XAT-ACTUAL PAPER 2007 Page 1 XAT Actual Paper -2007(Memory Based Questions) QUANTITATIVE ABILITY DIRECTIONS for Questions 1 to 4: Seven integers A, B, C, D, E, F and G are to be arranged in an increasing order such that-A. First four numbers are in arithmetic progression. B. Last four numbers are in geometric progression. C. There exists one number between E and G. D. There exist no numbers between A and B. E. D is the smallest number and E is the greatest. F. A G F 1 D C A = = > G. E = 960 1. The common difference in the A. P. is a. 20 b. 22 c. 25 d. 30 e. None of these 2. D = a. 30 b. 25 c. 22 d. 20 e. None of these 3. EA= a. 2 b. 3 c. 4 d. 5 e. None of these 4. The position and value of A is a. 5th highest and 100 b. 4th highest and 100 c. 4th highest and 100 d. 3rd highest and 180 e. None of these DIRECTIONS for Questions 5 and 6: It is possible to arrange eight of the nine numbers 2, 3, 4, 5, 7, 10, 11, 12, 13 in the vacant squares of the 3 by 4 array shown below so that the arithmetic average of the numbers in each row and column is the same integer. 1 15 9 14 5. Which one of the nine numbers must be left out when completing the array? a. 4 b. 5 c. 7 d. 10 e. 11 6. The arithmetic average is -a. 6 b. 7 c. 8 d. 9 e. 12 7. For which value of non-negative ‘a’ will the system x2 – y2 = 0, (x – a)2 + y2 = 1 have exactly three real solutions? a. – 2 b. 1 c. 2 d. 2 e. No such ‘a’ exists.XAT-ACTUAL PAPER 2007 Page 2 8. Let {An} be a unique sequence of positive integers satisfying the following properties: A1 = 1, A2 =2, A4 = 12. and An+1.An–1= An2 ± 1 for n = 2, 3, 4 ... Then, A7 is a. 60 b. 120 c. 149 d. 169 e. 187 9. If the polynomial x3 + px + q has three distinct roots, then which of the following is a possible value of p ? a. –1 b. 0 c. 1 d. 2 e. 3 10. In a certain factory, each day the expected number of accidents is related to the number of overtime hours by a linear equation. Suppose that on one day there were 1000 overtime hours logged and 8 accidents reported, and on another day there were 400 overtime hours logged and 5 accidents. What are the expected numbers of accidents when no overtime hours are logged? a. 2 b. 3 c. 4 d. 5 e. None of these DIRECTIONS for Questions 11 to 13: Each questions is followed by two statements labeled as (1) and (2). You have to decide if these statements are sufficient to conclusively answer the questions. Choose a. If statements (1) alone is sufficient to answer the question. b. If statement (2) alone is sufficient to answer the question. c. If statement (1) and statement (2) together are sufficient but neither of the two alone is sufficient to answer the question. d. If either statement (1) or statement (2) alone is sufficient to answer the question. e. If both statement (1) and statement (2) together are insufficient to answer the question. 11. Five integers A, B, C, D, and E are arranged in such a way that there are two integers between B and C and B is not the greatest. There exists one integer between D and E and D is smaller than E. A is not the smallest integer. Which one is the smallest? 1. E is the greatest 2. There exists no integer between B and E. 12. ABC is a triangle with B = 90°. Ð What is the length of the side AC? 1. D is the midpoint of BC and E is the midpoint of AB. 2. AD = 7 and CE = 5. 13. What is the maximum value of a/b? 1. a, a + b and a + 2 b are three sides of a triangle. 2. a and b both are positive. 