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Re: HCL Recruitment Papers
As you want old recruitment papers of HCL, so I am providing old papers here for your reference: HCL Recruitment Paper 1 1. Which of the following about the following two declaration is true i ) int *F() ii) int (*F)() Choice : a) Both are identical b) The first is a correct declaration and the second is wrong c) The first declaraion is a function returning a pointer to an integer and the second is a pointer to function returning int d) Both are different ways of declarin pointer to a function Answer : c) 2. What are the values printed by the following program? #define dprint(expr) printf(#expr "=%d\n",expr) main() { int x=7; int y=3; dprintf(x/y); } Choice: a) #2 = 2 b) expr=2 c) x/y=2 d) none Answer: c)x/y=2 3. Which of the following is true of the following program main() { char *c; int *ip; c =(char *)malloc(100); ip=(int *)c; free(ip); } ans: The code functions properly releasing all the memory allocated 4.output of the following. main() { int i; char *p; i=0X89; p=(char *)i; p++; printf("%x\n",p); } ans:0X8A 5.which of the following is not a ANSI C language keyword? a) Incorrect definition b) structures cannot refer to other structure c) Structures can ref er to themselves. Hence the statement is OK d) Structures can refer to maximum of one other structure Answer :c) 6. What is the size of the following union. Assume that the size of int =2, size of float =4 andsize of char =1. Union Tag{int a;flaot b;char c;}; a)2 b)4 c)1 d) 7 ans : b.) 7. What is the output of the following program? (.has been used to indicate a space) main() { char s[]="Hello,.world"; printf(%15.10s",s); } a)Hello,.World... b)....Hello,.Wor c)Hello,.Wor.... d)None of the above ans: b.) total 15 spaces and print only 10 characters. 8. If taxi fares were Rs 1.00 for the first 1/5 mile and Rs 0.20 for each 1/5 miles thereafter. The taxi fare for a 3mile ride was (A)Rs 1.56 (B)Rs 2.40 (C)RS 3.00 (D)Rs 3.80 (E)Rs 4.20 Answer :d)Rs 3.80 9. A computer routine was developed to generate two numbers (x,y) the first being a random number between 0 and 100 inclusive, and the second being less than or equal to the square root of the first.Each of the following pair satisfies the routine EXCEPT (A) (99.10) (B) (85.9) (C) (50.7) (D) (1.1) (E)(1.0) Answer : A) (99.10) 10. A warehouse had a square floor with area 10,000 sq.meters. A rectangular addition was built along one entire side of the warehouse that increased the floor by one  half as much as the original floor.How many metersdid the addition extend beyond the original buildings ? (A)10 (B)20 (C)50 (D)200 (E)500 Answer: c)50 11. A digital wristwatch was set accurately at 8.30 a.m and then lost 2 seconds every 5 minutes. What time was indicated on the watch at 6.30 p.m of the same day if the watch operated continuously that time ? (A)5:56 (B)5:58 (C)6.00 (D)6.23 (E)6.26 Answer :E) 6.26 12. A 5 liter jug contains 4 liters of a salt water solution that is 15 percent salt. If 1.5 litres of the solution spills out of the jug, and the jug is then filled to capacity with water, approximately what percent of the resulting solution in the jug is salt? (A)7.5% (B)9.5% (C) 10.5% (D)12% (E)15% Answer :A)7.5% 13. A plane traveled K miles in the first 96 miles off light time.If it completed the remaining 300 miles of the trip in 1 minute,what was its average speed in miles per hour for the entire trip ? Answer 300+k)/97 * 60 14. A merchant sells an item at a 20 percent discount, but still makes a gross profit of 20percent of the cost. What percent of cost would be gross profit on the item have been if it had been sold without the