Physics beat its target for the first time all year. 125 against a floor of 120, with only three wrong answers in 45 questions. The AC and EMI work held, and the two mirror blocks were clean. The ten blanks are all Ray Optics and all of them are two to four lines long — they were a time problem, not a knowledge problem.
Chemistry and Botany came in within a few marks of target (151 and 152). Six losses each, and in both sections two of those six were questions she has answered correctly on her own practice papers within the last ten days.
Zoology is the whole story: 106 / 180, fourteen wrong out of fifteen lost. That is 59 marks below target and it accounts for the entire gap to 600. Evolution was her best chapter on 8 October (150/160) and it was her worst section here, three days later.
3. Why Evolution fell apart
These were not hard questions and they were not unfamiliar. Five of the fourteen test ideas she got right on her own paper on 8 October:
AITS question
Same idea on 8 Oct
Then
Now
Q151 Haeckel vs von Baer
Q56, Q57
both correct
reversed
Q171 stabilising selection
Q47 curve panels
correct
blank
Q177 first cellular life
Q69
2000 mya ✓
reference missed
Q179 analogous organs
Q52, Q55
both correct
one pair swapped
Q169 / Q173 divergent vs convergent
Q51, Q54
correct
both wrong
The common thread is that every one was reworded. Her practice paper asked them in NCERT's own phrasing; the AITS asked the same facts sideways — "how is Haeckel different from von Baer", "how many years after the formation of Earth", "favoured" expressed as mortality. The recall is solid; the recall is phrasing-bound.
Three more (Q159, Q178, and arguably Q160) turn on a negative stem — NOT, INCORRECT, necessarily. In each case she picked a true statement. That is four marks each time and it is the cheapest thing on this list to fix.
4. The fourteen blanks
Ten in Physics, one in Chemistry, two in Botany, one in Zoology — 56 marks of opportunity. Every Physics blank is solved below in four or five lines; none needed more than one formula. Q28, Q31, Q33, Q34, Q36 and Q44 are two-line questions.
All ten Physics blanks sit in Ray Optics, which came last in the paper. Two readings are possible: the section ran out of time, or Ray Optics felt unfamiliar enough to skip on sight. The scores say the second — because the three Ray Optics questions she did attempt, she got two wrong (Q27, Q41). Ray Optics is the chapter to rebuild before the next AITS, and the blanks make it look better than it is.
Physics — 125 / 180 · 3 wrong · 10 blank
Q9WrongCLimiting caseAlternating current — limiting behaviour
At very high frequencies, the effective impedance of the given circuit in ohm will be
2 ← correct
0.5 ← her answer
1
5
The three stages of the reduction. Highlight cycles through them; every stage is drawn in full.
Solution
At very high f, XC = 1/(2πfC) → 0, so every capacitor becomes a plain wire.
At very high f, XL = 2πfL → ∞, so the 20 H coil becomes a break in the wire — that whole lower-left path is dead.
The upper branch is now 1 Ω and 2 Ω in series with the source node between them; the vertical capacitor shorts the midpoint across to the lower 2 Ω.
From the midpoint onward there are two 2 Ω resistors in parallel = 1 Ω.
Z = 1 + 1 = 2 Ω.
Where it went wrong · fixA shorted capacitor removes the capacitor, not the resistors sitting beside it in that branch. Write the simplified circuit out before computing — on this one the whole question is the redraw.
Q27WrongCFormula set-upSilvered glass plate
One face of a rectangular glass plate 6 cm thick is silvered. An object held 8 cm in front of the first face forms an image 12 cm behind the silvered face. The refractive index of the glass is
1.8 ← her answer
2.1
1.2 ← correct
1.6
Light crosses the glass twice, so the thickness enters as 2t/μ. The dot traces the actual path.
Solution
Light crosses the glass, reflects at the silvered back, and crosses the glass again — so the thickness is traversed twice.
Standard result: image lies at d + 2t/μ measured from the FRONT face, where d = 8 cm and t = 6 cm.
Behind the silvered face that is d + 2t/μ − t = 12.
8 + 12/μ − 6 = 12 ⟹ 12/μ = 10.
μ = 1.2.
Where it went wrong · fixμ appears in exactly one term, 2t/μ. Write that equation first and solve it; 1.8 comes from mixing the 8 cm and 6 cm into the μ term.
If the distance between an object and its three-times magnified virtual image produced by a curved mirror is 20 cm, the focal length of the mirror must be
10/3 cm
−5 cm
−7.5 cm ← correct
15 cm
Solution
Virtual and magnified ⟹ m = +3, so −v/u = 3 and v = −3u.
Put u = −x (object in front). Then v = +3x (image behind the mirror).
Separation = 3x − (−x) = 4x = 20 ⟹ x = 5, so u = −5 cm and v = +15 cm.
1/f = 1/v + 1/u = 1/15 − 1/5 = −2/15.
f = −7.5 cm — a concave mirror.