14. Triangle ABC has vertices A (0, 0), B (0, 6) and C (9, 0). The points P and Q lie on side AC such that AP = PQ = QC. Similarly, the points R and S lie on side AB such that AR = RS = SB. If the line segments PB and RC intersect at X, then the slope of the line AX is a. 23 b. –23 c. 32 d. –32 e. None of theseXAT-ACTUAL PAPER 2007 Page 3 15. ABCD is a rectangle. The points P and Q lie on AD and AB respectively. If the triangles PAQ, QBC and PCD all have the same area and BQ = 2 then AQ = a. 1 5 + b.1 5 - c. 7 d. 2 7 e. Not uniquely determined DIRECTIONS for Questions 16 to 18: Substitute different digits (0, 1, 2,....9) for different letters in the problem below, so that the corresponding addition is correct and it results in the maximum possible value of MONEY. P A Y M E R E A L M O N E Y 16. The letter ‘Y’ should be a. 0 b. 2 c. 3 d. 7 e. None of the above 17. The resulting value ‘MONEY’ is: a. 10364 b. 10563 c. 10978 d. 19627 e. None of the above 18. There are nine letters and ten digits. The digit that remains unutilized is: a. 4 b. 3 c. 2 d. 1 e. None of the above DIRECTIONS for Questions 19 and 20: Let ‘g’ be a function defined on the set of integers. Assume that ‘f’ satisfies the following properties: 1. f(0) ¹ 0; 2. f(1) = 3; and 3. f(x) f(y) = f(x + y) + f(x – y) for all integers x and y. 19. f(7) a. 123 b. 322 c. 123 d. 1126 e. None of the above 20. f(3) a. 7 b. 18 c. 123 d. 322 e. None of the above 21. For how many integers n, n 20 n - is the square of an integer? a. 0 b. 1 c. 2 d. 3 e. 4XAT-ACTUAL PAPER 2007 Page 4 22. Which equation can be graphically represented as follows? 2 6 10 14 18 –2 –6 –10 –14 –18 – 20 20 20 – 20 a. 8x2 – 15y2 = 169 b. 9x2 – 16y2 = 144 c. |(x – 8) (y – 15)| = 12 d. |(x – 9) (y – 16)| = 13 e. None of the above DIRECTIONS for Questions 23 and 24: A truck traveled from town A to town B over several days. During the first day, it covered 1p of the total distance, where p is a natural number. During the second day, it traveled 1q of the remaining distance, where q is a natural number. During the third day, it traveled 1p of the distance remaining after the second day, and during the fourth day, 1q of the distance remaining after third day. By the end of the fourth day the truck and traveled 34 of the distance between A and B. 23. If the total distance is 100 kilometers, the minimum distance that can be covered on day 1 is _____ kilometers. a. 25 b. 30 c. 33 d. 35 e. 40 24. The value of the sum of p and q is a. 4 b. 5 c. 6 d. 7 e. 8 25. ABC is a triangle with ÐCAB = 15° ÐABC = 30°. If M is the midpoint of AB then ÐACM = a. 15° b. 30° c. 45° d. 60° e. None of the above 26. Let p be any positive integer and 2x + p = 2y, p + y = x and x + y = z. For what value of p would x + y + z attain its maximum value? a. 0 b. 1 c. 2 d. 3 e. None of the aboveXAT-ACTUAL PAPER 2007 Page 5 27. Consider the system of linear equations 2x + 3y + 4z = 16 4x + 4y + 5z = 26 ax + by + cz = r For r = 5 and a = 1 then the system of linear equation will have infinite number of solutions if c = a. 32 b. 1 c. 12 d. 0 e. None of the above 28. ABC is a triangle with ÐBAC = 60°. A point P lies on one-third of the way from B to C, and AP bisects ÐBAC, ÐAPC is a. 30° b. 45° c. 60° d. 90° e. 120° 29. A management institute has six senior professors and four junior professors. Three professors are selected at random for a Government project. The probability that at least one of the junior professors would get selected is: a. 56 b. 23 c. 15 d. 16 e. None of the above 30. We define a function f on the integers f(x) = x10, if x is divisible by 10, and f(x) = x + 1 if x is not divisible by 10. If A0 =1994 and An + 1 = f(An). What is the smallest n such that An = 2? a. 9 b. 