Perimeter of semi circular = πr+2r
Where r is radius , π = 22/7
πr+2r= 144 (given)
(22/7)r +2r = 144
22r+14r = 144*7 (multiply both sides by 7)
36r = 144*7
r= 144*7/36 = 28
radius =28 cm
Area =( 1/2)πr^2
Area=( 1/2 )*(22/7)*28*28
= 1232cm^2
Ans : area is 1232 cm^2
12(1/3) = 4 so women = 4
==>men = 12-4 = 8
if team needs 20% of women then men Will be 80%
consider u want to hire “x” men
so,
(12+x)(80/100) = 8 + x
so, x= 8
Smart Question. What was the total gain by both of them, not Just Krishan
Its 2 times faster than the other train
v1*t=v2
v2*t=4*v1
solving these two,we get
v2/v1=2
Let d = 7r. And use distance is = rate × time
7r= ( r+12) 5
7r= 5r + 60
Subtract 5r from both sides
2r = 60
Divide out 2
Rate = 30 km/h
Original question 30 km/h × 7 = 210
30 +12 = 42× 5 = 210
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
1920
1.5 hr
144
9000
4%
125