# Top 100 questions on QA & DI ## Top 100 questions on QA & DI

In absence of any prescribed syllabus aspirants get intrigued on exams and it is pertinent to note that QA & DI are the two main sections which overlap with all exams and since these two sections are most difficult hence, our experts have prepared important question bank with solutions for you who aspire to crack forthcoming CAT/XAT/IIFT/NMAT/CMAT/SNAP and MAT exams

## SECTION – I : Quantitative Aptitude

1. Of five letters mailed today, each have a 2/5 probability of arriving in three days or sooner. What is the probability that exactly two of the five letters will arrive three days or sooner?

 (a) 216/125 (b) 216/625 (c) 125/216 (d) 64/125

2. Given: Min(a, b) = Least of a, b
Max(a, b) = Larger of a, b
G(a, b) = GCD of a, b
L(a, b) = LCM of a, b
P(a, b) = Product of a, b
Then the value of P(7, Max(3, Min(L(3, 5), P(17, 1)))) is

 (a) 17 (b) 105 (c) 119 (d) None of these

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#### 1.Quantitative Aptitude 2. Data Interpretation 3. Solutions

3. If a, b and c are real numbers, then the minimum value of ab + bc + ca is

 (a) 0 (b) -1/2 (c) -3/2 (d) None of these

4. Find the total distance covered by an elastic ball before it comes to rest, if it is dropped from a height of 20 m, and after each fall, it rebounds to 50% of the height from which it falls.

 (a) 120m (b) 100m (c) 60m (d) None of these

5. Given N = 11 X 13 X 17. What will be the number of factors of 12N, excluding 1 and 12N?

 (a) 5 (b) 48 (c) 24 (d) 46

6. The perimeter of a triangle is 8 cm and one of the sides is 3 cm. If the triangle has the maximum area then the other two sides are

 (a) (5/2, 5/2) (b) (3, 2) (c) (3/2, 7/2) (d) None of these

7. Two watermelons of the same quality are being sold. The first is 60 cm in circumference and the other is 50 cm. The first is one and half times as costly as the second. Which of the two is more profitable to buy?

 (a) First (b) Second (c) Both are same (d) Depend on the weight

8. A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time (in hours) required by the first pipe is

 (a) 10 (b) 12 (c) 15 (d) 20

9. A company could advertise about its new products in 4 magazines, 3 newspapers and 2 television channels. But in later move it decided to give advertisements in only 2 of the magazines one of the newspapers and one of the television channels. In how many ways can they advertise their products?

 (a) 3 (b) 36 (c) 44 (d) None of these

10. There are two positive integers a and b. What is the probability that a + b is odd?

 (a) 1/4 (b) 1/3 (c) 1/2 (d) 1/5

11. The sum of three fractions is 59/24. When the largest fraction is divided by the smallest fraction thus obtained is 7/6 which is 1/3 more than the middle one. Find the fractions.

 (a) 1/2, 2/3, 3/4 (b) 4/3, 8/7, 5/6 (c) 2, 4, 5 (d) 3/4, 5/6, 7/8

12. If log10 5 + log10 (5x + 1) = log10 (x + 5) + 1, then the value of x is

 (a) 1 (b) 2 (c) 3 (d) 4

13. In a âˆ† ABC, the median drawn from the vertex B is perpendicular to the side BA, and then which of the following is correct?

 (a) tanA – tanB = 0 (b) 2tanA – tanB = 0 (c) 2tanA + tanB = 0 (d) None of these

14. A and B can do a work in 30 days. B and C can do the same work in 24 days whereas C and A can do in 20 days. A, B and C started the work together, but B and C left after 10 days. How many days more will A take to complete the work?

 (a) 18 (b) 21 (c) 22 (d) 24

15. If f(x) = x-3, g(x) = x2, and h(x) = 1/x, then for -1 < x < 1,

 (a) f(g(h(x))) = h(f(g(x))) (b) f(g(h(x))) Ëƒ h(f(g(x))) (c) f(g(h(x))) Ëƒ f(h(g(x))) (d) None of these

16. If six bells start ringing together and ring at intervals of 4, 8, 10, 12, 15 and 20 seconds respectively then how many times will they ring together in 60 minutes?

 (a) 28 (b) 31 (c) 36 (d) None of these

17. If all the terms of a geometric series are positive, the first term of the series is 2 and the third term is 8, then how many digits are there in the 40th term?

 (a) 10 (b) 11 (c) 12 (d) 13

18. When (6767 +67) is divided by 68 then the remainder is

 (a) 65 (b) 66 (c) 67 (d) 1

19. In a row of 40 boys, Ram was shifted 10 places to the right of Shyam and Hari was shifted 10 places to the left of Manu. If Manu was 26th from the left and there were 3 boys between Hari and Ram after shifting, what was the position of Shyam in the row?

 (a) 11th from the right end (b) 13th from the left end (c) 30th from the right end (d) Insufficient Data

20. In two alloys A and B, copper and zinc are in the ratios of 4:1 and 1:3 respectively. A new alloy C is obtained by melting together 10 kg of A, 16 kg of B and some of pure copper. If in C the ratio of copper to zinc is 3:2, find the weight of the alloy C in Kg.

 (a) 34 (b) 35 (c) 42 (d) 45

21. Three boys A, B and C start running at constant speeds from the same point P along the circumference of a circular track. The speeds of A, B and C are in the ratio 5:1:1. A and B run clockwise while C runs in the anticlockwise direction. Each time A meets B or C on the track he gives them a card. What is the difference in the number of cards received by B and C if A distributes 33 cards in all?

 (a) 3 (b) 7 (c) 5 (d) 11

22. If  S = 1/100 + 1/101 + 1/102  .... 1/1000 ,  then

 (a) S ≤ 1/3 (b) 1/3 < S ≤ 2/3 (c) 2/3 < S ≤1 (d) S > 1

23. A and B are two points on a co-ordinate plane. All the points, in the same plane as A and B, whose distance from B is twice that from A lie on

 (a)  a straight line intersecting AB at a point O such that 2AO = BO. (b) a circle with center at a point O on AB such that AO = 2BO. (c) a circle with center at a point O on AB extended such that 4AO = BO. (d)  None of these

24. 677 has exactly 5 digits when converted into base ‘n’ from the decimal system. What is the minimum possible value of ‘n’?

 (a) 3 (b)  5 (c) 4 (d)  6

25. a1 = 1, a2 = 2 and an+2 = an(an + 1 – 1), where n is a natural number. What is the value of a1000

 (a) 2 (b) – 1 (c) 0 (d)  1

26.

A faulty clock gains 10 minutes every hour. If the time is set correctly at 12 Noon on 1st Jan 2010, then how many times will its minute-hand and hour-hand meet in the next 24 hours ?

 (a) 22 (b) 26 (c) 24 (d)  25

27. India and Brazil play the Soccer World Cup final in which India defeats Brazil 4 – 2. In how many different ways could the goals have been scored given that Brazil never had a lead over India during the match?

 (a) 9 (b) 10 (c) 8 (d)  None of these

28. How many real values of ‘x’ satisfy the given equation?

 (a) 4 (b) 3 (c) 2 (d)  None of these

29. ‘M’ and ‘N’ are natural numbers such that by M = (5N – 4) (5N + 1). If 1 ≤ N ≤ 200,   what is the harmonic mean of all the possible values of M?

 (a) 2002/1001 (b) 1001 (c) 1001/2002 (d)  1/1001

30.

In the figure given below, ABCDEFGH is a regular octagon of side ‘a’ units inscribed in a circle of radius ‘r’ units. If a2 what is the area (in square units) of the shaded region? 31.

In the given figure AB is the diameter of a circle with center ‘O’. C is any point in the circle such that ∠ACB = 120° . OD is perpendicular to BC and the length of OD is 1.5 cm. What is the length of AC (in cm)?â€‹

 (a) 3√3/2 (b) 3√3 (c) 2√3 (d)  6

32. Fourteen fruits and twenty two flowers are to be distributed among 10 people in such a way that each person gets something. Anyone who gets more than two flowers cannot get more than one fruit and anyone who gets more than one fruit cannot get more than three flowers. What is the maximum number of flowers that one can get?

 (a) 3 (b) 5 (c) 19 (d)  22

33.

The cost price of four articles A, B, C and D are ‘a’, ‘b’, ‘c’ and ‘d’ respectively. A, B, C and D are sold at profits of 10%, 20%, 30% and 40% respectively. If the net profit on the sale of these four articles is 25%, ‘a’, ‘b’, ‘c’ and ‘d’ cannot be in the ratio

 (a) 4 : 1 : 4 : 3 (b) 1 : 2 : 2 : 1 (c) 2 : 3 : 6 : 1 (d)  5 : 2 : 7 : 3

34.  x2 – 3y2 = 1376 How many integer solutions exist for the given equation?

 (a) One (b) Two (c) Four (d)  Zero

35.

