CA Foundation Maths Question Paper May 26 With Solution

∫ x²⁵ / x²⁶ dx = ∫ 1/x dx

= log x + C

2. If x(m) = am², y(m) = a / m², then find the value of dy/dx.

(A) 1 / m

(B) -1 / m2

(D) -1 / m4

Solution: (D)

x = am^2, so dx/dm = 2am

y = a/m2 = am^-2, so dy/dm = -2am^-3

dy/dx = (dy/dm) / (dx/dm) = -1/m⁴

3. If x^y = y^x, then dy/dx = __________.

(A) y(x log y − y) / x(y log x − x)

(B) y(x log y + y) / x(y log x − x)

(D) x(x log y − y) / y(y log x − x)

Solution: (A)

x^y = y^x

Taking log: y log x = x log y

dy/dx = y(x log y − y) / x(y log x − x)

4. For a given Revenue function R(x) = 100x − 2x², the maximum revenue occurs at x = __________.

(A) 20

(B) 25

(D) 50

Solution:

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5. Find the next term in the series 11, 12, 14, 18, 22, 30, __________.

(A) 32

(B) 43

(D) 41

Solution: (B)

Series: 11, 12, 14, 18, 22, 30

The pattern follows alternate increasing differences:

11 + 1 = 12

12 + 2 = 14

14 + 4 = 18

18 + 4 = 22

22 + 8 = 30

30 + 13 = 43

Hence, the next term is 43.

6. Two dice are rolled. What will be the probability that one dice have multiple of 3 and other dice have multiple of 2?

(A) 1/2

(B) 1/3

(D) 1/5

Solution: (B)

Multiples of 3 on a dice = {3, 6} → 2 outcomes

Multiples of 2 on a dice = {2, 4, 6} → 3 outcomes

Required favourable outcomes = 2 × 3 = 6

Total outcomes when two dice are rolled = 6 × 6 = 36

Since either dice can satisfy either condition, favourable outcomes = 12

Probability = 12 / 36 = 1 / 3

7. The probability of getting pass in an examination is x/3. If the probability of getting fail is 2/3, then the value of x is __________.

(A) 2

(B) 3

(D) 0

Solution:

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8. Which of the followings is a property of Arithmetic Mean?

(A) Sum of deviations from Arithmetic Mean is always positive.

(B) Arithmetic Mean is not affected by extreme values.

(D) Arithmetic Mean cannot be calculated for grouped data.

Solution:

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9. The mean of 10 observations is 15. If one observation 12 is replaced by 22, what will be the new mean?

(A) 16

(B) 14

(D) 18

Solution:

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10. If f(x) = kx², 0 ≤ x ≤ 1 is a probability density function of a random variable x, then the value of k is __________.

(A) 0

(B) 3

(D) 2

Solution: (B)

Since f(x) is a probability density function,

∫₀¹ kx² dx = 1

k [x³ / 3]₀¹ = 1

k × 1/3 = 1

k = 3

11. Mr. Arun bought a car for ₹ 3,00,000 by making a down payment ₹ 50,000 and decided to pay equal annual payment for 10 years. How much would be each payment if the interest on unpaid amount is 18% per annum, compound annually? (Given that (1.10, 0.18) = 4.49409)

(A) ₹ 55,628.61

(B) ₹ 55,266.86

(D) ₹ 50,000.00

Solution: (A)

Unpaid amount = ₹ 3,00,000 - ₹ 50,000 = ₹ 2,50,000

Equal annual payment = Loan Amount / Present Value Annuity Factor

= ₹ 2,50,000 / 4.49409
= ₹ 55,628.61

Therefore, each annual payment will be ₹ 55,628.61.

12. The compound interest for ₹ 15,000 at 20% per annum for 2 years compounded semi-annually is __________.

(A) ₹ 9,661.50

(B) ₹ 6,961.50

(D) ₹ 6,696.15

Solution: (B)

Principal = ₹ 15,000

Rate = 20% per annum compounded semi-annually

So, half-yearly rate = 10%

Number of periods in 2 years = 4

Amount = 15000(1 + 10/100)4

= 15000(1.1)4

= 15000 × 1.4641

= ₹ 21,961.50

Compound Interest = 21,961.50 − 15,000

= ₹ 6,961.50

13. The simple interest at the rate of p% per annum for p years will be ₹ p. Then, the principal is __________.

(A) ₹ p

(B) ₹ 100p

(D) ₹ 100 / p

Solution: (D)

