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\(\frac{2}{9} = 0.222222222...........2\)

\(\frac{3}{11}= 0.272727.............. 27\)

Addition (Does not have carries)

\(\frac{2}{9} + \frac{3}{11} = 0.4949494949....... 49\)

Odd's have 4

99th would have 4

Answer = C
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\(\frac{2}{9}\) = 0.22222.... ==> 99th digit is 2

\(\frac{3}{11}\) = 0.27272727.... ==> Every odd digit is 2. So, 99th digit will be 2.

2 +2 = 4 ( Ans C)
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\(\frac{2}{9}\) + \(\frac{3}{11}\) = \(\frac{49}{99}\)

\(\frac{49}{99}\) = 0.49494949..............

so the digits are getting repeated in order of two digits 49

so the 99 th digit is 4

correct answer - C
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What is the 99th digit after the decimal point in the decimal expansion of 2/9 + 3/11?

A. 1
B. 2
C. 4
D. 7
E. 9

\(\frac{2}{9} + \frac{3}{11}=\frac{22}{99} + \frac{27}{99}=\frac{49}{99}=0.494949...\). So, 99th (odd placed after the decimal) digit will be 4.

Answer: C.

For more on converting a recurring decimal to fraction check this: https://gmatclub.com/forum/math-number-theory-88376.html

Hope it helps.


HI what do you mean by 99th digit could you explain, I understand the theory of repeating no.'s but what do you mean by 99th Digit ?
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What is the 99th digit after the decimal point in the decimal expansion of 2/9 + 3/11?

A. 1
B. 2
C. 4
D. 7
E. 9

\(\frac{2}{9} + \frac{3}{11}=\frac{22}{99} + \frac{27}{99}=\frac{49}{99}=0.494949...\). So, 99th (odd placed after the decimal) digit will be 4.

Answer: C.

For more on converting a recurring decimal to fraction check this: https://gmatclub.com/forum/math-number-theory-88376.html

Hope it helps.


HI what do you mean by 99th digit could you explain, I understand the theory of repeating no.'s but what do you mean by 99th Digit ?

We need 99th digit after the decimal point of 0.494949...

The first digit is 4, the second digit is 9, the third digit is 4 again, the fourth digit is 9 again, ... 99th (odd placed after the decimal) digit will be 4.
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What is the 99th digit after the decimal point in the decimal expansion of 2/9 + 3/11?

A. 1
B. 2
C. 4
D. 7
E. 9


2/9 + 3/11 = 22/99 + 27/99 = 49/99 = 0.494949….

SInce each odd-positioned digit to the right of the decimal point has a value of 4, the 99th digit is 4.

Alternate Solution:

We see that 2/9 = 0.222… and 3/11 = 0.272727…

If we add the two decimal expressions, we see that we obtain 0.494949… Since all the odd-positioned digit to the right of the decimal point has a value of of 4, the 99th digit is 4.

Answer: C
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\(\frac{2}{9} + \frac{3}{11}=\frac{22}{99} + \frac{27}{99}=\frac{49}{99}=0.494949...\). So, 99th (odd placed after the decimal) digit will be 4.

Answer: C.

For more on converting a recurring decimal to fraction check this: https://gmatclub.com/forum/math-number- ... 88376.html

Hope it helps.
I was able to get to 49/99, but do you just need to memorize that anything divided by 99 is itself repeated? And why is the 9 the odd place? It looks to me that the first (odd place) after after the decimal is the 4 and the 9 would be the 2nd place so even? Edit: realize that 9 is the even place and 4 is the odd
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This is an aeasy sum
2/9+3/11=49/99

Now break 99 into int's prime factorization 99=3*3*11

According to the cyclicity rule of numbers, the number with the highest cyclicity or higher in value will determine the last digit, here 11 and has a cyclicity 2.

49/11= 4.454545... goes on. Here, 98th term will 5, then plus 1= 99th term will be 4.

Answer is C
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Hi analystpgh,

Good news on both counts. First, your edit is spot on: the block is 49, so the 1st digit is 4, the 2nd is 9, the 3rd is 4 again, and so on. Odd positions are 4, even positions are 9, and since 99 is odd, the 99th digit is 4. Nothing more to fix there.

Now the real question: you don't need to memorize the /99 rule - you can derive it in seconds, which is safer than trusting memory.

Why x/99 repeats the block

Take your result and let d = 0.494949.... The block is two digits long, so multiply by 100 to shift it one full block:

- 100d = 49.494949...
- d = 0.494949...
- Subtract: 99d = 49, so d = 49/99.

Run that backward and you see why49/99 = 0.494949...: dividing by 99 is exactly what reproduces the two-digit block. The 100 you multiplied by is what forces a two-digit repeat - and 100 pairs with the 99.

The pattern behind it

The number of 9s in the denominator tells you the block length:

- 7/9 = 0.7777... - one 9, so a 1-digit block.
- 7/99 = 0.070707... - two 9s, so a 2-digit block (write 7 as "07").
- 7/999 = 0.007007... - three 9s, so a 3-digit block.

So 49/99 gives the 2-digit block 49 repeating - no memorizing required, just the shift-and-subtract idea. Once you trust that, the only remaining job is matching the position (odd/even) to the right digit, which you already nailed.

Answer: C

analystpgh

I was able to get to 49/99, but do you just need to memorize that anything divided by 99 is itself repeated? And why is the 9 the odd place? It looks to me that the first (odd place) after after the decimal is the 4 and the 9 would be the 2nd place so even? Edit: realize that 9 is the even place and 4 is the odd
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