Where it went wrong · fixThree lines of algebra and no figure needed. 'Virtual and magnified' fixes the sign of m immediately; the only trap is forgetting that the separation is 3x + x, not 3x − x.
Q30BlankTBlank — repeated formulaTwo mirrors in succession
One concave mirror M₁ (f₁ = 20 cm) faces a convex mirror M₂ (f₂ = 10 cm) 40 cm away. An object Q sits 30 cm from M₁. Find the total magnification after reflection first at M₁ and then at M₂.
+4
−2
+2 ← correct
−4
Solution
At M₁ (concave): u = −30, f = −20 ⟹ 1/v = −1/20 + 1/30 = −1/60, so v = −60 cm.
m₁ = −v/u = −(−60)/(−30) = −2.
That image lies 60 cm from M₁, i.e. 20 cm behind M₂ — a virtual object for M₂, u = +20.
At M₂ (convex): 1/v = 1/10 − 1/20 = 1/20, so v = +20 cm, and m₂ = −20/20 = −1.
m = m₁ × m₂ = (−2)(−1) = +2.
Where it went wrong · fixTwo-mirror questions are just the mirror formula twice, with the first image becoming the second object. The only new idea is the virtual object when the image falls behind the second mirror.
Q31BlankTBlank — looked harder than it isConvex mirror — speed of the image
An observer in a parked car watches a jogger approaching in a side-view mirror of R = 2 m. The jogger runs at 5 m s⁻¹ and is 39 m away. The average speed of the image over the next one second is
(1/280) m s⁻¹ ← correct
(1/200) m s⁻¹
(1/150) m s⁻¹
5 m s⁻¹
Solution
Convex mirror: f = R/2 = 1 m.
At u = −39 m: 1/v = 1/f − 1/u = 1 + 1/39 = 40/39 ⟹ v = 39/40 m.
One second later the jogger is 34 m away: 1/v = 1 + 1/34 = 35/34 ⟹ v = 34/35 m.
Where it went wrong · fixThe word average is the instruction: compute two positions and subtract. No differentiation is needed — which is why this question is shorter than it looks.
Q32BlankTBlank — unfamiliar wrapperRefraction written as vectors
A ray along the unit vector (−î − 2ĵ)/√5 refracts at the xz-plane. For y > 0, μ = 2; for y < 0, μ = √5/2. The unit vector along the refracted ray is
−(4î + 3ĵ)/5 ← correct
−(3î − 4ĵ)/5
−(3î + 4ĵ)/5
(4î + 3ĵ)/5
Solution
The normal is ĵ. The ray's y-component is negative, so it travels from y > 0 (μ₁ = 2) into y < 0 (μ₂ = √5/2).
Component ⟂ to the normal is 1/√5, so sin i = 1/√5.
Snell: 2 × (1/√5) = (√5/2) sin r ⟹ sin r = 4/5, hence cos r = 3/5.
The refracted ray keeps the same sideways sense (x negative) and continues into y < 0.
Unit vector = −(4/5)î − (3/5)ĵ = −(4î + 3ĵ)/5.
Where it went wrong · fixA vector dressed up as something exotic. Split the unit vector into 'along the normal' and 'across the normal' and it becomes ordinary Snell's law.
Q33BlankTBlank — one-line recallAir bubble in a slab
An air bubble in a glass slab of μ = 1.6 appears 3 cm deep from one face and 7 cm deep from the opposite face. The thickness of the slab is
10 cm
25 cm
27 cm
16 cm ← correct
Looking INTO glass, real depth = apparent × μ. The two real depths add to the thickness.
Solution
Looking from air into glass, real depth = apparent depth × μ.
From the first face: 3 × 1.6 = 4.8 cm.
From the second face: 7 × 1.6 = 11.2 cm.
The bubble is one point, so the two real depths meet: t = 4.8 + 11.2.
t = 16 cm.
Where it went wrong · fixNote the direction of the rule. Apparent = real/μ when looking into the denser medium, so going the other way you multiply. Getting 6.25 cm means the division was done the wrong way round.
Q34BlankTBlank — series stepA stack of slabs
n transparent slabs of μ = 1.5, of thicknesses 1, 2, 3 … n cm, are stacked. A point object viewed from the top through the stack is shifted by 5 cm. Then n is
5 ← correct
4
3
6
Five slabs: the thicknesses are an arithmetic series, so Σt = n(n+1)/2.
Solution
One slab shifts the object by t(1 − 1/μ).
μ = 1.5 ⟹ 1 − 1/1.5 = 1/3, the same factor for every slab.
Where it went wrong · fixThe sum 1 + 2 + … + n is the only step that is not already a standing formula. Recognise n(n+1)/2 and the question is two lines.
Q36BlankTBlank — direct formulaLateral displacement
A ray meets a parallel-sided slab 9 cm thick at i = 60°; the angle of refraction is 30°. The lateral displacement is
√3 cm
3√3 cm ← correct
3 cm
2/√3 cm
Lateral displacement is the gap between the emergent ray and the undeviated path.
