s1(1+x/100)+s2(1+x/100)+s3(1+x/100)
The phrase is usually ‘like looking for a needle in a haystack’. It describes a task that is virtually impossible because you would have to search a huge area
7000
5:6::7:10::6:5
=((5/6)/(7/10)/(6/5))
=((5/6)*(6/5)/(7/10))
=(1/(7/10))
=10/7
1^1,2^2,3^3,4:^4,5^5,6^6
1,4,27,256,3125,46656
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
ans: 40
6/3*12/3*15/3=40
600
16.25
(p*r*t)/100 = I ………………..(p*3*10)/100 = 840 …..p = 84000/30 = 2800
let t = total no of students.. then
students who passed one or both subjects,
n(e U h) = n(e) + n(h) – n(e intersection h)
=> t = 0.8t + 0.7t – 144
=> t = 1.5t – 144
students who failed both subjects is 10% i.e. 0.1t
=>t-n(e U h) = 0.1t,
=>t -(1.5t – 144) = 0.1t
=>t- 1.5t- 0.1t = -144
=> -0.6t = -144
=>t = 240
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
ans: 250 rs exactly
1
In a cube all the diagonal and sides are equal, we can go diagonally.
441=21^2
Dice