Find probability of the natural no. between 101 and 999 which are divisible by both 2 and 5.
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Answer:89 natural number
explanation:-
no. divisible by 2&5 must be divisible by 10
Form an AP of natural number present between 101 & 999
AP = 110, 120, 130,............, 990
from the AP we concluded
a = 110
d = 10
nth term of the AP = 990
a+(n-1)d=990
110+(n-1)10=990
(n-1)10=990-110
(n-1)= 880/10
n-1=88
n=88+1
n=89
natural number present between 101 & 109 = 89
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