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If a, b, and c are positive, is a>(b+c)/2

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If a, b, and c are positive, is a>(b+c)/2  [#permalink]

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14 Dec 2014, 02:22
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60% (01:55) correct 40% (01:47) wrong based on 121 sessions

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If a, b, and c are positive, is a>(b+c)/2
(1) On the number line, a is closer to b than it is to c.
(2) b > c

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Re: If a, b, and c are positive, is a>(b+c)/2  [#permalink]

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14 Dec 2014, 04:54
1. a is closer to b than c. so in this case assume b=2 c=3 a=1 then we have b+c/2=2.5 > a(=1)
also , say b=4 a=5 c=3 then we have b+c/2=3.5 < a(=5)
Insufficient.

2. b>c insufficient as there is no relationship mentioned between a ,b and c.

Together 1 &2 , since a,b,c>0 and b>c and a is closer to b than c , we have a midpoint between b and c which is always less than a . i.e
b+c/2<a since a is much closer to b than c.

Eg: assume c =0.1 ,b=0.2 and a =0.06 (closer to b) or 0.3 which is > b+c/2 =0.05

Sufficient.

So Ans is C
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Re: If a, b, and c are positive, is a>(b+c)/2  [#permalink]

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10 Oct 2017, 06:36
so the question is asking: 2a> b+c?

based on statement 1:

possible scenarios:

---a--b-c (scenario 1)
or
--c---a--b (scenario 2)

if (b and c lies on the right side of a) then 2a> b+c
if (b lies on right side of a, while c on left side of a) then 2a< b+c

thus statement 1 INSUFF

statement 2:

b> c , INSUFF
we don't know the placement of a

Thus, combined statement 1&2, we know scenario (2) works and a is between c and b. Answer C
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Re: If a, b, and c are positive, is a>(b+c)/2  [#permalink]

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11 Oct 2017, 10:24
If a, b, and c are positive, is $$a>\frac{(b+c)}{2}$$

(1) On the number line, a is closer to b than it is to c.
let a=1 , b=2 and c=10 => a is closer to b but average of b and c = $$\frac{(b+c)}{2}$$ will be > from a

Let a=10, = and c =1 => => a is closer to b, average of b and c = $$\frac{(b+c)}{2}$$ will be < from a

Two conditions
Insufficient

(2) b > c
Clearly insufficient as we don't know anything about a

(1)+(2)
On the number line, a is closer to b than it is to c. and b >c

=> Here two situations possible
1st : a > b => average of b and c = $$\frac{(b+c)}{2}$$ will always be < from a
2nd: a < b ... in this case for a to be closer to b than it is to c ... a need to be greater than average of b and c => $$\frac{(b+c)}{2}$$ < a
both equations give same and definite answer
Sufficient

Re: If a, b, and c are positive, is a>(b+c)/2 &nbs [#permalink] 11 Oct 2017, 10:24
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