ACET Jun 2026 Solution

FOR ACET JUN EXAM PREPARATION

ACET Jun 2026
Solution
EXAM YEAR TYPE

ACET Jun 2026 Solution

Notes · Sample Papers · Previous Year Papers · Mock Tests

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Institute of Actuaries aof
ACET June 2026 Solutions
Q.1 Minimum value of f(x) = |3x² – 6x + 5|

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The expression inside is g(x) = 3x² – 6x + 5 = 3(x–1)² + 2. Since 3(x–1)² ≥ 0, the minimum of g(x) is 2 at x = 1. Since g(1) = 2

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> 0, the absolute value does not change it. So min f(x) = 2.

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Answer: A (2)
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Q.2 f(x) = x³ – 6x² + 9xl+ 2 strictly decreasing on: a
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f'(x) = 3x² – 12x + 9 = 3(x–1)(x–3). f'(x) < 0 when 1 < x < 3. So f is strictly decreasing on (1, 3).

Answer: A (1, 3)

Q.3 Number of real roots of 2x³ + 3x + 5 = 0

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Let f(x) = 2x³ + 3x + 5. Then f'(x) = 6x² + 3 > 0 for all x. So f is strictly increasing, and a strictly monotone cubic has exactly

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one real root. Check: f(0) = 5 > 0, f(–1) = –2 + (–3) + 5 = 0. So x = –1 is the only real root.

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Answer: C (Exactly 1 real root)

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Q.4 General term: T(r+1) = C(5,r) · x^(5−r) · 2^r
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For x³ → 5 − r = 3 → r = 2

Coefficient = C(5,2) · 2² = 10 · 4 = 40

Answer: B) 40

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Q.5 lim(x→0) [sin(5x) – sin(3x)] / x

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Using the standard limit lim(x→0) sin(kx)/x = k: [sin(5x) – sin(3x)] / x → 5 – 3 = 2.

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laQ.6 Equation with roots a², b² when a, b are roots of x² – 8x + 15 = 0
Answer:
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From the original: a + b = 8, ab = 15. New sum = a² + b² = (a+b)² – 2ab = 64 – 30 = 34. New product = a²b² = (ab)² = 225.
New equation: x² – 34x + 225 = 0.

Answer: A (x² – 34x + 225 = 0)

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Q.7 Unit vector in direction of v⃗ = 5i + 12j

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|v| = √(25 + 144) = √169 = 13. Unit vector = (5/13)i + (12/13)j.

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Answer: A ((5/13)i + (12/13)j)

Q.8 Distributive laws for dot and cross products

Both the dot product and cross product are distributive over addition.

Statement X: a · (b + c) = a·b + a·c — TRUE.

Statement Y: a × (b + c) = a×b + a×c — TRUE.

So both hold.

Answer: C (Both X and Y)

Q.9 |2A| for 3×3 matrix A with |A| = 4

For an n×n matrix, |kA| = k^n · |A|. Here n = 3, k = 2: |2A| = 2³ · 4 = 8 · 4 = 32.

Answer: C (32)

Q.10 adj(adj(M)) for 3×3 matrix M

For an n×n invertible matrix, adj(adj(M)) = |M|^(n–2) · M. For n = 3: adj(adj(M)) = |M|^1 · M = |M| · M.

Answer: B (|M| · M)

Q.11 1 + ω + ω² for cube root of unity ω ≠ 1

A fundamental property of the cube roots of unity: 1 + ω + ω² = 0.

Answer: B (0)

Q.12 |z| where z = (2 + 3i)(1 – i)

Using |z₁z₂| = |z₁||z₂|: |2 + 3i| = √(4+9) = √13. |1 – i| = √(1+1) = √2. So |z| = √13 · √2 = √26.

Answer: B (√26)

Q.13 Trapezoidal rule with h = 0.5 for ∫₀² (x³ + 1) dx

Points: 0, 0.5, 1, 1.5, 2. f(x) = x³+1 values: f(0)=1, f(0.5)=1.125, f(1)=2, f(1.5)=4.375, f(2)=9. T = (0.5/2)[1 +
2(1.125+2+4.375) + 9] = 0.25[1 + 2(7.5) + 9] = 0.25[1+15+9] = 0.25·25 = 6.25. B matches (exact = [x⁴/4+x]₀² = 4+2 = 6).
The computed value is 6.25.

