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HCL Technologies Previous year Placement Papers
Any buddy, please provide me the HCL Technologies Previous year Placement Papers…..

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Re: HCL Technologies Previous year Placement Papers
Yes sure, here I am uploading a document file that contains the HCL Technologies Previous year Placement Papers. There are objective types of the questions available. I have taken following questions from the attachment: In the following question select the word which is OPPOSITE in the meaning of the given word. Q1. INDISCREET a.reliable b. honest c.prudent d. stupid Q2. SOLICITUDE a.insouciance b. ingenuity c.propriety d. austerity Q3. In the sentence there is a bold word or phrase. One of the words or phrases given in the options conveys almost the samemeaning as the bold word or phrase in the sentence. Select that option which is nearest in meaning. It is preposterous on your part to look for a job without first completing your education. a.Wise b. Imperative c.Advisable d. Most admirable e. Very absurd In the following questiones, fill in the blank space. Q4. The success that he has gained, though striking enough, does not, however, commensurate . . . . the efforts made by him. a.About b. From c.With d. Beside e. Over Q5. Vinod took his meals after he . . . . a.Had completed his work b. Had been completing his work c.Was completing his work d. Had been completed his work e. Had got completed his work In the following questions, select the word or phrase that is similar in meaning to the given word. Rests of the questions are in the attachment, click on it…….
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Re: HCL Technologies Previous year Placement Papers
Ok, as you want the Previous year Placement Papers of HCL Technologies Limited so here I am providing you. HCL Technologies Limited placement paper Aptitude Questions 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. 21. The hour hand lies between 3 and 4. Tthe difference between hour and minute hand is 50 degree.What are the two possible timings? Sol: The angle between the hour hand and minute hand at a given time H:MM is given by θ = 30×H – 211×MM The time after H hours, hour hand and minute hand are at MM =  211×((30×H)±θ)  given H = 3, MM = 50 Substituting the above values in the formula θ = 8011, 28011 22. Jack and Jill went up and down a hill. They started from the bottom and Jack met Jill again 20 miles from the top while returning. Jack completed the race 1 min a head of Jill. If the hill is 440 miles high and their speed while down journey is 1.5 times the up journey. How long it took for the Jack to complete the race ? Sol: Assume that height of the hill is 440 miles. Let speed of Jack when going up = x miles/minute and speed of Jill when going up = y miles/minute Then speed of Jack when going down = 1.5x miles/minute and speed of Jill wen going up = 1.5y miles/minute Case 1 : Jack met jill 20 miles from the top. So Jill travelled 440 – 20 = 420 miles. Time taken for Jack to travel 440 miles up and 20 miles down = Time taken for Jill to travel 420 miles up 440x+201.5x=420y 681.5x=420y 68y = 63x y = 63x68 (1) Case 2 : Time taken for Jack to travel 440 miles up and 440 miles down = Time taken for Jill to travel 440 miles up and 440 miles down – 1 440x+4401.5x=440y+4401.5y – 1 440×53(1y−1x)=1(2) Substitute (2) in (1) we get x = 440×5×53×63 t = 440×53(1x) t = 12.6min 23. Data Sufficiency question: A, B, C, D have to stand in a queue in descending order of their heights. Who stands first? I. D was not the last, A was not the first. II. The first is not C and B was not the tallest. Sol: D because A is not first neither C and B is not the tallest person. The only person will be first is D. So option (C). We can answer this question using both the statements together. 24. One of the longest sides of the triangle is 20 m. The other side is 10 m. Area of the triangle is 80 m2. What is the another side of the triangle? Sol: If a,b,c are the three sides of the triangle. Then formula for Area = (s(s–a)×(s–b)×(s–c))−−−−−−−−−−−−−−−−−−−−√ Where s = (a+b+c)2=12×(30+c) [Assume a = 20 ,b = 10] Now, Check the options. 25. Data Sufficiency Question: a and b are two positive numbers. How many of them are odd? I. Multiplication of b with an odd number gives an even number. II.a2 – b is even. Sol: From the 1st statement b is even, as when multiplied by odd it gives even a2 – b = even ⇒ a is even Here none of a and b are odd 26. Mr. T has a wrong weighing pan. One arm is lengthier than other. 1 kilogram on left balances 8 melons on right, 1 kilogram on right balances 2 melons on left. If all melons are equal in weight, what is the weight of a single melon. Sol: Let additional weight on left arm be x. Weight of melon be m x + 1 = 8 x m       (1) x + 2 x m = 1       (2) Solving 1 & 2 we get. Weight of a single Melon = 200 gm.
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