18 c. 128 d. 1993 d. An never equals 2XAT-ACTUAL PAPER 2007 Page 6 DATA INTERPRETATION DIRECTIONS for Questions 31 and 32: The graph given below contains data pertaining to number of electronic commerce transactions that have taken place in the last six months of the financial year 2005. This graph contains data related to private consumption and does not include corporate electronic commerce activities. Numbers mentioned above the bar graphs are in millions and average price per unit is mentioned in the brackets. Online Consumer Trends 37 32 28 31 44 17 24 22 21 25 25 16 37 23 22 9 6 21 16 20 0 10 20 30 40 50 Asia Pacific Europe North America Latin America South Africa Regions Number of Transactions Books($5) Video/DVDs /Games ($10) Airlines Tickets /Reservation ($20) Clothing /Accessories /Shoes ($15) 31. If the airline ticket purchase made through internet increase by 20% and the average price of the airline ticket 25% then the net increase in revenues from the ecommerce activities world wide will be ___________ percent of the corresponding pre-price-increase revenues. a. 18 b. 19 c. 20 d. 21 e. 22 32. For which product category is the revenue contribution of Asia Pacific region the maximum? a. Books b. Video /DVDs /Games c. Airline Tickets /Reservation and Clothing /Accessories /Shoes d. Clothing /Accessories /shoes e. Airline Tickets /ReservationXAT-ACTUAL PAPER 2007 Page 7 DIRECTIOSNS for Questions 33 and 34: Study the aggregate financial ratios of all registered Indian manufacturing companies in the table below to answer the questions that follow: 2000 2001 2002 2003 2004 2005 PBDIT 13.1 11.7 12.3 13.3 14.4 14.7 PBDT 8.1 7.1 8 9.9 11.8 12.7 PBIT 9.4 8.4 8.7 9.9 11 11.6 PAT 3.2 2.8 2.7 4.4 6 6.9 Raw Material expense 41 40.6 43.1 45.5 45.7 47.1 Salaries and wages 5.9 5.7 5.6 5.3 4.9 4.4 Interest payments 4.6 4.3 4 3.1 2.3 1.7 Operating profit 5.2 4.2 4.9 6.7 8 8.7 Net sales (% Growth Over Previous Year) 18.4 19.3 2.6 15.7 15.2 19.9 All figures are as % of net sales unless otherwise mentioned 33. In which year the annual growth rate in the aggregate Salaries and Wages expense was maximum? a. 2005 b. 2004 c. 2003 d. 2002 e. 2001 34. What is the annual growth rate in aggregate PAT of the Indian manufacturing companies in the financial year 2005 as compared to that in the financial year 2004? a. 15.0 Percent b. 5.7 Percent c. 88.6 Percent d. 37.9 Percent e. None of these DIRECTIONS for Questions 35 to 39: Study the tables of the Indian foreign trade given below to answer the questions. Principal Commodities Export COMMODITIES 2003-04 2004-05 2005-06 Plantations 0.92 0.78 0.71 Agri & allied prdts 8.39 7.61 7.21 marine Products 2.08 1.6 1.4 ores & minerals 3.69 5.29 6.02 leather & mfrs. 3.19 2.89 2.56 gems & jewellery 16.56 17.29 15.13 sports goods 0.15 0.12 0.13 Chemicals & related products 15.43 16 15.1 engineering goods 16.41 18.41 18.66 electronic goods 2.74 2.28 2.18 project goods 0.09 0.06 0.13 Textiles 18.86 15.16 14.8 Handicrafts 0.7 0.43 0.4 Carpets 0.9 0.75 0.81 Cotton raw incl. waste 0.28 0.1 0.61 petroleum products 5.54 8.57 11.21 unclassified exports 4.07 2.66 0.94 GRAND TOTAL 100.00 100.00 100.00 Total Exports in Rupees Crore 293,366.75 375,339.53 454,799.97 US Dollar Exchange Rate 45.9513 44.9315 44.2735 Weight (%)XAT-ACTUAL PAPER 2007 Page 8 Principal Commodities' Import COMMODITIES 2003-2004 2004-2005 2005-2006 bulk imports 37.87 39.09 42.56 pearls, precious & semiprecious stones 9.25 8.8 6.42 Machinery 10.63 10 10.94 Project goods 0.49 0.54 0.57 Others 41.76 41.57 39.51 TOTAL IMPORTS 100 100 100 Total