ACET 2023 Answer Key (Mar)

Institute of Actuaries of India
ACET March 2023 Solutions
Mathematics
1. C. By definition of greatest integer less than or equal to 𝑥,−1 < ⌊𝑥⌋ − 𝑥 ≤ 0 , for x
real. Since |⌊𝑥⌋ − 𝑥| = −( ⌊𝑥⌋ − 𝑥), it follows that
0 ≤ |⌊𝑥⌋ − 𝑥| < 1 .

𝑥 𝑥 2 𝑥2
2. A. 𝑓°𝑔(𝑥) = 𝑓(𝑔(𝑥)) = 𝑓 (𝑥−2) = [𝑥−2] + 4 = (𝑥−2)2 + 4.
Hence, 𝑓°𝑔(3) = 9 + 4 = 13.

3. D. Let 𝑥 = √72 + 𝑥. Since 𝑥 is positive, 𝑥 2 − 𝑥 − 49 = 0 giving
1+√1−4(1)(−49) 1+√197
𝑥= = .
2 2

4. B. The 𝑟 𝑡ℎ term in (𝑎 − 3𝑏)12 is 𝑇𝑟+1 = (12
𝑟
)𝑎𝑟 (−3𝑏)12−𝑟 =
(12
𝑟
)𝑎𝑟 (−3)12−𝑟 𝑏12−𝑟 .
Thus, the coefficient of 𝑎5 𝑏 7 is obtained when 𝑟 = 5 in 𝑇𝑟+1 .
It is given by (12
5
)(−3)7 .

7 7
5. D. log 𝑚 7 − 3 log 𝑚 2 = 2 ⇒ log 𝑚 7 − log 𝑚 23 = 2 ⇒ log 𝑚 =2, giving 𝑚2 = .
8 8
7
Thus, 𝑚 = √8 since 𝑚 is positive.

6. A. Since sin 𝜃 and cos 𝜃 are the roots of the equation 𝑎𝑥 2 + 𝑏𝑥 + 1 = 0, we have
𝑏 1
sin 𝜃 + cos 𝜃 = − 𝑎 and sin 𝜃 cos 𝜃 = 𝑎.
𝑏2
Hence (sin 𝜃 + cos 𝜃) 2 = 𝑎2 .
𝑏2 1 𝑏2
Thus,sin2 𝜃 + cos2 𝜃 + 2 sin 𝜃 cos 𝜃 = 𝑎2 ⇒ 1 + 2 𝑎 = 𝑎2 ⇒ 𝑏 2 − 𝑎2 = 2𝑎.

7. C. Given that 𝑛 th term of an AP: 𝑎𝑛 = 3 + 4𝑛. The first term 𝑎1 = 7 and 𝑑 = 4.
15
Hence, the sum of the first 15 terms 𝑆15 = 2 [(2 × 7) + 14 × 4] = 525.

𝑖(𝑖+1) 𝑖 2 +𝑖 1 𝑛(𝑛+1)(2𝑛+1) 𝑛(𝑛+1)
8. B. ∑𝑛𝑖=1 ∑𝑖𝑗=1 𝑗 = ∑𝑛𝑖=1 = ∑𝑛𝑖=1 2 = 2 [ + ].
2 6 2
𝑛(𝑛+1) 2𝑛+1 2𝑛(𝑛+1)(𝑛+2)
= [ + 1] = = (𝑛+2
3
).
4 3 12

3+𝑖 3+𝑖 2+𝑖 6+5𝑖+𝑖 2 5𝑖+5
9. C. Given that 𝑧 = 2−𝑖, 𝑖 2 = −1. 𝑧 = 2−𝑖 . 2+𝑖 = = = 𝑖 + 1.
4−𝑖 2 5

𝑧 2 = (𝑖 + 1)2 = 𝑖 2 + 1 + 2𝑖 = 2 𝑖; 𝑧16 = (2𝑖)8 = 256.