Simple Interest formula:

SI = (P × R × T) / 100

Given: SI = p, R = p%, T = p years

So, p = (P × p × p) / 100

p = Pp2 / 100

P = 100 / p

14. The future value of an annuity of ₹ 2000 made annually for 8 years at interest rate of 14% per annum, compound annually is __________. (Given that (1.14)⁸ = 2.8526)

(A) ₹ 26,465.71

(B) ₹ 26,646.57

(D) ₹ 18,564.52

Solution: (A)

Future value of annuity formula:

S = R [(1 + i)n − 1] / i

Here, R = ₹ 2000, i = 14% = 0.14, n = 8

S = 2000 [(2.8526 − 1) / 0.14]

S = 2000 × (1.8526 / 0.14)

S = 2000 × 13.232857

S = ₹ 26,465.71

15. Ravi deposits some amount in bank for 5 1/2 years at the simple interest rate of 7% per annum. Ravi receives ₹ 66,480 at the end of term. Compute the amount of initial deposit by Ravi in the Bank.

(A) ₹ 48,000

(B) ₹ 50,000

(D) ₹ 51,000

Solution: (A)

Amount = ₹ 66,480

Rate = 7% p.a.

Time = 5 1/2 years = 11/2 years

Using simple interest formula:

A = P(1 + RT/100)

66,480 = P [1 + (7 × 11)/(100 × 2)]

66,480 = P (1 + 77/200)

66,480 = P × 277/200

P = 66,480 × 200 / 277

P = ₹ 48,000

16. Six friends A, B, C, D, E and F sit in a row facing north.

A and B have two persons between them.

A who is at right end, is immediate right of A.

D is not neighbour of B.

E sits at left end.

Who sits third from left?

(A) F

(B) A

(D) B

Solution:

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17. Five persons A, B, C, D & E are sitting on a bench. A is immediately to the left of C. E is immediately to the left of D and right of A. D is to the left of B. Which person is sitting in the middle of the bench?

(A) B

(B) E

(D) D

Solution:

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18. Six persons, A, B, C, D, E, and F are seated at a round table facing outside the centre but not necessarily in the same order. A sits at the immediate right of E. C sits after one person to the right of A. B sits beside A and F sits at the immediate right of D. How many persons are sitting between A & D?

(A) 1

(B) 2

(D) 4

Solution:

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19. In a circular arrangement of 5 persons facing centre, A is sitting between B and C, D is sitting immediate right of C, and E is sitting immediate left of B. Who is sitting between D and B?

(A) A

(B) B

(D) E

Solution:

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20. The product of the price index and the quantity index are equal to the corresponding value index in __________.

(A) Time reversal test

(B) Factor reversal test

(D) Unit test

Solution:

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21. For a set of observations on variables x and y, the following summary statistics are given:

n = 5, ∑x = 10, ∑y = 25, ∑xy = 70, ∑x² = 30.

The regression equation of y on x is expressed as:

y = a + bx.

What is the value of the slope b?

(A) 0.5

(B) 1.0

(D) 2.0

Solution: (B)

b = [n∑xy - ∑x∑y] / [n∑x² - (∑x)²]
= [5 × 70 - 10 × 25] / [5 × 30 - 10²]
= 100 / 50
= 2.0

22. The Spearman’s Rank correlation coefficient between Economics and Accountancy marks for a class student is 75/99 and the sum of Square of differences in ranks for Economics and Accountancy marks is 40. What is the number of students in the class?

(A) 10

(B) 15

(D) 20

Solution: (B)

r = 1 - [6∑D² / n(n² - 1)]

75/99 = 1 - [6 × 40 / n(n² - 1)]

240 / n(n² - 1) = 24/99

n(n² - 1) = 990

n = 10

23. When the two regression coefficients are given as b_xy = 0.6 and b_yx = 0.9, determine the value of the coefficient of correlation.

(A) 0.73

(B) 0.90

(D) 0.54

Solution:

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24. The sum of the first n terms of an arithmetic progression (A.P.) is 4n² + 3n. The 10^th term of the A.P. is __________.

(A) 77

(B) 83

(D) 79

Solution: (D)

Given sum of first n terms:

S_n = 4n² + 3n

n^th term of an A.P. is:

a_n = S_n − S_n−1

a₁₀ = S₁₀ − S₉

S₁₀ = 4(10)² + 3(10) = 400 + 30 = 430

S₉ = 4(9)² + 3(9) = 324 + 27 = 351

a₁₀ = 430 − 351 = 79

25. If the first term of a geometric progression exceeds the second term by 4 and the sum of its terms till infinity is 100, then the common ratio is __________.