Function f(x) is a continuous function defined for all real values of x, such that f(x) = 0 only for two distinct real values of x. It is also known that
f(6) + f(8) = 0
f(7).f(9) > 0
f(6).f(10) < 0
f(0) > 0 and f(1) < 0

How many of the following statements must be true? I.

f(1).f(2).f(3) < 0 II. f(3).f(5).f(7).f(9) > 0 III. f(7).f(8) < 0 IV. f(0) + f(1) + f(9) + f(10) > 0â€‹

 (a) 1 (b) 2 (c) 3 (d) 4

36. When working alone A, B and C can complete a piece of work in 8, 12 and 30 days respectively. At the most only two people can work on each day and nobody works for more than two consecutive days. What is the minimum number of days that they will take to finish the work? 37. ABCD is a rectangle with BC = a units and DC = √3a units. The perpendicular dropped from point A meets BD at point F. The diagonals AC and BD intersect at point G. What is the area (in square units) of âˆ†AFG?

 (a) √3a2/12 (b) √3a2/ 6 (c) √3a2/10 (d) √3a2/ 8

38.

The question given below is followed by two statements, A and B. Mark the answer using the following instructions:
Mark (a) if the question can be answered by using one of the statements alone, but cannot be answered by using the other statement alone.
Mark (b) if the question can be answered by using both the statements together, but cannot be answered by using either statement alone.
Mark (c) if the question can be answered by using either statement alone.
Mark (d) if the question cannot be answered even by using both the statements together

K is an integer and a, b are positive real numbers. Is (a+K)/(b+K) > a/ b ?â€‹

 (a) a > b (b) a > 7, b > 10

39.  f(x) = x2 + bx + c
The equation f(x) = 0 has two distinct roots which are from the set {– 3, – 2, – 1, 0, 1, 2, 3}. How
many different expressions of f(x) are possible such that f(0) is non-negative?

 (a) 3 (b) 6 (c) 12 (d) Infinite

40. From a solution that has milk and water in the ratio 5 : 3, ‘x’ percent is removed and replaced with water. The concentration of milk in the resulting soution lies between 30% and 50%. Which of the following best describes the value of ‘x’?

 (a)  25 < x < 50 (b)  20 < x < 52 (c)  20 < x < 48 (d) 25 < x < 60

41. Four boxes are labeled as A, B, C and D. Each box contains three balls - one red, one blue and one green. In how many ways can a person pick 2 red and 3 blue balls?

 (a) 48 (b)  24 (c)  8 (d) 16

42. Kamla fasted on 1st, 3rd, 6th, 10th, 15th...... day of the year 2010 and she continued fasting in this pattern thereafter. Bimla fasted on 1st, 4th, 8th, 13th, 19th..... day of the year 2010 and maintained the pattern thereafter. When would they fast together again?

 a)  23rd December, 2010 (b)  13th September, 2010 (c)  21st December, 2010 (d) Never

43. In the X-Y plane two distinct lines are drawn parallel to the line 3y – 4x = 15, each at a distance of 3 units from the given straight line. What are the lengths of the line segments of these two lines lying inside the circle x2 + y2 = 25?

 (a) 6 and 8 (b)  0 and 8 (c)  0 and 10 (d) 8 and 10

44.  How many numbers are there between 0 and 1000 which on division by 2, 4, 6, 8 leave remainders 1, 3, 5, 7 respectively?

 (a) 21 (b)  40 (c)  41 (d) 39

45. If ‘a’ is one of the roots of x5 – 1 = 0 and a 1, ≠ then what is the value of a15 + a16 + a17 +.......a50 ?

 (a) 1 (b)  5a (c)  35 (d) None of these

46.

If ‘n’ is a natural number then the greatest integer less than or equal to (5 + √19 )n is

 (a) even (b)  odd (c)  even when ‘n’ is even and odd when ‘n’ is odd. (d)  even when ‘n’ is odd and odd when ‘n’ is even.

47. A welder can finish job A and job C in 10 days and 20 days respectively. A blacksmith can finish job B and job C in 15 days and 10 days respectively. On the first day, two welders begin work on job A and a day later three blacksmiths begin work on job B. What is the least time required to complete all the three jobs?

 (a) 33/4 days (b)  37/4 days (c)  31/4 days (d)   None of these

48.

In the given figure ABCD is a quadrilateral with BC = 4 cm and AD = 2 cm. What is the length of AB (in cm)?â€‹

 (a) 4 – 1/√3 (b)  (4√3 – 2)/√3 (c)  (2√3 + 1)2 (d) (2√3 - 1)2

49. Vijay, Saral and Himanshu visited a shop to buy a certain article and each one of them bought atleast one unit of the article. The shopkeeper gave a discount of 20% to Vijay, two successive discounts of 20% and 25% to Saral and three successive discounts of 20%, 25% and 40% to Himanshu. After selling ‘x’ units of the article he calculated that he had given an overall discount of 50%. If 8 ≤ x ≤ 20, then how many values of ‘x’ are possible?

 (a) 5 (b)  7 (c)  0 (d)   None of these

50.  ABCD is a parallelogram. E is a point on AB such that AE : BE = 2 : 3. A line EF is drawn parallel to AD and it meets CD at F. G is a point on BC such that GB : GC = 1 : 4. What is the ratio of the area of âˆ†DEC to the area of âˆ†EFG?

 (a)  3 : 5 (b)  10 : 3 (c)  25 : 12 (d)   None of these

51.  In how many ways can 4 distinct balls be distributed into 3 identical boxes?

 (a)  14 (b)  17 (c)  11 (d)  6

52.  The sum of three numbers in A.P. is 45. If the sum of their squares is 683, what is the largest of the three numbers?

 (a)  16 (b)  19 (c)  17 (d)  18

## SECTION – II : Data Interpretation

DIRECTIONS FOR QUESTIONS 53 to 57 :
Refer to the data below and answer the questions that follow. 53. What was the percentage increase in imports between 1995-96 and 1999-2000?

 (a) 80 (b) 90 (c) 100 (d) None of these

54. What was the approximate percentage change in trade gap between 1998-99 and 1999-2000?

 (a) 4.6 (b) 4.9 (c) 5.1 (d) 6.4

55. What was the approximate percentage increase in trade gap between 1996-97 and 1997-98?

 (a) 160 (b) 175 (c) 190 (d) 200

56. Which of the following is true?

 (a) In all the years exports are greater than imports. (b) Trade gap is continuously increasing. (c) Trade gap is continuously decreasing. (d) The trade gap first increases and then exhibits a decreasing trend.

57. If oil imports constituted 20% of the total imports in 1997-98, then what percent of the trade gap was due to oil (assuming that no oil is exported)?

 (a) 30 (b) 40 (c) 50 (d) 60

DIRECTIONS FOR QUESTIONS 58 to 62 :
RAILWAY TIMETABLE : NEW DELHI – BHUBANESWAR RAJDHANI EXPRESS

 Station Name Arrival Time Departure Time Distance(In KM) New Delhi -- 17:05 0 Kanpur Central 21:48 21:53 440 Mughal Sarai JN 01:55 02:05 786 Gaya JN 04:18 04:21 989 Koderma 05:22 05:24 1066 Bokaro Steel City 07:35 07:40 1191 Tatanagar JN 10:35 10:40 1342 Kharagpur JN 12:25 12:40 1476 Balasore 14:02 14:04 1595 Bhadrak 15:12 15:14 1657 Cuttack 16:33 16:35 1772 Bhubaneswar 17:30 -- 1800

58. The longest run for the train between the two successive halts is

 (a) Mughal Sarai JN - Gaya JN (b) New Delhi - Kanpur Central (c) Kanpur Central - Mughal Sarai JN (d) Balasore - Balasore

59. The average speed that the train maintained between two successive stations was the highest between

 (a) Kanpur Central - Mughal Sarai JN (b) Mughal Sarai JN - Gaya JN (c) New Delhi - Kanpur Central (d) Bokaro Steel City - Tatanagar JN

60. The average speed that the train maintained between New Delhi and Bhubaneswar was nearly equal to

 (a) 72 kmph (b) 74 kmph (c) 75 kmph (d) 82 kmph

61. If we consider a journey that begins in New Delhi and ends in Bhubaneswar, the train has the longest halt at

 (a) Kanpur Central (b) Mughal Sarai JN (c) Tatanagar JN (d) Kharagpur JN

62. The train begins its return journey from Bhubaneswar to New Delhi Seventeen hours after it has arrived at Bhubaneswar. If the train left New Delhi on Tuesday on what day will it have returned to New Delhi? (Assume that on the return journey that train maintains the same average speed as on the onward journey).

