Set-2 Quant

[[email protected]]

If positive integer x is a multiple of 6 and positive integer y is a multiple of 14, is xy a multiple of 105?
(1) x is a multiple of 9.
(2) y is a multiple of 25.
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone 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.
----------------------------------------------------------------------------------------------------------------------

[samirpandeyit62]

x is a mulitple of 6 i.e its factors include prime factors of 6 i.e 2,3

y is mulitple of 14 so its factors include 7,2 (14=7X2)

now xy factors will therefore include 3 & 7 ( 3 PF of 6, 7 PF of 7)

Now 105 has prime factors (7,3,5) (105 =7X3X5)

so in order for xy to be a multiple of 105 , it must have 5 as a PF as it already has 3 & 7

now stm1 says x is divisible by 9 so 9=3X3 ,so this does not add 5 to the list of PF's of x

but still mulitples of 6 & 9 such as 6X5X9 & of 14 like 14X5 etc will add 5 as PF of
x or y therby xy

hence IN SUFF

stmt 2 says that y is a multiple of 25 ie.e 5X5 hence cf 5 is added to the prime factors of y, thereby xy hence with stmt we can say that now xy will have PF's 3,7,5 whose multiple is 105 ,hence xy will be divisible by 105

SUFF

ans should be B

pls check if this is correct

[krishnamurthyu]

Q-type: Y/N ;

Given:
x = 6a
y =14b
105 = 3 * 5 * 7

is xy/105 = (1,2,.....n)

xy = 6 * 14 * a * b
= 2 * 3 * 2 * 7 * ab
=4 * 3 * 7 * ab

1. x = 6a = 9c
= 3 * 2 ( 3 * c ) ; (where c =1 ...n )
c = 1; xy = 4 * 3 * 7 * ab * 3 = Not Divisible by 3*5*7 : N
c=5; xy = 4 * 3 * 7 * ab * 5 = Divisible by 3*5*7 Y

Insufficient;

2. y is multiple of 25 i.e y = 14b = 14 * 5 * 5
xy = * 3 * 7 * ab * 5 * 5 Divisible by 3*5*7

Hence Ans. B