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1 ball – 4 runs (1st batter – 94+4=98)
2 ball – 3 runs and one run is short run during running between the wicket (1st batter- 98+3-1 =100)and strike changes for 2nd batter
3 ball – 6 runs ( 2nd batter 94+6=100)
( c ) 10
23 Teams
3hour 45 min = 13500 sec
In 1 sec he covers = 12 m
In 13500 sec he covers = 12×13500 m = (12×13500)/1000 km = 162 km
Ans : 162 km
2*3(15+4)=114
114*1/2=57 rs
Pfull = 4 Hrs; Qfull = 5 Hrs
Assume velocity Vp = x then Vq = 4/5 x –> avg velocity = (5/5 + 4/5) / 2 = 0.9
If fastest time = 4hrs then the time it would take to fill up tank by alternating is 4/0.9 = 4.44 Hrs
30+2=32
The answer is 42
A
4
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.
Solution:
As given, we have,
The cost of one pen = 36 Rs.
So, the cost of 15 pens = 36 × 15 = 540 Rs.
The cost of one book = 45 Rs.
So, the cost of 12 books = 45 × 12 = 540 Rs.
The cost of one pencil = 8 Rs.
So, the cost of 10 pencils = 8 × 10 = 80 Rs.
Now,
the cost of each eraser is 40 Rs. less than the combined costs of pen and pencil.
So,
Combined costs of pen and pencil = 36 + 8 = 44 Rs.
Cost of one eraser = 44 – 40 = 4 Rs.
So, the cost of 5 erasers = 4 × 5 = 20 Rs.
Hence,
The total amount spent is
Hence, the total amount spent is 1180 Rs.
7 min clock:|——-7——-|
4 min clock:|—–4—-|—-4—–|—–4—-|—–4—-|
you got 9 min: |—————9————–|
3800