66.66%
Required Ratio=3:2
2, 5, 10, 50, 500, 5000
5*2=10
10*5=50
50*10=500
500*50=25000
So odd one is 5000
4
3606
F
Matches played: 60.
Matches won: 30% of 60 => (60*(30/100)) = 18 matches.
Iterative approach:
On adding 1 to matches played and matches won, on every iteration until the win percentage gets to 50. So
19 / 61 = 0.3114754098360656
20 / 62 = 0.3225806451612903
21 / 63 = 0.3333333333333333
22 / 64 = 0.34375
…
…
…
…
Similarly,
41 / 83 = 0.4939759036144578
42 / 84 = 0.5
So, after 60th match 24 more matches has to be played and won to get 50% average winning rate.
30+2=32
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. let x is distance from A to B
and y is initial speed.
30/y+(x-30)5/4y -x/y = 3/4
=> 4x-12y=120 —-(1).
45/y + (x-45)5/4y -x/y=3/5
=> 5x-12y=225 ——-(2).
From equ (1) and equ (2) we will get.
x=25 and y=105
so initial speed is 25 km/hr
and Distance From A to B is 105 km
area is given by Area of Rectangle = Length of Rectangle*(sqrt((Diagonal of Rectangle^2)-(Length of Rectangle^2)))
so substituting we get area=4*(sqrt((25-16)))
area=4*(sqrt(9))
since area cannot be negative hence we take sqrt of 9 as 3 (as sqrt of 9 is 3 and -3 as well)
area=4*3=12m^2
NO but i hope to you hire me
Let x and y be the two numbers. And x > y.
Then,
x + y = 80—–(1)
y=80-y
Also,
x = 4y – 5
Solving the two equations,
80 – y = 4y – 5
85 = 5y
Therefore, y = 17
put the value of y in equation(1)
then
x+y=80
x+17=80
x=63
the numbers are 17&63
15km/hr