Solution
d = t · sin(i − r)/cos r.
i − r = 60° − 30° = 30°, so sin(i − r) = 1/2.
cos r = cos 30° = √3/2.
d = 9 × (1/2) ÷ (√3/2) = 9/√3.
d = 3√3 cm.
Where it went wrong · fixStraight substitution into a formula on the sheet. Both the sine and the cosine must be taken of the right angle — sin of the difference, cos of r alone.
Q38BlankTBlank — needed the figureTotal internal reflection in a block
A ray enters air → a denser medium of μ = 4/3 and then suffers total internal reflection at the adjacent (side) face. The maximum value of θ is
sin⁻¹(√7/3) ← correct
sin⁻¹(√5/4)
sin⁻¹(√7/4)
sin⁻¹(√5/3)
The two angles inside the block are complementary — that is the whole question.
Solution
Inside the block the refracted ray makes θ′ with the top normal and strikes the side face at θ″, and θ″ = 90° − θ′.
TIR at the side face needs θ″ ≥ C; the largest θ is when θ″ = C exactly.
sin C = 1/μ = 3/4, and sin θ″ = cos θ′, so cos θ′ = 3/4 and sin θ′ = √7/4.
Snell at the top face: sin θ = μ sin θ′ = (4/3)(√7/4).
θ(max) = sin⁻¹(√7/3).
Where it went wrong · fixThe one idea is that the two internal angles are complementary. Mark θ′ and θ″ on the figure and write θ″ = 90° − θ′ before any algebra.
Q39BlankTBlank — standard resultThe circle of light on a water surface
A small bulb lies 70 cm below the surface of water (μ = 4/3). The area of the surface through which light emerges is
91800 cm²
19800 cm² ← correct
81900 cm²
2825 cm²
Only rays inside the critical cone get out; the lit patch is a circle of radius h·tan C.
Solution
Only rays meeting the surface inside the critical angle get out; the rest are totally internally reflected.
So the lit patch is a circle of radius r = h tan C.
sin C = 1/μ = 3/4 ⟹ tan C = 3/√7.
A = πr² = π h² tan²C = π × 70² × 9/7 = π × 6300.
A ≈ 19 800 cm².
Where it went wrong · fixStandard result worth memorising: A = π h² tan²C. 91800 and 81900 are digit-shuffles of the right answer, so the arithmetic matters as much as the formula.
Q41WrongCAlgebra stepPrism at minimum deviation
If the refractive index of the material of a prism is cot(A/2), where A is the angle of the prism, the angle of minimum deviation is
π − 2A ← correct
π/2 − A
π − A
π/2 − 2A ← her answer
At minimum deviation the ray inside runs parallel to the base — that symmetry gives the formula.
Solution
Prism formula: μ = sin((A + δm)/2) / sin(A/2).
Given μ = cot(A/2) = cos(A/2)/sin(A/2).
Write cos(A/2) as sin(90° − A/2); the denominators match, so the numerators must: sin((A + δm)/2) = sin(90° − A/2).
(A + δm)/2 = 90° − A/2 ⟹ A + δm = 180° − A.
δm = 180° − 2A = π − 2A.
Where it went wrong · fixThe step that decides it is turning cos into sin so both sides are sines of an angle. Her π/2 − 2A keeps the 90° instead of doubling it back to 180° when clearing the halves.
Q44BlankTBlank — conceptual, not numericalA slab moving towards the observer
A slab of μ = 3/2 and thickness 3 cm is moved towards a stationary observer at 3 m/s while a point O is viewed through it by paraxial rays. The velocity of the image is
2 m/s towards left
4/3 m/s towards left
3 m/s towards left
Zero ← correct
The slab slides across the figure and the green image stays put — the shift depends only on t and μ.
Solution
A slab shifts the image by t(1 − 1/μ) towards the observer.
That expression contains only t and μ — the distance of the slab from either the object or the eye does not appear.
Sliding the slab along the axis therefore changes nothing about where the image sits.
The shift stays 3(1 − 2/3) = 1 cm throughout the motion.
Velocity of the image = zero.
Where it went wrong · fixThe question is testing whether she knows what the shift formula depends on. Any answer with a number in m/s assumes position matters; it does not.
Chemistry — 151 / 180 · 5 wrong · 1 blank
Q61WrongNNaming suffixCoordination — IUPAC naming
The correct IUPAC name of the complex [Cr(NH₃)₅(NCS)][ZnCl₄] is
Where it went wrong · fixShe had the cation, the ligands and the oxidation state right and lost it on the last word. If the complex is the anion, its metal always ends in -ate. Check that one suffix before anything else.
Q65WrongCTwo-condition filterHomoleptic vs heteroleptic; d-electron count
Identify the homoleptic complexes with an ODD number of d electrons in the central metal: (A) [Fe(CN)₆]⁴⁻ (B) [Fe(CN)₆]³⁻ (C) [Fe(CN)₅NO]²⁻ (D) [CoCl₄]²⁻ (E) [Co(H₂O)₃F₃]
(B) and (D) only ← correct
(C) and (E) only
(A), (B) and (D) only ← her answer
(A), (C) and (E) only
Solution
Homoleptic = one kind of ligand only. That removes (C) — CN⁻ plus NO — and (E) — H₂O plus F⁻.
