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  #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
Super Moderator
 
Join Date: Mar 2013
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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