- prachi18oct
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If m and q are divisors of c, is mq a divisor of c that is not c or 1?
(1) mq < c
(2) m and q have no common divisors.
A)Statement (1) ALONE is sufficient, but statement (2) is not sufficient.
B)Statement (2) ALONE is sufficient, but statement (1) is not sufficient.
C)BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D)EACH statement ALONE is sufficient.
E)Statements (1) and (2) TOGETHER are NOT sufficient.
Lets take numbers c = 12, m , q can be from (1,2,3,4,6,12)
1) mq < c => m,q cannot be 2,6/3,4/4,6/6,12
m q can be 1,2;1,3;1,4;1,6;2,3 or 2,4
From all the possible combinations,
if m,q = 2,3 then it is divisor of 12
if m,q = 2,4 then it is not divisor of 12
SO INSUFFICIENT
2)m , q no common divisors
so we can eliminate some sets from previous selection.
m,q can be 3,4; 1,2; 1,3 ;4,1; 6,1; 2,3
All of them have mq as common divisor of c but if m,q = 3,4 then mq = 12 = c. SO NO
INSUFFICIENT.
Together,
we have only 1,2; 1,3 ;4,1; 6,1; 2,3 . ALl of these combinations satisfy mq < c and mq divisor of c.
HENCE SUFFICIENT.
Please let me know if I missed something or assumed anything.
(1) mq < c
(2) m and q have no common divisors.
A)Statement (1) ALONE is sufficient, but statement (2) is not sufficient.
B)Statement (2) ALONE is sufficient, but statement (1) is not sufficient.
C)BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D)EACH statement ALONE is sufficient.
E)Statements (1) and (2) TOGETHER are NOT sufficient.
Lets take numbers c = 12, m , q can be from (1,2,3,4,6,12)
1) mq < c => m,q cannot be 2,6/3,4/4,6/6,12
m q can be 1,2;1,3;1,4;1,6;2,3 or 2,4
From all the possible combinations,
if m,q = 2,3 then it is divisor of 12
if m,q = 2,4 then it is not divisor of 12
SO INSUFFICIENT
2)m , q no common divisors
so we can eliminate some sets from previous selection.
m,q can be 3,4; 1,2; 1,3 ;4,1; 6,1; 2,3
All of them have mq as common divisor of c but if m,q = 3,4 then mq = 12 = c. SO NO
INSUFFICIENT.
Together,
we have only 1,2; 1,3 ;4,1; 6,1; 2,3 . ALl of these combinations satisfy mq < c and mq divisor of c.
HENCE SUFFICIENT.
Please let me know if I missed something or assumed anything.