 (a) Thursday (b) Friday (c) Saturday (d) Sunday

DIRECTIONS FOR QUESTIONS 63 to 64 :

Answer the questions on the basis of the information given below. Two stock traders – Hasan and Sajid – trade in the shares of Pineapple Group only. Each of them follows a different strategy for selling and buying the stocks. Sajid sells an equal number of shares at 9 a.m. and 10 a.m. and buys them back in equal numbers at 1 p.m. and 2 p.m. Hasan sells some shares at 11 a.m. and the rest at 12 noon such that the amount he gets on the two occasions is the same. He buys back the whole lot at 3 p.m. The chart given below displays the performance of Pineapple Group stock on two particular days – Day 1 and Day 2. The profit/loss made by a trader on a particular day is the difference between the amount that he obtains by selling the shares and the amount that he spends in buying the shares. “Margin” for a day is the profit/loss expressed as a percentage of the total amount obtained by selling the shares on that day. 63.

If Sajid and Hasan sold an equal number of shares on Day 1, then which of the following statements would definitely be true?

I. Hasan made more profit than Sajid on Day 1.
I. Hasan made less profit than Sajid on Day 1.
III. Hasan’s Margin was more than Sajid’s Margin on Day 1.
IV. Hasan’s Margin was less than Sajid’s Margin on Day

 (a)  I only (b)  I and IV (c) II and IV (d) I and III

64. What was the ratio of Hasan’s Margin to Sajid’s Margin on Day 2?

 (a)  13:20 (b)  11:20 (c) 20:13 (d) Cannot be determined

DIRECTIONS FOR QUESTIONS 65 to 67 :

Answer the questions on the basis of the information given below. The graph given below shows the statistics of five companies – Perfitti VM, UB Group, ITC, Unilever and Wipro – in India. Each point on the graph indicates the Total Revenue generated by a different department of one of these companies and the Average Revenue generated per Unit Energy Consumed by that particular department. 65. If the departments represented in the graph are the only departments in the five companies where energy is consumed, then for which company is the Energy Consumption the highest?

 (a)  Wipro (b)  UB Group (c) ITC Ltd. (d) None of these

66. How many of the represented departments across the five companies have consumed less than 100W-hrs of energy?

 (a)  Seven (b) Three (c) Five (d) Six

67. Which of the five companies has the highest number of departments that generate more than Rs. 600 crores as the Total Revenue and consume less than 200W-hrs of energy?

 (a)  Perfitti (b)  Unilever (c) Wipro (d) ITC Ltd.

DIRECTIONS FOR QUESTIONS 68 to 69 :

Answer the questions on the basis of the information given below. The graph given below shows the total Exports and Imports of a country for four different years (i) Total Exports is the sum of Oil Exports and Non-Oil Exports.
(ii) Total Imports is the sum of Oil Imports and Non-Oil Imports.
(iii) Oil Trade Balance is the surplus of Oil Exports over Oil Imports.
(iv) Non-Oil Trade Balance is the surplus of Non-Oil Exports over Non-Oil Imports.

68. Which of the following experiences the highest increase?

 (a)  Oil Exports from 2000-01 to 2010-11 (b)  Oil Imports from 1990-91 to 2000-01 (c) Oil Trade Balance from 1990-91 to 2000-01 (d) Total Trade Balance from 1990-91 to 2000-01

69. Which of the following experiences the lowest percentage change?

 (a)  Non-Oil Exports from 2000-01 to 2010-11 (b)  Total Exports from 1990-91 to 2000-01 (c) Oil Imports from 1980-81 to 1990-91 (d) Total Imports from 2000-01 to 2010-11.

DIRECTIONS FOR QUESTIONS 70 to 71 :

Answer the questions on the basis of the information given below.

The performance appraisal of the employees of Hondai Motors Pvt Ltd. was done three times in the year 2010. The first appraisal was done in January, the second in July and the third in November. Only the employees who were appraised in January were eligible for appraisal in July and only those who were appraised in July were eligible for appraisal in November.
During an appraisal, an employee was appraised on exactly one of the three performance areas – Individual Performance, Team Performance and Moral Conduct. An employee already appraised on a particular performance area was not appraised on the same performance area in subsequent appraisal(s) during the year.

The table given below shows the number of employees appraised in each of the three appraisal months in 2010 for different performance areasâ€‹

 Performance Area Appraisal Month January July November Individual Performance 71 30 9 Team Performance 67 22 13 Moral Conduct 97 29 11

70. How many employees of Hondai Motors Pvt Ltd. were appraised on exactly one performance area in 2010?

 (a)  235 (b)  121 (c) 154 (d) None of these

71.  Among the employees of Hondai Motors Pvt Ltd. who were appraised in 2010, how many were not appraised on Individual Performance?

 (a)  125 (b)  134 (c) 113 (d) 165

DIRECTIONS FOR QUESTIONS 72 to 74 :

Answer the following questions on the basis of information given below: The graph given below shows the statistics of 12 Cricket players. Each point on the graph indicates the average score per match of a player and the number of matches played by that player. Each of the players plays for one of the four teams - Team 1, Team 2, Team 3 and Team 4. 72. If only the runs scored by the above mentioned 12 players are considered then which team has got the maximum aggregate score?

 (a)  Team 1 (b)  Team 2 (c) Team 3 (d) Team 4

73. How many players are there who have played more than 200 matches and have scored less than 9,000 runs?

 (a) 2 (b)  3 (c) 4 (d) 5

74. What is the overall average score of those players of Team 4, whose averages are better than the average of that player who has played the second highest number of matches for Team 3?

 (a)  41.16 (b)  42.50 (c)  42.07 (d) 43.40

DIRECTIONS FOR QUESTIONS 75 to 77 :

Answer the following questions on the basis of information given below: The following table shows the marks obtained by four students in three Mock CAT tests. The four students Abhishek, Akshay, Amitabh and Aamir are disguised as A, B, C and D in no particular order.

 Mock CAT A B C D Mock 1 47 51 53 Mock 2 62 59 10 Mock 3 72 25 28 Total score a b c d

• The maximum possible marks in each Mock CAT were 75.
• In each Mock CAT, a correct answer carried +3 marks and an incorrect answer carried -1 mark.
• Abhishek scored the lowest marks in Mock 1 and Aamir scored the highest marks in Mock 2 among the four students.
• Abhishek's total score is more than Amitabh's total score.
• Akshay's total score is less than Amitabh's total score.
• A’s total score is a multiple of 6
75. How many different total scores are possible for Akshay?

 (a)  101 (b)  97 (c)  98 (d) 100

76. What is the name of the person disguised as B?

 (a)  Aamir (b)  Abhishek (c)  Akshay (d) Either (a) or (b)

77. How many different values are possible for ‘a’?

 (a)  7 (b)  8 (c)  9 (d) 15

DIRECTIONS FOR QUESTIONS 78 to 80 :

Answer the following questions on the basis of information given below: Five actresses Careena, Catrina, Carishma, Coena and Cashmira are disguised as U, V, X, Y and Z, not necessarily in this order. Director Bhakti Kapoor interviewed the actresses in the order Z, Y, X, U and V for his film “Bhoot Aunty”.

• Catrina was interviewed after Cashmira.
• Careena was interviewed before Carishma.
• The fee (in Rs.) demanded by X, U, V were 2397 lakhs, 2379 lakhs and 1213 lakhs respectively.
• One of the remaining two actress demanded Rs. 1123 lakhs and the other demanded a sum (in Rs.) between 1201 lakhs and 2288 lakhs.
• The sum of the fees demanded by Careena and Carishma is the same as the sum of the fees demanded by Cashmira and Catrina.
78.  Who is disguised as Y?

 (a)  Coena or Cashmira (b)  Coena or Careena (c)  Coena or Cashmira or Careena (d) Coena or Cashmira or Catrina

79. What is the fee (in Rs.) demanded by Coena?

 (a)   2397 or 1213 lakhs (b)  1213 lakhs (c)  1213 or 2379 lakhs (d) 1123 lakhs

80.  Who was interviewed last ?

 (a)  Coena or Catrina (b)  Coena or Carishma (c)  Carishma or Catrina (d) None of these

DIRECTIONS FOR QUESTIONS 81 to 83 :

Answer the questions on the basis of the information given below.

In a country called Khelabad, each sportsman plays either Cricket or Hockey, but not both. Each of them plays for one of the two teams – Under-19 or Above-19. All the sportsmen of Khelabad who had passed in a Physical Fitness Examination conducted by the Sports Ministry recently are certified as “fit” and the rest are called “unfit”. It is also known that:
(i) The ratio of the number of fit sportsmen to the number of unfit sportsmen is 3 : 2.
(ii) Fifty percent of the sportsmen who play Hockey are fit.
(iii) The number of unfit sportsmen who play Hockey for the Under-19 team is 2000.
(iv) Eighty percent of the Hockey players play for the Under-19 team.
(v) All the sportsmen who play Cricket are fit.
(vi) The number of unfit sportsmen who play Hockey for the Above-19 team is equal to the total number of sportsmen who play Cricket for the Above-19 team.
(vii) The total number of sportsmen who play Cricket is 1250

81.  What is the total number of sportsmen in Khelabad?

 (a)  5000 (b)  6250 (c)  6000 (d) 5500

82.  What is the ratio of the number of unfit sportsmen who play Hockey for Above-19 team to the number of unfit sportsmen who play Hockey for Under-19 team?

