使用し、ローカルチェックは公式ライティングスコアを割り当てません。