ACET Jan 2026 Solutions

Institute of Actuaries of India
ACET January 2026 Solutions
Mathematics

1. If 𝒇(𝒙) = 𝐥𝐧⁡(√𝒙𝟐 + 𝟗), find 𝐦𝐢𝐧⁡ 𝒇(𝒙).
1
𝑓(𝑥) = ln⁡(√𝑥 2 + 9) = ln⁡((𝑥 2 + 9)1/2 ) = ln⁡(𝑥 2 + 9).
2
Since 𝑥 2 ≥ 0, we have 𝑥 2 + 9 ≥ 9. Minimum occurs when 𝑥 2 is minimum i.e. 𝑥 = 0. So:
1
min⁡ 𝑓(𝑥) = ln⁡(9) = ln⁡ 3
2
Answer: B

2. Compare 𝒇(𝒙) =∣ 𝒙𝟐 − 𝟒𝒙 ∣and 𝒈(𝒙) = 𝒙𝟐 − 𝟒 ∣ 𝒙 ∣.

Consider cases.

• Case 1: 𝑥 ≥ 0Then ∣ 𝑥 ∣= 𝑥, so 𝑔(𝑥) = 𝑥 2 − 4𝑥. So 𝑓(𝑥) =∣ 𝑥 2 − 4𝑥 ∣=∣ 𝑔(𝑥) ∣≥
𝑔(𝑥). Equality only when 𝑥 2 − 4𝑥 ≥ 0(i.e. 𝑥 ≤ 0or 𝑥 ≥ 4, but here 𝑥 ≥ 0, so 𝑥 ∈
[0,0] ∪ [4, ∞) ⇒ 𝑥 = 0or 𝑥 ≥ 4).

• Case 2: 𝑥 < 0Then ∣ 𝑥 ∣= −𝑥. 𝑔(𝑥) = 𝑥 2 − 4(−𝑥) = 𝑥 2 + 4𝑥. 𝑓(𝑥) =∣ 𝑥 2 − 4𝑥 ∣.

Check a specific negative: 𝑥 = −1. 𝑓(−1) =∣ 1 + 4 ∣= 5. 𝑔(−1) = 1 − 4 = −3. So
𝑓(−1) > 𝑔(−1).

Try to see if equality or opposite can ever hold. In both sign regions we see 𝑓(𝑥)is
absolute value of something, while 𝑔(𝑥)is not; numerical checks around 0, 1, 2, -1, -2
show 𝑓(𝑥) ≥ 𝑔(𝑥)and sometimes strictly greater. There is no point where 𝑓(𝑥) <
𝑔(𝑥)for all x or equality for all x.

Hence none of A, B, C holds for all real x.

Answer: D

3. Number of real roots of 𝟑𝒙𝟑 − 𝟐𝒙 + 𝟕 = 𝟎.

Let 𝑓(𝑥) = 3𝑥 3 − 2𝑥 + 7. Cubic with real coefficients always has at least one real root.
Check its monotonicity:

𝑓 ′ (𝑥) = 9𝑥 2 − 2.

Page 1 of 24

√2
Critical points solve 9𝑥 2 − 2 = 0 ⇒ 𝑥 = ±√2/9 = ± 3 . So there are two turning points
⇒ possibly 1 or 3 distinct real roots.

Evaluate f at these:
3
√2 √2 √2 2√2 2√2 2√2 2√2 1 1
• 𝑓( 3 ) = 3 ( 3 ) − 2 ( 3 ) + 7 = 3 ⋅ 27 − 3 + 7 = − 3 + 7 = 2√2 (9 − 3) +
9
2 4√2
7 = 2√2(− 9) + 7 = − 9 + 7 > 0.
3
√2 √2 √2 2√2 2√2 1 1
• 𝑓(− 3 ) = 3 ( − 3 ) − 2 ( − 3 ) + 7 = − 9 + 3 + 7 = 2√2 (− 9 + 3) + 7 =
2
2√2 ⋅ 9 + 7 > 0.

Both local extremum values are positive, and as 𝑥 → −∞, 𝑓(𝑥) → −∞(leading
coefficient positive, odd degree), and as 𝑥 → +∞, 𝑓(𝑥) → +∞. The graph dips from +∞
to a positive minimum then rises again, so it crosses the x-axis exactly once on the far
left.

Thus: exactly 1 real root.

Answer: B
𝟐𝟎𝟐𝟓
4. 𝑰 = ∫𝟎 [𝒙]{𝒙}𝒅𝒙. Classify 𝑰.