(A) [Fe(CN)₆]⁴⁻ → Fe(II) → d⁶, even.
(B) [Fe(CN)₆]³⁻ → Fe(III) → d⁵, odd ✓
(D) [CoCl₄]²⁻ → Co(II) → d⁷, odd ✓
Answer: (B) and (D) only.
Where it went wrong · fixTwo filters, applied in order: first strike out the heteroleptic ones, then count d electrons. She kept (A), whose d⁶ count is even — the second filter was never applied to it.
Q66WrongCRead the reagent firstQualitative analysis — the medium names the group
Both sulphides are black. The medium, not the colour, decides the cation.
Solution
Read the reagent, not the colour: H₂S with NH₄Cl and NH₄OH is the Group IV reagent (ammoniacal, basic medium).
Group II (Cu²⁺, Pb²⁺, Cd²⁺) precipitates with H₂S in dilute HCl — acidic medium — and would already have been removed before this stage.
Group IV cations are Co²⁺, Ni²⁺, Mn²⁺, Zn²⁺.
Of these, NiS and CoS are black; MnS is buff, ZnS is white.
M²⁺ = Ni²⁺.
Where it went wrong · fixCuS and NiS are both black, so the colour cannot decide it — only the medium can. Underline the reagent line before reading the options on every salt-analysis question.
Q69WrongNNaming suffixCoordination — spotting the wrong name
Wrong match about the IUPAC name of the complex is
The question asks for the name that is wrong, so each option has to be checked against the rules.
Options 1–3 are all correctly named, including the bis( ) prefix in option 3 because 'diamine' already contains a numeral.
Option 4 fails twice: the ligand CN⁻ is cyanido, not cyanato (cyanato is OCN⁻).
And [Ag(CN)₂]⁻ is the anion, so its metal must take -ate: argentate.
Correct name: diamminesilver(I) dicyanidoargentate(I) — so option 4 is the wrong match.
Where it went wrong · fixSame suffix rule as Q61, one question later, and it was missed again in the opposite direction — she flagged a correct name as wrong. Scan every option for an anionic complex and check its -ate ending first; that single pass settles both questions.
Q79BlankTBlank — four quick checksCrystal field — which electron goes where
About octahedral complexes: (A) If Δ₀ < P the fourth electron enters e(g), giving t₂g³e(g)¹. (B) If Δ₀ > P the fourth electron enters t₂g, giving t₂g⁴e(g)⁰. (C) If Δ₀ > P the sixth electron enters t₂g, giving t₂g⁴e(g)². (D) If Δ₀ < P the sixth electron enters t₂g, giving t₂g⁴e(g)².
A, B & C only
A, C & D only
A, B & D only ← correct
B, C & D only
Solution
Δ₀ < P is the weak field, high-spin case: electrons spread out rather than pair.
(A) d⁴ high spin → t₂g³e(g)¹ ✓
(B) Δ₀ > P, strong field, d⁴ → all four pair into t₂g → t₂g⁴e(g)⁰ ✓
(C) Strong-field d⁶ must be t₂g⁶e(g)⁰, not t₂g⁴e(g)² ✗
(D) Weak-field d⁶ → t₂g⁴e(g)² ✓ — so A, B and D.
Where it went wrong · fixFour statements, each a one-line check against 'does Δ₀ beat P?'. Left blank — but the same reasoning was done correctly on Q88's distractors, so the knowledge was there and the time was not.
Q88WrongCTwo halves crossed overCFSE of a d⁷ octahedral complex
The crystal field stabilisation energy of an octahedral complex with d⁷ configuration can show: (A) −1.8Δ₀ with one unpaired electron (B) −1.8Δ₀ with three unpaired electrons (C) −0.8Δ₀ with one unpaired electron (D) −0.8Δ₀ with three unpaired electrons
A and D ← correct
A and B
C and D
B and C ← her answer
Fill t2g first only when Δo beats the pairing energy P. Highlight alternates between the two cases.
Solution
d⁷ has two possible fillings, so the question wants both.
Strong field (low spin): t₂g⁶e(g)¹ → CFSE = 6(−0.4) + 1(+0.6) = −1.8Δ₀, with 1 unpaired electron → statement A.
Weak field (high spin): t₂g⁵e(g)² → CFSE = 5(−0.4) + 2(+0.6) = −0.8Δ₀, with 3 unpaired electrons → statement D.
B pairs −1.8 with three unpaired; C pairs −0.8 with one. Both cross the two cases over.
A and D.
Where it went wrong · fixHer 'B and C' is exactly the two correct numbers with the two electron counts swapped. Draw both boxes, count the arrows, and only then read the options — the arrows carry the unpaired count for free.
