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Answer: z and u
Explanation:
Y is to the right of U and exactly in front of V. Therefore,
U Y
V
Z is behind W and W and X are at extreme ends. So, W has to be to the right of Y. The final arrangement is as follows.
U Y W
X V Z Therefore, Z and U are at extreme ends is true.
Statements :
All pens are roads. All roads are houses.
Conclusions :
I. All houses are pens.
II. Some houses are pens.
A.
Prosperity for all
Answer: Option D
Explanation:
As Bob is sitting between Charles and Elena and Douglas is sitting to the left of Elena. We get the following arrangement.
Andy is opposite Bob. Therefore, the final arrangement is as follows.
Fred will be sitting right of Charles if each person interchange his/her place with the person sitting opposite them.
Balls- B1, B2, B3, B4, B5, B6, B7, B8, B9.
Group1 – (B1, B2, B3), Group2 – (B4, B5, B6), Group3 – (B7, B8, B9)
Now weigh any two groups. Group1 on left side of the scale and Group2 on the right side.
When weighing scale tilts left – Group1 has a heavy ball or right – Group2 has a heavy ball or balanced – Group3 has a heavy ball.
Lets assume Group 1 has a heavy ball.
Now weigh any two balls from Group1. B1 on left side of the scale and B2 on right side.
When weighing scale tilts left – B1 is the heavy or tilts right – B2 is the heavy or balanced – B3 is the heavy.
Here is the solution to the given version of the puzzle (9 balls, one is heavier, need to identify oddball), where we label the balls A, B, …, I:
1. Weigh ABC versus DEF.
Scenario a: If these (1) balance, then we know the oddball is one of G, H, I.
2. Weigh G versus H.
Scenario a.i: If these (2) balance, the oddball is I.
Scenario a.ii: If these (2) do not balance, the heavier one is the oddball.
Scenario b: If these (1) do not balance, then the oddball is on the heavier side. For simplicity, assume the ABC side is heavier, so the oddball is one of A, B, C.
2. Weigh A versus B.
Scenario b.i: If these (2) balance, the oddball is C.
Scenario b.ii: If these (2) do not balance, the heavier one is the oddball.
answer is maximum of 2.
A can give B (5 min – 4 1/2 min) = 30 sec start.
The distance covered by B in 5 min = 1000 m.
Distance covered in 30 sec = (1000 * 30)/300 = 100 m.
A can give B 100m start.
28
to arrange m objects in n places => nCm
i.e. for this example m=2(+,- sign) and n = 8(places between two number from 1 to 9)
SO, answer is : 8C2 = (8*7) / 2 = 28
Yes, technology give me better efficiency to my work. Just we have take a work piece of sheet metal to convert in product like is box, different size, where manually it’s complete on 3 to 4 days but with the help of technology I have done in 3 hours that’s I said the technology give better efficiency to my better work.
40
x – 30 = 1/4 x || *4
4x – 120 = x || -x + 120
3x = 120 || /3
x = 40
Speed Ratio = 1:7/6 = 6:7
Time Ratio = 7:6
1 ——– 7
4 ——— ? 28 m