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2 2
10. B. |𝑎⃗ + 𝑏⃗⃗ + 𝑐⃗| = |𝑎⃗|2 + |𝑏⃗⃗| + |𝑐⃗|2 + 2(𝑎⃗ . 𝑏⃗⃗ + 𝑎⃗. 𝑐⃗+𝑐⃗. 𝑎⃗) = 1 + 1 + 1 = 3 .
|𝑎⃗ + 𝑏⃗⃗ + 𝑐⃗| = √3.

11. B. ⃗⃗⃗⃗ − 2𝑖⃗ + 3𝑗⃗ − 4𝑘
In order that the vectors 𝑖⃗ − 𝑥𝑗⃗ + 3𝑘, ⃗⃗ and −𝑗⃗ + 2𝑘
⃗⃗ are coplanar,
1 −𝑥 3
we must have |−2 3 −4| = 0.
0 −1 2
1 −𝑥 3
Thus, |−2 3 −4| = 2 − 4𝑥 + 6 = 0 implying 𝑥 = 2.
0 −1 2

12. D. As there are three values,𝑦0 , 𝑦2 , 𝑦4 the function 𝑦 can be represented by a second
degree polynomial. Hence, ∆3 𝑦 = 0 for all 𝑦. The difference table is given below.

𝑦 ∆𝑦 ∆2 𝑦 ∆3 𝑦
𝑥0 3 𝑦1 − 3
𝑥1 𝑦1 5 − 2𝑦1
2 − 𝑦1 𝑦3 + 3𝑦1 − 9
𝑥2 2 𝑦3 − 4 + 𝑦1
𝑦3 − 2 8.4 − 3𝑦3 − 𝑦1
𝑥3 𝑦3 4. 4 − 2𝑦3
2.4 − 𝑦3
𝑥4 2.4
This gives 𝑦3 + 3𝑦1 − 9 = 0 and 8.4 − 3𝑦3 − 𝑦1 = 0. Solving these equations,
we have 𝑦1 = 2.325 and 𝑦3 = 2.025.

13. A. Towards computing the value of the integral, we need the following information.
0 1 2

𝑦0 = log e √1 + 0 = 0 𝑦1 = log e √1 + 1 𝑦2 = log e √1 + 2
= 0.3466 = 0.5493
2 1 1
∫0 log 𝑒 √1 + 𝑥 𝑑𝑥 = 3 [(𝑦0 + 𝑦2 ) + 4𝑦1 ] = 3 [0 + 0.5493) + 4 × 0.3466] =
0.6452 .
𝑥 6 −2 26 −2 31
14. B. . lim 4 = 4 = . (l'Hospital's rule is not applicable, as the expression does
𝑥→2 𝑥 −2 2 −2 7
0
not have the form 0 as 𝑥 → 2.)

sin 𝑥
15. C. Let 𝑦 = tan−1 (1+cos 𝑥). Then
𝑑𝑦 1 (1 + cos 𝑥) cos 𝑥 − sin 𝑥 (− sin 𝑥)
= 2 ×
𝑑𝑥 sin 𝑥 (1 + cos 𝑥)2
1+( )
1 + cos 𝑥
(1 + cos 𝑥)2 cos 𝑥 + cos 2 𝑥 + sin2 𝑥
= ×
1 + cos2 𝑥 + 2 cos 𝑥 + sin2 𝑥 (1 + cos 𝑥)2
1 + cos 𝑥 1
= = .
2 + 2 cos 𝑥 2

𝑥 𝑥2 𝑥𝑛 𝑑𝑦 1 2𝑥 3𝑥 2 𝑛𝑥 𝑛−1
16. D. Let 𝑦 = 1 + 1! + 2! + ⋯ + 𝑛! . = 1! + 2! + 3! + ⋯ + .
𝑑𝑥 𝑛!

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𝑑2 𝑦 𝑥 𝑥2 𝑥 𝑛−2
= 1 + 1! + 2! + ⋯ (𝑛−2)!.
𝑑𝑥 2

𝑑2 𝑦 𝑥 𝑛−1 𝑥𝑛
Hence, 𝑑𝑥 2 + (𝑛−1)! + 𝑛! = 𝑦.