Botany — 152 / 180 · 4 wrong · 2 blank
Q98WrongNTwo facts mergedHuman Genome Project — salient features
Statements on the human genome: A. The largest known human gene is on chromosome 1. B. Repeated sequences make up a very small portion of the genome. C. The Y chromosome has 231 genes. D. The average gene consists of 3000 bases. Which are correct?
C and D only ← correct
A and C only ← her answer
B, C and D only
A and D only
Solution
A — the largest known human gene is dystrophin, 2.4 million bases, on the X chromosome, not chromosome 1. ✗ (Chromosome 1 has the most genes, 2968 — a different fact.)
B — repeated sequences make up a very large portion of the genome. ✗
C — chromosome 1 has 2968 genes and the Y has 231, the fewest. ✓
D — the average gene is about 3000 bases. ✓
C and D only.
Where it went wrong · fixStatement A blends two true NCERT facts about chromosome 1 into one false sentence. The chapter's numbers need to be held as pairs: most genes = chromosome 1 · fewest = Y · largest gene = dystrophin on X.
Q100WrongNSecond blank not checkedTranscription — substrates and template
Transcription utilises ______ as substrates and polymerises in a ______ dependent manner, following the rule of complementarity.
ribonucleoside triphosphates, non-template ← her answer
deoxyribonucleoside triphosphates, template
deoxyribonucleoside triphosphates, non-template
ribonucleoside triphosphates, template ← correct
Solution
Transcription makes RNA, so the substrates are ribonucleoside triphosphates — she had this half right.
Complementarity is always read against the strand being copied.
The strand that is copied is the template strand (3′→5′).
The coding (non-template) strand is simply displayed, never read by the polymerase.
Ribonucleoside triphosphates, template.
Where it went wrong · fixHalf-right answers are the most expensive kind: the first blank pulled her towards the option whose second blank was wrong. Fill both blanks on the paper before scanning the options, then match the pair.
Q123BlankTBlank — eliminate on codon 2Reading the genetic code
In prokaryotes the mRNA reads 5′ AUG CCC AAG UCU GGG UUU AAA UAA 3′. The sequence of amino acids in the polypeptide is
f-met, Pro, Lys, Leu, Gly, Phe, Lys
f-met, Gly, Lys, Ser, Pro, Phe, Lys
f-met, Pro, Lys, Ser, Gly, Phe, Lys ← correct
f-met, Gly, Lys, Ser, Gly, Phe, Lys
Solution
AUG → f-met (formyl methionine in prokaryotes).
CCC → Pro · AAG → Lys · UCU → Ser.
GGG → Gly · UUU → Phe · AAA → Lys.
UAA → stop, contributing no amino acid.
f-met, Pro, Lys, Ser, Gly, Phe, Lys.
Where it went wrong · fixThree of the four options can be struck out on the second codon alone (CCC is proline, not glycine). Decode one codon and eliminate rather than decoding all seven — that turns a 90-second question into a 20-second one.
In a gene encoding a polypeptide of 50 amino acids, the 25th codon (UAU) is mutated to UAA in the mRNA. What happens?
A polypeptide of 24 amino acids will be formed ← correct
Two polypeptides of 24 and 25 amino acids will be formed
A polypeptide of 49 amino acids will be formed
A polypeptide of 25 amino acids will be formed
The ribosome runs along the mRNA and halts at the new stop codon.
Solution
UAU codes for tyrosine; UAA is a stop codon — this is a nonsense mutation.
Translation proceeds normally through codons 1 to 24.
At codon 25 a release factor binds instead of a tRNA, and the chain is released.
Codon 25 therefore contributes no amino acid.
A polypeptide of 24 amino acids.
Where it went wrong · fixThe trap is answering 25 — the stop codon is at position 25, so the chain stops one short of it. Count what is made, not where the mutation sits.
Q125WrongNWrong processTermination of translation
The process of translation is terminated by
sigma factor
release factor ← correct
rho factor ← her answer
sense codon
Solution
Translation happens on the ribosome and makes protein; transcription happens on DNA and makes RNA.
Sigma factor initiates transcription.
Rho factor terminates transcription — her answer belongs to the wrong process.
A stop codon is read by a release factor, which ends translation.
Release factor.
Where it went wrong · fixThe four-word card: sigma starts transcription · rho ends transcription · release factor ends translation. Rho and release both mean 'the end', which is exactly why they get swapped.
Q132WrongNOver-rejectionTemplate polarity and polymerase direction
Statement I: The polarity of the template strand is always 3′ to 5′. Statement II: DNA polymerase adds new deoxynucleoside phosphates to the 3′ end of the terminal nucleotide of the existing strand.
Statement I is incorrect, but statement II is correct
Both statements are incorrect
Statement I is correct, but statement II is incorrect ← her answer
Both statements are correct ← correct
Solution
Statement I: the strand that is read must run 3′→5′, because the new strand is built 5′→3′ antiparallel to it. Correct.
Statement II: a new nucleotide can only be joined to a free 3′-OH. Correct.
Both are unchanged NCERT lines, and they are two sides of the same rule.
Both statements are correct.