discount? (A)20% (B)40% (C)50% (D)60% (E)66.6% Answer :c) 50% 15. A millionaire bought a job lot of hats 1/4 of which were brown.The millionaire sold 2/3 of the hats including 4/5 of the brown hats. What fraction of the unsold hats were brown. (A)1/60 (B)1/15 (C)3/20 (D)3/5 (E)3/4 Answer :c)3/20 16. How many integers n greater than10 and less than 100 are there such that, if the digits of n are reversed, the resulting integer is n+9 ? (A)5 (B)6 (C)7 (D)8 (E)9 Answer )8 17. An investor purchased a shares of stock at acertain price.If the stock increased in price Rs 0.25 per share and the total increase for the x shares was Rs 12.50, how many shares of stock had been purchased ? (A)25 (B)50 (C)75 (D)100 (E)125 Answer :B)50 18. At a special sale, 5 tickets can be purchased for the price of 3 tickets. If 5 tickets are purchased at the sale,the amount saved will be what percent of the original price of the 5 tickets? (A)20% (B)33.3% (C)40% (D)60% (E)66.6% Answer :c)40% 19.Working independently, Tina can do a certain job in 12 hours.Working independently, Ann can do the same job in 9 hours. If Tina works independently at the job for 8 hours and then Ann works independently, how many hours will it take Ann to complete the remainder of the jobs? (A)2/3 (B)3/4 (C)1 (D)2 (E)3 Answer :E)3 20. A decorator bought a bolt of d m number of red chips in any one stack ? (A)7 (B)6 (C)5 (D)4 (E)3 Answer :C) 5 21.Statistics indicate that men drivers are involved in more accidents than women drivers.Hence it may be concluded that... a) sufficiently information is not there to conclude anything b) Men are actually better drivers but drive more frequently c) Women Certainly drive more cautiously than Men d) Men chauvinists are wrong about women's abilties. e) Statistics sometimes present a wrong picture of things 22. What does the hex number E78 correspond to in radix 7 ? a) 12455 b) 14153 c) 14256 d) 13541 e) 13112 ans:d 23. Given that A,B,C,D,E each represent one of the digits between 1 and 9 and that the following multiplication holds: A B C D E X 4  E D C B A  what digit does E represent ? a) 4 b) 6 c) 8 d) 7 Ans: c 24. HCL prototyping machine can make 10 copies every 4 seconds. At this rate, How many copies can the machine make in 6 min.? a) 900 b) 600 c) 360 d) 240 e) 150 ans: a 25. 10^2(10^8+10^8) =10^4 a) 2(10)^4 b) 2(10)^6 c) 10^8 d) 2(10)^8 e) 10^10 ans: b HCL Recruitment Paper 2 1. What is the 8th term in the series 1,4, 9, 25, 35, 63, . . . Sol: 1, 4, 9, 18, 35, 68, . . . The pattern is 1 = 21 – 1 4 = 22 – 0 9 = 23 + 1 18 = 24 + 2 35 = 25 + 3 68 = 26 + 4 So 8th term is 28 + 6 = 262 2. USA + USSR = PEACE ; P + E + A + C + E = ? Sol: 3 Digit number + 4 digit number = 5 digit number. So P is 1 and U is 9, E is 0. Now S repeated three times, A repeated 2 times. Just give values for S. We can easily get the following table. USA = 932 USSR = 9338 PEACE = 10270 P + E + A + C + E = 1 + 0 + 2 + 7 + 0 = 10 3. In a cycle race there are 5 persons named as J, K, L, M, N participated for 5 positions so that in how many number of ways can M make always before N? Sol: Say M came first. The remaining 4 positions can be filled in 4! = 24 ways. Now M came in second. N can finish the race in 3rd, 4th or 5th position. So total ways are 3 x 3! = 18. M came in third. N can finish the race in 2 positions. 2 x 3! = 12. M came in second. N can finish in only one way. 1 x 3! = 6 Total ways are 24 + 18 + 12 + 6 = 60. Shortcut: Total ways of finishing the race = 5! = 120. Of which, M comes before N in half of the races, N comes before M in half of the races. So 120 / 2 = 60. 4. If POINT + ZERO = ENERGY, then E + N + E + R + G + Y = ? Sol: 4 digit number + 5 digit number = 6 digit number. So E = 1, P = 9, N = 0 Observe R + 0 = G. But R = G not possible. 