ACET 2019 Answer Key (Sep)

Institute of Actuaries of India
ACET September 2019 Solutions

Mathematics
1. C. 1.23425001 + 2.03049912 = 3.26474913. After rounding this number 4
places after decimal, we obtain 3.2647.

2. A. The determinant of the matrix is easily computed as 𝑏(𝑏 − 2)(𝑏 + 1). When
𝑏 ≠ −1,0,2, this determinant is non-zero. Hence rank of the given matrix is 3.

3. C. The system has no solution, as the lines are parallel, i.e., they never intersect.

4. B. Given limit is equivalent to

𝑡=𝑔(𝑥)
𝑡2 𝑔2 (𝑥) − 4 𝑔(𝑥) − 2
lim | = lim = lim × lim (𝑔(𝑥) + 2)
𝑥 → 1⁄2 1 𝑥 → 1⁄2 1 𝑥 → 1⁄2 1 𝑥 → 1 ⁄2
𝑥−2 𝑥−2 𝑥−2
𝑡=2

1
𝑔(𝑥) − 𝑔 (2) 1 1 1
= lim × [𝑔 ( ) + 2] = 𝑔′ ( ) × (2 + 2) = 4𝑔′ ( ).
𝑥 → 1⁄2 1 2 2 2
𝑥−2

−3 𝑢−1 −3 𝑢
5. D. Given limit = lim (1 + 𝑢 ) = lim (1 + 𝑢 ) = 𝑒 −3 .
𝑢→∞ 𝑢→∞

6. D. There is jump discontinuity at every integer value of 𝑥.
1/𝑒 log𝑒 𝑦 𝑒 log𝑒 𝑧
7. C. ∫1 𝑑𝑦 = ∫1 𝑑𝑧. Therefore,
𝑦+1 𝑧(𝑧+1)

𝑒 1
𝑒
log 𝑒 𝑦 log 𝑒 𝑦 log 𝑒 𝑦 1
ℎ(𝑒) = ∫ [ + ] 𝑑𝑦 = ∫ 𝑑𝑦 = ∫ 𝑡 𝑑𝑡 = .
𝑦 + 1 𝑦(𝑦 + 1) 1 𝑦 2
1 0

1
Here 𝑧 = and log 𝑒 𝑦 = 𝑡.
𝑦

8. C. 𝑔 ∘ 𝑓: [0, ∞) → (−∞, ∞), 𝑔 ∘ 𝑓(𝑥) = 𝑔(𝑓(𝑥)) = 𝑔(𝑥) = |𝑥| = 𝑥, 𝑥 ∈ [0, ∞). Hence
𝑔 ∘ 𝑓 is one-to-one. But 𝑔: (−∞, ∞) → (−∞, ∞) and 𝑔(−3) = 𝑔(3) = 3. So, 𝑔 is not one-to-
one.

9. D. 4! = 24 is divisible by 8. Hence 5!, 6!, 7!, … ,99! are all divisible by 8.
Thus the required remainder is obtained when 1! + 2! + 3! = 9 is divided by 8,
which is 1.
1
10. B. |𝑥𝑛 | = |2 − | has the least upper bound 2.
𝑛

11. C. 𝑓 ′ (𝑥) does not exist at 𝑥 = 0, since lim 𝑓′(𝑥) = −∞, lim 𝑓′(𝑥) = ∞.
𝑥↓0 𝑥↑0

1

Moreover, 𝑓(0) = 1 and 𝑓(𝑥) < 1, for all other 𝑥. Thus, 𝑓(𝑥) is maximum at 𝑥 = 0,
although derivative does not exist there.
2 cos 2𝑥+𝜏 cos 𝑥
12. A. By the use of L’Hôpital’s rule, the limit is reduced to lim .
𝑥→0 2𝑥
Since there is 0 in the denominator, for finiteness of the limit as 𝑥 → 0 through
L’Hôpital’s rule, we must have lim (2cos2𝑥 + 𝜏cos𝑥) = 0. Thus 2 + 𝜏 = 0, i.e., 𝜏 =
𝑥→0
−2.
sin𝑥cos𝑥
13. C. Here 𝑓 ′ (𝑥) = |sin𝑥|
, exists only when 𝑥 ≠ 𝑛𝜋, 𝑛 = 0, ±1, ±2, …..

