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(60/100)*60 + (40/100)*40
=52%
(a) 29.12 (actual)
first year rate of paper (inflation rate) : 25 + 25 *
6.5/100 = 26.625
paper rate in first year is 1.5 % more . ..thr4
inflated cost in 1 yr = 26.625 + 1.5 / 100 * 26.625 =
27.024.
second year : 27.024 + 27.024 * 6.5 / 100 = 28.78
paper cost : 1.5 /100 * 28.78 + 28.78 = 29.12 ( approx..)
Mr. Brown painted the first whole house in 6 days plus 1/3 of the second house in the next 2 days. Mr. Black can paint a whole house in 8 days and had 8 days to work, so he painted the equivalent of 1 whole house. That accounts for 2-1/3 houses painted. Mr. Blue only needs to paint 2/3 of the last house, so 2/3 times 12 days equals 8 days.
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12
23. 15s squre is 225and 8 squre is 64 addition is 289 product also 120 addition 15+8=23
6/11 or 54.54%
LCM of 4,6,8 and 10 are 120
So, the least number of four digits is 120*9=1080 ( <1000 )
-Total /n=80
– x/5=40
x=200
Total/n=y/n
y-x/n-5=90
80n-200=90(n-5)
80n-200=90n-450
450-200=10n
n =25 (students) write exam
the answer is 45 km/h
Define x:
Let the total distance be x.
Find the time needed in terms of x for speed at 30 km/h:
Find the time needed in term of x for 40 km./h:
Find the difference in time:
Solve x:
Find the total time needed to for the whole journey:
Given that he was 40 mins late:
Find the minimum speed needed:
Answer: The minimum speed is 45 km/h
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