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16 Sep 2014, 01:51
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C
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Difficulty:
25%
(medium)
Question Stats:
80% (01:26) correct
20%
(01:56)
wrong
based on 61
sessions
History
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What is the 99th digit after the decimal point in the decimal expansion of \(\frac{2}{9}\) + \(\frac{3}{11}\)?
A. 1
B. 2
C. 4
D. 7
E. 9
A. 1
B. 2
C. 4
D. 7
E. 9
16 Sep 2014, 01:51
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Official Solution:
What is the 99th digit after the decimal point in the decimal expansion of \(\frac{2}{9}\) + \(\frac{3}{11}\)?
A. 1
B. 2
C. 4
D. 7
E. 9
There are two ways to approach this problem. We'll go through them in turn.
(1) Figure out the decimal expansions of the two fractions separately (using long division), then add.
\(\frac{2}{9} = 0.2222(2)\)
\(\frac{3}{11} = 0.2727(27)\)
Thus, the sum is 0.4949… The first digit is a 4, as is the third digit, the fifth digit, and every digit after that in an odd-numbered position. (The digits in even-numbered positions are all 9’s.) Thus, the 99th digit must also be a 4.
(2) Add the fractions together, then figure out the decimal expansion.
\(\frac{2}{9} + \frac{3}{11} = \frac{22}{99} + \frac{27}{99} = \frac{49}{99}\)
Now, you can either figure out that the decimal expansion of \(\frac{49}{99}\) is 0.4949(49), or you can simply know a shortcut: any two-digit number divided by 99 becomes a decimal with that two-digit number repeating. For instance, \(\frac{17}{99} = 0.1717(17)\), \(\frac{91}{99} = 0.9191(91)\), and so on. (This pattern generalizes to any number of digits, as long as the denominator is composed of only 9’s. For instance, \(\frac{125}{999 = 0.125125(125)\)).
The final analysis is the same: every digit in an odd-numbered position is a 4, so the 99th digit after the decimal point is a 4 as well.
Answer: C
What is the 99th digit after the decimal point in the decimal expansion of \(\frac{2}{9}\) + \(\frac{3}{11}\)?
A. 1
B. 2
C. 4
D. 7
E. 9
There are two ways to approach this problem. We'll go through them in turn.
(1) Figure out the decimal expansions of the two fractions separately (using long division), then add.
\(\frac{2}{9} = 0.2222(2)\)
\(\frac{3}{11} = 0.2727(27)\)
Thus, the sum is 0.4949… The first digit is a 4, as is the third digit, the fifth digit, and every digit after that in an odd-numbered position. (The digits in even-numbered positions are all 9’s.) Thus, the 99th digit must also be a 4.
(2) Add the fractions together, then figure out the decimal expansion.
\(\frac{2}{9} + \frac{3}{11} = \frac{22}{99} + \frac{27}{99} = \frac{49}{99}\)
Now, you can either figure out that the decimal expansion of \(\frac{49}{99}\) is 0.4949(49), or you can simply know a shortcut: any two-digit number divided by 99 becomes a decimal with that two-digit number repeating. For instance, \(\frac{17}{99} = 0.1717(17)\), \(\frac{91}{99} = 0.9191(91)\), and so on. (This pattern generalizes to any number of digits, as long as the denominator is composed of only 9’s. For instance, \(\frac{125}{999 = 0.125125(125)\)).
The final analysis is the same: every digit in an odd-numbered position is a 4, so the 99th digit after the decimal point is a 4 as well.
Answer: C
06 Jun 2017, 11:36
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I think it can solved like this.
2/9 = 0.2222 (till 99 and 100th position).
3/11 = 0.2727272...
so for 98th, 99th and 100th position
for 2/9 it would be 2-2-2
for 3/11 it would be 7-2-7 (ending in 7 since there is repetition of two digits 2 and 7)
Hence the addition would give
2/9 + 3/11 = 0.22.......222 + 0.27.....727 = 0.49.......949
Hence the answer is 4.
Smash that C option
2/9 = 0.2222 (till 99 and 100th position).
3/11 = 0.2727272...
so for 98th, 99th and 100th position
for 2/9 it would be 2-2-2
for 3/11 it would be 7-2-7 (ending in 7 since there is repetition of two digits 2 and 7)
Hence the addition would give
2/9 + 3/11 = 0.22.......222 + 0.27.....727 = 0.49.......949
Hence the answer is 4.
Smash that C option
