@ScareCrow28 said:You can check my post, if you have time!
i am not convinced about the solution sir..
@ all OA!
Solution:-
@pavimai said:can you share your approach
speed be n
now when m has travelled 27 n have covered
27- (27n/45)
now as n makes her speed twice so n catches m this means 2n>m
(27- (27n/45))/2n-45 = 6
n comes out as 23.5xx
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Suppose that a andb are digits, not both nine and not both zero, and the repeating decimal 0.abababababab...... is expressed as a fraction in lowest terms. How many different denominators are possible?
a)3 b)4 c)5 d)8 e)9
@albiesriram said:Suppose that a andb are digits, not both nine and not both zero, and the repeating decimal 0.abababababab...... is expressed as a fraction in lowest terms. How many different denominators are possible?
a)3 b)4 c)5 d)8 e)9
4
@albiesriram said:Suppose that a and b are digits, not both nine and not both zero, and the repeating decimal 0.abababababab...... is expressed as a fraction in lowest terms. How many different denominators are possible?a)3 b)4 c)5 d)8 e)9
N = 0.abababababa..
99N = ab
N = ab/99
Denom = 99 = 3^2*11
No of diff ways =3, 9, 11, 33 and 99
5 ways ?..
@albiesriram said:Suppose that a andb are digits, not both nine and not both zero, and the repeating decimal 0.abababababab...... is expressed as a fraction in lowest terms. How many different denominators are possible?a)3 b)4 c)5 d)8 e)9
5
x= 0.abab...
100x = ab.abab...
subtract both eqns
99x = ab
x= ab/99
5 values can be.. 9,11,3,33,99
x= 0.abab...
100x = ab.abab...
subtract both eqns
99x = ab
x= ab/99
5 values can be.. 9,11,3,33,99
@ScareCrow28 said:
N = ab/99Denom = 99 = 3^2*11No of diff ways =1, 3, 9, 33 and 99 5 ways ?
why is 11 not getting counted ?