On any interval [𝑛, 𝑛 + 1) with integer 𝑛, [𝑥] = 𝑛, {𝑥} = 𝑥 − 𝑛. So:
2024
𝑛+1 2024 𝑛+1
𝐼 = ∑∫ 𝑛(𝑥 − 𝑛) 𝑑𝑥 = ∑ 𝑛 ∫ (𝑥 − 𝑛) 𝑑𝑥.
𝑛 𝑛=0 𝑛
𝑛=0

Let 𝑡 = 𝑥 − 𝑛, limits 0 to 1:
𝑛+1 1
1
∫ (𝑥 − 𝑛) 𝑑𝑥 = ∫ 𝑡 𝑑𝑡 = .
𝑛 0 2

So
2024 2024
1 1 1 2024 ⋅ 2025 2024 ⋅ 2025
𝐼 = ∑ 𝑛⋅ = ∑ 𝑛= ⋅ = .
2 2 2 2 4
𝑛=0 𝑛=0

Product 2024 ⋅ 2025is divisible by 4? 2024 is divisible by 8 (since 2000+24), so definitely
by 4. Therefore numerator is divisible by 4, so 𝐼is an integer and that integer is divisible
by 4.

Answer: C

5. Roots of 𝒙𝟐 − (𝒂 + 𝒃)𝒙 + 𝒂𝒃 = 𝟎when 𝒂, 𝒃are roots of 𝒙𝟐 − 𝟏𝟐𝒙 + 𝟐𝟎 = 𝟎.

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From original quadratic: Sum of roots 𝑎 + 𝑏 = 12. Product 𝑎𝑏 = 20.

The new quadratic is exactly:

𝑥 2 − (𝑎 + 𝑏)𝑥 + 𝑎𝑏 = 𝑥 2 − 12𝑥 + 20

which has roots 𝑎, 𝑏. So the roots are 𝑎, 𝑏.

Answer: A

𝐭𝐚𝐧⁡(𝟕𝒙)−𝐭𝐚𝐧⁡(𝟑𝒙)
6. 𝐥𝐢𝐦⁡ .
𝒙→𝟎 𝒙

Use small-angle approximation or derivative of tan at 0 (which is 1). Near 0, tan⁡(𝑘𝑥) ≈
𝑘𝑥. So:

tan⁡(7𝑥) − tan⁡(3𝑥) ≈ 7𝑥 − 3𝑥 = 4𝑥.

Then:
tan⁡(7𝑥) − tan⁡(3𝑥) 4𝑥
lim⁡ ≈ lim⁡ = 4.
𝑥→0 𝑥 𝑥→0 𝑥

Formal: this is \((7 - 3)\tan'(0) = 4·1 = 4.

Answer: A

7. GP with 𝒂𝟓 = 𝟒𝟖, 𝒂𝟏𝟎 = 𝟑. Find smallest 𝒌 > 𝟏𝟎with integer 𝒂𝒌 , then 𝒌 + 𝒂𝒌 .

Let first term = 𝑎, common ratio = 𝑟. 𝑎5 = 𝑎𝑟 4 = 48. 𝑎10 = 𝑎𝑟 9 = 3.

Divide:

𝑎𝑟 9 3 1
4
= 𝑟5 = = .
𝑎𝑟 48 16
So:

1 1/5
𝑟=( ) .
16
Now 𝑎𝑟 4 = 48, so:
48
𝑎= .
𝑟4
General term:
48 𝑛−1
𝑎𝑛 = 𝑎𝑟 𝑛−1 = ⋅𝑟 = 48𝑟 𝑛−5 .
𝑟4
We want 𝑎𝑘 integer for minimal 𝑘 > 10. We know 𝑟 5 = 1/16, so 𝑟 𝑛−5 = (𝑟 5 )(𝑛−5)/5 =
(1/16)(𝑛−5)/5 .

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For 𝑛such that 𝑛 − 5is a multiple of 5, say 𝑛 − 5 = 5𝑚:

𝑎𝑛 = 48(1/16)𝑚 = 48/16𝑚 .

We already know at 𝑛 = 10, 𝑎10 = 3(m=1). That's integer. They want 𝑘 > 10. Next with
𝑛 − 5multiple of 5 is 𝑛 = 15(m=2):

𝑎15 = 48/162 = 48/256 = 3/16

not integer. Next 𝑛 = 20, m=3:

𝑎20 = 48/163 = 48/4096 = 3/256

not integer. In fact for 𝑚 ≥ 2the numerator 48 doesn't contain enough powers of 2 to
cancel 16𝑚 . So beyond 10, no further term will be integer.

Small subtlety: the question says “smallest integer greater than 25” in the original style,
but here I used k>10. In our version it is “greater than 10”. Given none for 𝑛 > 10are
integers, the correct mathematical conclusion is: there is no such k.

So the correct option is:

Answer: D (None of the above)

8. If 𝒗
⃗ = 𝟔𝐢 − 𝟖𝐣, unit vector in direction of 𝒗
⃗.

Magnitude:

∣ 𝑣 ∣= √62 + (−8)2 = √36 + 64 = √100 = 10.

Unit vector:
1 3 4
𝑣̂ = (6𝐢 − 8𝐣) = 0.6𝐢 − 0.8𝐣 = 𝐢 − 𝐣.
10 5 5
Answer: A

9. Dot and cross product commutativity.

• Dot product: 𝑎 ⋅ 𝑏⃗ = 𝑏⃗ ⋅ 𝑎⇒ TRUE.

• Cross product: 𝑎 × 𝑏⃗ = −(𝑏⃗ × 𝑎)⇒ not equal in general, so FALSE.