(A) 1/5

(B) 4/5

(D) 2/5

Solution: (B)

Let first term = a and common ratio = r.

Second term = ar

Given, first term exceeds second term by 4:

a − ar = 4

a(1 − r) = 4 ...(1)

Sum to infinity = a / (1 − r) = 100

a = 100(1 − r) ...(2)

Substituting (2) in (1):

100(1 − r)2 = 4

(1 − r)2 = 1/25

1 − r = 1/5

r = 4/5

26. Find out the number of 5-digit even numbers that can be formed from digits 1 to 7 without repetition of any digit.

(A) 720

(B) 360

(D) 840

Solution:

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27. If the sum of 4^th and 8^th term of an arithmetic progression (A.P.) is 120, then the 6^th term of the A.P. is __________.

(A) 10

(B) 70

(D) 100

Solution:

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28. The sum of infinite terms of the geometric series 1 − 1/5 + 1/25 − 1/125 + ... will be __________.

(A) 1

(B) 5/6

(D) 4/5

Solution:

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29. The standard deviation is zero only if all the observations assumed by a variable are __________.

(A) different

(B) equal

(D) square root of natural numbers

Solution:

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30. If the mean 40, k, 6k, 4k², 8k − 4k² is 20, then the value of k is __________.

(A) 15

(B) 10

(D) 4

Solution: (D)

Mean = Sum of observations / Number of observations

(40 + k + 6k + 4k² + 8k − 4k²) / 5 = 20

(40 + 15k) / 5 = 20

40 + 15k = 100

15k = 60

k = 4

Therefore, the value of k is 4.

31. An ogive represents cumulative frequencies on a graph. By what other name is this graphical form commonly known?

(A) Frequency Histogram

(B) Cumulative Frequency Curve

(D) Area Diagram

Solution:

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32. To add flexibility to the sampling process, the __________ is preferred.

(A) Simple Random Sampling

(B) Stratified Sampling

(D) Judgment Sampling

Solution:

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33. In a sample survey, sampling error mainly arises because:

(A) Enumerators record some responses incorrectly

(B) A part of the selected sample does not respond

(D) Some questions in the schedule are ambiguously worded

Solution:

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34. The mean of five observations is 28. Among the five observations, three observations are 10, 23 and 62. The difference between the remaining two observations is 13. Then the remaining two observations are __________.

(A) 30 and 17

(B) 38 and 15

(D) 35 and 22

Solution: (C)

Total of 5 observations = 28 × 5 = 140

Given observations total = 10 + 23 + 62 = 95

Remaining two observations total = 140 - 95 = 45

Let the two observations be x and y.

x + y = 45 and x - y = 13

2x = 58

x = 29 and y = 16

Therefore, the remaining two observations are 29 and 16.

35. If a:b = 3:4, then the value of (2a + 3b) / (3a + 2b) is __________.

(A) 18/17

(B) 17/18

(D) 7/6

Solution:

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36. The simplified value of 4√(a6b x4) × √(a3 x−4)−b will be __________.

(A) x^1+2b

(B) x^2b

(D) a^3b/2xb

Solution: (A)

Given expression:

⁴√(a^6bx⁴) × [√(a³x⁻⁴)]−b

= a^3b/2x × a^−3b/2x2b

a^3b/2 and a^−3b/2 cancel each other.

Therefore, simplified value = x^1+2b

37. The product of two numbers is 7644 and their ratio is 12:13. Then, the smaller of two numbers is __________.

(A) 91

(B) 84

(D) 90

Solution:

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38. The value of (log₃16 × log_√23) / (log₈3 × log₉4) is __________.

(A) 16

(B) 24

(D) 1/12

Solution: (B)

Given expression:

(log₃16 × log_√23) / (log₈3 × log₉4)

log₃16 = 4log₃2

log_√23 = 2log₂3

So, numerator = 4 × 2 = 8

log₈3 × log₉4 = 1/3

Required value = 8 / (1/3) = 24

39. If x = 3 + √8, then the value of x + 1/x is __________.

(A) 3√8

(B) 6

(D) 2√8

Solution: (B)

Given, x = 3 + √8

1/x = 1 / (3 + √8)

Rationalizing:

1/x = (3 − √8) / [(3 + √8)(3 − √8)]

1/x = (3 − √8) / (9 − 8)

1/x = 3 − √8

x + 1/x = (3 + √8) + (3 − √8)

x + 1/x = 6

40. Ram started walking 2 kms towards the Sun in the morning. He turned right and walked 2 kms. Then, he turned right and walked 2 kms. Finally he turned left and walked 2 kms. In which direction is he walking now?

