4/9
B.25
NTFS — New Technology File System
Fat — file allocation table
NTFS having a quota, commpress system with securites base as
administrator giving, multipal user, groups, to set permission
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.
3×3=9
435
13
(10×1000) / (2×3.1416×1.75) = 909 times (about)
Total Pages = 42+28 = 70 pages
for 70 pages we give Rs.20/-
and for 42 pages we give x
thus x = (42×20)/70
x = 12
The Answer = 12/-
A
Let the age of the man be x
Then age of his son becomes (x−24)
2 years later from now,
Age of man will be = x+2
and age of his son will be =(x−24+2)=x−22
According to question,
2(x−22)=x+2
i.e., 2x−44=x+2
i.e., x=46
Therefore ,
Present age of man=46 years
And present age of his son =46−24=22 years
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
Dear friends i ofcourse confused like you when i come
across to solve this problem. Really friends its very
simple if you understand the question clearly. First thing
is what is mean by “as many as” means its called ‘idiom and
phrase’ in english and it means “the same number of”. now
read the question “how many pairs of letters in STAINLESS
which has same number of letters between them in the word
as they have in english alphabet”.
In the alphabetical order, A-Z can be numbered as 1-26.
In A(INL)E which is same as in the alphabetical order A
(BCD)E. In both the cases E is in the Fourth position. so
we got one pair.
And in ST, there are no letters between them in the word
stainless. In alphabetical orer from A-Z also there is no
letters between them..so we got the second pair…
In STAINLESS it has two pairs ST and AE