1 1 1
17. A. Put √5 + log 𝑥 = 𝑡. Then 2𝑡 . 𝑥 𝑑𝑥 = 𝑑𝑡 ⇒ 𝑥 𝑑𝑥 = 2𝑡𝑑𝑡. Hence,
3
√5+log 𝑥 𝑡3 2
∫ 𝑑𝑥 = ∫ 2𝑡 2 𝑑𝑡 = 2 3 + 𝑐 = 3 (5 + log 𝑥)2 + 𝑐.
𝑥

18. B. Let 𝑓(𝑥) = 𝑥 15 cos6 𝑥 . 𝑓(−𝑥) = (−𝑥)15 cos6 (−𝑥) = − 𝑥15 cos 6 𝑥 = −𝑓(𝑥).
1
Hence 𝑓(𝑥)is an odd function. Thus ∫−1 𝑥15 cos 6 𝑥 𝑑𝑥 = 0.

4 2 1 2 5 4
19. A. Let 𝐴 = [ ] and 𝐵 = [ ]. Then, 𝐴 + 𝐵 = [ ].
1 1 0 1 1 2
The characteristic roots of 𝐴 + 𝐵 is the solution of the equation.
5−𝜆 4
| | = 0 ⇒ (5 − 𝜆)(2 − 𝜆) − 4 = 0 ⇒ 𝜆2 − 7𝜆 + 6 = 0.
1 2−𝜆
Thus, 𝜆 = 1,6. The characteristic roots are 1, 6.

20. C. In a diagonal matrix, if the diagonal elements are not zero, then its inverse is
also a diagonal matrix with entries on its diagonal being the reciprocal of the
corresponding diagonal elements of the given matrix. Thus, the inverse of the
1
0 0
3 0 0 3
1
matrix 𝐴 = [0 5 0] is 𝐴−1 0 5 0 .
0 0 7 1
[0 0 7 ]
35 0 0
Alternatively, |𝐴|=105. 𝐴𝑑𝑗 𝐴 = | 0 21 0 |.
0 0 15
1
0 0
35 0 0 3
1 1
Hence 𝐴−1 = 105 [ 0 21 0]= 0 5 0 .
0 0 15 1
[0 0 7 ]

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Statistics
21. D. Twenty books can be arranged among themselves in 20! ways.
The number of ways the books can be placed so that two particular books always be
together is 19! × 2!.
19!×2! 1
P(a particular pair of books is always together) = 20! = 10
P(a particular pair of books shall never be together)
= 1 – P(a particular pair of books is always together)
1 9
= 1 − 10 = 10.

22. A. Probability of drawing a red ball from the box containing 24 balls out which 𝑥 are
𝑥
red balls is 𝑝 = 24.
If 12 more red balls are put in the box, number of red balls = 𝑥 + 12 and total
number of balls = 24 + 12 = 36.
𝑥+12
The probability of drawing a red ball = = 2𝑝.
36
𝑥+12 2𝑥
So = 2𝑝 = 24. This implies 𝑥 = 6.
36