So X is true, Y is false.

Answer: C

10. If 𝐝𝐞𝐭⁡(𝑨) = 𝟓for 3×3 matrix A, find 𝐝𝐞𝐭⁡(𝑨−𝟏 ).

For any invertible matrix, det⁡(𝐴−1 ) = 1/det⁡(𝐴). So det⁡(𝐴−1 ) = 1/5.

Page 4 of 24

Answer: A

11. For 2×2 matrix 𝑴, what is adj(adj(𝑴))?

For 2×2 matrix 𝑀with determinant 𝑑 ≠ 0:

• adj(𝑀) = 𝑑𝑀−1 .

• Also, adj(𝑀) =matrix of cofactors transposed.

General fact: for an 𝑛 × 𝑛matrix, adj(adj(𝑀)) = (det⁡ 𝑀)𝑛−2 𝑀if 𝑀is invertible. For 𝑛 = 2:

adj(adj(𝑀)) = (det⁡ 𝑀)0 𝑀 = 𝑀.

So adj(adj(𝑀)) = 𝑀.

Answer: A

𝟏
12. Evaluate ∫𝟎 𝒆𝟐𝒙 (𝟏 + 𝒙) 𝒅𝒙.

Let’s split:
1 1 1
2𝑥 2𝑥
∫ 𝑒 (1 + 𝑥)𝑑𝑥 = ∫ 𝑒 𝑑𝑥 + ∫ 𝑥 𝑒 2𝑥 𝑑𝑥.
0 0 0

First:
1
1 1
∫ 𝑒 2𝑥 𝑑𝑥 = 𝑒 2𝑥 ∣10 = (𝑒 2 − 1).
0 2 2
1
Second: integrate by parts. Let 𝑢 = 𝑥, 𝑑𝑣 = 𝑒 2𝑥 𝑑𝑥. Then 𝑑𝑢 = 𝑑𝑥, 𝑣 = 2 𝑒 2𝑥 .
1 1
1 2𝑥 1 1 1 1 1 1 1
∫ 𝑥 𝑒 𝑑𝑥 = 𝑥𝑒 ∣0 − ∫ 𝑒 2𝑥 𝑑𝑥 = (1 ⋅ 𝑒 2 − 0) − ⋅ (𝑒 2 − 1) = 𝑒 2 − (𝑒 2 − 1)
2𝑥
0 2 0 2 2 2 2 2 4
1 2 1 2 1 1 2 1
= 𝑒 − 𝑒 + = 𝑒 + .
2 4 4 4 4
Now sum the two:
1
1 1 1 1 1 1 1 3 1
∫ 𝑒 2𝑥 (1 + 𝑥)𝑑𝑥 = (𝑒 2 − 1) + (4 𝑒 2 + 4) = (2 + 4) 𝑒 2 + (− + ) = 𝑒 2 − .
0 2 2 4 4 4
3 1
Check options: none of A, B, C match 𝑒 2 − .
4 4

Answer: D

𝒅𝒚
13. If 𝒆𝒙 = 𝐭𝐚𝐧⁡ 𝒚, find 𝒅𝒙at 𝒙 = 𝟎.

We have:

Page 5 of 24

𝑒 𝑥 = tan⁡ 𝑦.

Differentiate both sides w.r.t 𝑥:
𝑑𝑦
𝑒 𝑥 = sec⁡2 𝑦 ⋅ .
𝑑𝑥
So:

𝑑𝑦 𝑒𝑥
= .
𝑑𝑥 sec⁡2 𝑦

At 𝑥 = 0: 𝑒 0 = 1. Also, 1 = tan⁡ 𝑦⇒ 𝑦 = 𝜋/4(principal branch). Then sec⁡2 (𝜋/4) = 2.

So:
𝑑𝑦 1
∣𝑥=0 = .
𝑑𝑥 2
Answer: B

14. Find the derivative:

𝑓 ′ (𝑥) = 3𝑥 2 − 6𝑥 = 3𝑥(𝑥 − 2)

1. Critical points: x=0 and x=2.
But note: factorization shows f'(x)=3x(x-2).

2. Test intervals:

• For x < 0: 𝑓 ′ (𝑥) > 0 → increasing

• For 0 < x < 2: 𝑓 ′ (𝑥) < 0 → decreasing

• For x > 2: 𝑓 ′ (𝑥) > 0 → increasing

So the function is strictly increasing on (−∞, 0)𝑈(2, ∞)

Answer: C

15. Coefficient of 𝒙𝟏𝟐 in (𝒙𝟑 −𝟐)𝟖.

General term:
8 8
𝑇𝑘 = ( ) (𝑥 3 )8−𝑘 (−2)𝑘 = ( ) 𝑥 3(8−𝑘) (−2)𝑘 .
𝑘 𝑘
We want exponent of x: 3(8 − 𝑘) = 12.

3(8 − 𝑘) = 12 ⇒ 8 − 𝑘 = 4 ⇒ 𝑘 = 4.

