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let x be the avg age of each person in the family ten years
back…
so total age=6x
after 10 years i.e,now the total age of the family =6x+60
let grand father age as y
then the average of the family=(6x+60-y)/5
(bcoz grand father has expired and the new born baby has no
age at present)
according to the given question (6x+60-y)/5=x
bcoz x is the avg age bfore 10 years.
while calc the equation according to the presnt age we have
to add 10 to the result….
so the answer is 70
9 because from 64 stubs 8 cigarettes and again from 8 stubs he’ll get 1 cigarette then his score will be 9
21
The formula to find number of diagonals (D) given total number of vertices or sides (N) is
N * (N – 3)
D = ———–
2
Using the formula, we get
1325 * 2 = N * (N – 3)
N2 – 3N – 2650 = 0
Solving the quadratic equation, we get N = 53 or -50
It is obvious that answer is 53 as number of vertices can not be negative.
Alternatively, you can derive the formula as triange has 0 diagonals, quadrangel has 2, pentagon has 5, hexagon has 9 and so on……
Hence the series is 0, 0, 0, 2, 5, 9, 14, …….. (as diagram with 1,2 or 3 vertices will have 0 diagonals).
Using the series one can arrive to the formula given above.
Sachin is 14 years old whereas Rahu is 18 years old.
325
30
4.224 days
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