Are x and y both positive ?
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- phanideepak
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range = maxheight - minheight
So maxheight(A) - minheight(A) = r
g-r = minheight(A)
Similarly minheight(B) = h-s
So the question is g-r > h-s ?
Statement 1 : r > s but here we know nothing about g or h So Insufficient.
Statement 2 : g > h but here we know nothing about r or s
1 and 2
g>h and r>s so g-h>r-s Sufficient so IMO answer is C
Q2.
X&Y > 0 ?
1: 2x-2y = 1. So here consider x= 2.5 and y = 2 x&y > 0 But consider x=-2 and y = -2.5 here X&Y<0
Insuff
2: x/y > 1. In this case either x&y > 0 or X&Y<0
1 and 2
x-y = 0.5 and x/y > 1 from these two we can rule out the case X&Y<0 since if x<0 then y < 0 but x-y is positive so y < x but this is not possible as both are -ve and x/y > 1
for example consider x = -2 to get 0.5 we should have y = -2.5 but then we have x/y > 1 and (-2/-2.5) < 1 so x&y < 0 is ruled out
and we are left with x&y>0
So IMO answer is
C
So maxheight(A) - minheight(A) = r
g-r = minheight(A)
Similarly minheight(B) = h-s
So the question is g-r > h-s ?
Statement 1 : r > s but here we know nothing about g or h So Insufficient.
Statement 2 : g > h but here we know nothing about r or s
1 and 2
g>h and r>s so g-h>r-s Sufficient so IMO answer is C
Q2.
X&Y > 0 ?
1: 2x-2y = 1. So here consider x= 2.5 and y = 2 x&y > 0 But consider x=-2 and y = -2.5 here X&Y<0
Insuff
2: x/y > 1. In this case either x&y > 0 or X&Y<0
1 and 2
x-y = 0.5 and x/y > 1 from these two we can rule out the case X&Y<0 since if x<0 then y < 0 but x-y is positive so y < x but this is not possible as both are -ve and x/y > 1
for example consider x = -2 to get 0.5 we should have y = -2.5 but then we have x/y > 1 and (-2/-2.5) < 1 so x&y < 0 is ruled out
and we are left with x&y>0
So IMO answer is
C
- Mom4MBA
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For class A let 'a' be minimum height, so r=g-a ; range = max - min
For class B let 'b' be minimum height, so s=h-b
statement 1:
r < s
g-a<h-b
but we do not know anything about g and h so insufficient
statement 2:
g > h
r+a > s+b
but we do not know anything about r and s so insufficient
taking both the statements together we get
r < s (from 1)
g-a < h-b
given g>h (from 2)
so we get a > b
We need both 1 and 2
For class B let 'b' be minimum height, so s=h-b
statement 1:
r < s
g-a<h-b
but we do not know anything about g and h so insufficient
statement 2:
g > h
r+a > s+b
but we do not know anything about r and s so insufficient
taking both the statements together we get
r < s (from 1)
g-a < h-b
given g>h (from 2)
so we get a > b
We need both 1 and 2
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