#1
June 8th, 2016, 03:25 PM
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HCL Network Engineer
Hello, I want the paper of the HCL for the placement of network engineer post, will you please provide me?
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#2
June 8th, 2016, 03:50 PM
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Re: HCL Network Engineer
Hello, here I m providing you the paper of HCL for placement of network engineer post as under: 1.Riya is twice as old as Priya now. Three years ago she was thrice as old as priya.How old is Riya now? a. 7 years b. 9 years c. 6 years d. 12 years Answer – d 2. Complete the series 81, 69, 58, 48, 39, ----- a. 7 b. 10 c. 22 d. 31 Answer – d 3. Riya and Priya set on a journey. Riya moves eastward at a speed of 20kmph and Priya moves westward at a speed of 30 kmph.How far will be priya from Riya after 30 minutes a. 25kms b. 10kms c. 50kms d. 30kms Answer – a 4. What percentage of 20 is .05? a. 0.5 b. 0.25 c. 0.1 d. 0.025 Answer – b 5. If a number cube is rolled once and coin is tossed once. What is the probability that the coin will show tails and composite number on the cube? a. 1/8 b. 1/16 c. 1/2 d. 1/4 Answer – a 6. A loan of Rs.700 is made at 9.5% simple interest for 4 months. How much interest is owed when the loan is due? a. RS.20 b. Rs 23 c. Rs 22 d. Rs 24 Answer – c 7. If ¼ of ¼ is subtracted from ½ of ½, the result will be a. 3/8 b. 1/16 c. 3/16 d. 1/4 Answer – c 8.If x(a) = a2 +2a-1, then x(8)-x(5) = a. 43 b. 45 c. 14 d. 3 Answer – b 9. How many pieces, each of length 4.5 meters, can be cut out of 225 meters of wires? a. 40 b. 45 c. 50 d. 55 Answer – c 10. During a stock clearance sale, a pair of shoes which costs Rs.500 was sold at a small profit of 5%. What was the selling price of the pair of shoes? a. 400 b. 425 c. 450 d. 475 Answer – d 11. 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 12. 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 13. 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. 14. 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 15. 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 |
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