x,y sides of rectangle
area=xy
increased by 100% =double length
hence
new area=4xy
increased area=4xy-xy=3xy
percentage of area will be increased
by 300%
7 1/2
2, 3, 6, 15, 52.5, 157.5, 630
52.5
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
sow
Time taken to fill repaired cistern = 12 mins
In 1 min, pipe will fill = 1/12
Time taken to fill cistern with leak = 12 + 18 = 30 mins
In 1 min, it will fill = 1/30
Consider the leak as an outlet pipe, so
1/30 = 1/12 – [ in 1 min how much the outlet pipe leaks ]
= 1/12 – 1/30
= 5/60 – 2/60 = 3/60 = 1/20
Therefore, it will take 20 mins for the leak to empty the cistern
1000 cu.m
1/3
Answer=8
obey