1 + R = G possible. So R and G are consecutive. G > R. 1 + I = R, So I and R are consecutive. R > I. i.e., G > R > I. and G, R, I are consecutive. Now O + T should give carry over and O + Z also give carry over. So O is bigger number. Now take values for G, R, I as 8, 7, 6 or 7, 6, 5 etc. and do trial and error. POINT = 98504, ZERO = 3168 and ENERGY = 101672. So E + N + E + R + G + Y = 1 + 0 + 1 + 6 + 7 +2 = 17 5. There are 1000 junior and 800 senior students in a class. And there are 60 sibling pairs where each pair has 1 junior and 1 senior.1 student is chosen from senior and 1 from junior randomly.What is the probability that the two selected students are from a sibling pair? Sol: Junior student = 1000 Senior student = 800 60 sibling pair = 2 x 60 = 120 student Probability that 1 student chosen from senior = 800 Probability that 1 student chosen from junior = 1000 Therefore,1 student chosen from senior and 1 student chosen from junior n(s) = 800 x 1000 = 800000 Two selected student are from a sibling pair n(E) = 120C2 = 7140 Therefore P(E) = n(E)/n(S) = 7140?800000 6. SEND + MORE = MONEY. Then what is the value of M + O + N + E + Y ? Sol: Observe the diagram. M = 1. S + 1 = a two digit number. So S = 1 and O cannot be 1 but 0. Also E and N are consecutive. Do trial and error. SEND = 9567, MORE = 1085, MONEY = 10652 SO M + O + N + E + Y = 1 + 0 + 6 + 5 + 2 = 14 7. A person went to shop and asked for change for 1.15 paise. But he said that he could not only give change for one rupee but also for 50p, 25p, 10p and 5p. What were the coins he had ? Sol: 50 p : 1 coin, 25 p : 2 coins, 10 p: 1 coin, 5 p : 1 coin, Total: 1.15 p 8. 1, 1, 2, 3, 6, 7, 10, 11, ? Sol: The given pattern is (Prime number  consecutive numbers starting with 1). 1 = 2 – 1 1 = 3 – 2 2 = 5 – 3 3 = 7 – 4 6 = 11 – 5 7 = 13 – 6 10 = 17 – 7 11 = 19 – 8 14 = 23 – 9 9. A Lorry starts from Banglore to Mysore At 6.00 a.m, 7.00 a.m, 8.00 a.m.....10 p.m. Similarly another Lorry on another side starts from Mysore to Banglore at 6.00 a.m, 7.00 a.m, 8.00 a.m.....10.00 p.m. A Lorry takes 9 hours to travel from Banglore to Mysore and vice versa. (I) A Lorry which has started At 6.00 a.m will cross how many Lorries. (II) A Lorry which has started At 6.00 p.m will cross how many Lorries. Sol: I. The Lorry reaches Mysore by 3 PM so it meets all the Lorries which starts after 6 a.m and before 3 p.m. So 9 lorries. Also the Lorry which starts at night 10 p.m on the previous day at Mysore reaches Bangalore in morning 7 a.m. So it also meets that Lorry. So the Lorry which starts at 6:00 am will cross 10 Lorries. II. The lorry which has started at 6 p.m reaches destination by 3 a.m. Lorries which start at the opposite destination at 10 am reaches its destination at 7 pm. So all the lorries which starts at 10 am to 10 pm meets this lorry . So in total 13. 10. GOOD is coded as 164 then BAD coded as 21.if ugly coded as 260 then JUMP? Sol: Coding = Sum of position of alphabets x Number of letters in the given word GOOD = (7 + 15 + 15 + 4 ) x 4 = 164 BAD = (2 + 1 + 4) x 3 = 21 UGLY = (21 + 7 + 12 + 25) x 4 = 260 So, JUMP = (10 + 21 + 13 + 16) x 4 = 240 11. If Ever + Since = Darwin then D + a + r + w + i + n is ? Sol: Tough one as it has 10 variables in total. 4 digit number + 5 digit number = 6 digit number. So left most digit in the answer be 1. and S = 9, a = 0. Now we have to use trial and error method. Here E appeared 3 times, I, R, N two times each. Now E + I or E + I + 1 is a two digit number with carry over. What could be the value of E and I here. 8 and 7 are possible. But from the second column, 8 + C = 7 or 17 not possible. Similarly with 7 and 6. If E = 5, then the remaining value can be filled like above. 