(A) South

(B) North

(D) West

Solution:

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41. If the word ‘CODE’ is coded as 1357 and ‘GAMER’ is coded as 24678, what word does the number 84178 represent?

(A) ROVER

(B) RCEAR

(D) RAECR

Solution:

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42. If ‘DOCTOR’ is coded as 423527 and ‘PATIENT’ is coded as 8651905, then how is ‘OPERATION’ coded?

(A) 289765120

(B) 297681205

(D) 286712065

Solution:

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43. In a certain Language, ‘SUBJECT’ is written as ‘UWDJGEV’ then ‘MENTION’ will be written as __________.

(A) OGKVPQP

(B) NOFUPOQ

(D) OGPVKQP

Solution:

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44. Find the next term of the series: 7, 8, 18, 57, 232, 1165, __________.

(A) 6996

(B) 5840

(D) 6990

Solution: (A)

7 × 1 + 1 = 8

8 × 2 + 2 = 18

18 × 3 + 3 = 57

57 × 4 + 4 = 232

232 × 5 + 5 = 1165

1165 × 6 + 6 = 6996

45. If the standard deviation of a Poisson distribution is 3, then P(X = 0) is __________.

(A) e⁻⁶

(B) e⁻³

(D) e⁻¹

Solution: (C)

For a Poisson distribution:

Mean = Variance = λ

Standard deviation = √λ = 3

Therefore, λ = 9

Probability function of Poisson distribution:

P(X = x) = (e^−λ λ^x) / x!

For X = 0:

P(X = 0) = e⁻⁹

46. A card is drawn at random from a well-shuffled deck of 52 cards. What is the probability that the card drawn is either a King or a Heart?

(A) 13/52

(B) 17/52

(D) 15/52

Solution:

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47. In a Shooting competition, A hit the target 6 out of 13 shots, and B hit 8 out of 11 shots. If they both try once, what is the probability that the target would be hit at least once?

(A) 21/143

(B) 56/143

(D) 122/143

Solution: (C)

Probability that A hits the target = 6/13

Probability that B hits the target = 8/11

Probability that neither hits the target:

= (1 − 6/13) × (1 − 8/11)

= (7/13) × (3/11)

= 21/143

Probability that target is hit at least once:

= 1 − 21/143

= 122/143

48. A bag contains 5 red, 4 blue, and 3 green balls. Two balls are drawn at random without replacement. What is the probability that both balls are of different colours?

(A) 45/66

(B) 47/66

(D) 45/132

Solution: (B)

Total balls = 5 + 4 + 3 = 12

Total ways of drawing 2 balls = 12C2 = 66

Ways of drawing balls of same colour:

2 red = 5C2 = 10

2 blue = 4C2 = 6

2 green = 3C2 = 3

Total same colour ways = 10 + 6 + 3 = 19

Different colour ways = 66 − 19 = 47

Probability = 47 / 66

49. A machine depreciates at a rate of 10% per year on its beginning-of-year value. If its value is ₹ 6,000 after 9 years, what was its original purchase price?

(A) ₹ 14,587

(B) ₹ 15,488

(D) ₹ 15,888

Solution: (B)

Let the original purchase price be P.

After 9 years:

P × (1 − 10/100)⁹ = 6000

P × (0.9)⁹ = 6000

P = 6000 / (0.9)⁹

P = 6000 / 0.3874

P = ₹ 15,488 approximately

Therefore, the original purchase price was ₹ 15,488.

50. What will be the approximate future value of an annuity of ₹ 1000 made annually for 5 years at interest rate of 7% per annum, compounded annually? (Given (1.07)⁵ = 1.40255)

(A) ₹ 8,025

(B) ₹ 5,750

(D) ₹ 6,412

Solution:

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51. If the difference between the annually compounded interest and simple interest on a certain sum of money at 8% per annum for 3 years is ₹ 788. Then the principle amount is __________.

