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14 Nov 2022, 20:14
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Bunuel
Bunuel
To clarify, when the question asks for "is z between x and y?" does that translate to just finding out is x<z<y OR also y<z<x? Essentially, I was confused as to which order x and y needed to be. Thank you for your help.
14 Nov 2022, 22:11
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woohoo921
It could be both but from each statement we get that x < y, so the question becomes is x < z < y.
26 Oct 2023, 10:41
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Visualizing it on a number line helped me the most. Don't try to come up with numbers (though in this case, it can be used) in such question but rely on intuition and number line.
x,y,z are +ve int.
Asked whether x<z<y .
Since all are +ve int. it can inferred ------0-------x-------z-------y .
S1: x<2z<y. i.e x/2<z<y/2
Now putting x and y in this case.
Case1 : x/2-------z------x---------y/2------y . Here x/2<z<y/2. But this case doesn't fall in case question asks us.
Case2 : x/2-------x------z---------y/2------y . This gives YES to our asked condition.
Case 1 and 2 both gives different answers. S1 insufficient.
S2: 2x<z<2y. Now putting this on number line with x and y.
Case1: x--------2x------z-------y--------2y. This satisfies our asked condition.
Case2: x------y------2x------z------2y. This doesn't satisfies our condition.
Case 1 and 2 both gives different answers. S2 insufficient.
Combining S1 and S2. lets just add eqn of S1 and S2.
x<2z<y ----- S1
2x<z<y ------S2
It gives 3x<3z<3y , dividing by 3 gives our desired eqn i.e x<z<y.
Thus C is answer.
x,y,z are +ve int.
Asked whether x<z<y .
Since all are +ve int. it can inferred ------0-------x-------z-------y .
S1: x<2z<y. i.e x/2<z<y/2
Now putting x and y in this case.
Case1 : x/2-------z------x---------y/2------y . Here x/2<z<y/2. But this case doesn't fall in case question asks us.
Case2 : x/2-------x------z---------y/2------y . This gives YES to our asked condition.
Case 1 and 2 both gives different answers. S1 insufficient.
S2: 2x<z<2y. Now putting this on number line with x and y.
Case1: x--------2x------z-------y--------2y. This satisfies our asked condition.
Case2: x------y------2x------z------2y. This doesn't satisfies our condition.
Case 1 and 2 both gives different answers. S2 insufficient.
Combining S1 and S2. lets just add eqn of S1 and S2.
x<2z<y ----- S1
2x<z<y ------S2
It gives 3x<3z<3y , dividing by 3 gives our desired eqn i.e x<z<y.
Thus C is answer.