Imports (in Crores fo Rupees) 359,107.66 501,064.54 630,526.77 Weight (%) 35. Growth of trade imbalance (exports less imports) in dollar terms in the year 2005 -06 as compared to the previous year was: a. 39.77 b. 41.85 c. 91.24 d. 95.98 e. None of these 36. Given that the weight (%) of Petroleum crude & products in the total imports of India is 26.70, 27.87, and 30.87 in the years 2003 -04, 2004-05, and 2005-06 respectively. What is the ratio of yearly difference in the export of Petroleum Products and Import of Petroleum crude & products in dollar terms, in the year 2005-06 versus 2004-05? a. 1.36 b. 1.38 c. 1.46 d. 1.48 e. None of these 37. The three commodities which had highest export growth rate in the year 2004-05 as compared to the previous year, arranged in descending order of growth rates are: a. Petroleum products, ores & minerals, engineering goods b. Ores & minerals, gems& jewellery, chemicals & related products c. Gems & jewellery, chemicals & related products, agri & allied products d. Ores & minerals, chemical and related products, arri & allied products. e. Ores & minerals, engineering goods, chemicals & related products. 38. In the year 2005-06 the commodity which witnessed maximum growth in exports (in Indian Rupees) as compared to the year 2004-05 is: a. Petroleum products b. Project goods c. Ores & minerals d. Sports goods e. None of these 39. In the two year period from 2004-2005 to 2005-2006, the average growth in import (in Indian Rupees) of which commodity to India was maximum? a. bulk imports b. pearls, precious & semi precious stones c. machinery d. project goods e. othersXAT-ACTUAL PAPER 2007 Page 9 DIRECTIONS for Questions 40 to 43: Read the data below and choose the correct option for the questions that follow. Queen Airlines offers the following Privilege program. There are 5 membership rises, each with its own set of enhanced tier-specific benefits. New members join at the Blue tier level, then upgrade to the Blue plus, with the added benefit of Tele check-in. The program also has three elite tiers-Silver Gold, and Platinum. The 6 months fast -track upgrade and the 12 months standard upgrade system operate in parallel implying that whenever a particular passengers satisfies conditions, either in terms of number of flights in the stipulated period or in terms of accumulated Queen-Miles in either of the two systems, they are automatically upgrade systems considered the data in rolling 6 months period. The same holds for the 12 months period. If conditions for more than one upgrade are satisfied, the passenger is given the higher of the two. Tier Upgrade 6-month Fast Track Upgrade System 12-month Standard Upgrade System Blue to Blue Plus 3/3,000 Not applicable Blue Plus to Silver Not Applicable 5/12,500 Silver to Gold 5/12,500 10/20,000 Gold to Platinum 10/20,000 20/30,000 Cummulative Queen Airways flights/Cummulative Status Queen-Miles Blue Blue Plus Silver Gold Platinum Tele Check-in Ö Ö Ö Ö Additional baggage allowance 10 kgs 20 kgs 35 kgs Confirmed upgrade vouchers 1 3 5 Guaranteed reservations up to 24 hours prior to departure Ö Ö Cancellation fees waived on published fares Ö Distance between cities in Queen-miles Kolkata 1461 Mumbai 1407 1987 Hyderabad 1499 1516 711 Bangalore 2061 1881 998 562 Coimbatore 2401 2167 1265 902 340 Guwahati 1959 1081 2746 2370 2932 3209 Chennai 2095 1676 1329 688 331 491 2718 Delhi Kolkata Mumbai Hudrabad Banglore Chennai GhyXAT-ACTUAL PAPER 2007 Page 10Mr. kakkar, a newly recruited MBA from a business school, started his career with the start of the