23. C. 𝑃(𝐸 ∩ 𝐹) = 𝑝, 𝑃(𝐸 ∩ 𝐹̅ ) = 𝑝, 𝑃(𝐸̅ ∩ 𝐹) = 𝑝.
𝑃(𝐸 ∪ 𝐹) = 𝑃(𝐸 ∩ 𝐹̅ ) + 𝑃(𝐸̅ ∩ 𝐹) + 𝑃(𝐸 ∩ 𝐹) = 3𝑝.

24. D. 𝐸 and 𝐹 are independent, so 𝑃(𝐸 ∩ 𝐹) = 𝑃(𝐸)𝑃(𝐹).
A. 𝑃(𝐸 ∩ 𝐹̅ ) = 𝑃(𝐸) − 𝑃(𝐸 ∩ 𝐹) = 𝑃(𝐸) − 𝑃(𝐸)𝑃(𝐹) = 𝑃(𝐸)(1 − 𝑃(𝐹)) =
𝑃(𝐸)𝑃(𝐹̅ ), so 𝐸 and 𝐹̅ are independent.
B. 𝑃(𝐸̅ ∩ 𝐹̅ ) = 1 − 𝑃(𝐸 ∪ 𝐹) = 1 − [𝑃(𝐸) + 𝑃(𝐹) − 𝑃(𝐸 ∩ 𝐹)] = 1 −
[𝑃(𝐸)(1 − 𝑃(𝐹)) + 𝑃(𝐹)] = 1 − 𝑃(𝐹) − 𝑃(𝐸)𝑃(𝐹̅ ) = 𝑃(𝐹̅ ) − 𝑃(𝐸)𝑃(𝐹̅ ) =
(1 − 𝑃(𝐸))𝑃(𝐹̅ ) = 𝑃(𝐸̅ )𝑃(𝐹̅ ).
C. 𝑃(𝐸 ∪ 𝐹) = 𝑃(𝐸) + 𝑃(𝐹) − 𝑃(𝐸 ∩ 𝐹) = 𝑃(𝐸) + 𝑃(𝐹) − 𝑃(𝐸)𝑃(𝐹) =
𝑃(𝐸) + 𝑃(𝐹)(1 − 𝑃(𝐸)) = 𝑃(𝐸) + 𝑃(𝐹)𝑃(𝐸̅ ).
D. 𝑃(𝐸|𝐹)𝑃(𝐹|𝐸) = 𝑃(𝐸)𝑃(𝐹).

25. B. Let 𝐸 be the event that number 5 appears at least once.
𝐹 be the event that the sum of the numbers appearing is 8.
𝐸 = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (1, 5), (2, 5), (3, 5), (4, 5), (6, 5)}
𝐹 = {(2, 6), (3, 5), (4, 4), (5, 3), (2, 6)}
11 5
𝑃(𝐸) = 36 , 𝑃(𝐹) = 36.
2
2 𝑃(𝐸∩𝐹) 36 2
𝐸 ∩ 𝐹 = {(3, 5), (5, 3)}. So 𝑃(𝐸 ∩ 𝐹) = . 𝑃(𝐸|𝐹) = = 5 = .
36 𝑃(𝐹) 5
36
0+1×30+2×25+3×12+4×11+5×2
26. D. The average number of accidents per day = = 1.7.
100
Median = (50th obs. + 51st obs. )/2 = (1 + 2)/2 = 1.5.
Mode = 1.

27. B. Mean = −1+1+(𝑛−2)×0 = 0.
𝑛
Mean absolute deviation about the mean
1 1+1 2
= 𝑛 (|−1 − 0| + |1 − 0| + (𝑛 − 2) × 0) = 𝑛 = 𝑛.
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28. B. Coefficient of variation = 𝑠 × 100. Standard deviation 𝑠 = 3.8.
𝑥̅
𝑠
Coefficient of variation cv = 𝑥̅ × 100 = 7.6. ⇒ 𝑥̅ = 50
If each observation is multiplied by 2, then new 𝑥̅ = 2 × 50 and 𝑠 = 2 × 3.8.
2×3.8
New CV = 2×50 × 100 = 7.6.

29. A. The probability distribution of 𝑋 is
1
𝑃(𝑋 = 𝑥) = 10 , 𝑥 = 21, 22, 23, 25, 27, 29, 30, 32, 35, 40.
𝑃(𝑋>25,𝑋≥23) 𝑃(𝑋>25)
𝑃(𝑋 > 25|𝑋 ≥ 23) = = 𝑃(𝑋≥23) .
𝑃(𝑋≥23)
6 8 6 3
𝑃(𝑋 > 25) = 10, 𝑃(𝑋 ≥ 23) = 10. 𝑃(𝑋 > 25|𝑋 ≥ 23) = 8 = 4.