 (a)  1:3 (b)  3:1 (c)  1:4 (d) 4:1

83. What is the total number of fit sportsmen who play for Above-19 team?

 (a) 1000 (b)  2000 (c) 1500 (d) Cannot be determined

DIRECTIONS FOR QUESTIONS 84 to 86 :

Answer the questions on the basis of the information given below.

The table given below shows the marks scored by six students of a School in Physics, Chemistry, Mathematics and English in an exam. Each subject is assigned a Credit as mentioned along with the subject name in bracket. E.g. the Credit assigned to Physics is 2.5.â€‹

 Subject Physics (2.5) Chemistry (1.5) Mathematics (4) English (3.5) Student Aman 40 67 77 38 Bipin 55 65 88 44 Cute 75 64 55 60 David 64 74 98 48 Esha 75 65 41 77 Fardeen 41 39 67 84

The Grade assigned to a student against the marks scored by him/her in a subject is calculated as per the table given below.

 Marks (X) Grade X< 40 E 40 ≤X < 50 D 50 ≤ X < 65 C 65 ≤ X < 75 B X ≥ 75 A

The points received by a student against the grade assigned to him/her in a subject are calculated as per the table given below.

 Grade Points A 10 B 8 C 6 D 4 E 2

The CGPA (Cumulative Grade Point Average) of a student is calculated using the formula given below. (‘Y’ represents one of the four subjects.)â€‹

84. Who among the six students got the highest CGPA?

 (a)  Aman (b)  Bipin (c)  Esha (d) Fardeen

85. Which two students got the same CGPA?

 (a) Cute and Fardeen (b)  Bipin and Esha (c)  Bipin and David (d) Aman and Fardeen

86. What was the CGPA of Fardeen?

 (a)  6.00 (b)  7.04 (c)  6.96 (d) 6.86

DIRECTIONS FOR QUESTIONS 87 to 89 :

Answer the questions on the basis of the information given below. The break-up of the financial budget of a country called Chaupatland for FY 2010-11 is represented by the pie charts given below. Pie chart A represents Budgeted Revenue and pie chart B represents Budgeted Expenditure. Budgeted Revenue was less than Budgeted Expenditure by Rs. 625 thousand crores. Interest Payment exceeded Non-Planned Capital Account by 12% of Budgeted Expenditure. Non-Tax Revenue was Rs. 285 thousand crores less than Tax Revenue.â€‹

87.  The difference between Budgeted Expenditure and Budgeted Revenue was what percentage of Budgeted Revenue?

 (a)  120% (b)  125% (c)  25% (d) 20%

88. By how much did the sum of Planned Revenue Account and Planned Capital Account exceed Non-Tax Revenue?

 (a) Rs. 281.25 thousand crores (b) Rs. 236.25 thousand crores (c) Rs. 186.25 thousand crores (d) Rs. 88.25 thousand crores

89. In FY 2011-12, if Budgeted Revenue is increased by 52% over that of the previous year, then what will be the new percentage share of Tax Revenue in Budgeted Revenue?

 (a)  50% (b)  76% (c) 56% (d) Cannot be determined

DIRECTIONS FOR QUESTIONS 91 to 92 :

 Year 1951 1961 1971 1981 1991 1. Full Time Child Workers Census data based estimates (0-14 Yrs.) 13387144 14469775 10664018 11195544 12669909 NSS data based estimates (5-14 Yrs.) 11339526 13777443 16330000 16166330 13950225 2. Non-workers & Non-students (5-14 Yrs.) Census data based estimates 49700129 64914609 89482123 89541313 97659410 MHRD & NSS data based estimates 52997224 61123492 86092259 77352410 94554833 3. Child Marginal workers Census estimates (1981 and 1991) NA NA NA 2445329 10498822 4. Estimates of total child workers Census data based estimates 13387144 15469775 16153985 18340873 23161013 NSS data based estimates 11339526 15299910 16330000 18611659 23449047

90.  In Census data based estimates (0-14 Yrs.), which year saw the maximum percentage change in Full Time Child Workers as compared to the previous Census?

 (a)  1961 (b)  1971 (c) 1981 (d) 1991

91. In Non-workers & Non-students (5-14 Yrs.) category, the two estimates are based on Census and MHRD & NSS respectively. For which year is the absolute difference between the two estimates the highest?

 (a)  1951 (b)  1961 (c) 1971 (d) 1981

DIRECTIONS FOR QUESTIONS 92 to 94 :

Answer the questions on the basis of the information given below.

The table given below shows the marks obtained by 20 students of a school in their annual exam. The Performance Score of a student is calculated as: Performance Score = Academic Score + 3 × Extra Curricular Score

 Name Gender Academic Score Extra Curricular Score Stage of Education Vikrant M 73 10 Secondary Saurabh M 70 12 Senior Secondary Kaushik M 46 7 Primary Meenal F 46 6 Primary Manmeet M 61 4 Primary Vikram M 78 10 Secondary Manjari F 30 16 Primary Asmita F 32 6 Primary Mohit M 68 15 Senior Secondary Pavan M 58 3 Secondary Lokesh F 67 6 Secondary Pranab M 32 8 Secondary Mudit M 54 12 Senior Secondary Amit F 74 11 Primary Maneet M 47 19 Primary Suresh M 49 19 Secondary Puneet M 33 12 Secondary Visakha F 57 12 Senior Secondary Renuka F 59 3 Primary Santosh M 40 9 Senior Secondary

The school calculates ‘Happy Growth Index’ for every student to determine the relative happiness of its students. Students are classified as either High or Low on Happy Growth Index based on the criteria shown in the table given below.

 Gender Stage of Education Performance Score Happy Growth Index Male Primary 70 ≤ PS ≤ 100 High Primary PS < 70 or PS > 100 Low Secondary 75 ≤ PS ≤ 107 High Secondary PS < 75 or PS > 107 Low Senior Secondary 85 ≤ PS ≤ 110 High Senior Secondary PS < 85 or PS > 110 Low Female Primary 60 ≤ PS ≤ 90 High Primary PS < 60 or PS > 90 Low Secondary 70 ≤ PS ≤ 95 High Secondary PS < 70 or PS > 95 Low Senior Secondary 75 ≤ PS ≤ 100 High Senior Secondary PS < 75 or PS > 100 Low
92.  How many students are Low on Happy Growth Index at Primary Stage of Education?

 (a)  2 (b)  3 (c) 4 (d) 5

93. What is the ratio of the number of male students at Senior Secondary Stage who are High on Happy Growth Index to the number of female students at Primary Stage who are High on Happy Growth Index?

 (a)  1 : 1 (b)  2 : 3 (c) 1 : 3 (d) 4 : 3

94.  How many female students whose Academic Score is less than 50 have the same Happy Growth Index as that of Renuka?

 (a)  1 (b)  4 (c) 3 (d) 2

DIRECTIONS FOR QUESTIONS 95 to 97 :

Answer the questions on the basis of the information given below. The political parties mentioned in the graph given below participated in a constituency election. The graph shows the number of votes received by the candidates of six major parties and other smaller parties in Round 1 of the election. The election is held as follows: In Round 1, if a candidate gets more than 50% of the total number of votes cast in that round, (s)he is declared the winner of the election, else, top 5 candidates based on the number of votes move to Round 2. In Round 2, if a candidate gets more than 50% of the total number of votes cast in that round, (s)he is declared the winner of the election, else, top 3 candidates based on the number of votes move to Round 3. In Round 3, the candidate who gets the maximum number of votes in that round is declared the winner of the election. Assume that there are no ties in deciding top 5, top 3 (if Round 2 happens) candidates or the winner (if Round 3 happens) in Round 3. 95. Among the five candidates who moved to Round 2, the candidate with the maximum number of votes in Round 1 got what percentage of the total number of votes in that round?

 (a)  17.4 % (b)  19.8% (c) 16.5% (d) 18.2%

96. If the votes received by the candidates of Samajik Party and Hindu Muslim Party in Round 1 are not taken into account, then by what percentage points does the share of “Others” increase?

 (a)  0.64 (b)  0.74 (c) 0.84 (d) 0.94

97. If 1 Million votes are cast in Round 2 and 3 Million votes are cast in Round 3, then what is the minimum total number of votes needed by a candidate to be declared the winner of the election?