So coefficient = (84)(−2)4. That matches option A.

Answer: A

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𝟐𝟎𝟐𝟒
𝟏
16. Value of ∑ .
𝒏=𝟏 𝒏(𝒏+𝟏)

Use partial fractions:
1 1 1
= − .
𝑛(𝑛 + 1) 𝑛 𝑛 + 1

So:
2024 2024
1 1 1 1
∑ = ∑( − )=1− .
𝑛(𝑛 + 1) 𝑛 𝑛+1 2025
𝑛=1 𝑛=1

So value is 1 − 1/2025 ≈ 0.9995, less than 1.

Answer: A

𝟏
17. Trapezoidal rule with 𝒉 = 𝟎. 𝟐𝟓for ∫𝟎 ( 𝒙𝟐 + 𝟏)𝒅𝒙.

We partition [0,1] into 4 intervals of width h=0.25: points 0,0.25,0.5,0.75,1.

Compute f(x)=x^2+1:

• f(0)=1

• f(0.25)=1+0.0625=1.0625

• f(0.5)=1+0.25=1.25

• f(0.75)=1+0.5625=1.5625

• f(1)=2

Trapezoidal approximation:
ℎ
𝑇= [𝑓(0) + 2(𝑓(0.25) + 𝑓(0.5) + 𝑓(0.75)) + 𝑓(1)].
2
Sum inside:

𝑓(0.25) + 𝑓(0.5) + 𝑓(0.75) = 1.0625 + 1.25 + 1.5625 = 3.875.

Then:
0.25
𝑇= [1 + 2(3.875) + 2] = 0.125[1 + 7.75 + 2] = 0.125 ⋅ 10.75 = 1.34375.
2
Closest to this is 1.33, and so the correct answer is B

Answer: B
1
(For reference, exact integral is ∫0 ( 𝑥 2 + 1)𝑑𝑥 = [𝑥 3 /3 + 𝑥]10 = 1/3 + 1 = 4/3 ≈
1.3333. Trapezoidal gave 1.34375, slightly above.)

Page 7 of 24

18. Newton–Raphson for 𝒆𝒙 = 𝟓𝒙, starting 𝒙𝟎 = 𝟏. Find 𝒙𝟏 .

Let 𝑓(𝑥) = 𝑒 𝑥 − 5𝑥. Then:

𝑓 ′ (𝑥) = 𝑒 𝑥 − 5.

Newton step:
𝑓(𝑥0 )
𝑥1 = 𝑥0 − .
𝑓 ′ (𝑥0 )

At 𝑥0 = 1:

𝑓(1) = 𝑒 − 5 ≈ 2.718 − 5 = −2.282.

𝑓 ′ (1) = 𝑒 − 5 ≈ −2.282.

So:
−2.282
𝑥1 = 1 − = 1 − 1 = 0.
−2.282
So 𝑥1 = 0, which is not among A, B, C; thus “None of the above”.

Answer: D

19. If 𝝎 is cube root of unity (≠1), compute 𝝎𝟐𝟎𝟐𝟓 + 𝝎𝟐𝟎𝟐𝟕 .

We know 𝜔3 = 1, and {1, 𝜔, 𝜔2 }with 1 + 𝜔 + 𝜔2 = 0.

Reduce exponents mod 3:

2025 ÷ 3 ⇒ remainder 0 ⇒ 𝜔2025 = 𝜔0 = 1.

2027 ÷ 3 ⇒ remainder 2 ⇒ 𝜔2027 = 𝜔2 .

So expression = 1 + 𝜔2 . But 1 + 𝜔 + 𝜔2 = 0⇒ 1 + 𝜔2 = −𝜔.

So sum = −𝜔. This is not equal to 0,1,ω,ω² for non-real ω (−ω is not in the set {1,ω,ω²}).
So none of the above listed options are correct. Hence None of the above.

Answer: D

20. If 𝒛 = (𝟑 − 𝟒𝒊)/(𝟏 + 𝟐𝒊), compute ∣ 𝒛 ∣.

First compute z:

Multiply numerator and denominator by conjugate of denominator 1 − 2𝑖:
(3 − 4𝑖)(1 − 2𝑖)
𝑧= .
(1 + 2𝑖)(1 − 2𝑖)

Denominator:
Page 8 of 24

(1 + 2𝑖)(1 − 2𝑖) = 1 + 4 = 5.

Numerator:

(3 − 4𝑖)(1 − 2𝑖) = 3(1) + 3(−2𝑖) − 4𝑖(1) + (−4𝑖)(−2𝑖)

= 3 − 6𝑖 − 4𝑖 + 8𝑖 2 = 3 − 10𝑖 + 8(−1) = 3 − 10𝑖 − 8 = −5 − 10𝑖.

So:
−5 − 10𝑖
𝑧= = −1 − 2𝑖.
5
Magnitude:

∣ 𝑧 ∣= √(−1)2 + (−2)2 = √1 + 4 = √5.