𝑥
30. B. Mean of 𝑋 is ∫∞ 𝑥 𝑒 −𝜃 𝑑𝑥 = 𝜃 ∫∞ 𝑢𝑒 −𝑢 𝑑𝑢 = 𝜃. 1 = 𝜃.
0 𝜃 0
25 1 𝑥
The median of 𝑋 = 25. So 𝑃(𝑋 ≤ 25) = 0.50. This implies ∫0 𝑒 −𝜃 𝑑𝑥 = 0.50
𝜃
25 25 25
⇒ 1 − exp (− 𝜃 ) = 0.5 ⇒ exp (− 𝜃 ) = 0.5 ⇒ 𝜃 = log 2 .
𝑒

31. D. 𝑋 ∼ 𝐵𝑖𝑛𝑜𝑚𝑖𝑎𝑙 (𝑛, 𝑝). 𝐸(𝑋) = 𝑛𝑝 and Var(𝑋) = 𝑛𝑝𝑞.
4 8 𝑛𝑝𝑞 2 2 1 4
So 𝑛𝑝 = 3 and 𝑛𝑝𝑞 = 9. 𝑛𝑝 = 3 ⇒ 𝑞 = 3 ⇒ 𝑝 = 3. 𝑛𝑝 = 3 ⇒ 𝑛 = 4.
2 4 1 2 3 16 32
𝑃(𝑋 ≥ 2) = 1 − 𝑃(𝑋 = 0) − 𝑃(𝑋 = 1) = 1 − (3) − 4 (3) (3) = 1 − 81 − 81 =
33
.
81

32. D. 𝑉𝑎𝑟(𝑋 − 𝑌) = 𝑉𝑎𝑟(𝑋) − 2 𝑐𝑜𝑣(𝑋, 𝑌) + 𝑉𝑎𝑟(𝑌) = 𝜎12 − 2𝜌𝜎1 𝜎2 + 𝜎22 .
𝑉𝑎𝑟(𝑋) + 𝑉𝑎𝑟(𝑌) − 𝑉𝑎𝑟(𝑋 − 𝑌)
𝑉𝑎𝑟(𝑋) + 𝑉𝑎𝑟(𝑌) − 𝑉𝑎𝑟(𝑋 − 𝑌) = 2𝜌𝜎1 𝜎2 ⇒ 𝜌 = .
2𝜎1 𝜎2

33. C. A. 𝑋 ∼ 𝑃𝑜𝑖𝑠𝑠𝑜𝑛(3). 𝑃(𝑋 ≤ 1) = 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) = 𝑒 −3 + 3𝑒 −3 = 4𝑒 −3
B. Suppose 𝑋 ∼ 𝑃𝑜𝑖𝑠𝑠𝑜𝑛 (𝜆). 𝑃(𝑋 ≥ 1) = 1 − 𝑒 −𝜆
𝑃(𝑋 ≥ 1) = 1 − 𝑒 −𝜆 = 1 − 𝑒 −3 ⇒ 𝜆 = 3. So variance = 3.

C. Suppose 𝑋 ∼ 𝑃𝑜𝑖𝑠𝑠𝑜𝑛 (𝜆).
𝜆2 𝜆
𝑃(𝑋 = 2) = 2 𝑃(𝑋 = 1) ⇒ 2! 𝑒 −𝜆 = 2 × 1! 𝑒 −𝜆 ⇒ 𝜆 = 4

Mean = 4, Variance =4. So standard deviation = 2.
1 1 𝑒 −1 𝑒 −1
D. 𝑋 ∼ 𝑃𝑜𝑖𝑠𝑠𝑜𝑛(1). 𝐸 (1+𝑋) = ∑∞ ∞ −1
𝑥=0 1+𝑥 . 𝑥! = ∑𝑥=0 (𝑥+1)! = 1 − 𝑒 .

34. C. 𝐸(𝑋 3 ) = 𝐸(3 + 𝑍)3 = 𝐸(33 + 3 × 32 𝑍 + (3) 3 × 𝑍 2 + 𝑍 3 )
2
= 𝐸(27 + 27𝑍 + 9𝑍 2 + 𝑍 3 ).
2
𝑍 ∼ 𝑁(0, 1). 𝐸(𝑍) = 0, 𝐸(𝑍 3 ) = 0. 𝐸(𝑍 2 ) = 𝑉𝑎𝑟(𝑍) + (𝐸(𝑍)) = 1 + 0 = 1.
𝐸(𝑋 3 ) = 27 + 27𝐸(𝑍) + 9𝐸(𝑍 2 ) + 𝐸(𝑍 3 ) = 27 + 27 × 0 + 9 × 1 + 0 = 36.

