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by shahdevine » Fri Aug 28, 2009 10:45 am
is x greater than 6?

1) x/y>1.4
2) average of x and y is 2

OA is c...but i got e... here was my reasoning:

s1) x/y>1.4

restate: x>1.4y, insufficient. two unknowns.

s2) (x+y)/2=2 therefore x+y=4. insufficient. two unknowns.

s1+s2)

use substitute method. y=4-x so

x>1.4(4-x)
x>5.6-1.4x
2.4x>5.6
x>3.5

so x can be greater 6 but also less than 6. insufficient. what am i missing?
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by scoobydooby » Fri Aug 28, 2009 10:57 am
x/y>1.4 => x>1.4y when y is positive
=> x<1.4y when y is negative.

so when you combine you would get:
x>1.4(4-x) and x<1.4(4-x)
x>2.33 and x<2.33
only possible solution is x=2.33 less than 6

hence, C
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by shahdevine » Fri Aug 28, 2009 11:10 am
scoobydooby wrote:x/y>1.4 => x>1.4y when y is positive
=> x<1.4y when y is negative.

so when you combine you would get:
x>1.4(4-x) and x<1.4(4-x)
x>2.33 and x<2.33
only possible solution is x=2.33 less than 6

hence, C
hey scoob, thx..but i still don't get it.

if x/y>1.4 then even although we don't know the signs of x or y we know that because the division of them is a positive greater than 1.4, either they are both positive or both negative. why would you isolate y as being possibly negative? its not a square, or absolute value hence there are no hidden signs.

i appreciate...
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by scoobydooby » Fri Aug 28, 2009 11:27 am
yes, either they are both positive or both negative.

when both positive: as the denominator is positive, one can safey cross multiply w/o altering the sign and get x>1.4y

when both negative: one has to alter the sign as the denominator is negative so we would get x<1.4y when we crossmultiply over the inequality

say x/y>3
when x=16, y=4 we have 16>12
when x=-16, y=-4 we have -16<-12
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