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# If set s={7,y,12,8,,x,9} is x+y less than 18? 1.range of set

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If set s={7,y,12,8,,x,9} is x+y less than 18? 1.range of set [#permalink]

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23 Mar 2011, 23:50
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If set s={7,y,12,8,,x,9} is x+y less than 18?

1.range of set s is less than 9
2.average of x and y is less than average of set s.
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Re: explain the concept ! [#permalink]

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24 Mar 2011, 01:19
If range is less than 9, then the difference between the largest and teh smallest members of the set should be ,9

The difference of all the given numbers < 9

So even if we were to assume that the difference between x and y is less than 18

then let us try to find values of x and y

So x = 7 + 10 = 17

y can be the same as well = 7 + 10 = 17

So x + y can be > 18,

Again, x+y can be any other numbers as 1 and 2, and the range is still < 9.

so (1) is not sufficient.

From (2)

(x+y)/2 < (7 + y + 12 + 8 + x + 9)/6

=> 6(x+y) < 2(x+y + 36)

=> 4(x+y) < 2*36

=> (x+y) < 18

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Re: explain the concept ! [#permalink]

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25 Mar 2011, 05:22
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AnkitK wrote:
If set s={7,y,12,8,,x,9} is x+y less than 18?

1.range of set s is less than 9
2.average of x and y is less than average of set s.

(1) x= 5, y= 11 range (12-5) is less than 9 and x+y <18
but x= 5, y= 13 here range is less than 8 but x+y=18 so INSUFFICIENT.

(2) (X+Y)/2<(7+y+12+8+x+9)/6
X+Y<18 SUFF.
ANS B
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Re: If set s={7,y,12,8,,x,9} is x+y less than 18? 1.range of set [#permalink]

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14 Apr 2016, 01:20
In order to prove that option B is the right answer we are multiplying the inequality by the LCM (6) without knowing if x+y is positive or negative.

Is there any reason for x+y to be assumed to be positive?
Re: If set s={7,y,12,8,,x,9} is x+y less than 18? 1.range of set   [#permalink] 14 Apr 2016, 01:20
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