35. A. 𝑃(𝑋 = 𝑌) = ∑𝑛𝑥=1 𝑝(𝑥, 𝑥) = ∑𝑛𝑥=1 2 2𝑥 = 2 2 ∑𝑛𝑥=1 𝑥 = 2 2 × 𝑛(𝑛+1) =
𝑛 (𝑛+1) 𝑛 (𝑛+1) 𝑛 (𝑛+1) 2
1
.
𝑛

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36. B. 𝐴1 : the event that a student has 100% attendance
𝐴2 : the event that a student is irregular
𝐸: the event that a student attains A grade
Given that 𝑃(𝐴1 ) = 0.40, 𝑃(𝐴2 ) = 0.60, 𝑃(𝐸|𝐴1 ) = 0.70, 𝑃(𝐸|𝐴2 ) = 0.10
𝑃(𝐴1 𝐸) 𝑃(𝐸|𝐴1 )𝑃(𝐴1 ) 0.7 × 0.4
𝑃(𝐴1 |𝐸) = = =
𝑃(𝐸) 𝑃(𝐸|𝐴1 )𝑃(𝐴1 ) + 𝑃(𝐸|𝐴2 )𝑃(𝐴2 ) 0.7 × 0.4 + 0.6 × 0.1
14
= .
17

37. C. 𝑥̅ = 1.2, 𝑦̅ = 12
Regression coefficient of 𝑦 on 𝑥 is 𝑏𝑦𝑥 = 3.2
Regression coefficient of 𝑥 on 𝑦 is 𝑏𝑥𝑦 = 0.2
A. 𝑟 2 = 𝑏𝑦𝑥 𝑏𝑥𝑦 = 3.2 × 0.2 = 0.64 ⇒ 𝑟 = ±0.8. So 𝑟 = 0.8, since 𝑏𝑦𝑥 and 𝑏𝑥𝑦 are
positive.
𝑠 𝑠
B. 𝑏𝑦𝑥 = 𝑟 𝑠𝑦 ⇒ 3.2 = 0.8 . 𝑠𝑦 ⇒ 𝑠𝑦 = 4𝑠𝑥 .
𝑥 𝑥
C. The regression of 𝑦 on 𝑥 is −𝑦̅ = 𝑏𝑦𝑥 (𝑥 − 𝑥̅ ) ⇒ 𝑦 − 12 = 3.2(𝑥 − 1.2) ⇒ 𝑦 −
3.2𝑥 − 8.16 = 0.
D. The regression of 𝑥 on 𝑦 is 𝑥 − 𝑥̅ = 𝑏𝑥𝑦 (𝑦 − 𝑦̅) ⇒ 𝑥 − 1.2 = 0.2 (𝑦 − 12) ⇒ 𝑥 −
0.2𝑦 + 1.2 = 0.

38. A. 2 1
2 1 𝑃 (0 < 𝑋 < , 0 < 𝑌 < )
𝑃 (0 < 𝑋 < |0 < 𝑌 < ) = 3 3
3 3 1
𝑃 (0 < 𝑌 < 3)
1
2 1 2 2
2 1 𝑦2 3 10
𝑃 (0 < 𝑋 < 3 , 0 < 𝑌 < 3) = ∫03 ∫03 4𝑥(1 − 𝑦)𝑑𝑦𝑑𝑥 = ∫03 4𝑥 [𝑦 − 2 ] 𝑑𝑥 = 9 ∫03 𝑥𝑑𝑥 =
0
20
81
.
1
1
1 3 10 1 5
𝑃 (0 < 𝑌 < ) = ∫ ∫ 4𝑥(1 − 𝑦)𝑑𝑦 𝑑𝑥 = ∫ 𝑥𝑑𝑥 = .
3 0 0 9 0 9

