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C
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Difficulty:
85%
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Question Stats:
44% (01:43) correct
56%
(01:24)
wrong
based on 489
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Which of the following is closest to 1/9+1/99+1/999
A. 1/10
B. 1/9
C. 1/8
D. 1/6
E. 1/5
A. 1/10
B. 1/9
C. 1/8
D. 1/6
E. 1/5
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02 Jan 2010, 08:47
Kudos
Bookmarks
Which of the following is closest to 1/9+1/99+1/999
A. 1/10
B. 1/9
C. 1/8
D. 1/6
E. 1/5
Probably the easiest way would be if we just add the fractions. Notice that these fractions would be repeated decimals:
1/9+1/99+1/999=0.(111)+0.(010)+0.(001)=~0.122=~0.125=1/8
BTW exact fraction would be: 0.(111111)+0.(010101)+0.(001001)=0.(122213).
Answer: C.
For more on repeated decimals check: math-number-theory-88376.html
Hope it helps.
A. 1/10
B. 1/9
C. 1/8
D. 1/6
E. 1/5
Probably the easiest way would be if we just add the fractions. Notice that these fractions would be repeated decimals:
1/9+1/99+1/999=0.(111)+0.(010)+0.(001)=~0.122=~0.125=1/8
BTW exact fraction would be: 0.(111111)+0.(010101)+0.(001001)=0.(122213).
Answer: C.
For more on repeated decimals check: math-number-theory-88376.html
Hope it helps.
02 Jan 2010, 07:15
Kudos
Bookmarks
Infact there is another way of getting the answer without even solving it...
logically u r adding \(\frac{1}{99}+\frac{1}{999}\) to \(\frac{1}{9}\), therefore the answer would be more than \(\frac{1}{9}\). Hence \(\frac{1}{10}\) is wrong for sure...
Now \(\frac{1}{8}\) is \(\frac{1}{72}\) more than \(\frac{1}{9}\). This implies the midpoint betwen \(\frac{1}{8}\) & \(\frac{1}{9}\) is \(\frac{1}{144}\)..
Since you are adding \(\frac{1}{99}\) (which is more than \(\frac{1}{144}\)) to \(\frac{1}{9}\), then the sum would be closer to \(\frac{1}{8}\) than \(\frac{1}{9}\) ..... Also note that the \(\frac{1}{99}\) is less than \(\frac{1}{72}\) and hence it would not exceed \(\frac{1}{8}\)
Hence answer would be C.
logically u r adding \(\frac{1}{99}+\frac{1}{999}\) to \(\frac{1}{9}\), therefore the answer would be more than \(\frac{1}{9}\). Hence \(\frac{1}{10}\) is wrong for sure...
Now \(\frac{1}{8}\) is \(\frac{1}{72}\) more than \(\frac{1}{9}\). This implies the midpoint betwen \(\frac{1}{8}\) & \(\frac{1}{9}\) is \(\frac{1}{144}\)..
Since you are adding \(\frac{1}{99}\) (which is more than \(\frac{1}{144}\)) to \(\frac{1}{9}\), then the sum would be closer to \(\frac{1}{8}\) than \(\frac{1}{9}\) ..... Also note that the \(\frac{1}{99}\) is less than \(\frac{1}{72}\) and hence it would not exceed \(\frac{1}{8}\)
Hence answer would be C.
