x, y, z are numbers on the number line. Is x < y < z ?
1). |y-x| + |y-z| = |z-x|
2). x<z
OA-E
Number line
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x, y, z are numbers on the number line. Is x < y < z ?
1). |y-x| + |y-z| = |z-x|
i.e distance between y & x + dis bet y & z = dis bet z & x
This has possibilties i.e xyz or zxy(y should be on the way from x to z or vice versa) NOT SUFF
2). x<z
this states that x < z but x may be greater than or less than y
INSUFF
combine we have
xyz or zxy & x<z
hence x<y<z
Ans should be C
Arocks can you confirm that the ans is E.
i.e xyz is the combination so x<y<z
1). |y-x| + |y-z| = |z-x|
i.e distance between y & x + dis bet y & z = dis bet z & x
This has possibilties i.e xyz or zxy(y should be on the way from x to z or vice versa) NOT SUFF
2). x<z
this states that x < z but x may be greater than or less than y
INSUFF
combine we have
xyz or zxy & x<z
hence x<y<z
Ans should be C
Arocks can you confirm that the ans is E.
i.e xyz is the combination so x<y<z
Regards
Samir
Samir