Why isn't the answer to TMUA-P2-ERR1-009 shown upfront?

The options are lettered without marking the correct one so you attempt TMUA-P2-ERR1-009 first — TMUA questions reward spotting the trap before you check the worked solution below.

Do I need a calculator for TMUA-P2-ERR1-009?

No — calculators are banned in the TMUA. TMUA-P2-ERR1-009 is set for a no-calculator target of about 270 seconds; the worked solution shows the fastest by-hand route.

Which TMUA specification point is TMUA-P2-ERR1-009?

TMUA-P2-ERR1-009 is Paper P2 · Identifying errors in proofs (specification code ERR1), level 3. Drill more of the same topic from this page.

TMUA-P2-ERR1-009 — Paper P2 · Identifying errors in proofs · Level 3

P2ERR1Level 3~270s target

Spec: ERR1 · No calculator · ~270s

A six-line algebra proof submitted for a scholarship problem claims to show that

x = 5

follows from the equation

(x - 2)(x - 5) = 3(x - 2)

. The lines are: line one states the equation; line two divides both sides by

(x - 2)

to get

x - 5 = 3

; line three adds five to obtain

x = 8

; line four substitutes

x = 8

back and verifies both sides equal

30

; line five concludes the only solution is

x = 8

. An examiner suspects a lost solution hidden by an illegal division. Which verdict correctly locates the first fault?

Options

Line three: adding five to both sides is invalid.
Line four: substitution cannot verify a solution.
Line two is the first fault: dividing by $(x - 2)$ assumes $x \ne 2$ and discards the solution $x = 2$ , which also satisfies the original equation.
Line one is false: the equation has no solutions.
Line five is the first fault; all algebra before it is fine and complete.

Letters only — the answer key is hidden. Work it out, then check the solution below.

Hint (check trope)

No written hint yet — check the trope: line-number-fault, division-by-zero-variable. Name the trap first, work the stem by hand, then open the solution below.

Show solution

Dividing by a variable expression needs the non-zero proviso: if

x = 2

both sides are

0

, so

x = 2

is a genuine solution lost at line two. The correct route factors:

(x-2)(x-8) = 0

. Trap: a successful check of

x = 8

masks the missing root — verification never recovers solutions destroyed upstream.

Keep practising

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Browse more

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Options

Line three: adding five to both sides is invalid.

Line four: substitution cannot verify a solution.

Line two is the first fault: dividing by

(x - 2)

assumes

x \ne 2

and discards the solution

x = 2

, which also satisfies the original equation.

Line one is false: the equation has no solutions.

Line five is the first fault; all algebra before it is fine and complete.

Letters only — the answer key is hidden. Work it out, then check the solution below.

Keep practising

Same topic · same level

P2ERR1Level 3

TMUA-P2-ERR1-010

A flawed lemma in a student journal claims that from the true statement “For every real x there exists a real y with y >…

Identifying errors in proofs

Open TMUA-P2-ERR1-010 →

Same trap: line-number-fault

P2ERR1Level 1

TMUA-P2-ERR1-001

A “proof” that 1 = 2 reads: (1) Let a = b. (2) Then a^2 = ab. (3) So a^2 - b^2 = ab - b^2. (4) Factorising gives (a-b)(a…

Identifying errors in proofs

Open TMUA-P2-ERR1-001 →

TMUA-P2-ERR1-009 — Paper P2 · Identifying errors in proofs · Level 3

Options

Keep practising

Master Identifying errors in proofs

TMUA-P2-ERR1-009 — Paper P2 · Identifying errors in proofs · Level 3

Options

Keep practising

Master Identifying errors in proofs