year 2007. His travel plans for the year of 2007 is given below in the table. Date From To 2.1.2007 Mumbai Hydrabad 8.1.2007 Hydrabad Mumbai 3.2.2007 Mumbai Delhi 8.3.2007 Delhi Guwahati 20.3.2007 Guwahati Kolkatta 11.4.2007 Kolkatta Guwahati 30.4.2007 Guwahati Chennai 4.7.2007 Chennai Guwahati 20.7.2007 Guwahati Kolkatta 2.9.2007 Kolkatta Hydrabad 11.9.2007 Hydrabad Guwahati 22.9.2007 Guwahati Delhi 1.10.2007 Chennai Banglore 11.10.2007 Guwahati Chennai 4.11.2007 Chennai Delhi 29.11.2007 Delhi Hydarabad 1.12.2007 Hydrabad Guwahati 31.12.2007 Guwahati Mumbai 40. The difference in complete calendar months, discounting any partial months, between the first tier upgrade and the last tier upgrade for Mr. Kakkar is: a. 7 b. 8 c. 9 d. 10 e. 11 41. In which month will Mr. Kakkar become eligible for guaranteed reservations up 24 hours prior to departure? a. January b. February c. March d. April e. May 42. The number of complete calendar months, discounting any partial months, for which Mr. Kakkar avails the gold tier membership is: a. 7 b. 8 c. 9 d. 10 e. 11 43. If 6 months upgrade had not been in operation, then Mer. Kakkar would have reached Gold tier in the months of a. June b. July c. August d. September e. OctoberXAT -ACTUAL PAPER 2007 Page 11 Answers and Explanations 1. d As per the condition giver in the question, following combination are possible, i.e. (i) DABCGFE (ii) DBACGFE (iii) DCBAGFE (iv) DCABGFE Since it is given that the greatest term is 960, that is ‘E’ and A F F 1 D C A = = > , we can assume that the common ratio is 2 and check the feasibility. Therefore E = 960, F = 480, G = 240 Also, it can easily observed that DCBAGFE is the only combination satisfying the condition. Now other values will be A = 120, B = 90, C = 60, D = 30 Now since none of the above values are against the condition given, the sequence will be 30 60 90 120 240 480 960 (D) (C) (B) (A) (G) (F) (E) Hence Option ‘d’ is the correct answer. 2. a 3. e 4. e For questions 5 and 6: 5. d There is no hard and fast rule for solving these kind of question. The only points to be considered are as follows (i) Every row has summation being multiple of 4 (ii) Every column sums up to be multiple of 3 (iii) First row, already has the number 15, the largest number. Hence, the blank boxes should contain the minimum possible number. Keeping in mind the above points, the complete arrangement is as follows. 1 13 3 15 11 9 7 5 12 2 14 4 Observe that rows sum up to ‘32’ and column sum up to ‘24’. 5. d 6. c 7. b Given that x2 – y2 = 0 ... (i) (x – a)2 + y2 = 1 ... (ii) Solving (i) and (ii), we get- = 2 a 2 –a 2 ± Now for all the real values of x 2 2 –a 0 ³ or – 2 a 2 £ £ For non-negative values of ‘a’ we have o a 2 £ £ Now, when we put a = 1, we get x = 2, 0. Þ y = 2, – 2, 0. Hence we get-exactly three real solutions (2, 2), (2, – 2) and (0, 0) ( ) y x =± Q . 8. d Putting n = 2 in the equation An+1 An–1 = 2n A ± 1, we get. 2 3 1 2 A A A 1 = ± 2 3 2 1 A 3,5 1± = = Putting n = 3.2 4 2 3 A A A 1 = ± 2 2 4 3 1 5 1 A or 2 2 ± ± Þ = But, A4 = 12, Therefore A3 = 5 Now putting n = 4, 5, 6 similarly as in above equation we get 2 6 7 5 A 1 4901 A 169 A 29 ± = = = . 9. a Given polynomial is x3 + px + q = 0 ...