(A) ₹ 39,175

(B) ₹ 39,475

(D) ₹ 40,475

Solution: (A)

Difference between CI and SI for 3 years = P × r²(300 + r) / 100³

788 = P × 8²(300 + 8) / 100³

788 = P × 64 × 308 / 10,00,000

P = ₹ 39,975 approximately

52. Rakesh requires ₹ 1,00,000 to buy a scooter after 4 years. What will be the approximate present value of ₹ 1,00,000 if the interest rate is 10% per annum?

(A) ₹ 67,800

(B) ₹ 68,300

(D) ₹ 71,430

Solution:

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53. Raju deposit ₹ 20,000 in a nationalized bank for 3 years at an annual interest rate of 8%, with the interest compounded every quarter. Find out how much interest Raju earns in the first year and the second year.

(A) ₹ 1,632 & ₹ 1,712

(B) ₹ 1,684 & ₹ 1,738

(D) ₹ 1,696 & ₹ 1,746

Solution: (C)

Quarterly rate = 8% ÷ 4 = 2%

Interest in 1st year
= 20,000 × [(1.02)⁴ − 1]
= 20,000 × 0.082432
= ₹ 1,648 approximately

Amount after 1st year
= 20,000 + 1,648
= ₹ 21,648

Interest in 2nd year
= 21,648 × [(1.02)⁴ − 1]
= 21,648 × 0.082432
= ₹ 1,784 approximately

Therefore, the correct answer is ₹ 1,648 & ₹ 1,784.

54. Find the effective interest rate, if nominal rate is 12% per annum, quarterly compounding.

(A) 12%

(B) 12.36%

(D) 13%

Solution:

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55. Dr. Madhu said, “The engineer who is my brother is the son of the only Professor in our family.” If the Professor is Mr. Murli, how is Mr. Murli related to Dr. Madhu?

(A) Uncle

(B) Father

(D) Cousin

Solution: (B)

The engineer is Dr. Madhu’s brother.

The engineer is the son of the only Professor in the family.

Therefore, the Professor Mr. Murli is also the father of Dr. Madhu.

Hence, Mr. Murli is related to Dr. Madhu as father.

56. P is the sister of Q. R is the mother of Q. S is the father of C. R is the sister of S. T is the mother of S. What is the relation of P with T?

(A) P is Mother of T

(B) T is P’s Grand Mother

(D) T is Brother of P

Solution: (B)

P is the sister of Q, so P and Q are children of the same parent.

R is the mother of Q, so R is also the mother of P.

R is the sister of S, and T is the mother of S, so T is also the mother of R.

Therefore, T is P’s grandmother.

57. Sharma points to a girl and says “She is the mother of my son’s wife’s daughter”. What is the relation of the girl with Sharma?

(A) Mother

(B) Daughter

(D) Daughter-in-law

Solution:

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58. A is brother of B. P is the sister of Q. B is the son of P. How is A related to P?

(A) Brother

(B) Son

(D) Uncle

Solution:

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59. Q’s mother is the sister of P and daughter of S. A is the son of P and Brother of H. G is the father of H. How is G related to S?

(A) Nephew

(B) Son-in-law

(D) Father-in-law

Solution: (B)

Q’s mother is sister of P and daughter of S, so P is also child of S.

A is son of P and brother of H, so H is also child of P.

G is father of H, therefore G is husband of P.

Hence, G is son-in-law of S.

60. Interview method is used for collection of __________.

(A) Primary data

(B) Secondary data

(D) Continuous data

Solution:

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61. Which formula correctly represents Fisher’s Price Index?

(A) √[(∑p₁q₀ / ∑p₀q₀) × (∑p₁q₁ / ∑p₀q₁)] × 100

(B) (∑p₁q₀ / ∑p₀q₀) × 100

(D) √[(∑p₀q₁ / ∑p₁q₀)] × 100

Solution: (A)

Fisher’s Price Index = √[(∑p₁q₀ / ∑p₀q₀) × (∑p₁q₁ / ∑p₀q₁)] × 100

62. A price index with base 2000 shows:

Index for 2010: 140

Index for 2020: 210

What is the index for 2010 when base is shifted to 2020?

(A) 57.14

(B) 66.67

(D) 133.33

Solution:

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63. If ∑p₀q₀ = 400 and ∑p₁q₀ = 720 and Paasche’s index number is 125, then the Fisher’s index number is __________.

(A) 125

(B) 250

(D) 180

Solution: (C)

Laspeyres Index = ∑p₁q₀ / ∑p₀q₀ × 100

= 720 / 400 × 100

= 180

Fisher’s Index = √(Laspeyres Index × Paasche’s Index)

= √(180 × 125)

= √22500

= 150

64. What is the approximate real wage of a worker earning ₹ 12,500 nominal wage, when the cost of living index is 250 (base 100)?

