a. (x+y)'=x'.y' b. (x'+y')'=x.y
c. (x'.y')'=x+y d. (x'+y')'=x'.y'
C
Answer: Pile II
here the 30% of 100 is 30
and 10% of the 30 = 3
so 20% = 6%
so ans = 6%
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
P+C+M=80
P+C=70
Therefore M=10
Given , P+M=90
If M=10
P=80
LCM=48
Then add remainder 3
ans =51
let original radius=100
original surface area lets say it x =4 * (22/7) * 100 *100
new radius=100 + 50% of 100=150
new surface area lets say it y = 4 * (22/7)* 150* 150
increase in surface area=((y-x)/x) * 100
i.e.
((4*(22/7)*150*150 – 4*(22/7)*100 *100) / (4*(22/7)*100*100))*100
=125
so the answer is 125%.
hope this helps you.
In 3 hours thinker candle becomes half
In the same time thinner candle becomes one fourth. Now the
required condition is satisfied. So 3 Hours
1×2×…100=100!
Number of zeros in product of n numbers =[5n]+[52n]+[53n]+…
Number of zeros in product of 100 numbers =[5100]+[52100]+[53100]
where [.] is greatest integer function
=[20]+[4]+[0.8]=20+4=24
B1
9 sec=4 times
x sec=12 times
x=(9*12)/4=27 sec