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multipe of 15

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multipe of 15

by ST » Fri May 15, 2009 5:56 am
is the integer n a multip eof 15?

1) n is a multiple of 20
2) n+6 is a multiple of 3


Answer is c
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Source: — Data Sufficiency |
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by ManSab » Fri May 15, 2009 9:18 am
1) n is a multiple of 20 => 40,60,80 , only number whose mutlitple is 15 is 60........... Insufficient
2) n+6 is a multiple of 3 ==> 15 + 6=21 -yes,, 21+6 = 27.. - No insufficient

1 and 2 together is also insufficient
E..

Please correct if I am wrong

Thanks
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by scoobydooby » Fri May 15, 2009 9:43 am
1 and 2 are insufficient (mansab)

together, n=20k and also n=3t-6
equating, 20k=3t-6
k=(3t-6)/20 must be an integer

for t=22, k=3=>n=60 multiple of 15
for t=42, k=6=>n=120 multiple of 15
for t=62, k=9=>n=180 multiple of 15
for t=82,102, 122,......n is a multiple of 15

hence C
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by aj5105 » Fri May 15, 2009 7:30 pm
Together, consider 30+6, 60+6, 90+6. Sufficient.

(C)
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by francopiccolo » Fri May 15, 2009 10:15 pm
For I, n is a multiple of 5, for II, n is a multiple of 3, hence n is a multiple of 15.

(C)
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by success1111 » Sun May 17, 2009 7:55 am
aj5105 wrote:Together, consider 30+6, 60+6, 90+6. Sufficient.

(C)
Your explanation is wrong. Together you will consider 60 +6,120+6,180+6.
Not 30, 90 as you stated because 30 and 90 violated the statement one' principle of multiple of 20.
Trust but verify.
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by gmt123 » Sun May 17, 2009 8:31 am
is the integer n a multip eof 15?

1) n is a multiple of 20
2) n+6 is a multiple of 3


Stmt 1 -
n is multiple of 20 -> so n is also multiple of 5 and 4

Stmt 2 -
n+6 is a multilpe of 3 -> so n is also multiple of 3

combine stmt1 and stmt2 - n is multiple of 5 and 3 , hence multiple of 15.

Ans C
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