14. A. Since the points 2 and 4 are closest to 3, the most suitable linear interpolation
for approximating 𝑓(3) is

3−2 3−2 1
(1 − ) 𝑓(2) + ( ) 𝑓(4) = [𝑓(2) + 𝑓(4)] = 45.
4−2 4−2 2

15. C. The given equation is equivalent to

𝑎 cos 𝑥 − 𝑏 sin 𝑥 = √𝑎2 + 𝑏 2 cos(𝑥 + 𝜃) = 𝑑,

where cos 𝜃 = 𝑎/√𝑎2 + 𝑏 2 and sin 𝜃 = 𝑏/√𝑎2 + 𝑏 2. It simplifies to cos(𝑥 + 𝜃) =
𝑑
𝑑⁄√𝑎2 + 𝑏 2 , which has a solution if and only if | | ≤ 1.
√𝑎 2 +𝑏2

16. D. The condition (𝑎⃗ × 𝑏⃗⃗) = 2(𝑎⃗ × 𝑐⃗) implies that the planes of the pairs of
vectors (𝑎⃗, 𝑏⃗⃗) and (𝑎⃗, 𝑐⃗) are identical, i.e., the three vectors 𝑎⃗, 𝑏⃗⃗ and 𝑐⃗ are co-planar.
Therefore, the vector 𝑏⃗⃗ × 𝑐⃗ is perpendicular to that plane, and in particular, to the
vector 𝑎⃗.

17. B. For the rank to be equal to 1, all the columns should be a multiple of the first
column. The second column is already a multiple. The only value of 𝑐 that makes the
third column a multiple also is −6.

18. A. Let 𝑦 = 0.272727 …, so 100𝑦 = 27.272727 … = 27 + 𝑦.
27 3
Thus 99𝑦 = 27, which implies 𝑦 = = .
99 11
8
Also, 0.727272 … = 1 − 𝑦 = 11.
2𝑦(1−𝑦) 48
As 𝑦, 𝑥, (1 − 𝑦) are H.P., we have 𝑥 = 𝑦+(1−𝑦) = 2𝑦(1 − 𝑦) = 121, a rational number
lying between 0 and 1.
log 𝑏 log 𝑐
19. B. It is given that log10 𝑎 = log10 𝑏. Hence, log10 𝑎, log10 𝑏 and log10 𝑐 are in G.P.
10 10

2

Statistics
𝑛
20. D. If 𝑛 be the number of sides, then ( ) − 𝑛 = 65 ⇒ 𝑛 = 13.
2

21. D. 7𝑃3 × 4 × 3! = 7!.

22. A. Alphabetical order of the letters 𝐴, 𝐾, 𝑁, 𝑅. Number of words starting with 𝐴 is
6 and same is true for 𝐾 and 𝑁. The first word starting with 𝑅 is 𝑅𝐴𝐾𝑁, next word will be
𝑅𝐴𝑁𝐾. Thus the position of the word 𝑅𝐴𝑁𝐾 is (6 × 3 + 1) + 1 = 20.

23. C. 𝐴 and 𝐵 are already mutually exclusive. If they have to be independent too, then
𝑃(𝐴) × 𝑃(𝐵) = 𝑃(𝐴 ∩ 𝐵) = 0, i. e. , both 𝑃(𝐴) and 𝑃(𝐵) cannot be positive.
Conversely, if 𝑃(𝐴) ≠ 0 and 𝑃(𝐵) ≠ 0, 𝐴 and 𝐵 cannot be independent.

24. D. 𝑃(𝐴𝑐 ) + 𝑃(𝐵𝑐 ) = 2 − 𝑃(𝐴) − 𝑃(𝐵) = 2 − 𝑃(𝐴 ∪ 𝐵) − 𝑃(𝐴 ∩ 𝐵) = 1.

25. B. 𝐴1 : coin drawn is two tailed
𝐴2 : coin drawn is fair
𝐵: the toss produces tail. Required probability 𝑃(𝐵) = 𝑃(𝐴1 )𝑃(𝐵|𝐴1 ) +
31 𝑚 𝑚+1 1 31
𝑃(𝐴2 )𝑃(𝐵|𝐴2 ) = ⇒ ×1+ × = ⇒ 𝑚 = 10.
42 2𝑚+1 2𝑚+1 2 42

26. B. Arranging the 10 integers in ascending order without the median 𝑥:
2,3,3,7,9,11,13,17,19,21
6th position is the median position, and hence the only three integer options for the median
are 9, 10 and 11.