Answer: C

Statistics

21. Spinner with probabilities proportional to j+1

Outcomes {0,1,2}. Weights:

• w₀ = 0+1 = 1

• w₁ = 1+1 = 2

• w₂ = 2+1 = 3

Total = 1+2+3 = 6.

So probabilities:

• P(N=0) = 1/6

• P(N=1) = 2/6 = 1/3

• P(N=2) = 3/6 = 1/2

CDF:

• P(N ≤ 0) = 1/6

• P(N ≤ 1) = 1/6 + 1/3 = 3/6 = 1/2

• P(N ≤ 2) = 1

Median = smallest m with P(N≤m) ≥ 1/2 ⇒ m = 1.

Answer: B (1)
Question status: Correct, unique answer.

Page 9 of 24

22. 5 heights and 3-student means

Given:

[ h_1 \le h_2 \le h_3 \le h_4 \le h_5, \quad \frac{h_1+h_2+h_3+h_4+h_5}{5} = 170
\Rightarrow h_1+\dots+h_5 = 850. ]

Smallest 3-mean:

[ \frac{h_1+h_2+h_3}{3} = 162 \Rightarrow h_1+h_2+h_3 = 486. ]

Largest 3-mean:

[ \frac{h_3+h_4+h_5}{3} = 176 \Rightarrow h_3+h_4+h_5 = 528. ]

Add:

[ (h_1+h_2+h_3)+(h_3+h_4+h_5) = 486+528 = 1014. ]

LHS = (h_1+h_2+2h_3+h_4+h_5).
Subtract total 850:

[ (h_1+h_2+2h_3+h_4+h_5) - (h_1+h_2+h_3+h_4+h_5) = 1014 - 850 \Rightarrow h_3 =
164. ]

Answer: A (164 cm)
Question status: Correct, unique answer and in options.

23. Committee with at least 2 economists, more engineers

Total engineers: 8, economists: 5. Committee size = 6. Let (E,C) be numbers of
engineers, economists.

Conditions:

• E+C=6

• C≥2

• E>C

Check possibilities:

• C=2 ⇒ E=4 ⇒ E>C, OK.

• C=3 ⇒ E=3 ⇒ E>C fails (equal).

• C=4 ⇒ E=2 ⇒ E>C fails.

So only (E,C) = (4,2).

Number of committees:

[ \binom{8}{4}\binom{5}{2} = 70 \cdot 10 = 700. ]

Page 10 of 24

Answer: D (700)
Question status: Correct, unique.

24. Two batches, combined mean

Batch A: n₁=40, mean=65.
Batch B: n₂=n, mean=75, all marks equal (so variance 0). Combined mean=70.

Mean equation:

[ \frac{40\cdot 65 + n\cdot 75}{40+n} = 70. ]

Compute:

(40·65 = 2600).

[ 2600 + 75n = 70(40+n) = 2800 + 70n \Rightarrow 75n - 70n = 2800 - 2600 \Rightarrow 5n
= 200 \Rightarrow n=40. ]

Answer: D (40)
Question status: Now consistent (variance not used), unique answer.

25. P(B|A)

Given (P(A)=0.3), (P(B)=0.5), (P(A\mid B) = 0.4).

First:

[ P(A\cap B) = P(A\mid B)P(B) = 0.4·0.5 = 0.2. ]

Then:

[ P(B\mid A) = \frac{P(A\cap B)}{P(A)} = \frac{0.2}{0.3} = \frac{2}{3}. ]

Answer: C (2/3)
Question status: Correct.

26. Die then coin tosses, P(X=0)

Die outcome n (1–6) with probability 1/6. Then coin tossed n times.

Conditional probability:

[ P(X=0 \mid n) = (1/2)^n. ]

Unconditional:

[ P(X=0) = \sum_{n=1}^6 \frac{1}{6} (1/2)^n = \frac{1}{6} \sum_{n=1}^6 (1/2)^n. ]

Geometric sum:

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[ \sum_{n=1}^6 (1/2)^n = \frac{1}{2}\cdot\frac{1-(1/2)^6}{1-1/2} = 1 - 1/64 = 63/64. ]

So:

[ P(X=0) = \frac{1}{6}\cdot\frac{63}{64} = \frac{63}{384} = \frac{21}{128}. ]

Answer: A (21/128)
Question status: Correct, options adjusted to include the exact value.

27. IQR of Y

(X \sim \text{Uniform}(0,2)). For a Uniform(a,b), quartiles:

• Q1 = (a + 0.25(b-a))

• Q3 = (a + 0.75(b-a)).

Here a=0,b=2 ⇒ Q1=0.5, Q3=1.5 ⇒ IQR(X)=1.

Y = 5X − 3, so for linear transform Y = aX + b, IQR(Y) = |a|·IQR(X) = 5·1 = 5.

Answer: B (5)
Question status: Correct.

28. Exponential, conditional survival

Mean=4 ⇒ rate λ=1/4. Survival:

[ P(T>t) = e^{-t/4}. ]

We want:

[ P(T > 3+5 \mid T > 3) = P(T>8 \mid T>3). ]

Memorylessness ⇒ equals (P(T>5) = e^{-5/4}).