Answer: B (6.25)

Q.14 Newton–Raphson: solve x² – 7 = 0 with x₀ = 3

f(x) = x² – 7, f'(x) = 2x. x₁ = x₀ – f(x₀)/f'(x₀) = 3 – (9–7)/6 = 3 – 2/6 = 3 – 0.333... = 2.667 ≈ 2.67.

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as Answer: A (2.67)
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Q.15 dy/dx at x = 0 given eˣ = cos y

Differentiate: eˣ = –sin(y) · dy/dx. So dy/dx = –eˣ / sin(y). At x = 0: e⁰ = 1, so cos y = 1, y = 0. sin(0) = 0. The derivative is
undefined (denominator is 0). None of the listed values apply.

Answer: D (None of the above — derivative is undefined at x = 0)

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Q.16 Common ratio of GP with a₄ = 24 and a₇ = 3

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a₇/a₄ = r³ = 3/24 = 1/8. So r = (1/8)^(1/3) = 1/2.

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Answer: A (1/2)

Q.17 Evaluate between the limits 0 and π.

Let I = , limits 0 to π.

Using integration by parts:

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Let u = x → du = dx

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Let dv = sin(x) dx → v = −cos(x)

So:
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I = , evaluated from 0 to π a g
I = [−x·cos(x) + sin(x)] from 0 to π

Substituting limits:

I = (−π·cos π + sin π) − (−0·cos 0 + sin 0)

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I = (−π·(−1) + 0) − (0 + 0)

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I=π

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laQ.18 Evaluate: 𝑥 ⋅ 𝑒 𝑑𝑥 ag
Answer:

ag ∫
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0
𝑥

Integration by parts: u = x, dv = eˣ dx.

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Then 𝑑𝑢 = 𝑑𝑥, 𝑣 = 𝑒 𝑥 . 𝑥𝑒 𝑥 𝑑𝑥 = [𝑥𝑒 𝑥 ]10 − ∫0 𝑒 𝑥 𝑑𝑥 = 𝑒 − [𝑒 𝑥 ]10 = 𝑒 − (𝑒 − 1) = 1.

So the answer is A (1)

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Q.19 Value of 1² + 2² + 3² + ... + 20² is:
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Using the standard formula: sum of squares of first n natural numbers = n(n+1)(2n+1) / 6

Here n = 20:

= 20 × 21 × 41 / 6

= 17220 / 6

= 2870

Answer: A

Q.20 Local minimum of f(x) = x⁴ – 4x³ + 6

f'(x) = 4x³ – 12x² = 4x²(x–3). Critical points: x = 0 and x = 3. f''(x) = 12x² – 24x. f''(0) = 0 (inconclusive — check sign change:
f'(x) = 4x²(x–3) is negative for x < 3 (x≠0), so x=0 is not a local min). f''(3) = 108 – 72 = 36 > 0, so x = 3 is a local minimum.

Answer: C (x = 3)

Q.21 Outcomes are 1, 2, 3, 4 each with probability 1/4. With an even number of equally likely outcomes, the median is
the average of the two middle values: (2+3)/2 = 2.5.

Answer: B (2.5)

Q.22 Distribution shape given mean = 50, median = 48, mode = 45

When mean > median > mode, the distribution is positively skewed (right-skewed). The longer tail is on the right.

Answer: C (Positively skewed)

Q.23 5-member team from 6 boys and 4 girls with at least 2 girls

Cases: (2 girls, 3 boys) = C(4,2)·C(6,3) = 6·20 = 120. (3 girls, 2 boys) = C(4,3)·C(6,2) = 4·15 = 60. (4 girls, 1 boy) =
C(4,4)·C(6,1) = 1·6 = 6. Total = 120 + 60 + 6 = 186.

Answer: C (186)

Q.24 Combined mean of Batch X (n=30, mean=60) and Batch Y (n=20, mean=75)

Combined mean = (30×60 + 20×75) / (30+20) = (1800 + 1500) / 50 = 3300/50 = 66.

Answer: B (66**)**

Q.25 P(A|B) given P(A)=0.4, P(B)=0.6, P(A∪B)=0.76

P(A∩B) = P(A) + P(B) – P(A∪B) = 0.4 + 0.6 – 0.76 = 0.24. P(A|B) = P(A∩B)/P(B) = 0.24/0.6 = 0.4. Since 0.4 = 2/5, this is not
in the given options A–C (1/6, 1/4, 1/3). None of the above.