5653 + 97825 = 103478 Answer is 23 12. There are 16 hockey teams. find : (1) Number of matches played when each team plays with each other twice. (2) Number of matches played when each team plays each other once. (3) Number of matches when knockout of 16 team is to be played Sol: 1. Number of ways that each team played once with other team = 16C2. To play with each team twice = 16 x 15 = 240 2. 16C2 = 120 3. Total 4 rounds will be played. Total number of matches required = 8 + 4 + 2 + 1 = 15 13. 15 tennis players take part in a tournament. Every player plays twice with each of his opponents. How many games are to be played? A. 190 B. 200 C. 210 D. 220 E. 225 Sol: Formula: 15C2 x 2. So 15 x (15  1) = 15 x 14 = 210 14. 1, 11, 21, 1211, 111221, 312211, . . . . . what is the next term in the series? Sol: We can understand it by writing in words One One time 1 that is = 11 Then two times 1 that is = 21 Then one time 2 and one time 1 that is = 1211 Then one time one, one time two and two time 1 that is = 111221 And last term is three time 1, two time 2, and one time 1 that is = 312211 So our next term will be one time 3 one time 1 two time 2 and two time 1 13112221 and so on 15. How many five digit numbers are there such that two left most digits are even and remaining are odd. Sol: N = 4 x 5 x 5 x 5 x 5 = 2375 Where 4 cases of first digit {2,4,6,8} 5 cases of second digit {0,2,4,6,8} 5 cases of third digit {1,3,5,7,9} 5 cases of fourth digit {1,3,5,7,9} 5 cases of fifth digit {1,3,5,7,9} 16. If a refrigerator contains 12 cans such that 7 blue cans and 5 red cans. In how many ways can we remove 8 cans so that atleast 1 blue can and 1 red can remains in the refrigerator. Sol: Possible ways of keeping atleast 1 blue and 1 red ball are drawing cans like (6,2) (5,3) (4,4) (6,2) ?7C6×5C2 ? 710 = 70 (5,3) ?7C5×5C3 ? 21 x 10 = 210 (4,4) ?7C4×5C4 ? 35 x 5 = 175 70 + 210 + 175 = 455 17. Find the 8th term in series? 2, 2, 12, 12, 30, 30,      Sol: 11 + 1 = 2 22 – 2 = 2 32 + 3 = 12 42 – 4 = 12 52 + 5 = 30 62 – 6 = 30 So 7th term = (72 + 7) = 56 and 8th term = ({82} – 8) = 56 Answer is 56 18. Rahul took a part in cycling game where 1/5 ahead of him and 5/6 behind him then total number of participants = Sol: Let x be the total number of participants including Rahul. Excluding rahul = (x – 1) 15(x–1)+56(x–1) = x 31x – 31 = 30x Total number of participants x = 31 19. Data sufficiency question: What are the speeds two trains travels with 80 yards and 85 yards long respectively? (Assume that former is faster than later) a) they take 75 seconds to pass each other in opposite direction. b) they take 37.5 seconds to pass each other in same direction Sol: Let the speeds be x and y When moves in same direction the relative speed, x – y = (85–80)37.5 = 0.13      (I) When moves in opposite direction the relative speed, x + y = 165/75 = 2.2     (II) Now, equation I + equation II gives, 2x = 0.13 + 2.2 = 2.33 ? x = 1.165 From equation l, x – y = 0.13 ? y = 1.165 – 0.13 = 1.035 Therefore the speeds are 1.165 yards/sec and 1.035 yards/sec. 20. Reversing the digits of father's age we get son's age. One year ago father was twice in age of that of his son? find their current ages? Sol: Let father's age = 10x + y Son's age = 10y + x (As, it is got by reversing digits of fathers age) At that point (10x + y) – 1 = 2{(10y + x) – 1} ? x = (19y – 1)/8 Let y = 3 then x = 7. For any other y value, x value combined with y value doesn't give a realistic age (like father's age 120 etc) So, this has to be solution.Hence father's age = 73. Son's age = 37. 
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