(i) Now let a, b, c be three distinct roots of above equation. So we can write- + b + c = 0 ab + bc + ca = p abc = – q Also, we can write here, Squring on both sides, we get (a2 + b2 + c2) = (a + b + c)2 – 2 (ab + bc ac) Þ (a2 + b2 + c2) = – 2p Here L. H. S cannot be negative as it is the sum of squares. So ‘p’ has to be negative which is option (a) only.Page 12 XAT -ACTUAL PAPER 2007 10. b Let the expected number of accidents be ‘A’ and number of overtime hours be ‘T’ As expected number. of accidents and number of overtime hours are related by a linear equation, we can write here A = K + K1 (T) Now, 8 = K + K1 (1000) ... (i) 5 = K + K1 (400) ... (ii) Solving (i) and (ii) we get 1 1 K 3,K 200 = = Hence, ( ) 1 A 3 0 200 = + A = 3. 11. c From the given, three different arrangements are possible. case (i) DBEAC case (ii) BADCE case (iii) CADBE From statement (1), either case (ii) or case (iii) could be valid so we cannot answer which one is the smallest. From statement (2), either case (i) or case (ii) is possible. If we combine the two statements; only case (iii) is possible. Hence (c) is correct option. 12. c AB C D Ey x y x Neither statement (i) nor (ii) alone is sufficient to determine the length of side AC. But when statement (i) and (ii) are combined, value of AC can be determined. As per the diagram above x2 + 4y2 = 49 ...(i) 4x2 + y2 = 25 ...(ii) Solving (i) and (ii) we can determine the value of 4 (x2 + y2) = AC2. Hence (c) is correct. 13. e Since ‘b’ is the common difference of three sides of a triangle, ab can take any values in the given range of real; numbers. Suppose a = 1000 b = 1 Then there sides will be 1000, 1001, 1002 Here a 1000 1000 b 1 = = So the values of ‘a’ and ‘b’ can be varied to any values making the ratio ab a variable or undeterminable. 14. a AB C SR P Q x y x (0, 6) (0, 4) (0, 2) (0, 0)(3, 0) (6, 0) (9, 0) From the above diagram Equation of line PB: y = – 2x + 6 ...(i) ( ) 2 1 1 1 2 1 y –y y – y x –x x –x æ ö ç ÷ = ç ÷ è ø Q Equation of line CR: 2 y – x 2 9 = + ...(ii) Solving equation (i) and (ii) for the co-ordinates of point ‘X’, we get 9 3 x , 4 2 æ ö =ç ÷ è ø Now slope of AX is 2 1 2 1 3–0 y –y 2 2 m 9 3 x –x –0 4 æ ö Þ = ç ÷ è ø Q 15. a A BC DP Q x x+2 2 Let AQ be ‘x’ and area of triangles PAQ, QBC and PCD be ‘A’ Now 1 2 BC A BC A 2´ ´ = Þ = ...(i) Similarly in D PDC ( ) 1 2A PD x 2 A PD 2 x 2 ´ ´ + = Þ = + ...(ii) Also in D PAQ 2A AP x = ...(iii) Now AD = PD + PA or, 2A 2A A x 2 x = + + Solving above equations, we get x 1 5 = ± As x cannot be negative. x 1 5 \ = + .XAT -ACTUAL PAPER 2007 Page 13 For questions 16 and 18: This question is a ‘hit and trial’ based question with little application of logic. Observe that-(i) Summation of ‘E’ and ‘L’ must give zero in the end so that ‘y’ comes as it is in the fourth row. (E + L = 10) (ii) Rest all is ‘hit and trial’ The correct assignment of digits is 7 2 5 1 6 9 6 2 4 1 0 3 6 5 Hence, Option (e) is the correct answer. 16. e 17. e 18. e 19. c Given that f(i) = 3 f(x) f(y) = f(x+y) + f(x–y) Putting x = 0, y = 0, we get f(o)2 = 2f(0) Þ f(0) = 2 Putting x=1, y = 1, we get f2(1) = f(2) + 2 ( ) ( ) f 0 2 = Q ( ) f 2 7 Þ = Similarly f(3) = 18 and f(4) = 47 Now putting x = 4, x = 3 f(4) f(3) = f(7) + f(1) Þ f(7) = 843 Therefore, option (c) is the correct answer. 20. b 21. e For n = 0, 10, 16, 18 the expression n 20 –n is the square for an integer. Therefore the correct option is (e). 22. b According to the graph at y = 0, x should be ‘4’. Option (b) satisfies this condition. 23. a According to the statement of the question ( ) p –1 q –1 p –1 q –1 1 p q p q 4 æ öæ öæ ö= ç ÷ç ÷ç ÷ è øè øè ø or 2(p–1) (q–1) = pq or (p – 2) (q – 2) = 2. Hence, p – 2 = 2 and q – 2 = 1 Þ p = 4 & q = 3 or p – 2 = 1 and q – 2 = 2 Þ p = 3 & q = 4 Which gives p = 3, 4 and q = 3, 4 Minimum distance is covered when p = 4, i.e. 1 100 25 4 ´ = Also in each of the above cases p + q = 7. 24. d 25. b A B C M x y 30° 15° 165–q q 135–q x Applying since rule to D ABC y 2x sin30 sin135 = ° ° ...