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At 130 years.
Out of 10 persons, 4 are graduates; so, (10 – 4) = 6 are under-graduates.
If there is no restriction, any three can be chosen from the ten in (10C3) = 120 ways.
Now, if all three chosen are under-graduates; it can take place in (6C3) = 20 ways.
Therefore, the probability that there will be no graduate among the three chosen = (20 / 120) = (1 / 6).
Therefore, the probability that there will be at least one graduate among the three chosen = {1 – (1 / 6)} = (5 / 6) = 0.8333.
copy cat
3*2-1=5
5*2-2=8
8*2-3=13
13*2-4=22
22*2-5=39
39*2-6=72
so series is 3,5,8,13,22,39,72
24 should be replaced by 22
24 is wrong…….
One day’s work of A, B and C = (1/24 + 1/30 + 1/40) = 1/10.
C leaves 4 days before completion of the work, which means only A and B work during the last 4 days.
Work done by A and B together in the last 4 days = 4 (1/24 + 1/30) = 3/10.
Remaining Work = 7/10, which was done by A,B and C in the initial number of days.
Number of days required for this initial work = 7 days.
Thus, the total numbers of days required = 4 + 7 = 11 days.
1×2×…100=100!
Number of zeros in product of n numbers =[5n]+[52n]+[53n]+…
Number of zeros in product of 100 numbers =[5100]+[52100]+[53100]
where [.] is greatest integer function
=[20]+[4]+[0.8]=20+4=24
1, 1, 2, 6, 24, 96, 720
C. 96
N O O N
S O O N
+ M O O N
----------
J U N E
4 1 1 4
5 1 1 4
+ 0 1 1 4
———-
9 3 4 2
As it is a right angled triangle so we know the hypotenuse must be 13 and 5 & 13 are base and height (in any order).
Area of triangle = 0.5* base * height = 0.5 * 5 * 12 = 30
Let the number of one rupee coins in the bag be x.
Number of 50 paise coins in the bag is 93 – x.
Total value of coins
[100x + 50(93 – x)]paise = 5600 paise
=> x = 74
ANS = 74
3 groups
both a and b is correct
35000