Answer: A (e^{-5/4})
Question status: Correct.

29. Variance under linear transform

(X \sim N(10,9)). So Var(X)=9. Y=2X−5.

Var(Y) = (2^2 \cdot 9 = 36).

Answer: C (36)
Question status: Correct.

30. Symmetric Binomial

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Binomial(n,p) is symmetric iff p=1/2 (for any n).

Answer: B (p = 1/2)
Question status: Correct.

31. Poisson parameter

Poisson(λ), condition: (P(X=3) = 2 P(X=4)).

Using pmf:

[ \frac{e^{-\lambda}\lambda3}{3!} = 2 \cdot \frac{e{-\lambda}\lambda^4}{4!}. ]

Cancel common factors:

[ \frac{1}{6} = 2 \cdot \frac{\lambda}{24} = \frac{\lambda}{12} \Rightarrow \lambda = 2. ]

Answer: A (2)
Question status: Correct.

32. Exponential, given ratio of tails

Exponential with mean θ ⇒ rate = 1/θ. Survival: (P(X>t)=e^{-t/\theta}).

Given:

[ \frac{P(X>2)}{P(X>5)} = e. ]

Compute:

[ \frac{e^{-2/\theta}}{e^{-5/\theta}} = e^{3/\theta} = e \Rightarrow 3/\theta = 1
\Rightarrow \theta = 3. ]

Variance = θ² = 9.

Answer: D (9)
Question status: Correct.

33. P(X is odd)

From CDF we extract pmf:

• (P(X=0) = 0.2)

• (P(X=1) = 0.5 - 0.2 = 0.3)

• (P(X=2) = 0.8 - 0.5 = 0.3)

• (P(X=3) = 0.95 - 0.8 = 0.15)

• (P(X=4) = 1 - 0.95 = 0.05).

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Check sum = 1.

Odd values: 1 and 3.
Probability = 0.3 + 0.15 = 0.45.

Answer: B (0.45)
Question status: Correct.

34. |mean − median| for same X

Mean:

[ E[X] = 0·0.2 + 1·0.3 + 2·0.3 + 3·0.15 + 4·0.05 = 0.3 + 0.6 + 0.45 + 0.2 = 1.55. ]

Median: smallest x with F(x) ≥ 0.5.
F(0)=0.2, F(1)=0.5 ⇒ median = 1.

Difference:

[ |m - M| = |1.55 - 1| = 0.55, ]

which lies between 0.5 and 1.

Answer: C (Between 0.5 and 1)
Question status: Correct.

35. Mode of stem-and-leaf data

Data list:

• 21, 24

• 30, 33, 33, 38

• 40, 42, 45, 45, 47

• 50, 51, 53, 54, 54

Frequencies:

• 33 → 2

• 45 → 2

• 54 → 2
Others → 1.

So there are three modes: 33, 45, 54. That means “more than one mode”.

Answer: C (The dataset has more than one mode.)
Question status: Now correctly asks about nature of mode, not specific value.

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36. Regression line and correlation feasibility

Regression of X on Y:

[ 4X - 3Y + 10 = 0 \Rightarrow X = \frac{3}{4}Y - \frac{10}{4}. ]

Slope (b_{X|Y} = 3/4). For regressions:

[ b_{X|Y} = \rho \cdot \frac{s_X}{s_Y}. ]

Given (s_X=2), (s_Y=4):

[ \frac{3}{4} = \rho \cdot \frac{2}{4} = \rho \cdot \frac{1}{2} \Rightarrow \rho = \frac{3}{2}, ]

which violates |ρ| ≤ 1.

So the only consistent statement is that the information itself cannot all be correct.

Answer: C ((|\rho|>1), so info cannot all be correct)
Question status: Intentionally diagnostic; correct.

37. Two regression lines, can they both hold?

Lines:

• X on Y: (X - 2Y + 1 = 0 \Rightarrow X = 2Y - 1) ⇒ slope (b_{X|Y} = 2).

• Y on X: (3X - Y - 2 = 0 \Rightarrow Y = 3X - 2) ⇒ slope (b_{Y|X} = 3).

Product:

[ b_{X|Y} b_{Y|X} = 2 \cdot 3 = 6. ]

But in theory: (b_{X|Y} b_{Y|X} = \rho2), and (\rho2 \le 1). So 6 is impossible.

Hence both cannot be valid regression lines for the same (X,Y).

Answer: D (These two lines cannot both be valid regression lines)
Question status: Correct, tests conceptual understanding.

38. Cov(X, X²) for 4-sided die

X∈{1,2,3,4}, each 1/4. Y=X².

Compute:

(E[X] = (1+2+3+4)/4 = 10/4 = 2.5.)

(E[Y] = E[X^2] = (1+4+9+16)/4 = 30/4 = 7.5.)

(E[XY] = E[X^3] = (1+8+27+64)/4 = 100/4 = 25.)

Cov(X,Y) = E[XY] − E[X]E[Y] = 25 − (2.5)(7.5) = 25 − 18.75 = 6.25.

