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saurabh9gupta
Bunuel
In the decimal representation of d, where 0 < d < 1, is the tenths’ digit of d greater than 0?

(1) 12d is an integer.
(2) 6d is an integer.
chetan2u - your approach plz

what is tenths digit
0.xy here x is tenths digit

(1) 12d is an integer.
so \(12d=x.....d=\frac{x}{12}\)
case I :- x is 1
\(d=\frac{1}{12}\) Now \(\frac{1}{10}>\frac{1}{12}\).......\(0.1>\frac{1}{12}\) there fore d = 0.0rst
yes, tenths digit is 0
case I :- x is 6
\(d=\frac{6}{12}=\frac{1}{2}=0.5\)
no, tenths digit is not 0 but 5
insuff

(2) 6d is an integer.
so \(6d=x.....d=\frac{x}{6}\)
Min value of d when :- x is 1
\(d=\frac{1}{6}\) Now \(\frac{1}{10}<\frac{1}{6}\).......\(0.1<\frac{1}{6}\) therefore d = 0.rst
yes, tenths digit is always>0
suff

B
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Bunuel
In the decimal representation of d, where 0 < d < 1, is the tenths’ digit of d greater than 0?

(1) 12d is an integer.
(2) 6d is an integer.

1) 12d is an integer that means d is in the form of 1/2, 1/3, 1/4, 1/6, 1/12

The tenths digit in \(\frac{1}{2}\), \(\frac{1}{3}\), \(\frac{1}{4}\) and \(\frac{1}{6}\) is >0

1/12 units digit = 0

Insufficient.

2)
6d is an integer means d is in the form of \(\frac{1}{2}\), \(\frac{1}{3}\), \(\frac{1}{4}\) and \(\frac{1}{6}\)

each of which the units digit is >0

Sufficient

B
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(1) 12d is an integer

Solving for d, d = 1/12, 2/12, 3/12, ...

1/12 is 0.0...

INSUFFICIENT

(2) 6d is an integer

d minimum is 1/6 = 0.1...

SUFFICIENT

The answer is B
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