(i) Also( ) y x sin 165– sin = ° q° q ...(ii) or ( ) sin 165– 2sin ° q= q Þ q= 30° Hence option (b) is the correct answer. 26. a 27. c Given equation are 2x + 3y + 4z = 16 ...(i) 4x + 4y + 5z = 26 ...(ii) ax + by + cz = r ...(iii) Also given that a = 1 and r = 5. Subtracting (i) from (ii), we get 2x + y + z = 10 ...(iv) x + by + cz = 5. ...(v) Now for above two lines to have infinite solution 2 1 1 10 1 b c 5 = = =1 b c 2 Þ = = Hence option (c) is correctPage 14 XAT -ACTUAL PAPER 2007 28. e A B C 3 0° q 120° –q 30 ° 2y y 3 2x x3 P From angle bisector theorem BP AB 1 PC AC 2 = = 3y x y sin . sin60 sin 2x = Þ q= ° q Now y 3 1 Area or ab sin 2x 2 æ ö = D= q ç ÷ è ø Also applying cosine rule to the D ABC ( ) 2 2 2 2 y 2y –x Cos 60 3y x 4y + ° = Þ = ...(ii) Putting his value of y in (i) we get q = 30° Therefore APC 120 Ð = °. 29. a Total number of cases 10 3 C Favorable (required) case = 10 6 3 3 C – C The required probability 10 6 3 3 10 3 C – C C = 56 =Hence option ‘a’ is the correct answer. 30. a It is clear with the definition of given function that if given ‘x’ is divisible by 10, Then ( ) x f x 10 = Also increment of ‘1’ will be given to x, i.e. f(x) = x + 1 Now, f(1994) = 1995 ( ) 1994 is not divisible by 10 Q ....f(2000) = 200 f(200) = 20 f(20) = 2 So in all ‘9’ steps required to get An = 2. n 9 \ = . For questions 31 and 32: 31. d The revenues from the e -commerce activities, world wide = 5 × (37 + 32 + 28 + 31 +44) + 10 × (17 + 24 + 22 + 21 + 25) + 20 × (25 + 21 + 16 + 16 + 37) + 15 × (20 + 23 + 22 + 9 + 6) = (860 + 1090 + 2300 +1200) = 5450 million dollars. The total number of e -commerce transactions made for airline ticket purchase = (25 + 21 + 16 + 16 + 37) = 115 million after 20% increase: = 115 × 1.20 = 138 million dollars The new average price = 20 × 1.25 = 25 $ Þnet increase in the worldwide e -commerce revenues = (138 × 25 – 115 × 20) million dollar = 1150 million dollars Þ The required percentage = 1150 100 5450´ = 21.1% » 21% Þ Option (d) is correct. 32. e The revenue contribution of the Asia Pacific region, for the four product categories is:-Books = 37 × 5 = 185 Mn $ Videos/DVDs/Games = 17 × 10 = 170 Mn $ Airline Tickets/Reservation = 25 × 20 = 500 Mn $ Clothings/Accessories/shoes =20 × 15 = 300 Mn $ Definitely (e) is the correct option. For questions 33 and 34: Let the net sales, in the year 2000 be 100. Hence, the net sales values in the after years are: YEAR NET SALES 2000 100 2001 119.3 2002 122.4 2003 141.6 2004 163.1 2005 195.6 33. e The absolute values of the aggregate salaries and wages expenses are, year wise given as :-YEAR SALERIS & WAGES 2000 5.9 2001 6.8 2002 6.85 2003 7.5 2004 7.99 2005 8.6 The maximum growth rate has been in the year 2001 when it increased to 6.80; from a previous value of 5.90. Hence (e) is correct.XAT -ACTUAL PAPER 2007 Page 15 34. d The absolute value of the aggregate PAT are In 2005 : 195.6 6.9 100´ = 13.49 & In 2004 163.1 6 9.78 100´ = Þ The annual growth rate = 13.49 – 9.78 100 9.78 ´ = 37.9% Þ Option (d) is correct. For questions 35 to 39: 35. a Trade imbalance in 2005 -06 is given by (630526.77) – 454799.97 = 175726.80 crore rupees & trade imbalance in the years 2004 -05 is given by (501064.54 – 375339.53) = 125725.01 crore rupees the growth in the trade imbalance is:-( ) 175726.80 – 125725.01 100 125725.01 ´ = 39.77 Hence answer is (a). 