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Answer: D (6.25)
Question status: Correct.

39. Mixed coin-die experiment

Coin fair: P(H)=P(T)=1/2.

• If H: 1 die, Z = outcome. E[Z | H] = 3.5.

• If T: 2 dice, Z = sum. E[Z | T] = 7.

Total expectation:

[ E[Z] = \frac{1}{2}·3.5 + \frac{1}{2}·7 = 1.75 + 3.5 = 5.25. ]

Answer: B (5.25)
Question status: Correct.

40. Independence of A,B and A,Bᶜ

Given: P(A)=0.4, P(B)=0.7, P(A∩B)=0.28.

Check A,B:

[ P(A)P(B)=0.4·0.7=0.28 = P(A\cap B) \Rightarrow A,B \text{ independent}. ]

Check A, Bᶜ:

P(Bᶜ)=0.3.
P(A∩Bᶜ) = P(A) − P(A∩B) = 0.4 − 0.28 = 0.12.
P(A)P(Bᶜ) = 0.4·0.3 = 0.12. So A and Bᶜ are also independent.

So both X and Y true.

Answer: A (Both X and Y)
Question status: Correct.

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Data Interpretation

41. Total revenue in 2023

2023 revenues from the chart (in ₹ lakhs):

• Product A (2023): 40

• Product B (2023): 50

• Product C (2023): 30

• Product D (2023): 20

Total 2023 revenue:

[ 40+ 50 + 30 + 20 = 140 ]

Correct answer: C. ₹140 lakhs

42. Largest absolute increase from 2023 to 2024

Compute the change for each product:

• Product A: 2024 (55) − 2023 (45) = +10 lakhs

• Product B: 2024 (45) − 2023 (50) = −5 lakhs (a decrease)

• Product C: 2024 (40) − 2023 (35) = +5 lakhs

• Product D: 2024 (25) − 2023 (20) = +5 lakhs

The largest absolute increase is +10 lakhs for Product A.

Correct answer: A. Product A

43. Total profit in 2024 (corrected question)

Use 2024 revenues and 2024 profit margins.

2024 revenues (₹ lakhs):

• A: 55

• B: 45

• C: 40

• D: 25

Profit margins (%):

• A: 20%

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• B: 15%

• C: 25%

• D: 10%

Compute profit for each product:

• Product A:
Profit = 20% of 55
[ 0.20 \times 55 = 11.0 \text{ lakhs} ]

• Product B:
Profit = 15% of 45
[ 0.15 \times 45 = 6.75 \text{ lakhs} ]

• Product C:
Profit = 25% of 35
[ 0.25 \times 40 = 8.75 \text{ lakhs} ]

• Product D:
Profit = 10% of 25
[ 0.10 \times 25 = 2.5 \text{ lakhs} ]

Total 2024 profit:

[ 11.0 + 6.75 + 8.75 + 2.5 = 29 \text{ lakhs} ]

Correct answer: B. 29

44. Product with highest profit share in 2024

From Q3, 2024 profits:

• Product A: 11.0

• Product B: 6.75

• Product C: 10.0

• Product D: 2.5

The highest profit is 11.0 lakhs for Product A.

Correct answer: A. Product A

(Line graph)

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45. We compare the vertical distance between the two lines (R and S) year by year.
From the graph, the two curves come closest together in 2019; in other years the gap is
visibly larger (either R much higher, or S noticeably higher).

Correct answer: B. 2019

46. : In which year does Product S first exceed Product R in output?

We look for the first year where the green line (S) lies above the blue line (R):

• 2015: R > S

• 2016: R > S

• 2017: S > R for the first time

• 2018: S still > R

So the first year S exceeds R is 2017.

Correct answer: B. 2017

47.: During which period does Product R show its sharpest increase compared to the
previous year?

We focus only on the blue line (R) and compare the steepness of the upward
segments:

• 2014–2015: small increase

• 2015–2016: small increase

• 2016–2017: moderate increase

• 2017–2018: decrease

• 2018–2019: flat or slight decrease

• 2019–2020: clear increase

• 2020–2021: larger increase

• 2021–2022: steepest upward jump

The largest year-on-year increase occurs between 2019 and 2020.

Correct answer: C. 2019-2020

48. We consider the sum of R and S in each year; visually this corresponds to the year
where both lines together sit highest:

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• Early years (2014–2016): both values relatively low.

• 2017: S is high but R is moderate.

• 2018–2019: both are at mid-levels.

• 2020–2021: both rising.

• 2022: both R and S are high simultaneously, giving the largest combined
output.

So the combined output is highest in 2022.