(A) ₹ 3,125

(B) ₹ 4,500

(D) ₹ 6,250

Solution:

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65. Relation on integers is defined by:

a ~ b ⇔ a − b is divisible by 5.

This relation is __________.

(A) Reflexive only

(B) Symmetric only

(D) Neither symmetric nor transitive

Solution: (C)

Given relation:

a ~ b ⇔ a − b is divisible by 5

It is reflexive because a − a = 0, and 0 is divisible by 5.

It is symmetric because if a − b is divisible by 5, then b − a is also divisible by 5.

It is transitive because if a − b and b − c are divisible by 5, then a − c is also divisible by 5.

Therefore, the relation is an equivalence relation.

66. In a class, 50 students have expressed that they like Mathematics, while 66 students like Accountancy. Among these, 36 students like both Mathematics and Accountancy. Based on this information, determine how many students in the class like either Mathematics or Accountancy?

(A) 14

(B) 20

(D) 80

Solution: (D)

Let M = students who like Mathematics

Let A = students who like Accountancy

Given:

n(M) = 50

n(A) = 66

n(M ∩ A) = 36

Students who like either Mathematics or Accountancy:

n(M ∪ A) = n(M) + n(A) − n(M ∩ A)

= 50 + 66 − 36

= 80

67. Let the function f : R → R is defined by f(x) = x² + 3, then f−1(12) is __________.

(A) √3

(B) 3

(D) 12

Solution: (B)

Given, f(x) = x2 + 3

To find f−1(12), solve:

x² + 3 = 12

x² = 9

x = ±3

From the given options, the value is 3.

68. Evaluate lim x→∞ (x² + 2x + 2) / (3x² + x + 1).

(A) 1/3

(B) 2/3

(D) 2

Solution: (A)

Given limit:

lim x→∞ (x2 + 2x + 2) / (3x2 + x + 1)

Divide numerator and denominator by x2.

= lim x→∞ (1 + 2/x + 2/x2) / (3 + 1/x + 1/x2)

As x → ∞, terms containing 1/x and 1/x2 become 0.

= 1 / 3

69. What will be the mean deviation for the numbers {2, 4, 7, 8, 9, 12} from the mean?

(A) 2.14

(B) 2.33

(D) 2.67

Solution:

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70. The given frequency distribution is:

X	1	2	3	4	5	6	7
Y	5	9	12	18	14	10	6

The arithmetic mean of the Frequency distribution is __________.

(A) 2.99

(B) 3.06

(D) 4.71

Solution: (C)

Arithmetic Mean = ∑XY / ∑Y

∑Y = 74 and ∑XY = 303

Mean = 303 / 74 = 4.10

71. If third quartile and first quartile are 64.5 and 22 respectively, then the quartile deviation is __________.

(A) 42.5

(B) 21.25

(D) 30.5

Solution:

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72. The arithmetic mean (A.M.) of two positive numbers exceeds their harmonic mean by 30. If their A.M. is 45, then their geometric mean will be __________.

(A) 27√2

(B) 25√3

(D) 15√3

Solution:

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73. Two groups of students have harmonic means of 50 and 30 for their test scores, with 10 students in the first group and 15 students in the second group. What will be the combined harmonic mean of the two groups?

(A) 15.35

(B) 35.71

(D) 42.50

Solution: (B)

Combined H.M. = (n₁ + n₂) / [(n₁/H₁) + (n₂/H₂)]

= 25 / [(10/50) + (15/30)]

= 25 / 0.7

= 35.71

74. The feasible region formed by linear inequalities is always a __________ region.

(A) Circular

(B) Concave

(D) Irregular

Solution:

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75. The set of inequalities 2x + 3y < 9, 5x − y > 3 has __________ region.

(A) Unbounded solution

(B) Bounded solution

(D) Exactly a point solution

Solution: (A)

Given inequalities:

2x + 3y < 9

5x − y > 3

Both inequalities represent half-planes extending infinitely.

The common feasible region is not enclosed from all sides.

Therefore, the solution region is unbounded.