Where it went wrong · fixShe answered this exact rule correctly as Q1 and Q2 of her own 8 October paper. Rejecting statement II here is the over-suspicion pattern: if you cannot name the word that was swapped, the statement is true.
Zoology — 106 / 180 · 14 wrong · 1 blank
Q141WrongSArrangement / orderChronology of the mammal line
Arrange in correct evolutionary order: A. Therapsids · B. Early reptiles (stem reptiles) · C. Pelycosaurs · D. Mammals
A → B → C → D
B → C → D → A ← her answer
B → A → C → D
B → C → A → D ← correct
Therapsids are named for being mammal-LIKE — they must sit immediately before mammals.
Solution
Stem reptiles (B) come first — every later group descends from them.
Pelycosaurs (C) are the earliest synapsids, the sail-backed forms.
Therapsids (A) are the mammal-like reptiles that arose from pelycosaurs.
Mammals (D) arose from therapsids, so D must be last.
B → C → A → D.
Where it went wrong · fixHer order puts mammals before the mammal-like reptiles they came from. On any sequence, fix the last item first — here the name 'mammal-like' settles it instantly.
Q151WrongNAttribution reversedHaeckel vs von Baer
How is Haeckel's view different from that of von Baer regarding embryological evidence for evolution?
Haeckel proposed that embryos pass through the adult stages of other animals ← correct
Haeckel proposed that embryos never pass through the adult stages of other animals ← her answer
von Baer proposed that embryos pass through the adult stages of other animals
von Baer disproved the presence of gill slits behind the head in human embryo
Solution
Haeckel observed the vestigial gill slits in all vertebrate embryos — that observation stands.
Haeckel's interpretation was that an embryo repeats the adult stages of its ancestors.
Von Baer disproved that interpretation: embryos never pass through the adult stages of other animals.
So the claim about passing through adult stages belongs to Haeckel.
First option.
Where it went wrong · fixShe answered this correctly twice on 8 October (Q56 and Q57, where she even underlined 'never'). Here the two names were swapped under time. Card: Haeckel claimed it · von Baer killed it.
The evolution of antibiotic-resistant bacteria is an example of
natural selection driven by anthropogenic action ← correct
artificial selection driven by anthropogenic action ← her answer
natural selection not involving anthropogenic action
artificial selection not involving anthropogenic action
Solution
Humans supply the antibiotic, so the pressure is anthropogenic — both of her candidate options agree on that half.
Artificial selection means humans deliberately choose which individuals breed, as in dogs or crop varieties.
Nobody selects which bacteria survive; the antibiotic kills the susceptible ones and the resistant ones simply reproduce.
The mechanism is therefore natural selection, merely with a human-made agent.
First option.
Where it went wrong · fixNCERT treats this and industrial melanism as the two standard examples of natural selection on a human-made pressure. Human-caused ≠ human-chosen — artificial selection needs deliberate breeding.
Q157WrongNNCERT listConvergent evolution — the Numbat
Which of the following best describes the Numbat?
A placental mammal adapted for ant-eating, found in Australia ← her answer
A marsupial adapted for ant-eating, found in Australia ← correct
A placental mammal adapted for aquatic life, found in Africa
A marsupial adapted for tree-climbing, found in South America
Solution
The Numbat appears in the NCERT table of Australian marsupials paired with their placental look-alikes.
Its placental counterpart is the anteater.
Australia's native mammals are overwhelmingly marsupials — that is the whole point of the convergence example.
Numbat = marsupial, ant-eating, Australian.
Second option.
Where it went wrong · fixThe convergence table is a memory list: marsupial mole / mole · marsupial mouse / mouse · numbat / anteater · bandicoot / rabbit · Tasmanian wolf / wolf. Every left-hand name is a marsupial.
Q159WrongSNegative stemDarwin's observations vs Lamarck's claim
Which of the following is NOT a factual observation on which the theory of natural selection is based?
Natural resources are limited. ← her answer
Population sizes are stable except for seasonal fluctuations.
Acquired characters are inherited by the offspring. ← correct
Most variations among individuals are inherited.
Solution
The question asks for the statement that is not one of Darwin's observations.
Limited resources, stable population sizes and heritable variation are the three standard observations behind natural selection.
'Acquired characters are inherited' is Lamarck's claim, and it was rejected.
It therefore cannot be a basis for natural selection.
Third option.
Where it went wrong · fixShe picked a statement that is one of Darwin's observations, which means the word NOT was read but not applied. Circle NOT / EXCEPT / INCORRECT in the stem and write the word 'false' at the top of the option list before choosing.
Q160WrongCOver-claim wordsWhat a fossil in a layer actually tells you
Fossils found in a particular sedimentary layer most likely represent
organisms whose species necessarily originated during the formation of that sediment ← her answer
organisms that died during the deposition of that sediment ← correct
organisms that died long after that sediment had hardened into rock
organisms whose species necessarily became extinct during the formation of that sediment
Solution
A fossil forms when a dead organism is buried by sediment being laid down.