 (a)  26,38,194 (b)  21,38,195 (c) 21,38,193 (d) 26,38,195

DIRECTIONS FOR QUESTIONS 98 to 100 :

Answer the questions on the basis of the information given below.
The two pie charts given below specify the percentage of men and women in the various age groups A, B, C, D and E residing in a town in the year 2010.
Group A: age ≤ 20
Group B: 20 < age ≤ 25
Group C: 25 < age ≤ 30
Group D: 30 < age ≤ 35
Group E: 35 < age
These men and women contribute to the income of their town and the ratio of men’s contribution to women’s contribution is 4 : 5. Assume that all the people residing in the town are represented in the pie charts. 98. What is the ratio of men’s contribution from Group D to women’s contribution from Group C to the town’s income?

 (a) 2 : 5 (b)  4 : 5 (c) 8 : 19 (d) Data Insufficient

99.  In 2011 the number of women in Group C increases by 15%. By what percentage will the total number of women in the town increase?

 (a) 15% (b)  3% (c) 5% (d) Cannot be determined

100.  If the average contribution of men to the town’s income is Rs 35,000, then what is the average contribution of women to the town’s income?

 (a) Rs 36,842 (b)  Rs 36,742 (c) Rs 36,942 (d) Rs 37,042

## Hints & Solutions for Top 40 Questions on QA & DI

### SECTION – I : Quantitative Aptitude

 1. Number of ways in which exactly 2 of 5 letters will arrive in three days or sooner = 5! ÷ [2! X (5 - 2)!] = 10. Now, the probability of a letter failing to arrive in 3 days or sooner = 1 – 2/5 = 3/5. The probability of the first 2 letters arriving in 3 days or sooner and the other 3 letters arriving later than that = (2/5)2 x (3/5)3 = 108/3125. So, probability of exactly two letters arriving in three days or sooner = (108/3125) x 10 = 216/625. 2. P(17, 1) = 17, L(3,5) = 15, Min(15, 17) = 15, Max(3, 15) = 15, So, P(7, 15) = 105 3. Let E = ab + bc + ca= ab + c(a + b), Taking a, b Ëƒ 0 and c = a very large negative value, E = A very large negative vale. Therefore as c decreases E decreases. Thus E tends to negative infinity. 4. Total distance travelled = 20 + 2[20(1/2) + 20(1/2)2 + ……. Up to infinity] = 20 + 2 x 1/2 x 20 x [1÷ (1 – 1/2)] = 60m 5. 12N = 22 X 31 X 111 X 131 X 171 = 2a X 3b X 11c X 13d X 17e Therefore, 0 ≤ a ≤ 2, 0 ≤ b ≤ 1, 0 ≤ c ≤ 1, 0 ≤ d ≤ 1, 0 ≤ e ≤ 1 So, total number of factors of 12N = 3 X 2 X 2 X 2 X 2 = 48, which includes 1 and 12N Thus the number of factors of 12N which does not include 1 and 12n = 48 – 2 = 46. 6. Let one side of the triangle is ‘a’ then the other sides are (5 – a) and 3. s = Semi perimeter = 8/2 = 4. Therefore, Area of Triangle, A = [ s(s – a)(s –b)(s – c)]1/2 ⇒ A = [4(4 – 3)(4 – a)(4 – 5 + a )]1/2 = 4(4 – a)(a – 1) ⇒ A2 = - 4a2 + 20a – 16 = - 4( a2 - 5a + 4) = - 4[ (a – 5/2)2 – 9/4] So, for maximum area(A), a = 5/2. Thus the other two sides are 5/2 and 5/2. 7. Since the ratio of their circumference is directly proportional to their diameter, therefore ratio of their diameter = 6 : 5. Hence ratio of their volume = (6/5)3 = 216/125 = 1.73 S0, if it were priced according to its size or weight, bigger one should cost 173% as much as the smaller one. ⇒ Bigger one is 73% more than the smaller one. Now as it is 50% costlier, it is more profitable to buy the bigger melon. 8. Let the first pipe alone can fill the tank in x hours Then the second pipe can fill the tank in (x-5) hours and the third pipe can fill the tank in (x-5)-4 = (x-9) hours Therefore, part filled by first pipe and second pipe together in 1 hr = part filled by third pipe in 1 hr ⇒ 1/ x + 1/ (x – 5) = 1/(x – 9) Solving we get, x = 15 or 3. But x cannot be 3 otherwise (x – 9) becomes negative. So, x = 15 9. Total number of ways of advertising = C(4, 2) X C(3, 1) X C(2, 1) = 36 10. Here we can have four cases: (i) a is even and b is even (ii) a is even and b is odd (iii) a is odd and b is even (iv) a is odd and b is odd Out of the four cases, in case (ii) and (iii), the sum will be odd. Therefore, the required probability = 2/4 = 1/2 11. Let the numbers in ascending orders be x, y and z. ⇒ x + y + z = 59/24 and z/x = 7/6 and y = 7/6 – 1/3 = 5/6 Since z/x = 7/6, so, x = 6z/7, ⇒ 6z/7 + 5/6 + z = 59/24, ⇒ z = 7/8, x = 3/4, y = 5/6 12. log10 5 + log10 (5x + 1) = log10 (x + 5) + 1 ⇒ log10 5 + log10 (5x + 1) = log10 (x + 5) + log10 10 ⇒ log10 [5(5x+1)] = log10 [10(x+5)] ⇒ 5(5x+1) = 10(x+5), => 5x+1 = 2(x+5) ⇒ 5x + 1 = 2x + 10, => 3x = 9, => x = 3 13. Draw an obtuse angled triangle ABC. Draw CE perpendicular to extended AB. In âˆ† CEA, tan A = CE/EA = CE/2EB = 1/2 tan CBE = 1/2 tan (π – B) = - 1/2 tan B ⇒ 2 tan A + tan B = 0 14. Let the work done by A, B and C in 1 day be x, y and z respectively. ⇒ x + y = 1/30, ïƒž y + z = 1/24, ïƒž z + x = 1/20 On adding we get, 2(x + y + z) = 1/30 + 1/24 + 1/20 ⇒ x + y + z = 1/16 ⇒ In 1 day work done by A, B and C = 1/16 So, in 10 day work done by A, B and C = 5/8 So, remaining work = 1 - 5/8 = 3/8 Now, work done by A in 1 day = 1/16 – 1/24 = 1/48 So, the number of days required by A complete the remaining work = 3/8 ÷ 1/48 = 18 15. f(g(h(x))) = 1/x2 – 3 …………….. (1) f(h(g(x))) = 1/x2 – 3 …………….. (2) h(f(g(x))) = 1/(x2 – 3) ……………(3) So, (a) and (c) are not true. Now, suppose f(g(h(x))) = h(f(g(x))), then x2 = 0.3 or 3.6 (approximately). The graphs of (1) and (3) intersect at x = 0.5 and x = 1.9. Hence, the answer is (d) 16. l.c.m.( 4, 8, 10, 12, 15, 20) = 120 Hence all the six bells will ring together in every 120 seconds or 2 minutes. So, the number of times they will ring together in 60 minutes = 60/2 + 1 = 31. 17. The first term is 21 and the third term is 23. So, the 40th term is 240. Now, Let x = 240, => log x = 40(log2), => log x = 40 x 0.3(Approx)= 12.0 (Approx) So, x = AL(12.0..), So, 240 has 13 digits. 18. For n = odd, (xn + 1) is always divisible by (x+1). ⇒ (6767 + 1) is divisible by (67 + 1) ⇒ (6767 + 1) is divisible by 68 ⇒ (6767 + 1) ÷ 68 gives a remainder of 0 ⇒ [(6767 + 1) + 66] ÷ 68 gives a remainder of 66 ⇒ (6767 + 67) ÷ 68 gives a remainder of 66. 19. Manu is 26th from the left and Hari is 10 places to the left of Manu. So, Hari is 16th from left. Now, there are 3 boys between Hari and Ram. Thus the position of Ram is either 12th or 20th from the left. Thus the exact position of Ram cannot be ascertained. Hence to find the position of Shyam in the row, given data is insufficient. 20. Let the quantity of pure copper be x kg. By given condition, the quantity of copper = 4/5 x 10 + 1/4 x 16 + x ⇒ 3/5(10 + 16 + x) = 8 + 4 + x ⇒ x = 9 Thus, the weight of the new alloy = 10 + 16 + 9 = 35 kg. 21. Let the speed of A, B and C be 5v, v and v respectively.  Let the circumference of the track be 12vt.  Instance when A meets:  B – 3t, 6t, 9t, 12t, 15t, 18t ....  C – 2t, 4t, 6t, 8t, 10t, 12t ....  In every 6t units of time A meets B twice while A meets C thrice.  So in 36t A would give: B – 2 × 6 = 12 cards C – 3 × 6 = 18 cards. (total 30) Now, at 38t A gives C a card (his 31st) at 39t A gives B a card (his 32nd) and at 40t A gives C a card (his last)  So the required difference = (18 + 2) – (12 + 1) = 7 22. S = 1/100 + 1/101 + 1/102  .... 1/998 + 1/999 + 1/1000 ,  then In the first 450 terms, the sum of the two numbers in the denominator is constant. Thus, their product will be maximum when they are equal 23. Let the co-ordinates of A and B be (0,0) and (a,0) respectively. Let the co-ordinates of point C be (h, k). Given: 2AC = BC The equation written above is that of a circle with center (say O) at  (-a/3 – 0)  and radius  2a/3, The center O  (-a/3 – 0)  divides AB externally such that 4AO = BO 24. We need to check for a base where 677 must have 5 digits if converted.We also know that in Base 3, the decimal value of a 5-digit number must lie between 80 and 243 (as ‘81 signifies 10000’ and ‘242 signifies 22222’ – the minimum and maximum possible 5-digit number in base 3).So, the base cannot be 3.Next we check for Base 4. The decimal value of a 5-digit number must lie between 255 and 1024. As 256 < 677  ≤ 1023, this must be the minimum possible value of n. 25. a1 = 1  a2 = 2  a3 = 1 (1) = 1  a4 = 2 (0) = 0  a5 = 1 (– 1) = – 1  a6 = 0 (– 2) = 0  a7 = – 1 (– 1) = 1  a8 = 0 (0) = 0  a9 = 1 (– 1) = – 1  After a1 and a2, there is a cyclicity of ‘4’ in the remaining terms a1000 = a2 + 4 × 249 + 2 = 0.â€‹ 26. The minute-hand of a normal clock covers 360/ 60 =6° per minute. The hour-hand of a normal clock covers 30/60 =1°/ 2 per minute. So once they are together, in every minute the minute hand gains 6 − 1/2 =11°/2  over the hour hand.  So, time between two meetings = 360/11/2 = 720/11 minutes.  So, in any clock the hour-hand and the minute-hand meet after every 720/11 minutes.  If 60 minutes have passed in a normal clock then time passed in the faulty clock is 70 minutes.  If 24 hrs (or 24 × 60 minutes) have passed in a normal clock then time passed in the faulty clock must be 24 × 70 = 1680 minutes.  Number of times the hands meet – 27. If Brazil never gets a lead over India, the first goal of the match should be scored by India. At this stage India leads by 1 – 0, and the only way in which Brazil can take a lead is by scoring the next two goals i.e. a scoring pattern like IBBIII. In all other cases Brazil would never be able to lead India.  Total cases where India scores the first goal 5!