76. If α & β are roots of the equations x² − 2x − 35 = 0, where (α ≥ β), then find the value of (α − β)² / (α + β)².

(A) 18

(B) 24

(D) 36

Solution: (D)

Given equation:

x2 − 2x − 35 = 0

For roots α and β:

α + β = 2

αβ = −35

(α − β)2 = (α + β)2 − 4αβ

= 22 − 4(−35)

= 4 + 140

= 144

Therefore,

(α − β)2 / (α + β)2 = 144 / 4 = 36

77. The system of equations

2x + y = 10

4x + 6y = 36

has solution with x = __________.

(A) 2

(B) 3

(D) 6

Solution: (B)

Given equations:

2x + y = 10 ...(1)

4x + 6y = 36 ...(2)

From equation (1):

y = 10 − 2x

Substitute in equation (2):

4x + 6(10 − 2x) = 36

4x + 60 − 12x = 36

−8x = −24

x = 3

78. One experienced person does 10 units of work per day, while a fresher does 5 units of work per day. The employer wants to maintain at least 50 units of work per day. This situation can be expressed as __________.

(A) 10x + 5y > 50, x ≥ 0, y ≤ 0

(B) 10x + 5y ≤ 50, x ≥ 0, y ≥ 0

(D) 10x + 5y = 50, x ≥ 0, y ≤ 0

Solution: (C)

Let x = number of experienced persons

Let y = number of freshers

Experienced person does 10 units of work per day.

Fresher does 5 units of work per day.

Total work done per day = 10x + 5y

Employer wants at least 50 units of work.

Therefore, 10x + 5y ≥ 50 with x ≥ 0 and y ≥ 0.

79. If A × B means A is to the south of B, A + B means A is to the north of B, A × B means A is to the east of B, A − B means A is to the west of B; then in P + Q + R − S, in which direction is S with respect to Q?

(A) South-West

(B) South-East

(D) North-West

Solution:

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80. Anil’s house faces east. From the back-side of the house, he walks straight 50 meters, then turns to the right and walks 50 meters. Finally he turns towards left and stops after walking 25 meters. Now, in which direction is Anil from the starting point?

(A) South-East

(B) North-East

(D) North-West

Solution: (D)

Anil’s house faces east, so the back-side direction is west.

From the back side, he first walks 50 m west.

Then he turns right, so he moves 50 m north.

Finally, he turns left, so he moves 25 m west.

Therefore, from the starting point, Anil is towards North-West.

81. Vishal started from his house toward west. After walking a distance 30 meters, he turned towards right and walk 20 meters. He then turned left and moved a distance of 10 meters, turned to his left again and walked 40 meters. He now turn to left and walk 5 meters, finally he turns to left. In which direction he is walking now?

(A) North

(B) South

(D) South-West

Solution: (A)

Vishal first walks west.

From west, right turn means north.

Then left from north means west.

Then left from west means south.

Then left from south means east.

Finally, left from east means north.

Therefore, he is walking towards North.

82. Punu started from his house towards west. After walking a distance of 25 m, he turned to the right and walked 10 m. He then again turned to the right and walked 15 m. After this, he is to turn right at 135° and to cover 30 m. In which direction should he go?

(A) West

(B) South

(D) South-East

Solution: (C)

Punu first walks west.

Then right turn from west means north.

Then right turn from north means east.

Now he is facing east.

A right turn of 135° from east will take him towards South-West.

Therefore, he should go towards South-West.

83. Six students A, B, C, D, E and F are sitting in a bench facing north. C is sitting second to the right of F. A is sitting extreme left end of the bench. E is sitting second to the right of C. D is sitting immediate left of E. Who is sitting between F and C?

(A) A

(B) B

(D) D

Solution: (B)

A is at the extreme left end.

C is second to the right of F, so one person must sit between F and C.

E is second to the right of C.

D is immediate left of E.

The correct arrangement is: A F B C D E.

So, B is sitting between F and C.

84. The covariance between two variables X and Y is 4. The standard deviation of X is 10 and the correlation coefficient between X and Y is 0.4. Find the standard deviation of Y.

(A) 1

(B) 2

(D) 4

Solution: (A)

Formula for correlation coefficient:

r = Cov(X,Y) / (σx × σy)

Given:

Cov(X,Y) = 4

σx = 10

r = 0.4

0.4 = 4 / (10 × σy)

4 = 4σy

σy = 1

85. For a Binomial Distribution with mean = 4 and variance = 3, what are the values of n and p?

(A) n = 16, p = 0.25

(B) n = 12, p = 1/3

(D) n = 16, p = 0.75

Solution:

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86. In a binomial distribution, the probability of success is 0.7. The variance for n = 15 is __________.

