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

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Math Expert
Joined: 02 Sep 2009
Posts: 56275
If a is a positive number less than 10, is c greater than the average  [#permalink]

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08 Jan 2018, 23:39
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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 (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.

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

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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
Intern
Joined: 27 Apr 2015
Posts: 40
GMAT 1: 370 Q29 V13
Re: If a is a positive number less than 10, is c greater than the average  [#permalink]

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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'

Regards
Dinesh
Math Expert
Joined: 02 Aug 2009
Posts: 7764
Re: If a is a positive number less than 10, is c greater than the average  [#permalink]

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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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