36. e In the year 2005 -06 export of petroleum products = 0.1121 × 454799.97 = 50983.08 crore rupees. & import of petroleum crude and products = 0.3087 × 63.526.77 = 194643.61 crore rupees. Þ yearly difference = (50983.08 – 194643.61) = 143660.53 crore rupees. In the year 2004 -05 export of petroleum products = 0.0857 × 375339.53 = 32166.59 crore rupees & import of petroleum crude and products = 0.2787 × 501064.54 = 139646.68 crore rupees Þ yearly difference = (32166.59 – 139646.68) = 107480.09 crore rupees. Þ the ratio = yearly difference in 2005 06 yearly differencein 2004 05 -- = 143660.53 1.336 107480.09= Þ (e) is the correct answer. 37. a The options discuss, 6 different commodities. Their export values in year 2003 -04 and 2004 -05 are given below. The last column shows the growth rate over the previous year. 2003 -04 2004 -05 Petroleum products 16252.517 32166.59 1.97 Ores and Minerals 10825.233 19855.46 1.83 Engineering Goods 48141.48 69100 1.43 Gems and Jewellery 48581.53 64896.2 1.33 Chemicals and related products 45266.48 60054.32 1.32 Agri and Allied Products 24613.47 28563.33 1.16 Expert values Commodities Growth Rate Clearly, option (a) has the correct descending order. 38. Growth exports = [(Weight percentage of 2005 -06) × (Total Exports value of 2005 -06)] – [(Weight percentage of 2004 -05) × (Total Exports value of 2004 -05)] The growth for “Petroleum products|” has been the maximum and is given below. 11.21 8.57 454799.97 – 375339.53 100 100 æ ö æ ö ´ ´ ç ÷ ç ÷ è ø è ø = 18816 crores rupees. 39. Average growth of import [Export value in 2005-06) – (Export value in 2004 -05)] +[(Export value in 2005 -05) – (Export value in 2003-04)] 2 = (Export value in 2005 06) – (Export value in 2003 04) 2 - - = bulk export has the maximum value of this as 0.4256 630526.77 – 0.3787 359107.66 2 ´ ´ = 66179.06 crore rupees. For questions 40 and 43: 40. d The first tier upgrade corresponds to Blue to Blue Plus and the last tier upgrade is Gold to Platium. A passenger gets the first tier upgrade when he fulfills either of the following two conditions: (i) He completes 3000 Queen Miles within the first 6 months of the first flight. (ii) He makes 3 journeys, within the first 6 months of the first flight. A passenger gets the last tier upgrade when he fulfills any of the following conditions:Page 16 XAT -ACTUAL PAPER 2007 (i) He completes 3000 Queen Miles within the first 6 months of the first flight. (ii) He makes 10 journeys, within the first 6 months of the first flight. (iii) He completes 20 journeys, within the first 6 months of the first flight. (iv) He completes 30000 Queen Miles within the first 12 months of the first flight Hence, Mr. Kakkar gets his first tier upgrade on 3.02.2007 and his last tier upgrade on 31.12.2007. The difference is 10 Months and 29 days. Discounting the partial month (the extra 29 days) the difference is of 10 months. Hence (d) is the correct answer. 41. c To avail this benefit, the passenger should at least be a Gold member. On 30.3.2007, he makes his 5th journey. This makes him eligible for Blue plus to Silver upgrade under the 12-month system. On the other hand, under the 6 month fast track system, he becomes eligible for Silver to Gold Upgrade, which is higher. Hence on 20.3.2007 his membership is upgraded to Gold and he becomes eligible for guaranteed reservations up to 24 hours prior to departure. Hence the correct answer is (c). 42. c Mr. Kakkar becomes a Gold member on 20.3.2007 and he becomes a Platinum member on 31.12.2007. Hence he enjoys the Gold membership for 9 months, discounting an partial months. 43. d Had there been no 6 – month system, then Mr. Kakkar membership, after completing 20,000 Queen Miles which happens in the month of September.