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8, 27, 64, 100, 125, 216, 343
A is travelling at 50kmph, B is travelling at 40
kmph……..according to the formula
time taken to meet = distance between them
———————-
relative speed of two vehicles
so, time taken to meet= 15/(50-40)=15/10=3/2hrs
sorry this question is irrelevant.
let A goes up and B comes down.
assume A takes 1 step in 1sec.
hence,B takes 5 steps in 1sec.
50 steps ll be moved by A in 50/1=50secs
125 steps ll be moved by B in 125/5=25 secs
hence speed of escalator=(difference of no of
steps)/(difference of time)
speed=(125-50)/(50-25)= 3 steps/sec
hence during upward journey of 50secs by A,total steps=50 by
A and 150 by escalator..total=150+50=200
during downward journey of 25 secs by B,total steps=125 by B
and 75 by escalator..total=125+75=200steps.
ANS:200 steps
430
5x5x6 = 150 cube
150 – 20( corner cube) = 130 cube
130 x 3 (side remaining) = 390
20 x 2 (side remaining) = 40
total side remaining = 390 + 40 = 430
Answers as expected
sorry the above ans was typed by mistake the correct ans is…..
1027.05 ? 314.005 + 112.25 = ?
ans:
1027.05 – 314.005 + 112.25 = 825.295
speed = 72 Kmph
Speed= 72 * 1000 / (60*60) m/s
speed= 72 * 5 / 18
speed= 20 m/s
time = 30 s
distance = 600 m
9
B
55 + 70 = 125
125/14hrs = 60 mph
14 hours
If the bell rung when they r started is also counted then it
will be 5 otherwise 4
To determine how many consecutive zeros the product of S will end with, we need to find the highest power of 10 that divides the product. This is equivalent to finding the highest power of 5 that divides the product, since the number of factors of 2 will always be greater than the number of factors of 5.
The primes in S are {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}.
There are 24 primes in S, so the product of S is:
2 x 3 x 5 x 7 x 11 x 13 x 17 x 19 x 23 x 29 x 31 x 37 x 41 x 43 x 47 x 53 x 59 x 61 x 67 x 71 x 73 x 79 x 83 x 89 x 97
We need to find the highest power of 5 that divides this product. To do this, we count the number of factors of 5 in the prime factorization of each number in S.
5 appears once: 5
5 appears once: 25
5 appears once: 35
5 appears once: 55
5 appears once: 65
5 appears once: 85
So, there are six factors of 5 in the product of S. However, we also need to consider the powers of 5 that arise from the factors 25, 35, 55, and 65.
25 = 5 x 5 appears once: 25
35 = 5 x 7 appears once: 35
55 = 5 x 11 appears once: 55
65 = 5 x 13 appears once: 65
Each of these numbers contributes an additional factor of 5 to the product of S. Therefore, there are 6 + 4 = 10 factors of 5 in the product of S.
Since each factor of 5 corresponds to a factor of 10, we know that the product of S will end with 10 zeros. Therefore, the product of S will end with 10 consecutive zeros
answer= 4:7
100