A wet stone
x and y can be equal to 1..
If so,then he gets 1/1 of Rs.10 which is equal to 10 and
again 1/1 of Rs.10=Rs.10..so,he gets total of 20..and
returns 20..so,no gain and no loss..
If x=1,y=2,he gets, 5+20=25..and returns 20..so he may not
lose..
So,whatever be the values of x and y,only these two answers
are possible..
so its a)He never loses..
b)Brother or Sister
Because Bis related to either Brother or Sister.
It’s choice A because you take the last to letters and move them to the front then the previous two letters go after them and so on.
26
let the for digit number be = pqrs
p=q/3 ==>q = 3p
r=p+q=p+3p =4p
s=3q = 3(3P)=9p
number:
p 3p 4p 9p
let p=1 answer is 1349
if it 2 answer 2 6 8 18
so it becomes five digit number so correct answer is 1349
When A runs 1000 meters, B runs 900 meters and when B runs 800 meters, C runs 700 meters.
Therefore, when B runs 900 meters, the distance that C runs = (900 x 700)/800 = 6300/8 = 787.5 meters.
So, in a race of 1000 meters, A beats C by (1000 – 787.5) = 212.5 meters to C.
So, in a race of 600 meters, the number of meters by Which A beats C = (600 x 212.5)/1000 = 127.5 meters.
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
4155