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If a is a positive number less than 10, is c greater than the average

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If a is a positive number less than 10, is c greater than the average  [#permalink]

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New post 08 Jan 2018, 23:39
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A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

53% (02:08) correct 47% (01:49) wrong based on 60 sessions

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If a is a positive number less than 10, is c greater than the average  [#permalink]

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New post 09 Jan 2018, 08:45
Bunuel wrote:
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.


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
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Re: If a is a positive number less than 10, is c greater than the average  [#permalink]

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New post 04 Mar 2018, 07:08
Bunuel wrote:
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.


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'

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Dinesh
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Re: If a is a positive number less than 10, is c greater than the average  [#permalink]

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New post 04 Mar 2018, 08:41
Bunuel wrote:
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.



a>10, is \(c>\frac{a+10}{2}\)

(1) On the number line, c is closer to 10 than it is to a.
since c is closer to 10 than a, MEANS c is more towards 10 than the midpoint or average of a and 10
suff

(2) 2c – 10 is greater than a.
2c-10>a...
you can simplify in two ways..
a) \(2c>a+10......c>\frac{a+10}{2}\)
exactly what we are looking for
b) \(2c>a+10....2c-a>10.......c-a>10-c\)..
this again tells us that c is closer to 10 than it is closer to a (same as statement I)
suff
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Re: If a is a positive number less than 10, is c greater than the average   [#permalink] 04 Mar 2018, 08:41
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