Answer: D (None of the above — P(A|B) = 2/5 = 0.4)

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as Q.26 E[Y] for coin-then-die experiment (4-sided die)
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If Heads: E[Y|H] = (1+2+3+4)/4 = 2.5. If Tails: E[Y|T] = 2×2.5 = 5. E[Y] = 0.5×2.5 + 0.5×5 = 1.25 + 2.5 = 3.75.

Answer: A (3.75)

Q.27 IQR of X ~ Uniform(–1, 3)

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For Uniform(a,b): Q1 = a + 0.25(b–a), Q3 = a + 0.75(b–a). Here a=–1, b=3, b–a=4. Q1=–1+1=0, Q3=–1+3=2. IQR = Q3–Q1

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= 2.

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Answer: B (2)
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Q.28 P(T > 10) for Exp(mean=5)

Rate λ = 1/5. P(T > t) = e^(–t/5). P(T > 10) = e^(–10/5) = e^(–2).

Answer: A (e^(–2))

Q.29 Std dev of Y = 3X + 2 where X ~ N(5, 16)

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Var(X) = 16, so SD(X) = 4. Var(Y) = 3²·Var(X) = 9·16 = 144. SD(Y) = 12.

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Answer: A (12)

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Q.30 A fair coin is tossed 8 times. The probability of getting

X ~ Binomial(n=8, p=1/2)

P(X = 5) = C(8,5) · (1/2)⁵ · (1/2)³

= C(8,5) · (1/2)⁸

= 56 · (1/256)

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= 56/256

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= 7/32
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laQ.31 λ for Poisson where P(X=2) = P(X=3) ag
Answer:

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e^(–λ)λ²/2! = e^(–λ)λ³/3!. Cancel common factor: 1/2 = λ/6 ⟹ λ = 3.

Answer: B (3)

Q.32 Variance of X (exponential) given P(X>3)/P(X>6) = e^(3/2)

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P(X>t) = e^(–t/θ). Ratio = e^(–3/θ)/e^(–6/θ) = e^(3/θ) = e^(3/2). So 3/θ = 3/2 ⟹ θ = 2. Variance = θ² = 4.

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Answer: B (4)

Q.33 P(X is even) from given CDF

PMF: P(0)=0.15, P(1)=0.30, P(2)=0.30, P(3)=0.20, P(4)=0.05. Even values: 0, 2, 4.

P(even) = 0.15+0.30+0.05 = 0.50.

Answer: B (0.50)

Q.34 Median of X from same CDF

Median = smallest x with F(x) ≥ 0.5. F(1) = 0.15+0.30 = 0.45 < 0.5. F(2) = 0.75 ≥ 0.5. So median = 2.

Answer: B (2)

Q.35 Mode of stem-and-leaf data

Data: 15, 18, 22, 25, 25, 29, 30, 33, 37, 37, 37, 41, 44, 46, 48, 48, 50, 50, 52. Frequencies: 25→2, 37→3, 48→2, 50→2.
The value 37 appears 3 times — it is the unique mode.

Answer: A (The dataset has a unique mode of 37)

Q.36 Correlation coefficient: 3Y – 2X + 4 = 0, sₓ=3, s_Y=2

Rewrite as Y = (2/3)X – 4/3. Slope b_{Y|X} = 2/3. Formula: b_{Y|X} = ρ · (sy/sₓ) = ρ · (2/3). So ρ = (2/3) ÷ (2/3) = 1. Since
|ρ|=1 ≤ 1, this is valid. ρ = 1.

Answer: B (ρ = 1)

Q.37 ρ² given two regression lines Y = 2X+1 and X = 0.4Y+0.5

b_{Y|X} = 2, b_{X|Y} = 0.4. ρ² = b_{Y|X} · b_{X|Y} = 2 × 0.4 = 0.8.

Answer: A (0.8)

Q.38 Cov(X, X²) for fair 6-sided die

E[X] = 3.5. E[X²] = (1+4+9+16+25+36)/6 = 91/6. E[X³] = (1+8+27+64+125+216)/6 = 441/6 = 73.5. Cov(X,X²) = E[X³] –
E[X]·E[X²] = 73.5 – 3.5×(91/6) = 73.5 – 318.5/6 = 73.5 – 53.083... = 20.417. Option D matches 20.42 (The question uses 6-
sided die so answer is D).