27. D.

2 2
6 6 6 6
1 1 1 1
√ ∑ ((𝑎𝑦𝑖 + 𝑏) − ∑(𝑎𝑦𝑗 + 𝑏)) = √ ∑ (𝑎𝑦𝑖 − ∑ 𝑎𝑦𝑗 )
5 6 5 6
𝑖=1 𝑗=1 𝑖=1 𝑗=1

2
6 6
𝑎2 1
= √ ∑ (𝑦𝑖 − ∑ 𝑦𝑗 ) = |𝑎|𝑚.
5 6
𝑖=1 𝑗=1

𝜎 𝜎
28. C. Let 𝐴 and 𝐵 represent the two distributions. Then 𝜇𝐴 = 0.4, 𝜇𝐵 = 0.5.
𝐴 𝐵

Therefore, 𝜎𝐴 = 0.4𝜇𝐴 = 0.4 × 25 = 10 and 𝜎𝐵 = 0.5𝜇𝐵 = 0.5 × 20 = 10,
i. e., 𝜎𝐴 − 𝜎𝐵 = 0.

29. A. The only two possible values of |𝑋 − 2| are 0 and 1, and these occur with probabilities 0.3
and 0.7, respectively. Therefore, 𝐸|𝑋 − 2| = 0.7.

30. C.
1 2 2/3 𝑐 3 2/3 8 1
𝑥
3 |1/3 27 − 27
2
2 1 𝑃 (𝑥 > , 𝑥 ≤ ) ∫ 𝑐𝑥 𝑑𝑥 7
𝑃 (𝑥 ≤ 3 |𝑥 > 3) = 3 3 1/3
= 1 = = = .
1 1 1
𝑃 (𝑥 > 3) ∫1/3 𝑐𝑥 2 𝑑𝑥 𝑐 𝑥 3 | 1 − 27 26
3 1/3

3

31. A. (𝑥 − 50)⁄10 = 1.645, i. e. , 𝑥 = 50 + 1.645 × 10 = 66.45.
32. C. The probability of an single person being female and smoker is 0.6 × 0.4 = 0.24. The
count of female smokers in a random sample of size 10 is a binomial random variable with
𝑛 = 10 and 𝑝 = 0.24. The probability of this count being 0 is 0.7610.
33. C. The sum of two independent Poisson distributed random variables with mean 5 is another
Poisson distributed random variable with mean 10. Therefore,
𝑒 −10 100 𝑒 −10 101
𝑃(𝑋 ≥ 2) = 1 − 𝑃(𝑋 = 0) − 𝑃(𝑋 = 1) = 1 − − = 1 − 11𝑒 −10 .
0! 1!

Alternative solution: Let 𝑋1 and 𝑋2 be the number of calls coming in the first and second
minute, respectively. Therefore,
𝑒 −5 50 𝑒 −5 50
𝑃(𝑋1 + 𝑋2 = 0) = 𝑃(𝑋1 = 0)𝑃(𝑋2 = 0) = × = 𝑒 −10 ;
0! 0!

𝑒 −5 51 𝑒 −5 50
𝑃(𝑋1 + 𝑋2 = 1) = 𝑃(𝑋1 = 1)𝑃(𝑋2 = 0) + 𝑃(𝑋1 = 0)𝑃(𝑋2 = 1) = 2 × ×
1! 0!
−10
= 10𝑒 ;

𝑃(𝑋1 + 𝑋2 ≥ 2) = 1 − 𝑃(𝑋1 + 𝑋2 = 0) − 𝑃(𝑋1 + 𝑋2 = 1) = 1 − 11𝑒 −10 .
1
34. A. Here, 𝑓(𝑥) = 2𝑎 over [𝑎, 3𝑎] and 0 elsewhere. Therefore,
3𝑎 3𝑎
2)
1 2 𝑥3 27𝑎3 − 𝑎3 13𝑎2
𝐸(𝑋 =∫ 𝑥 𝑑𝑥 = | = = .
𝑎 2𝑎 6𝑎 𝑎 6𝑎 3

