28
to arrange m objects in n places => nCm
i.e. for this example m=2(+,- sign) and n = 8(places between two number from 1 to 9)
SO, answer is : 8C2 = (8*7) / 2 = 28
we know,
area=b*h
b gets increased by 20% i.e (b+0.20b)
h gets decreased by 20% i.e (h-0.20h)
rewriting the equation(area=b*h),
area=(b+0.20b)*(h-0.20h)
area=b(1+0.20)*h(1-0.20)
area=b(1.20)*h(0.80)
area=b*h*(1.20)*(0.80)
area=b*h*(0.96)
i.e new area=0.96 times the original area
if 100% was the original area,it has decreased to 96%
so,100%-96%= 4%
1.7% loss
It is About 35 Billion
Prob(choosing two people who are a couple from 6 couples) = 1/6
We have 5 couple left (10 people)
Prob(choosing a person A from 10 people) = 1/10
Prob(choosing 1 person who is not a couple of A) = 1/8
Prob(choosing 4 people that has exactly one couple) = 1/6*1/8*1/10 = 0.0021
( a ) 28
15 litre water contains 20% alcohal ie. 12 (water)+ 3
(alcohol).
after adding 5 liter water it becomes 12+5=17 (water)& 3
(alc.).
ie. 17:3 and answer will be 85:15. ie. 15%.
C.1175
d / (x+1/2) = 4/5 * t
and d / (x-1/2) = 5/2 + t
solving we get d = 15.
1440 * 3/4 = 1080
1080 / 30 = 36
1440 * 1/4 = 360
360 / 24 = 15
36 * 2 + 15 * 1 = 87
6/(2+3)-(2+3)/6
=6/5-5/6
=(36-25)/30
=11/30
ans is 11/30
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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
Friday
6 pairs of earrings. So 12 earrings.
The total has to be 26, so let’s say there’s 6 pairs of earrings and the number of pairs has to be 4 less than the number of rings. So that would make 10 rings. The 10 rings has to be 2 1/2 greater than the number of pins. So you would divide 10/2.5 = 4 Pins. That would equate to a total of 12 earrings, 10 rings, and 4 pins. A total of 26 pieces.
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