Answer: D (Cov = 245/12 ≈ 20.42 for 6-sided die)

Q.39 Independence: P(A)=0.5, P(B)=0.6, P(A∩B)=0.30

Check X: P(A)P(B) = 0.5×0.6 = 0.30 = P(A∩B) ✓ Independent. Check Y: P(Bᶜ)=0.4. P(A∩Bᶜ) = P(A)–P(A∩B)=0.5–0.30=0.20.
P(A)P(Bᶜ)=0.5×0.4=0.20 ✓ Independent. Both X and Y are true.

Answer: A (Both X and Y)

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as Q.40 Expected winnings on biased 4-sided spinner
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Weights: k=1→1, k=2→2, k=3→3, k=4→4. Total = 10. P(k) = k/10. E[winnings] = Σ k·(k/10) = Σ k²/10 = (1+4+9+16)/10 =
30/10 = 3.

Answer: B (3)

Q.41 Total units sold in March

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March sales: North=55, South=65, East=45, West=50. Total = 55+65+45+50 = 215 (thousands).
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Answer: C (215)
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Q.42 Largest absolute increase in sales from January to April

North: 60–45=+15. South:70 –60=+10. East: 40–35=+5. West: 55–50=+5. North shows the highest with + 15. North is A
and is the correct choice here.

Answer: A (North — increase of 15 thousand units)

Q.43 Total profit in May using profit margins

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May sales: North=65, South=60, East=50, West=70 (thousands). Margins: N=18%, S=22%, E=15%, W=20%. Profits:

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N=65×0.18=11.7, S=60×0.22=13.2, E=50×0.15=7.5, W=70×0.20=14.0. Total=11.7+13.2+7.5+14.0=46.4.

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Answer: D (Total = 46.4 thousand)

Q.44 Month with highest combined sales a g
Jan:45+60+35+50=190. Feb:50+55+40+60=205. Mar:55+65+45+50=215. Apr:60+70+40+55=225. May:65+60+50+70=245.
Highest is May.

Answer: D (May)

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Q.45 Month with smallest temperature gap between City P and Q

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Differences: Jan:20–12=8, Feb:21–14=7, Mar:24–18=6, Apr:26–23=3, May:27–28=–1 (Q < P), Jun:29–32=–3. The absolute

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gap is smallest in April (3°C) or May (1°C). May has the smallest gap of 1°C.

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Answer:

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a Q.46 First month City P exceeds City Q
Temperatures: Jan P=12<Q=20, Feb P=14<Q=21, Mar P=18<Q=24, Apr P=23<Q=26, May P=28>Q=27. City P first exceeds
City Q in May.

Answer: D (May)

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Q.47 Sharpest month-on-month increase for City P

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City P changes: Jan→Feb: +2, Feb→Mar: +4, Mar→Apr: +5, Apr→May: +5, May→Jun: +4. The sharpest increase is +5°C,
occurring in both Mar–Apr and Apr–May. The first occurrence is Mar–Apr.

Answer: C (Mar–Apr)

Q.48 Month with highest combined temperature

Combined: Jan=32, Feb=35, Mar=42, Apr=49, May=55, Jun=61. Highest is June.

Answer: C (June)

Q.49 Department with highest average attendance

Commerce: (88+85+87+90)/4=350/4=87.5. Science: (76+80+78+82)/4=316/4=79.0. Arts: (91+89+93+92)/4=365/4=91.25.
Law: (84+88+86+85)/4=343/4=85.75. Highest is Arts.

Answer: C (Arts)

Q.50 Department with lowest attendance

Commerce : 85% in February

Science : 76% in January

Arts : 89% in February

Law : 84% in January

Science has the lowest with 76% in January

Q.51 Department with highest gap in attendance (largest range)

Ranges: Commerce: 90–85=5. Science: 82–76=6. Arts: 93–89=4. Law: 88–84=4.

Science is highest with 6% range difference

Answer: B (Science)

Q.52 Best word: 'The scientist's ______ research led to a breakthrough.'

'Diligent' means hardworking and careful — perfectly fitting for research leading to a breakthrough. The other options
(superficial, erratic, negligent) imply carelessness.

Answer: B (diligent)

Q.53 Synonym of 'verbose'

'Verbose' means using more words than necessary — i.e., wordy.

Answer: B (wordy)

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as Q.54 Opposite of 'benevolent'
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'Benevolent' means kind and generous. Its antonym is 'malevolent' — wishing evil upon others.