/3!2! =10  Different scoring patterns possible = 10 – 1 = 9 28. Now, three cases arise:  Case–1  P(x) = 1 and Q(x) may be anything.  ∴  (x−7) = 1 or x = 8 But, for x = 8, Q(x) is not defined.  Case–2  P(x) = – 1 and Q(x) is an even exponent. (x – 7) = – 1 or x = 6 For x = 6, Q(x) = – 4, which satisfies the given equation. Case–3  Q(x) = 0 and P(x) 0 ⇒ x = 7 or x = 22 But, for x = 7, P(x) = 0, for which the given equation is not defined. 29. When N = 1 M = 1 × 6 N = 2 M = 6 × 11 N = 200 M = 996 × 1001 30.  31. Construction: Extend BC to cut the circle at E and join AE. âˆ†AEB is right angled at E. (Angle in a semicircle is a right angle). Hence, both AE and OD are perpendicular to EB. By AA property: âˆ† AEB ~ âˆ†ODB  Hence, OB/AB = OD/AE or OD/AE = 1/2  ⇒ AE = 2OD =  3cm. In right âˆ†AEC :  ∠ACE = 180o −∠ACB = 180o − 120o = 60â€‹o 32. All the flowers can be given to one person and then the fruits distributed among all others in such a way that all of them get at least one fruit. 33. As per the information given in the question, we can conclude that ⇒ 0.05c + 0.15d = 0.05b + 0.15a ⇒ c + 3d= b + 3a The ratio 5 : 2 : 7 : 3 does not satisfy the given relation. 34. 3y2 = x2 – 1376 As we can see L.H.S. is definitely a multiple of 3 and in R.H.S. 1376 leaves a remainder of 2 when divided by 3. There are three possibilities for x in R.H.S: (i) If x is multiple of 3, so is x2, and R.H.S. will leave a remainder of 1 when divided by 3. (ii) If x is of the form 3m + 1, x2 will be of the form 3n + 1 and R.H.S will leave a remainder of 2. m, n ∈ N  (iii) If x is of the form 3m + 2, x2 will be of the form 3n + 1 and R.H.S. will leave a remainder of 2. m, n ∈ N  So R.H.S. can never be a multiple of 3, while L.H.S. is always a multiple of 3. Hence no real solution exists. 35. f(0) > 0 and f(1) < 0 implies that one root for f(x) = 0 lies between x = 0 and x = 1. f(6) + f(8) = 0 implies that f(6) and f(8) are of opposite sign but same absolute value. Hence another root for f(x) = 0 must lie between x = 6 and x = 8. As f(1) < 0, f(6) must also be less than zero, otherwise we’ll have more than 2 roots for f(x) = 0. Hence f(8) > 0 and f(6) < 0. Further f(7).f(9) > 0 implies that both f(7) and f(9) are greater than zero. So the second root for f(x) = 0 must lie between x = 6 and x = 7. So f(x) would look like :â€‹ As f(1), f(2) and f(3) are less than zero, f(1).f(2).f(3) < 0 is true. As f(3), f(5) < 0 and f(7), f(9) > 0, f(3).f(5).f(7).f(9) > 0 is true. As f(7), f(8) > 0, f(7).f(8) < 0 is false. f(0), f(9), f(10) > 0 and f(1) < 0, but since we don’t know the magnitude of any of these four we cannot judge if f(0) + f(1) + f(9) + f(10) is greater than zero or not.â€‹ 36. The maximum total work done for three consecutive days will be when A, B and C work in distinct pairs (for example ‘A and B’, ‘B and C’ and ‘C and A’). Hence on these three days fraction of total work done will be The remaining 4/120 fraction of work can be done in least amount of time if A and B work together on the 7th day. A and B together can do 1/8 +1/12 = 5/24 + = fraction of work in a day. So, they will complete the 4/120 fraction of work in 4/120 ÷ 5/24  = 4/25 day 37. By Pythagoras theorem: Since diagonals of a rectangle are equal and bisect each other: AG = DG = AC/2  = a Hence, âˆ†DGA is equilateral and since AF is perpendicular to the base DG we can say that AF must divide âˆ†DGA into two equal halves. Area of equilateral  âˆ†DGA = √3/4 a2 square units 38. From Statement A: This statement is clearly insufficient. The answer can given only when exact values of a, b and K are known. For example let (a, b) = (4, 2) or a/b = 2  Case 1: K = 0 (a + K)/(b + K) = a/b Case 2: K = 2 (a + K)/(b + K) = 6/4 = 3/2 > a/b Case 3: K = –1 (a + K)/(b + K) = 3/1 = 3 > a/b From Statement B: As explained above for Statement A, this statement is also insufficient to answer without knowing the exact values of a, b and K. Combining Statement A and Statement B: Even the two statments combined cannot give a unique answer. 39. If f(0) is non-negative, roots of f(x) = 0 cannot be of opposite signs. So the possibilities for the two roots are (0, 3), (0, 2), (0, 1), (1, 2), (1, 3), (2, 3), (0, –3), (0, –2), (0, –1), (–1, –2), (–1, –3) and (–2, –3). Hence the required answer is 12. 40. Let the quantity of total solution be 800 units. So, quantity of milk is 500 units and water is 300 units. Now x percent of 800 = 8x units mixture is removed. The quantity (in units) of milk and water removed is 5x and 3x respectively. Final concentration of milk (in percentage) is: (500 − 5x)/800 × 100   So, 30 < (500 − 5x)/8 < 50 ⇒  240 < 500 − 5x < 400 ⇒ − 260 < −5x < −100 ⇒ 20 < x < 52. 41. Required number of ways = 4C2 X 4C3 = 6 X 4 = 24 42. Days of the year on which Kamla fasted: 1, 3, 6, 10, 15, 21, 28…. Days of the year on which Bimla fasted: 1, 4, 8, 13, 19, 26, 34…. It can be analysed that if Kamla fasted on the Kth day, Bimla must have fasted on the (K – 2)th day. Hence for both of them to fast on the same day, Kamla must fast on some (K – 2)th day and Kth day as well. This is impossible as in the first series no two consecutive terms will have a difference of 2 after the first two days. 43. Distance of origin (0, 0) from the line 3y – 4x – 15 = 0: Let the new lines drawn parallel to 3y – 4x – 15 = 0 be L1 and L2. Distance of L1 from origin = 3 + 3 = 6 units Distance of L2 from origin = 3 – 3 = 0 units The circle x2 + y2 = 25 has a radius of 5 units. Hence line segment of L1 lying inside the circle will be of zero length (L1 does not cut the circle). Chord cut by L2 will be diameter = 10 units 44. We can see that the difference between the divisor and the respective remainder is the same in each division i.e. 2 – 1 = 4 – 3 = 6 – 5 = 8 – 7 = 1 Hence the general form of such numbers will be LCM(2, 4, 6 and 8).K – 1 = 24K – 1, where ‘K’ is any natural number. Hence the numbers are 23, 23 + 24, 23 + 2 × 24, ......, 23 + 40 × 24 A total of 41 such numbers are there between 0 and 1000. 45. a15 + a16 + a17 + ....... +a50 Sum = a15 {1 + a + a2 + ------- + a35} Since a is the root of equation x5 – 1 = 0,â€‹ a5 – 1 = 0 ⇒ a5 = 1 46. Let’s assume that ‘I’ denotes the integral part and ‘F’ denotes the fraction part of   (5 + √19)n. ⇒ I + F = 5n = nC1.5n-1.√19 +nC2.5n-2.19 + nC3.5n-3.19√19 + ........â€‹  Nowâ€‹ ( 5 − √19 )n  is a proper fraction as ( 5 − √19 ) < 1 Let’s assume that F′ = ( 5 − √19 )n ⇒ F′ = 5n − nC1.5n-1.√19 + nC2.5n-2.19 + nC3.5n-3.19√19 + ........â€‹  Hence,â€‹ I + F + F′ = 2 [ 5n + nC2.5n-2.19 + nC4.5n-4.192+ ........â€‹ ] Evidently, F + F = 1 ′ as the sum of two proper fractions is always less than 2 and greater than 0. Also the right hand side is an integer and hence F + F′ should also be an integer. Therefore I + 1 is an even number and I is an odd number. Alternative method: Putting n = 1, we get 5+ √19 whose integral part is 9. Putting n = 2, we get 25 + √19 + 10 19 whose integral part is 25 + 19 + 43 which is again an odd number. Now, through the options it can be judged that the greatest integer must always be an odd number. 47. Let A involves 30 units, B involves 15 units and C involves 20 units of work. One welder does 3 units of A and 1 unit of C in a day. One blacksmith does 1 unit of B and 2 units of C in a day. Two welders would take 5 days to complete A. They finish A at the end of day 5. On day 6 they finish 2 units of C. Three blacksmiths finish B at the end of day 6. After days 6, two welders and three blacksmiths together would take 18/8 days to finish the rest of C. So in all it takes 33/4 days to complete all the jobs. 48. ∠BAD = 90° Extend BA and CD to meet at E. âˆ†EBC becomes an equilateral triangle. So BE = 4 cm In âˆ†ADE, ∠ADE  = 30° tan30° = AE/2 = 1/√3 ⇒ AE = 2/√3 cm ∴â€‹ AB = BE − AE = 4 − 2/√3 = (4√3 − 2)/√3  cm 49. Discount to Saral 20 + 25 –(20 × 25)/100 = 40% Discount to Himanshu = 40 + 40 – (40 × 40)/100 = 64%  Let a, b, c be the number of units sold at 20%, 40% and 64% respectively ⇒ (a × 20)/100 + (b × 40)/100 + (c × 64)/100 = {(a+b+c)50}/100 ⇒ 7c/50 – b/10 – 3a/10 = 0 ⇒ 7c – 5b – 15a = 0 So ‘c’ has to be a multiple of 5. Values of a, b, c satisfying the equation are (2, 1, 5); (1, 4, 5); (4, 2, 10); (3, 5, 10); (2, 8, 10). Hence the possible values for x are 8, 10, 16, 18 and 20.â€‹ 50. Area of âˆ†DEC = Area of âˆ†DBC ……..( âˆ†'s between same parallel line and same base) = 1/2 × Area of parallelogram ABCDâ€‹ Area of âˆ†EFG = 1/2 × Area of parallelogram EBCFâ€‹ = 1/2 × 3/5 × Area of parallelogram ABCDâ€‹ 51. Case 1: All four balls in the same box: 1 way Since all the boxes are identical. Case 2: Two balls in 1 box and the remaining two in another box: 4C2/2! = 3 ways It is divided by 2! because the boxes are identical. Case 3: Two balls in a box and one ball in each of the remaining two boxes: 4C2 = 6 ways Note: We need not select a box for the remaining two balls as they will go one each in the remaining two boxes in one way only. Case 4: Three balls in one box and one remaining ball in another box:â€‹ 4C3 = 4 ways Since there is only one way of selecting the other box as the boxes are identical. Total ways = 1 + 3 + 6 + 4 = 14 52. The average of the three numbers will be 15. Let the numbers be 15 – d, 15 and 15 + d. ⇒ (15 – d)2  + 152  + (15 + d)2 = 683 ⇒ d = 2 So the numbers are 13, 15 and 17.

