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08 Jan 2018, 23:39
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
50% (02:04) correct
50%
(01:47)
wrong
based on 98
sessions
History
Date
Time
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Not Attempted Yet
If a is a positive number less than 10, is c greater than the average (arithmetic mean) of a and 10?
(1) On the number line, c is closer to 10 than it is to a.
(2) 2c – 10 is greater than a.
(1) On the number line, c is closer to 10 than it is to a.
(2) 2c – 10 is greater than a.
09 Jan 2018, 08:45
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Bunuel
Given \(a<10\) and to find is \(c>\)mid point of \(a\) & \(10\) i.e. \(c>\frac{a+10}{2}=>2c>a+10\)
Statement 1: This implies \(c\) is more than the mid-point of \(a\) & \(10\) because mid-point will be equidistant from both \(a\) & \(10\) and given \(c\) is closer to \(10\) than \(a\). Sufficient
Statement 2: implies \(2c-10>a=>2c>a+10\). Sufficient
Option D
04 Mar 2018, 07:08
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Bunuel
Given \(0<a<10\)
To find :- is \(c>\frac{(a+10)}{2}\)
Stat 1 On the number line, c is closer to 10 than it is to a. Thus following 2 cases
Case 1 \(c>10\), so 'c' will ALWAYS be closer to 10 than 'a' on the number line
=> Since \(a<10\). Therefore avg of (a, 10) i.e \(\frac{(a+10)}{2}<10\)
=> thus \(c>10>\frac{(a+10)}{2}\)
=> \(c>\frac{(a+10)}{2}\)
Case 2 \(c<10\)
=> Since 'c' NEEDS to be closer to 10 than to 'a' it has to be > than the midpoint .i.e (avg) of (a, 10) on the number line.(If 'c' is < than the midpoint of (a, 10) it will be closer to 'a' than 10)
=> Therefore \(\frac{(a+10)}{2}<c<10\)
=> OR \(c>\frac{(a+10)}{2}\)
Thus SUFFICIENT in both cases
Stat 2 2c – 10 is greater than a
=> OR (2c – 10)>a
=> OR 2c>10+a
=> OR \(c>\frac{(a+10)}{2}\)
SUFFICIENT
Option 'D'
Regards
Dinesh
