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

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If a is a positive number less than 10, is c greater than [#permalink]  02 Dec 2010, 02:27
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55% (hard)

Question Stats:

60% (02:05) correct 40% (01:04) wrong based on 105 sessions
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.
[Reveal] Spoiler: OA

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Re: DS Algebra [#permalink]  02 Dec 2010, 02:43
Expert's post
rxs0005 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: $$0<a<10$$. Question: is $$c$$ greater than the average (arithmetic mean) of $$a$$ and 10? --> or is $$c>\frac{a+10}{2}=average$$? --> or is $$2c>a+10$$?

(1) On the number line, c is closer to 10 than it is to a.

Number line approach:
a-----average-----10----- (average of a and 10 is halfway between a and 10). So the question ask whether c is either in the BLUE or GREEN area.

As, c is closer to 10 than it (c) is to a then this statement directly tells us that c is either in the BLUE or GREEN area. Sufficient.

Algebraic approach:
c is closer to 10 than it is to a, means that the distance between c and 10 is less than the distance between c and a. So, $$|10-c|<|c-a|$$. Now, as c is closer to 10 than it is to a, then c>a, so $$|c-a|=c-a$$ --> two cases for 10-z:

A. $$c\leq{10}$$ --> $$|10-c|=10-c$$ --> $$|10-c|<|c-a|$$ becomes: $$10-c<c-a$$ --> $$2c>10+a$$. Answer to the question YES.

B. $$c>{10}$$ --> in this case $$2c>20$$ and as $$a<10$$, then $$a+10<20$$, hence $$2c>10+a$$. Answer to the question YES.

(2) 2c – 10 is greater than a --> $$2c-10>a$$ --> $$c>\frac{a+10}{2}=average$$, again directly tells us that c is greater than the average of a and 10. Sufficient.

Similar questions:
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x-is-a-positive-number-less-than-86563.html
600-level-question-95138.html?hilit=halfway%20between%20greater%20closer

Hope it helps.
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Re: DS Algebra [#permalink]  05 Dec 2010, 06:08
q is c > (a+10)/2
from statement 1 c is closer to 10 than a
let us take a=9.4 c= 9.5
in this case c < (a+10)/2
if a=7 c=9 then c > (a+10)/2

so statement 1 not sufficient
can some one explain is there any wrong in this
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Re: DS Algebra [#permalink]  05 Dec 2010, 06:23
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Expert's post
anilnandyala wrote:
q is c > (a+10)/2
from statement 1 c is closer to 10 than a
let us take a=9.4 c= 9.5
in this case c < (a+10)/2
if a=7 c=9 then c > (a+10)/2

so statement 1 not sufficient
can some one explain is there any wrong in this

Statement (1) says: on the number line, c is closer to 10 than it is to a --> means that the distance between c and 10 is less than the distance between c and a.

Now, your example $$a=9.4$$ and $$c=9.5$$ is not valid as in this case $$c$$ is obviously closer to $$a$$ than to 10 (c-a=0.1 and 10-c=0.6).

There are 2 different approaches in my previous post shoving why is this statement sufficient.

Hope it's clear.
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Re: If a is a positive number less than 10, is c greater than [#permalink]  23 Jan 2014, 23:28
I did it with numbers.

From (1) I know that C is closer to 10 than to a. Pluggin numbers gives me e.g. c= 9, a = 7 average = 8,5 so true....continuing I figured that since c is ALWAYS closer (even if you take 9.99995) to 10, it will be always greater than the average. SUFF.

(2) 2c -10 > a --> c > 10 +a --> c > (10+a)/2 which is the average. SUFF.
Also: Since a is < 10, C is at least 10. which would give us the same answer.
Re: If a is a positive number less than 10, is c greater than   [#permalink] 23 Jan 2014, 23:28
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