So the layer dates the death and burial of that individual.
It says nothing about when the species first appeared — it may have existed long before.
It says nothing about extinction either — the species may still be alive.
Second option.
Where it went wrong · fixBoth of the options she had to choose between contain the word necessarily. In NEET, 'necessarily' and 'only' almost always mark an over-claim — the safer option is the plain descriptive one.
Q162WrongCTrue sentence, wrong theoryTheory of special creation
Statement I: According to the theory of special creation, all present-day species were created in their existing forms. Statement II: The theory holds that the Earth is billions of years old, allowing sufficient time for species to evolve.
Both statements are correct ← her answer
Both statements are incorrect
Statement I is correct, but statement II is incorrect ← correct
Statement I is incorrect, but statement II is correct
Solution
Statement I is the core of special creation: species were created as they are and have not changed. Correct.
Special creation dates the Earth at about 4000 years, not billions.
'Billions of years, allowing time for species to evolve' is the opposite of special creation — it is the evolutionary view.
Statement II is therefore incorrect.
I correct, II incorrect.
Where it went wrong · fixStatement II describes the rival theory. On a two-statement question about a named theory, check each statement against that theory specifically — a sentence can be perfectly true in general and still be wrong as a description of the theory being asked about.
Q167WrongCIntuition over evidenceDinosaurs and their living relatives
Dinosaurs are most closely related to which of the following living groups?
Lizards and snakes ← her answer
Turtles and mammals
Crocodiles and birds ← correct
Amphibians and lizards
Solution
Dinosaurs belong to the archosaur line of reptiles.
The two surviving archosaur groups are the crocodiles and the birds.
Birds in particular descend directly from theropod dinosaurs — Archaeopteryx is the NCERT link fossil.
Lizards and snakes are squamates, a separate reptile branch.
Crocodiles and birds.
Where it went wrong · fix'Dinosaur looks like a big lizard' is the intuition the question is testing. The chapter's own evidence — Archaeopteryx, the reptile-to-bird link — points at birds, so the answer is in the syllabus even if the relationship feels odd.
Q169WrongCDivergent vs convergentDarwin's finches — which phenomena
The evolution of Darwin's finches with different beak forms illustrates which phenomena? A. Adaptive radiation · B. Convergent evolution · C. Divergent evolution · D. Natural selection · E. Artificial selection
A, B and D only ← her answer
A, C and E only
B, D and E only
A, C and D only ← correct
Solution
A — one ancestral seed-eating stock radiating into many forms on one island group is the definition of adaptive radiation ✓
C — one ancestor diverging into several forms is divergent evolution ✓
D — the beak forms were selected by the available food, so natural selection ✓
B — convergent evolution is unrelated forms coming to resemble each other; these finches are closely related, so it is the opposite ✗
E — nobody bred them, so not artificial selection ✗ ⟹ A, C and D.
Where it went wrong · fixDivergent and convergent are the pair she has now crossed twice in one paper (also Q173). Fix it with the ancestor: one ancestor splitting = divergent · different ancestors meeting = convergent.
In a population of human babies, birth weight follows stabilising selection. Which correctly describes the outcome observed?
Babies with birthweight around the mean showed higher mortality
Babies with birthweight around the mean showed lower mortality ← correct
Babies with higher birthweight than the mean showed lower mortality
Babies with birthweight at both extremes showed lower mortality
The curve narrows around the mean. 'Lower mortality at the mean' is the same statement read from the survival side.
Solution
Stabilising selection favours the mean and removes both extremes.
'Favoured' expressed as mortality means the mean has the lowest mortality.
Both the very light and the very heavy babies die more often.
The curve therefore becomes taller and narrower around the mean.
Second option.
Where it went wrong · fixHuman birth weight is NCERT's own example of stabilising selection, and she answered the curve-shape version of this correctly on 8 October (Q47). The only new step is translating 'favoured' into 'lower mortality'.
Q173WrongCDivergent vs convergentDivergent evolution
The process by which organisms with a common evolutionary history evolve different phenotypic adaptations in response to different environmental challenges is called
genetic drift
genetic equilibrium
convergent evolution ← her answer
divergent evolution ← correct
Solution
The stem says common evolutionary history — one ancestor.
It then says they evolve different adaptations — they move apart.
One ancestor + moving apart = divergent evolution (homologous organs).
Convergent evolution is the reverse: different ancestors arriving at similar forms (analogous organs).
Divergent evolution.
Where it went wrong · fixThe phrase 'common evolutionary history' is the giveaway and it is in the first line of the question. Underline the ancestry phrase first, then pick the direction word.
Q176WrongNLamarck vs DarwinLamarck's explanation
Which sequence best represents LAMARCK's explanation for the evolution of long necks in giraffes?
Random appearance of short- and long-necked individuals → inheritance of the variation → survival of all variants
Pre-existing variation in neck length → greater reproductive success of long-necked giraffes → natural selection ← her answer
Short-necked giraffes → single large mutation → long-necked giraffes
Repeated use of the neck to reach high leaves → acquired elongation → inheritance by succeeding generations ← correct
Solution
Lamarck's mechanism is use and disuse followed by the inheritance of acquired characters.
