Out of 10 persons, 4 are graduates; so, (10 – 4) = 6 are under-graduates.
If there is no restriction, any three can be chosen from the ten in (10C3) = 120 ways.
Now, if all three chosen are under-graduates; it can take place in (6C3) = 20 ways.
Therefore, the probability that there will be no graduate among the three chosen = (20 / 120) = (1 / 6).
Therefore, the probability that there will be at least one graduate among the three chosen = {1 – (1 / 6)} = (5 / 6) = 0.8333.
France — D
trick
(r/10)*(r/10)
(20/10)*(20/10)
4% decrese
habeas corpus
Hog
5/9
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.
Let us consider previous salary as a ‘X’ then
0.12*X=0.10*(X+1200)
0.02X=120
X=120/.02=6000
X=6000
His previous salaray was 6000
17and+-12=5-48=-43
10
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
I solve it as the 280m train being stationary… and the 120m train moving at (42+30) = 72kph…. or 20mps
The front of the train will have to go 280m to pass the stationary train plus another 120m for the backs to clear… that is 400m in all.
The time to travel 400m at 20mps is 20secs
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
23
2hr 30min
Therefore, the number of possible 13 digit numbers using 1, 2, 3, 4, 5 which are divisible by 4
= x 3 + x 2
= (3 + 2)
=
= 244140625 ways
30%