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04 Dec 2018, 00:03
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B
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Difficulty:
45%
(medium)
Question Stats:
69% (02:07) correct
31%
(02:09)
wrong
based on 724
sessions
History
Date
Time
Result
Not Attempted Yet
When the decimal point of a certain positive decimal number is moved six places to the right, the resulting number is 9 times the reciprocal of the original number. What is the original number?
A. 0.0003
B. 0.003
C. 0.03
D. 0.3
E. 0.9
A. 0.0003
B. 0.003
C. 0.03
D. 0.3
E. 0.9
05 Dec 2018, 00:51
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Bunuel
IMO, Difficulty level should be low (around 15%).
Let's assume that the number is x. Moving decimal point six places means multiplying it by \(10^6\); So, we can get this equation:
\(10^6 x= 9(1/x)\); multiplying both sides by x and dividing them by \(10^6\), we get:
\(x^2=9/10^6\); and hence:
\(x=3*10^-3=0.003\)
General Discussion
04 Dec 2018, 00:25
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Bunuel
9/0.003 = 3*10^3
this would be equal to when 0.003 * 10^6' moved six places to right'= 3*10^3 IMO B..