(A) 4.50

(B) 3.15

(D) 2.10

Solution:

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87. The mean and mean deviation of a normal distribution are 13.5 and 4.8 respectively. Then the third quartile of the normal distribution is __________.

(A) 6.25

(B) 17.55

(D) 1.75

Solution:

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88. The mean of a binomial distribution is 4 and the variance is 3.2. If q < 0.5, find the value of p.

(A) 0.2

(B) 0.4

(D) 0.3

Solution: (A)

For binomial distribution:

Mean = np = 4

Variance = npq = 3.2

So, 4q = 3.2

q = 3.2 / 4 = 0.8

Since p + q = 1

p = 1 − 0.8 = 0.2

89. If 2x + 3y + 2 = 0 is the regression equation of x on y and the arithmetic mean of y is −2, then arithmetic mean of x is __________.

(A) 4

(B) 6

(D) 2

Solution: (A)

The regression line always passes through the point (x̄, ȳ).

Given regression equation:

2x + 3y + 2 = 0

Mean of y, ȳ = −2

Substitute y = −2 in the equation:

2x + 3(−2) + 2 = 0

2x − 6 + 2 = 0

2x − 4 = 0

x = 2

Therefore, arithmetic mean of x is 2.

90. In how many ways 5 Indians and 5 Americans people are seated around a table so that no two Indians are in adjacent positions?

(A) 3! × 4!

(B) 3! × 5!

(D) 4! × 6!

Solution: (C)

First arrange 5 Americans around the circular table.

Number of ways = (5 − 1)! = 4!

Now, there are 5 gaps between Americans.

To ensure no two Indians are adjacent, place 5 Indians in these 5 gaps.

Number of ways = 5!

Total ways = 4! × 5!

91. If ₹ 50,000 grows to ₹ 80,525.5 in 5 years, the compound annual growth rate (CAGR) is __________.

(A) 9%

(B) 10%

(D) 12%

Solution:

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92. In how many years will ₹ 50,000 become ₹ 75,000 at 8% per annum compound interest? (Given log(1.5) = 0.1761 and log(1.08) = 0.0334)

(A) 4.8 years

(B) 5.1 years

(D) 5.6 years

Solution: (C)

75,000 = 50,000 × (1.08)ⁿ

(1.08)ⁿ = 1.5

n = log(1.5) / log(1.08)

n = 0.1761 / 0.0334 = 5.27 years

Therefore, n = 5.3 years approximately.

93. A loan of ₹ 1,00,000 is repaid in 3 equal annual instalments at 10% per annum interest, compounded annually. Find the amount of each instalment. (Given P(3,0.1) = 2.48685)

(A) ₹ 41,211

(B) ₹ 41,311

(D) ₹ 40,212

Solution:

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94. A group consists of 7 men and 5 women. In how many ways can a group of 4 members be selected if the group has no women?

(A) 70

(B) 30

(D) 35

Solution:

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95. Four cards are drawn at random from a standard deck of 52 playing cards without replacement. In how many ways it can be done such that the selected cards consist of exactly one Jack and three Aces?

(A) 2304

(B) 2440

(D) 2164

Solution:

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96. A researcher wants to study the effectiveness of a new treatment by surveying the most experienced doctors in a city. He selects 50 doctors based on their reputation and years of experience. Which type of sampling method is used in this survey?

(A) Simple Random Sampling

(B) Stratified Sampling

(D) Systematic Sampling

Solution: (C)

The researcher selected doctors based on personal judgment, reputation and experience.

Selection is not random and depends on the investigator’s opinion.

Therefore, the sampling method used is Judgment Sampling.

97. Data that is classified according to an attribute or characteristic of the items under study is referred to as __________.

(A) Qualitative data

(B) Chronological data

(D) Quantitative data

Solution:

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98. Which one of the following parts is used to show the units of measurement?

(A) Body

(B) Caption

(D) Stub

Solution:

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99. Which of the following statement is true?

(A) Mean of the statistic is known as Standard Error

(B) Mean deviation of the statistic is known as Standard Error

(D) Geometric mean of the statistic is known as Standard Error

Solution: (C)

Standard Error is defined as the standard deviation of the sampling distribution of a statistic.

It measures the variability of the statistic from sample to sample.

Hence, Standard deviation of the statistic is known as Standard Error.

100. Mutually Exclusive classification of class intervals __________.

(A) Excludes both the class limits

(B) Excludes the upper class limit but includes the lower class limit

(D) Include both the class limits

Solution: (B)