Answer: C (malevolent)

Q.55 Error in: 'Neither of the students have submitted their assignment on time.'

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'Neither' is singular and takes a singular verb. The error is 'have submitted' — it should be 'has submitted'.

Answer: B (have submitted)
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Q.56 Fill in: 'The committee reached a ______ after hours of debate.'
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'Consensus' means aggeneral agreement reached by a group — exactly what a committee achieves after debate.
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Answer: A (consensus)

Q.57 person who hates mankind is called a ______.

A 'misanthrope' is a person who dislikes humankind and avoids human society.

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Answer: B (misanthrope)

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Q.58 Meaning of 'beats around the bush'
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This idiom means avoiding the main point or topic — talking around it without addressing it directly.

Answer: B (Avoiding the main topic)

Q.59 Doctor : Stethoscope :: Carpenter : ?

A doctor uses a stethoscope as their primary professional tool. A carpenter's primary tool is a hammer.

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Answer: A (Hammer)

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Q.60 Which is NOT a valid inference from the renewable energy passage?

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passage says critics argue renewable energy 'cannot yet reliably meet peak demand', which

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a Answer: C ('Renewable energy is already fully capable of meeting all energy demands' — not supported)
Q.61 Correct sequence of jumbled sentences

Logical order: II (customers complained about cost) → III (sales dropped) → IV (management ordered review) → I (this
led to revision of pricing strategy). Sequence: II, III, IV, I.

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Answer: A (II, III, IV, I)

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Q.62 Which sentences are erroneous?

I: 'more smarter' is doubly comparative — should be 'smarter'. II: 'refused accepting' should be 'refused to accept'. III:
'worked hardly' should be 'worked hard' (hardly means 'barely'). All three are erroneous.

Answer: D (I, II and III — all three)

Q.63 Year when 15th August again falls on Thursday (given 2024 = Thursday)

Normal year: +1 day. Leap year: +2 days (for dates after Feb 29). Leap years: 2028, 2032, 2036. 2024=Thu. 2025:+1=Fri.
2026:+1=Sat. 2027:+1=Sun. 2028(L):+2=Tue. 2029:+1=Wed. 2030:+1=Thu ✓. The answer is 2030.

Answer: A (2030)

Q.64 How many times do hands form a straight line between 6:00 and 7:00 p.m.?

The hands are in a straight line (180°) once per hour. Between 6:00 and 7:00 p.m., the hands form a straight line exactly
once (at approximately 6:32).

Answer: B (1)

Q.65 Cubes with exactly 2 painted faces in a 4×4×4 cube cut into 1×1×1 pieces

Cubes with paint on exactly 2 faces are the edge cubes (excluding corners). A 4×4×4 cube has 12 edges. Each edge of a 4-
cube has 4–2=2 non-corner cubes. So total = 12×2 = 24.

Answer: B (24)

Q.66 If A is C's mother's brother's daughter, then C is A's:

C's mother's brother is C's maternal uncle. His daughter is C's cousin. So A is C's cousin, and C is also A's cousin.

Answer: B (Cousin)

Q.67 Seating arrangement: 6 students, F at left end, D immediately right of C, A right of D

F is at position 1. C–D must be consecutive. A is to the right of D. B is to the right of A. In all valid arrangements, D's only
left neighbour is C, B is to the right of A, which is to the right of D. So B cannot sit next to D (B must be placed
elsewhere). Option C ('B does not sit next to D') must be true.

Answer: C (B does not sit next to D

Q.68 Syllogism: All pilots are engineers; Some engineers are doctors; No doctor is a teacher

From (2): Some engineers are doctors. From (3): No doctor is a teacher. Therefore, those engineers who are doctors are
not teachers. So 'Some engineers are not teachers' is definitely true.

Answer: C (Some engineers are not teachers)

Q.69 Logical sequence: Recruitment, Training, Appraisal, Promotion

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Natural order of employment cycle: Interview (I) → Recruitment (R) → Training (T) → Promotion (P).

Answer: A (I, R, T, P)

Q.70 Wrong number in series: 2, 5, 11, 23, 47, 96, 191

Pattern: ×2+1 each time. 2→5(×2+1)✓, 5→11(×2+1)✓, 11→23(×2+1)✓, 23→47(×2+1)✓, 47→95(×2+1)=95, but series
shows 96 ✗. Then 95→191(×2+1)✓. So the wrong number is 96 (should be 95).

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Answer: C (96 — should be 95)

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