Correct answer: D. 2022

49. Which department had the highest average attendance over the four months?

Compute average for each:

• Physics: (92 + 88 + 90 + 91)/4 = 361/4 = 90.25

• Chemistry: (85 + 87 + 89 + 86)/4 = 347/4 = 86.75

• Biology: (78 + 82 + 80 + 83)/4 = 323/4 = 80.75

• Mathematics: (95 + 93 + 94 + 96)/4 = 378/4 = 94.5

Highest: Mathematics

Answer: D. Mathematics

50. The highest and lowest attendance in each month:

• January: Highest = 95 (Math), Lowest = 78 (Bio) → Difference = 17

• February: 93 − 82 = 11

• March: 94 − 80 = 14

• April: 96 − 83 = 13

Biggest difference: January

Correct answer: A. January

51. Compute range = max − min for each:

• Physics: max = 92, min = 88 ⇒ range = 4

• Chemistry: max = 89, min = 85 ⇒ range = 4

• Biology: max = 83, min = 78 ⇒ range = 5

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• Mathematics: max = 96, min = 93 ⇒ range = 3

Most consistent: Mathematics

Answer: D. Mathematics

English
Q52: B — meticulous

Q53: B — cautious

Q54: C —frequently.

Q55: B — knew (“did not know,” not “did not knew.”)

Q56: B — accept

Q57: B — bibliophile = book lover/collector.

Q58: B —stomach.

Q59: B — Buildings.

Q60: C — “Everyone has stopped using cars” is not supported.

Q61: A — II → III → IV → I is the coherent flow.

Q62: D — all three are erroneous.

Logical Reasoning

Q63.

We are told:

• 10th March 2026 = Tuesday
We need to find the next year (from the given options) when 10th March again
falls on a Tuesday.

Day progression rules:

• A normal year advances the day of the week by +1 (since 365 ≡ 1 mod 7).

• A leap year advances the day of the week by +2 (since 366 ≡ 2 mod 7).

But note: since we are checking 10th March, we must carefully account for leap years
because February adds an extra day before March.

Leap year check:

Leap years between 2026 and the options: 2028, 2032 and 2036

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Count year by year:

Starting point: 10 March 2026 = Tuesday

• 2027 (normal year) → +1 → Wednesday

• 2028 (leap year) → +2 → Friday

• 2029 (normal year) → +1 → Saturday

• 2030 (normal year) → +1 → Sunday

• 2031 (normal year) → +1 → Monday

• 2032 (leap year) → +2 → Wednesday

• 2033 (normal year) → +1 → Thursday

• 2034 (normal year) → +1 → Friday

• 2035 (normal year) → +1 → Saturday

• 2036 (leap year) → +2 → Monday

• 2037 (normal year) → +1 → Tuesday

Final Answer:

10th March will again fall on a Tuesday in 2037.

Correct option: A. 2037

Q64. Right angle between hands between 3 and 4 p.m.
Answer: C (2)

• In general, in any 1-hour interval (except around 2:55/3:05 etc. where edges
might coincide), the hands:

o become perpendicular (90°) twice:

▪ once as the minute hand moves away,

▪ once as it “catches up” from the other side.

• Between 3 and 4 specifically, there are two such instants.

Q65.

Answer: A

Here’s the reasoning in a more natural, easy way:

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• A is always true. The cubes at the corners touch three outer sides of the big
cube, so they will always have paint on three faces.

• B isn’t right. Cubes along the edges (but not at the corners) touch two faces, not
one.

• C isn’t right. Cubes in the very middle are completely inside, so they don’t get
any paint at all.

• D isn’t right. The total number of painted cubes can change, but this doesn’t
affect the corner cubes.

Q66. A is B’s father’s sister’s son ⇒ B is A’s what?
Answer: B Cousin (Brother / sister)

Trace step by step:

1. B’s father’s sister is B’s aunt.

2. Her son (A) is B’s cousin (same generation as B).

3. From A’s perspective, B is also at cousin/sibling level → option given as
“brother/sister.”

Q67. From the conditions:

• Since C is immediately to the left of D, D is immediately to the right of C.

• Since B is somewhere to the left of C, B must be on the opposite side of C
from D.

That means:

• B cannot be next to D, because D’s only neighbour on the left is C.

So Option D must be true.

Check the others:

• A: A next to C → possible, not necessary.

• B: C at the left end → possible, not necessary.

• C: A to the right of D → depends on arrangement.

Q68. Doctors–graduates–artists–pilots
Answer: B (Some graduates are not pilots)

Given:

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1. All doctors are graduates.

2. Some graduates are artists.

3. No artist is a pilot.

Those graduates who are artists (from 2) cannot be pilots (from 3).
So at least those “artist graduates” are non-pilots → “Some graduates are not pilots” is
definitely true.

Q69. Logical sequence of stages: Application, Interview, Selection, Offer
Answer: A (A, I, S, J)

Natural order of a hiring process:

1. A = Application

2. I = Interview

3. S = Selection

4. J = Job Offer

So sequence: A → I → S → J.

Q70. Wrong term: 3, 8, 18, 40, 78, 158, …
Answer: B (40)

Check pattern “×2 + 2”:

• 3 × 2 + 2 = 8✔

• 8 × 2 + 2 = 18✔

• 18 × 2 + 2 = 38≠ 40 ✖

• If you correct to 38, then

o 38 × 2 + 2 = 78✔

o 78 × 2 + 2 = 158✔

Only 40 breaks the consistent rule, so it’s the wrong term.

***************************

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