## SECTION – II

DIRECTIONS FOR QUESTIONS 53 to 57 :
53. Imports in 1995-96 = 7000 crore
Imports in 1999-2000 = 14000 crore
Percentage increase in imports = (14000 – 7000)/7000x100 = 100 %

54. Trade gap in 1998-99 = 13500 - 7000 = 6500
Trade gap in 1999-2000 = 14000 – 7800 = 6200
Percentage decrease in trade gap = (6500 – 6200)/6500x100 = 4.6 %

55. Trade gap in 1996-97 = 9000 - 6600 = 2400
Trade gap in 1997-98 = 13000 – 6400 = 6600
Percentage increase in trade gap = (6600 – 2400)/2400x100 = 175 %

56. The trade gap first increases and then exhibits a decreasing trend.

57. Oil Imports in 1997-98 = 20/100 x 13000 = Rs. 2600 crore
Trade gap in 1997-98 = Rs. 6600 crore
Hence, percentage trade gap due to oil = 2600/6600 X 100 = 39.39 % ~ 40 %

DIRECTIONS FOR QUESTIONS 58 to 62 :
 Station Name Distance(In Km) Time Taken(In hours) Speed(In kmph) New Delhi - Kanpur Central 440 4.72 93.22 Kanpur Central - Mughal Sarai JN 346 4.03 85.86 Mughal Sarai JN - Gaya JN 203 2.22 91.44 Gaya JN - Koderma 77 1.98 38.89 Koderma - Bokaro Steel City 125 2.18 57.34 Bokaro Steel City - Tatanagar JN 151 2.92 51.71 Tatanagar JN - Kharagpur JN 134 1.75 76.57 Kharagpur JN- Balasore 119 1.37 86.86 Balasore - Bhadrak 62 1.13 54.87 Bhadrak - Cuttack 115 1.32 87.12 Cuttack - Bhubaneswar 28 0.92 30.43

58. With the run is of 440 km, the longest run is between New Delhi - Kanpur Central.

59. The average speed of 93.22 kmph is the highest between New Delhi - Kanpur Central.

60. The average speed that the train maintained between New Delhi and Bhubaneswar = 1800 km/25 hrs and 25 min = 70.82 kmph.

61. The train has the longest halt of 15 minutes at Kharagpur JN.

62. Friday

DIRECTIONS FOR QUESTIONS 63 to 64 :
63.

Let the number of shares sold by Sajid and Hasan on Day 1 be 36x each.

= 375 × 18x + 250 × 18x = Rs. 11,250x.
= 750 × 18x + 625 × 18x – 375 × 18x – 250 × 18x = Rs. 13,500x.
Sajid’s Margin on Day 1 ≈ 54.5%.

= 225 × 36x = Rs. 8,100x.
= 375 × 16x + 300 × 20x – 225 × 36x = Rs. 3,900x. Hasan’s Margin on Day 1 = 32.5%.

64.

For Hasan:
As the share price at 11:00 a.m. and 12:00 noon was Rs. 500 and Rs. 400 respectively, the number of shares sold by Hasan at 11:00 a.m. and 12:00 noon must be in the ratio 4 : 5 respectively. Let the number of shares sold by Hasan at 11:00 a.m. and 12:00 noon be 4x and 5x respectively.
Total sales amount = 500 × 4x + 400 × 5x = Rs. 4,000x Total investment in purchase = 750 × 9x = Rs. 6,750x

Margin (loss) = (6750 − 4000)/4000 = 68.75%

For Sajid:
Let the total number of shares sold by Sajid at 9 a.m. and 10 a.m. be 2y. Total sales amount = 200 × y + 300 × y = Rs. 500y
Total investment in purchase = 500 × y + 625y = Rs. 1,125y

Margin (loss) = (1125 − 500)500 = 125% Required ratio = 68.75 : 125 = 11 : 20

DIRECTIONS FOR QUESTIONS 65 to 67 :
65. The Energy Consumption of a department can be obtained by dividing the Total Revenue of that department by the Average Revenue per Unit Energy Consumed by that department. Among the five companies, the Energy Consumption is the highest for Perfitti VM at approximately 1900W-hrs.

66. There are six departments in all whose Energy Consumption is less than 100W-hrs. They include two departments of UB Group, one of Unilever and three of Wipro.

67. Unilever has two departments whose Total Revenue is more than Rs. 600 crores and Energy Consumption is less than 200W-hrs. The only other department that satisfies the given criteria is of ITC Ltd.