The giraffe stretches its neck to reach high leaves — that is the use.
The elongation acquired in its lifetime is then passed to the offspring.
Her chosen option is Darwin's explanation: pre-existing variation plus natural selection.
Fourth option.
Where it went wrong · fixThe two explanations of the same giraffe are the classic paired question. Lamarck = acquired during life, then inherited. Darwin = already there, then selected. Check the first step of the chain: 'pre-existing' can never be Lamarck.
Q177WrongSReference point missedReading the time reference
About how many years AFTER THE FORMATION OF EARTH did the first cellular form of life possibly originate?
500 million years
2000 million years ← her answer
2500 million years ← correct
4500 million years
Solution
NCERT: the Earth formed about 4500 million years ago.
The first cellular forms of life appeared about 2000 million years ago.
The question does not ask when — it asks how long after the Earth formed.
4500 − 2000 = 2500 million years.
Third option.
Where it went wrong · fixBoth numbers she needed were in her head — she answered '2000 million years ago' correctly on her own 8 October paper (Q69). The question changed the reference point, and 2000 was sitting there as the familiar-looking option.
Q178WrongSNegative stemMutation theory — the incorrect statement
Select the INCORRECT statement regarding mutation theory.
Hugo de Vries proposed the mutation theory.
Mutations arise suddenly in a population. ← her answer
Mutations are random and directionless.
Mutations cause continuous variation over generations. ← correct
Solution
De Vries proposed mutation theory from his work on evening primrose. ✓ true
He held that mutations are sudden, single-step and large — saltation. ✓ true
He held they are random and directionless, unlike Darwin's small directional variations. ✓ true
Continuous variation is Darwin's; de Vries's mutations are explicitly discontinuous. ✗ false
Fourth option is the incorrect statement.
Where it went wrong · fixShe picked a true statement, so again the word INCORRECT was not applied (see Q159). The contrast to hold: Darwin small, continuous, directional · de Vries large, discontinuous, random.
Q179WrongCHomologous vs analogousAnalogous organs
Pairs of organs: A. Wings of bats and wings of beetles · B. Forelimbs of horse and forelimbs of bat · C. Eye of octopus and eye of mammals · D. Heart of man and heart of frog · E. Flippers of penguins and dolphins. Select the correct set of ANALOGOUS organs.
A, C, E only ← correct
A, B, D only
B, C, D only
A, C, D only ← her answer
Sort every pair by asking about the BUILD first, never the function.
Solution
Analogous = same function, different structure (convergent evolution).
A — bat wing is a bony forelimb, beetle wing is a chitinous flap. Analogous ✓
C — octopus and mammal eyes are built completely differently. Analogous ✓
E — penguin and dolphin flippers: bird and mammal, independently streamlined. Analogous ✓
B and D share the vertebrate bone plan and the vertebrate heart plan — those are homologous. ⟹ A, C and E.
Where it went wrong · fixShe swapped E for D. The heart of man and frog are the same organ in two vertebrates — different chamber counts, but one plan. Ask 'can I name a shared plan?' before 'do they do the same job?'
5. Fix list
Ray Optics is the single biggest number on this page. Ten blanks plus two wrong = 48 marks in one chapter. Re-solve Q27, 28, 30, 31, 32, 33, 34, 36, 38, 39, 41 and 44 cold, then one 30-question timed set on mirrors, slabs, TIR and prisms. Nothing else in Physics needs attention this week.
Circle the negative word in every stem. NOT · INCORRECT · EXCEPT · necessarily · only. Three questions on this paper (Q159, Q160, Q178) were lost to true statements being chosen. Write "false wanted" at the top of the option list.
Re-learn Evolution from reworded questions, not from NCERT sentences. Her recall is fine in NCERT's own phrasing and breaks when the same fact is asked sideways. Every revision question this week should be a paraphrase.
One card for the pairs that keep crossing: divergent (one ancestor splitting) vs convergent (different ancestors meeting) · Lamarck (acquired, then inherited) vs Darwin (pre-existing, then selected) · Haeckel (claimed it) vs von Baer (disproved it) · homologous (shared plan) vs analogous (shared job).
Chemistry: the -ate suffix. Q61 and Q69 are the same rule eight questions apart. Any complex that is the anion takes zincate, argentate, ferrate, cuprate. Scan all four options for an anionic complex before reading anything else.
Botany: both blanks before you match. Q100 was lost because the first blank was right and the second was never checked.
Carry forward what worked: Physics cleared its target, every Assertion–Reason and match-the-following question on the paper came back correct, and the three sections written before Zoology all finished within reach of target.
Error tags: C concept or reasoning · N NCERT recall · S stem or arrangement slip · T not attempted. Marks: correct × 4, wrong × (−1), blank 0. All numerical answers independently verified. Figures animate; the static layer of each carries the full message, and animation is suppressed when the device asks for reduced motion.