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555-605 (Medium)|   Fractions and Ratios|   Inequalities|                           
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woohoo921
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If a and b are positive integers, is a/b < 9/11 ? 27 May 2022, 19:07
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Bunuel
If a and b are positive integers, is a/b < 9/11 ?

(1) a/b < 0.818
(2) b/a > 1.223
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avigutman

Do you discuss this problem in video form somewhere on your Youtube channel? Thank you!
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If a and b are positive integers, is a/b < 9/11 ? 30 May 2022, 15:26
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woohoo921

avigutman
Do you discuss this problem in video form somewhere on your Youtube channel?
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Video solution from Quant Reasoning published this morning, woohoo921 :)
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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If a and b are positive integers, is a/b < 9/11 ? 07 Dec 2022, 09:24
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Hadrienlbb
Same question as people above. This question is really easy with a calculator. A little less so without it, under stress and under 2 minutes.

Anyone willing to break down the mental math on this one? Because clearly approximation wither an option here.

Thanks,

Posted from my mobile device
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I think this may help:
with a simplified fraction, if its denominator contains factors other than 2 & 5 => It is a repeating fraction
You can transform a repeating fraction => ratio by converting its denominator into 9, 99, 999, or other numbers of "9" that are equivalent to the number of digits in the Numerator. Now the value in the Numerator is the repeating part.
You will see what I'm trying to say by looking at the example below:
9/11 is clearly a repeating fraction. 9/11 = 81/99 = 0.81818181...
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If a and b are positive integers, is a/b < 9/11 ? 07 Dec 2022, 12:09
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Bunuel
If a and b are positive integers, is a/b < 9/11 ?

(1) a/b < 0.818
(2) b/a > 1.223
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To compare two fractions:
Multiply the numerator in each fraction by the denominator in the other.
The numerator that yields the bigger product belongs to the bigger fraction.

Alternate solution:

Is a/b < 9/11?

Statement 1: a/b < 818/1000
Comparing \(\frac{818}{1000}\) and \(\frac{9}{11}\), we get:
818*11 = 818(10+1) = 8180 + 818 = 8998
9*1000 = 9000

Since the numerator for 9/11 yields the greater product:
818/1000 < 9/11

Since a/b < 818/1000 < 9/11, we get:
a/b < 9/11
Thus, the answer to the question stem is YES.
SUFFICIENT.

b/a > 1.223 --> \(\frac{b}{a}\) > \(\frac{1223}{1000}\) -->\(1000b > 1223a\) --> \(\frac{1000}{1223}\) > \(\frac{a}{b}\) --> \(\frac{a}{b}\)< \(\frac{1000}{1223}\)

Statement 2: a/b < 1000/1223
Comparing \(\frac{1000}{1223}\) and \(\frac{9}{11}\), we get:
1000*11 = 11000
9*1223 = (10-1)1223 = 12230-1223 = 11007

Since the numerator for 9/11 yields the greater product:
1000/1223 < 9/11

Since a/b < 1000/1223 < 9/11, we get:
a/b < 9/11
Thus, the answer to the question stem is YES.
SUFFICIENT.

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D
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If a and b are positive integers, is a/b < 9/11 ? 19 Aug 2023, 04:55
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Bunuel, can we solve in this way?

a/b < 9/11, cross multiply them, it becomes 11a < 9b, if we can find out between a and b which variable is bigger, we can find the answer, right?
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If a and b are positive integers, is a/b < 9/11 ? 19 Aug 2023, 05:47
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AnkurGMAT20
Bunuel, can we solve in this way?

a/b < 9/11, cross multiply them, it becomes 11a < 9b, if we can find out between a and b which variable is bigger, we can find the answer, right?
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a can be less that b but we still can have a/b > 9/11 as well as a/b < 9/11. For example:

(a = 10) < (b = 11) but a/b > 9/11

(a = 8) < (b = 11) but a/b < 9/11
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If a and b are positive integers, is a/b < 9/11 ? 29 Aug 2024, 09:14
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Salsanousi

Bunuel
If a and b are positive integers, is a/b < 9/11 ?

(1) a/b < 0.818
(2) b/a > 1.223

Kudos for a correct solution.
1/10 = 0.10
1/11 = 0.0909090 and so it goes.

so 9/11 = 0.818181 etc.. now the question asks whether a/b is < 0.818181...

From statement 1) we could tell that a/b < 0.818 so a/b would then be less than 9/11. We have a clear definitive answer.

For statement 2)

b/a > 1.223

1/9 = 0.1111

so having 10/9 = 1.11111 and 11/9 would be 1.22222

Now since b/a > 1.2223 then we know that b/a is larger than 11/9. Then we could tell that a/b is < 9/11

Hence sufficient.

Both statements are sufficient, thus, the answer is D.
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­Hey I understand the solution. I actually selected the answer B since they rounded that off but they didnt indicate that the first one was non-terminating cause I was under the impression that the GMAT will add the ... for non terminating decimal. like for the 0.8181.... . Cause they rounded off the above answer to 0.222.... to 0.223. Do let me know if GMAT provides this segregration ?
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If a and b are positive integers, is a/b < 9/11 ? 29 Aug 2024, 10:18
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s3sanghvi

Salsanousi

Bunuel
If a and b are positive integers, is a/b < 9/11 ?

(1) a/b < 0.818
(2) b/a > 1.223

Kudos for a correct solution.
1/10 = 0.10
1/11 = 0.0909090 and so it goes.

so 9/11 = 0.818181 etc.. now the question asks whether a/b is < 0.818181...

From statement 1) we could tell that a/b < 0.818 so a/b would then be less than 9/11. We have a clear definitive answer.

For statement 2)

b/a > 1.223

1/9 = 0.1111

so having 10/9 = 1.11111 and 11/9 would be 1.22222

Now since b/a > 1.2223 then we know that b/a is larger than 11/9. Then we could tell that a/b is < 9/11

Hence sufficient.

Both statements are sufficient, thus, the answer is D.
­Hey I understand the solution. I actually selected the answer B since they rounded that off but they didnt indicate that the first one was non-terminating cause I was under the impression that the GMAT will add the ... for non terminating decimal. like for the 0.8181.... . Cause they rounded off the above answer to 0.222.... to 0.223. Do let me know if GMAT provides this segregration ?
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­
There are terminating decimals 0.818 and 0.223. Please review the discussion again.
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If a and b are positive integers, is a/b < 9/11 ? 06 Dec 2025, 07:49
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