DIRECTIONS FOR QUESTIONS 68 to 69 :
68. Both Oil Trade Balance and Total Trade Balance decreased from 1990-91 to 2000-01. The increase in Oil Imports from 1990-91 to 2000-01 was more as compared to the increase in Oil Exports from 2000-01 to 2010-11.

69. Non-Oil Exports increased by approximately 273% from 2000-01 to 2010-11.
Total Exports increased by approximately 377% from 1990-91 to 2000-01.
Oil Imports increased by 175% from 1980-81 to 1990-91.
Total Imports increased by approximately 55% from 2000-01 to 2010-11.

DIRECTIONS FOR QUESTIONS 70 to 71 :
70. The total number of employees who were appraised in January was 71 + 67 + 97 i.e. 235. These were the employees who were appraised on at least one performance area.
The total number of employees who were appraised in July was 30 + 22 + 29 i.e. 81. These were the employees who were appraised on at least two performance areas.
The number of employees who were appraised on exactly one performance area is 235 – 81 i.e. 154

71. The number of employees who were not appraised on Individual Performance in January was 67 + 97 i.e. 164. The employees who were appraised on Individual Performance in July and November were among these 164 employees only. So the number of employees who were not appraised on Individual Performance in 2010 was 164 – (30 + 9) = 125.

DIRECTIONS FOR QUESTIONS 72 to 74 :
72. Team 4 has scored the maximum runs through the mentioned players.â€‹

73. Only 3 players have scored less than 9,000 runs even after playing more than 200 matches

74.

Player 1 and Player 2 of Team 4 have better averages than Player 2 of Team 3.

Runs scored by Player 1 and Player 2 of Team 4 = 13600 + 10800 = 24400

Matches played by them = 340 + 240 = 580

Average score = 24400/580 = 42.07

FOR DIRECTIONS FOR QUESTIONS 75 to 77 :

Aamir is either A or B. Accordingly, Abhishek would be either B or A. Therefore, Amitabh and Akshay are disguised as C and D. As the maximum possible marks in each Mock Cat were 75, the total marks obtained by C are definitely more than the total marks obtained by D. Therefore, C is Amitabh and D is Akshay.

Let us assume that Abhishek is disguised as B and the marks scored by Abhishek in Mock 1 are x.

∴  x + 87 > 138 or x > 51 which would mean that Abhishek could not score the lowest in Mock 1. Hence, the assumption is incorrect and so Abhishek must be disguised as A and Akshay as Bâ€‹

75.

Let’s assume that Akshay, who is disguised as D, score p marks in Mock 3.

So his total score d = 63 + p

As we know that d < c

63 + p < 138

or p < 75

All the scores from –25 to +75 are possible in any of the tests, except 70, 73 and 74.

Therefore, different scores possible for Akshay = 101 – 3 – 1 = 97

76. a

77.

Abhishek is disguised as A. Let’s assume that Abhishek scores q marks in Mock 2. As Aamir scored the highest in Mock 2, q < 62 and 47 + 72 + q > 51 + 59 + 28.

∴ 19 < q < 62

Abhishek’s score is a multiple of 6.

∴ 119 + q needs to be a multiple of 6.

∴ q = 6K + 1, where K is a whole number.

19 < 6K + 1 < 62

∴ 4 ≤ K ≤ 10 Therefore, 7 different values are possible for a.

FOR DIRECTIONS FOR QUESTIONS 78 to 80 :

Let the fee (in Rs. lakhs) demanded by Careena, Carishma, Cashmira and Catrina be a, b, c and d (not necessarily in that order).
Let the fee (in Rs. lakhs) demanded by Coena be e.
Also, either Z or Y must have demanded Rs. 1123 lakhs as the prices demanded by X, U and V are already given.

Let e = 1123.
a + b = c + d.
It can be seen that 2397 – 2379 = 18
and 18 + 1213 = 1231.
So, 1231 + 2379 = 2397 + 1213.
Also, 1201 < 1231 < 2288.
So, 1231 is one such acceptable value.
Coena must have demanded the fee of Rs. 1123 lakhs and so she must be either Z or Y (as concluded above).â€‹

The following cases are possible:â€‹

 I II III IV Z Coena (1123) Cashmira (1231) Coena (1123) Careena (1231) Y Cashmira (1231) Coena (1123) Careena (1231) Coena (1123) X Careena (2397) Careena (2397) Cashmira (2397) Cashmira (2397) U Catrina (2379) Catrina (2379) Carishma (2379) Carishma (2379) V Carishma (1213) Carishma (1213) Catrina (1213) Catrina (1213)

Note:
e cannot take any value other than 1123.
E.g. let e = 2397.
In such a case no equation of the type ‘a + b = c + d’ is possible, which satisfies the given conditions. A possible equation is ‘1123 + 2289 = 1213 + 2379’ but since 2289 > 2288, it cannot be taken.â€‹

78. câ€‹

79.  dâ€‹

80.  c

FOR DIRECTIONS FOR QUESTIONS 81 to 83 :

Let the number of fit sportsmen who play Hockey be x. From statement (ii), the number of unfit sportsmen who play Hockey is also x. From statements (v) and (vii), the number of fit sportsmen who play Cricket is 1250 and the number of unfit sportsmen who play Cricket is 0. From statement (i), we can deduce that x is equal to 2500. From statement (iv), the number of sportsmen who play Hockey for Under-19 team is 4000 and the number of sportsmen who play Hockey for Above-19 team is 1000. Further analysis leads to the tables given below.â€‹

 Hockey Under-19 team Above-19 team Fit Unfit Fit Unfit 2000 2000 500 500

 Cricketâ€‹ Under-19 team Above-19 team Fit Unfit Fit Unfit 750 0 500 0
81. The total number of sportsmen in Khelabad is 6250.â€‹

82. The required ratio is 1 : 4.â€‹

83.

The total number of fit sportsmen who play for Above-19 team is 1000.â€‹

DIRECTIONS FOR QUESTIONS 84 to 86 :

The table given below shows the grades received by the students in different subjects.â€‹

 Subject Physics (2.5) Chemistry (1.5) Mathematics (4) English (3.5) Student Aman D B A E Bipin C B A D Cute A C C C David C B A D Esha A B D A Fardeen D E B A 84. Esha got the highest CGPA.

85. Bipin and David got the same CGPA

86. The CGPA of Fardeen was 6.96

DIRECTIONS FOR QUESTIONS 87 to 89 :

Let Interest Payment be x% and Non-Planned Capital Account be y% of Budgeted Expenditure. From pie chart B, x + y = 22% and x – y = 12%. Thus, x = 17 and y = 5.
Now 76% – 19% i.e. 57% of Budgeted Revenue is given as Rs. 285 thousand crores. Thus Budgeted Revenue = Rs. 500 thousand crores.
Budgeted Expenditure = Budgeted Revenue + Rs. 625 thousand crores = Rs. 1125 thousand crores

87.

The difference between Budgeted Expenditure and Budgeted Revenue when expressed as a percentage of Budgeted Revenue

= 625/500 × 100 125%

88. The sum of Planned Revenue Account and Planned Capital Account (Rs. 281.25 thousand crores) exceeded the Non-Tax Revenue (Rs. 95 thousand crores) by Rs. 186.25 thousand crores.

89. Since nothing is mentioned about the break-up of Budgeted Revenue in FY 2011-12, the percentage share of Tax Revenue cannot be determined.

DIRECTIONS FOR QUESTIONS 90 to 91 :
90.

Percentage change for the year 1971 is the highest and the value is 91. The absolute difference for the year 1981 is clearly much more than that for the other four years and the value is 12188903.

DIRECTIONS FOR QUESTIONS 92 to 94 :
92. Among males, Performance Scores of Kaushik and Maneet don’t fall in the interval [70, 100] and so they are Low on Happy Growth Index. Among females, Performance Scores of Asmita and Amit don’t fall in the interval [60, 90] and so they are Low on Happy Growth Index. So 4 students in all are Low on Happy Growth Index at Primary Stage of Education.

93. Among males at Senior Secondary Stage, Saurabh and Mudit are High on Happy Growth Index. Among females at Primary Stage, Meenal, Manjari and Renuka are High on Happy Growth Index. So the required ratio is 2 : 3.

94. Renuka is High on Happy Growth Index. Meenal and Manjari are the only female students whose Academic Score is less than 50 and who are High on Happy Growth Index.

DIRECTIONS FOR QUESTIONS 95 to 97 :
95. In Round 1, the total number of votes cast is 1,00,00,000. The candidate with the maximum number of votes gets 16,52,754 votes or 16.5% of the total number of votes.

96. The total number of votes becomes 67,23,323. Then “Others” will have 1,51,325/67,23,323 = 2.25% share. This means an increase of 0.74

97. Among the candidates who enter Round 2, the Kongress candidat