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18 Sep 2021, 10:04
Kudos
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
23% (02:51) correct
77%
(02:33)
wrong
based on 71
sessions
History
Date
Time
Result
Not Attempted Yet
18 Sep 2021, 11:57
Kudos
Bookmarks
I think the question, especially statement I, is too calculation intensive to find it’s way in actuals
Is x<1.2?
I. \(|x-\frac{13}{11}|< |x-\frac{11}{9}|\)
Basically the expression tells us that x is closer to 13/11 as compared to 11/9.
Thus, x has to be less than middle of 13/11 and 11/9.
You can also find this relation algebraically.
When x is between 13/11 and 11/9, \(x-\frac{13}{11}>0 \ \ and \ \ x-\frac{11}{9}<0\).
\(x-\frac{13}{11}< -x+\frac{11}{9}\)
\(\frac{11}{9}+\frac{13}{11}> 2x\)
\(2x<\frac{121+117}{99}………x<\frac{238}{198}\)
Now we are looking at x<1.2 or x<\(\frac{240}{200}\)
If we increase both the numerator and the denominator by same number, the original number reduces in value. Check the link in signature below on the same.
So \(\frac{238}{198}>\frac{238+2}{198+2}=\frac{240}{200}\)
Thus, x can be <1.2 or between 240/200 and 238/198.
Insufficient
II. \(|x-\frac{31}{30}|< |x-\frac{34}{25}|\)
Basically the expression tells us that x is closer to 31/30 as compared to 34/25.
Thus, x has to be less than middle of 31/30 and 34/25.
You can also find this relation algebraically.
When x is between 31/30 and 34/25, \(x-\frac{31}{30}>0 \ \ and \ \ x-\frac{34}{25}<0\).
\(x-\frac{31}{30}< -x+\frac{34}{25}\)
\(\frac{31}{30}+\frac{34}{25}> 2x\)
\(2x<\frac{155+204}{150}………x<\frac{359}{300}\)
Now we are looking at x<1.2 or x<\(\frac{360}{300}\)
Now, \(x<\frac{359}{300}<\frac{360}{300}\)
Answer is always yes.
Sufficient
B
Is x<1.2?
I. \(|x-\frac{13}{11}|< |x-\frac{11}{9}|\)
Basically the expression tells us that x is closer to 13/11 as compared to 11/9.
Thus, x has to be less than middle of 13/11 and 11/9.
You can also find this relation algebraically.
When x is between 13/11 and 11/9, \(x-\frac{13}{11}>0 \ \ and \ \ x-\frac{11}{9}<0\).
\(x-\frac{13}{11}< -x+\frac{11}{9}\)
\(\frac{11}{9}+\frac{13}{11}> 2x\)
\(2x<\frac{121+117}{99}………x<\frac{238}{198}\)
Now we are looking at x<1.2 or x<\(\frac{240}{200}\)
If we increase both the numerator and the denominator by same number, the original number reduces in value. Check the link in signature below on the same.
So \(\frac{238}{198}>\frac{238+2}{198+2}=\frac{240}{200}\)
Thus, x can be <1.2 or between 240/200 and 238/198.
Insufficient
II. \(|x-\frac{31}{30}|< |x-\frac{34}{25}|\)
Basically the expression tells us that x is closer to 31/30 as compared to 34/25.
Thus, x has to be less than middle of 31/30 and 34/25.
You can also find this relation algebraically.
When x is between 31/30 and 34/25, \(x-\frac{31}{30}>0 \ \ and \ \ x-\frac{34}{25}<0\).
\(x-\frac{31}{30}< -x+\frac{34}{25}\)
\(\frac{31}{30}+\frac{34}{25}> 2x\)
\(2x<\frac{155+204}{150}………x<\frac{359}{300}\)
Now we are looking at x<1.2 or x<\(\frac{360}{300}\)
Now, \(x<\frac{359}{300}<\frac{360}{300}\)
Answer is always yes.
Sufficient
B
20 Sep 2021, 19:24
Kudos
Bookmarks
sjuniv32
Statement 1:
|x - a| means the distance between x and a.
We may interpret this as "the distance between x and \(\frac{13}{11}\) is less than the distance between x and \(\frac{11}{9}\)."
Hence, x is closer to \(\frac{13}{11}\) than \(\frac{11}{9}\). The midpoint of these two points is \(\frac{117 + 121}{2*99} = \frac{119}{99}\) which is more than \(\frac{120}{100} = 1.2\).
Recall that x is closer to smaller value \(\frac{13}{11}\), thus x is less than something greater than 1.2. Then x is not always less than 1.2. Insufficient.
Statement 2:
Using the logic above, x is closer to \(\frac{31}{30}\) than \(\frac{34}{25}\). Their midpoint is \(\frac{186 + 170}{300} = \frac{356}{300}\), which is less than \(\frac{360}{300} = 1.2\)
Since x is closer to the smaller value, we have \(x < \frac{356}{300} < 1.2\). Sufficient.
Ans: B