2 1 20 9 4
𝑃 (0 < 𝑋 < <𝑌< )= × = .
3 |0 3 81 5 9

𝜆𝑥 𝜆 𝜆𝑥−1 𝜆 𝜆𝑒 𝜆
39. D. 𝐸(𝑋) = ∑∞ ∞ 𝜆
𝑥=1 𝑥. 𝑥!(𝑒 𝜆 −1) = 𝑒 𝜆 −1 ∑𝑥=1 (𝑥−1)! = 𝑒 𝜆 −1 𝑒 = 𝑒 𝜆 −1.
𝜆𝑥 𝜆𝑥
𝐸(𝑋 2 ) = 𝐸(𝑋(𝑋 − 1) + 𝑋) = ∑∞ ∞
𝑥=1 𝑥(𝑥 − 1). 𝑥!(𝑒 𝜆−1 ) + 𝐸(𝑋)= ∑𝑥=2 (𝑥−2)!(𝑒 𝜆 −1)
𝜆𝑒 𝜆 𝜆2 𝑒 𝜆 𝜆𝑒 𝜆 𝜆(1+𝜆)𝑒 𝜆
+ 𝑒 𝜆 −1= 𝑒 𝜆−1 + 𝑒 𝜆 −1 = .
𝑒 𝜆 −1
2
2 𝜆(1+𝜆)𝑒 𝜆 𝜆𝑒 𝜆 𝜆𝑒 𝜆 𝜆𝑒 𝜆
𝑉𝑎𝑟(𝑋) = 𝐸(𝑋 2 ) − (𝐸(𝑋)) = − (𝑒 𝜆 −1) = 𝑒 𝜆−1 (1 + 𝜆 − 𝑒 𝜆−1) =
𝑒 𝜆 −1
𝜆𝑒 𝜆
2 (𝑒 𝜆 − 𝜆 − 1).
(𝑒 𝜆 −1)

3 4 1 9
40. A. 𝐸(𝑋) = ∑2𝑥=0 ∑2𝑦=0 𝑥 𝑝(𝑥, 𝑦) = 1 (12 + 12) + 2 × 12 = 12.
3 4 1 11
𝐸(𝑋 2 ) = ∑2𝑥=0 ∑2𝑦=0 𝑥 2 𝑝(𝑥, 𝑦) = 12 (12 + 12) + 22 × 12 = 12.
11 81 51
Var(𝑋) = 12 − 144 = 144.

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2 3 1 4 1 17
𝐸(𝑌) = ∑2𝑥=0 ∑2𝑦=0 𝑦 𝑝(𝑥, 𝑦) = 1 (12 + 12) + 2 (12 + 12 + 12) = 12.
2 3 1 4 1 29
𝐸(𝑌 2 ) = ∑2𝑥=0 ∑2𝑦=0 𝑦 2 𝑝(𝑥, 𝑦) = 12 ( + ) + 22 ( + + )= .
12 12 12 12 12 12
29 289 59
Var(𝑌) = 12 − 144 = 144.
3 4 1 15
𝐸(𝑋𝑌) = ∑2𝑥=0 ∑2𝑦=0 𝑥𝑦 𝑝(𝑥, 𝑦) = 1 × 1 × 12 + 1 × 2 × 12 + 2 × 2 × 12 = 12.
15 9 17 27 27
Cov(𝑋, 𝑌) = 𝐸(𝑋𝑌) − 𝐸(𝑋)𝐸(𝑌) = 12 − 12 × 12 = 122 = 144.
𝑐𝑜𝑣(𝑋,𝑌) 27 51 59 27
Corr(𝑋, 𝑌) = = 144⁄√144 × 144 = .
√𝑉𝑎𝑟(𝑋)𝑉𝑎𝑟(𝑌) √3009

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Data Interpretation
41. C. Number of families in 125000 - 150000 group owning at least 2 vehicles = 30 +
2 = 32.
Number of families in 150000 - 175000 group owning at least 2 vehicles = 62 +
5 = 67.
Number of families in 175000 - 200000 group owning at least 2 vehicles = 75 +
11 = 86.
Number of families in 200000 or more group owning at least 2 vehicles = 62 +
20 = 82.