35. D. The given condition implies that the correlation between 𝑋 and 𝑌 is −0.4. Therefore, the
correlation between 5 − 𝑋 and 2𝑌 + 3 is 0.4.
𝜎 3.5
36. C. The expected price at Mumbai is 𝜇𝑌 + 𝜌 𝜎𝑌 (𝑋 − 𝜇𝑋 ) = 80 + 0.9 × 2.5 (85 − 60) =
𝑋
111.5.
52
37. B. Number of all possible ways to drop four cards from the pack is ( ). Number of ways to
4
drop one card from a particular suit is 13. Therefore, number of selections producing one card
from each suit is 134 .
38. B. Clearly, 𝑆 = {2,3,5,7,11,13,17,19,23,29} contains 10 elements. Thus both numerator and
denominator can be selected in 10 ways. Hence required number is 10 × 10 − 10 + 1 = 91,
as out of 10 × 10 selections, 10 numbers are just 1.
4
39. A. Using the first row sum and the second column sum, we have 𝑃(𝑌 = 1) = 𝑎 + 9 and
1
𝑃(𝑋 = 2) = 𝑎 + 9. For independence, we must have
4 1
𝑎 = 𝑃(𝑋 = 2, 𝑌 = 1) = (𝑎 + ) (𝑎 + )
9 9
2 1 2 1 1 2
which yields 𝑎 = 9. Sum over all probabilities gives 𝑏 + 𝑐 = 1 − 3 − 9 − 9 − 9 = 9. Second
row and first column sums give
1 1 1 1 1
𝑏 = 𝑃(𝑋 = 1, 𝑌 = 2) = (𝑏 + 𝑐 + ) (𝑏 + ) = (𝑏 + ) , 𝑖. 𝑒. , 𝑏 = .
9 3 3 3 6
2 1 1
Hence, 𝑐 = 9 − 6 = 18.

4

Data Interpretation
40. A. Fruits and vegetables account for 33%, or about one-third of the total expenditure. If that
amount is Rs. 2000, the total expenditure would be Rs. 6000 and the expenditure on Rice
would be 25% of that, i.e., Rs. 1500.
16%
41. C. 84% ≈ 1/5.

42. D. Schools A and B both add up to 25.
43. D. The difference is 6 in both Schools C and E.
44. C. 2014, 2016 and 2017.
45. C. Smallest slope is visible between 2015 and 2016. Closer examination reveals that this
period had smaller growth than the other period of small slope (2016 to 2017).
46. A. 1.22 time FY 2016-17 profits is 10 Crores. The answer is 1000/1.44 = 694.44.
47. B. It is easy to visually identify that the ratio of the profit share to sales share is largest for
Hyundai; total profits and sales do not matter.
48. B. The ratio is 0.17/0.13 ≈ 1.307.
49. D. The given information is shown in the Venn diagram. Additionally it is known that 𝑎 +
40 = 114 and 𝑏 + 𝑐 + 40 + 388 = 500.

Java C
404
𝑎 222

40
𝑏 𝑐

388
C++

The number of students studying exactly two programming languages including C++ is 𝑏 +
𝑐 = 500 − 40 − 388 = 72.
50. B. The number of students studying exactly two languages is 𝑎 + 𝑏 + 𝑐 = (114 − 40) +
72 = 146.
51. C. If the number of students studying the Java language is 568, then 𝑏 = 568 − 404 − 114 =
50, i.e., 𝑐 = 72 − 50 = 22. Therefore, the number of students studying the C language is
222 + 114 + 𝑐 = 358.

5

English
52. D

53. B

54. C

55. B

56. D

57. A

58. A

59. A

60. C

61. B

62. A

6

Logical Reasoning
63. A. A and E are husband and wife, with daughters-in-law D and F; H is a son of F. Therefore
A is the grandfather of H.

64. C. Number of backward people who are educated = 11 + 3 = 14.

65. C. N is the second play to be staged.

O should be immediately followed by M, i.e., the order OM should be followed. N should
be immediately followed by L, i.e., the order NL should be followed. N or O should not be
the first or last play. Therefore, either of the orders NLOM or OMNL should be followed
from Tuesday to Friday. One play is staged between K and L. So, the order is

Monday K

Tuesday N

Wednesday L

Thursday O

Friday M

66. B. Leap years are generally separated by 4 years, the only exceptions being some years
divisible by 100 (e.g., the year 1900). In such cases, the separation between consecutive leap
years is 8 years. A longer separation is not possible.

67. B.

68. B. The relevant small cubes lie along the three edges common to the single corner
(mentioned in the question). Adjusting for the triple counting of the corner cube, the requisite
number is 3 × 4 − 2 = 10.

69. C. The gap between the hour hand and the minute hands reduces steadily starting from a gap
of 45 minute spaces at 9 o’clock. The gap reduces at the rate of 55 minute spaces per hour (60
60 1
minutes). Therefore, they will converge in × 45 = 49 minutes.
55 11

70. D.

7