42. D. Number of families owning exactly 1 vehicle = 22 + 90 + 165 + 290 + 258 +
307 + 159 + 65 = 1356.
Number of families owning exactly 2 vehicles = 0 + 3 + 7 + 14 + 30 + 62 +
75 + 62 = 253.
Number of families owning more than 2 vehicles = 0 + 0 + 0 + 0 + 2 + 5 + 11 +
20 = 38.
Number of families owning at least 1 vehicle = 1356 + 253 + 38 = 1647.

43. A. Number of families whose earning is Rs. 100000 or more per month owning exactly
1 vehicle = 290 + 258 + 307 + 159 + 65 = 1079.

44. A. S4: From about 2% in 2019 to about 95% in 2022.
S5: From about 6% in 2019 to about 75% in 2022.
S9: From about 3% in 2019 to about 55% in 2022.
S10: From about 4% in 2019 to about 54% in 2022.

45. A. It is evident from the graph that S1 has 100% households with tap water
connections in 2022.

46. C. It is evident from the graph that S4 has minimum and S2 has maximum share in
2019.

47. B. The states having more than 65% of the households with tap water connections are
S1, S2, S3, S4, S5, S6, S7 and S8.

48. D. A. The minimum number of frauds occurred in FY13.
B. The maximum number of frauds occurred in FY22.
C. This is not true. The number of frauds decreased in FY 21 over FY 20.
D. The number of frauds decreased over the previous year is in FY21.

49. C. Budget allocation on Centrally sponsored Schemes and Central Sector Schemes are
9% and 17%.
Total budget allocation on these sectors is 40 × 0.26 lakh crore = 10.4 lakh crore.

50. B. The sectors with less than 10% allocation are
Centrally Sponsored Schemes – 9%.
Subsidies – 7%.

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Defence – 8%.
Finance Commission and other transfers – 9%.
Pensions – 4%.
Other expenditure – 8%.
Percentage allocation in these sectors – 45%.
Total allocation in these sectors = 40 × 0.45 lakh crore = 18 lakh crore.

51. B. Maximum allocation – 20%.
Minimum allocation – 4%.
The difference between maximum and minimum allocation is 40 × 0.16 = 6.4 lakh
crore.

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English

52. A.
53. C.
54. B.
55. A.
56. D.
57. D.
58. C.
59. B.
60. D.
61. D.
62. A.

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Logical reasoning

63. A. Assuming 𝐴 is the number of students who play exactly 1 game, 𝐵 who play exactly
2 games and 𝐶 be those who play exactly 3 games.
𝐴 + 𝐵 + 𝐶 = 100.
𝐴 + 2𝐵 + 3𝐶 = 60 + 84 + 72 = 216.
Second Equation– 2 × First Equation = 𝐶– 𝐴 = 16.
Thus, 𝐶 = 𝐴 + 16. Since the minimum value of 𝐴 is 0, the minimum value of 𝐶 is
16.

64. B. Barfis, Tikkis, Jalebis and Samosas are successive subsets.

65. D. Before Alexa starts walking towards her house, she is 50 − 30 = 20 meters to the
South and 20 meters to the East of her house.

66. D. (360/12) × 6 = 180.

67. C. From Friday to Wednesday we need 5 odd days. There will be 1 odd day in 2017,
2018 and 2019 and 2 odd days in 2020 which is a leap year. So 5 odd days will be
completed in 2020.

68. D. He shot balloons in the order of 2,4,6,8,10,12,14,1,5,9,13,3,11 and 7. At the tenth
shot it was balloon numbered 9.

69. A. No cube is painted black on more than three faces.

70. C. There is no requirement for drawing a family tree to solve this question. Since Nimrit
is the grandmother of Shiv and Shiv and Ankit are cousins (from the first statement).
Ankit should be a grandson to Nimrit.
Alternative logic: Since Kishore is the grandfather of Ankit and Nimrit is the wife
of Kishore, Ankit should be a grandson to Nimrit.

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