GPrep Question

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Source: — Data Sufficiency |

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by srivas » Mon Aug 24, 2009 7:02 am
n/3 remainder 2 t/5 remainder 3

statement I

n-2 divisble by 5 so we can assume n = 17

we don't know about t

staement II

t is divisible by 3 so we can take t = 18

from both statements nt/15 remainer is 6

answer is C
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by srivas » Mon Aug 24, 2009 7:03 am
n/3 remainder 2 t/5 remainder 3

statement I

n-2 divisble by 5 so we can assume n = 17

we don't know about t

staement II

t is divisible by 3 so we can take t = 18

from both statements nt/15 remainer is 6

answer is C
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by navalpike » Mon Aug 24, 2009 9:17 am
Can someone show how to solve this using the remainder equation as well?

Thanks,

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by sreak1089 » Mon Aug 24, 2009 9:52 am
Given: n = 3p + 2
t = 5q + 3

stmt # 1: n-2 is divisible by 5
=> n-2 = 5r
=> n = 5r + 2

In (n*t), we don't know yet whether there is a factor of 3 or not. Hence stmt # 1 is NOT SUFFICIENT.

stmt # 2: t is divisible by 3
=> t = 3s

In (n*t), we don't know whether there is a factor of 3 or not. Hence stmt # 2 is NOT SUFFICIENT.

Combining # 1 and # 2, we have (n*t) = 15rs + 6s

(n*t)/15 = (15rs + 6s)/15 => remainder depends on 6s/15. Hence, even after combining both stmts, its NOT SUFFICIENT.

Hence answer is E IMO.

OA Please.....

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by tarun'susername » Mon Aug 24, 2009 10:30 am
OA is [spoiler]'C'[/spoiler].

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by kaulnikhil » Mon Aug 24, 2009 12:09 pm
take numbers
for the first set
5
8
11
14
17
and second
8
13
18
23
28
pick and choose them according to conditions you get C
try out additional case of 47 * 18 for C ..you will see it will also give remainder 6
the only problem is it will take more than two minutes.. :(

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by kbharadwaj.1987 » Tue Aug 25, 2009 6:05 am
But then consider this situation:
stmnt 1:
n-2 is divisible by 3=> n=5,8,11,...
let's say n is 5.
and thus n=5.
stmnt 1 not suff because t is not mentioned.
stmnt 2:
t is divisble by 3.
and also we know that t when divided by 5 yields a remainder 3.
=> t could be 18,33,...
But it is insuff because n isn't provided.
Considering both stmnts we get
n=5, t=18, thus nt when divided by 15 would give us a remainder of 0.
and when n=5 and t=33, we would get a remainder of 0.
But when n=8, and t=18,we would get a remainder of 9.
and also when n=8, and t=33, we would get a remainder of 9.
So it's E.

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by lazyfox » Tue Aug 25, 2009 10:45 am
Posted: Tue Aug 25, 2009 9:05 am Post subject:

--------------------------------------------------------------------------------

But then consider this situation:
stmnt 1:
n-2 is divisible by 3=> n=5,8,11,...
let's say n is 5.
and thus n=5.
stmnt 1 not suff because t is not mentioned.
stmnt 2:
t is divisble by 3.
and also we know that t when divided by 5 yields a remainder 3.
=> t could be 18,33,...
But it is insuff because n isn't provided.
Considering both stmnts we get
n=5, t=18, thus nt when divided by 15 would give us a remainder of 0.
and when n=5 and t=33, we would get a remainder of 0.
But when n=8, and t=18,we would get a remainder of 9.
and also when n=8, and t=33, we would get a remainder of 9.
So it's E.
****************************************************
IMO C is the correct answer.

You cannnot pick n=5 as the condition is n-2 is divisible by 5; the possible values of n are 17, 47 etc.. and the possible values for t are 3, 18 etc... if you mutiple whichever combination of n and t we will always get remainder of 6 when nt/15.

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by kbharadwaj.1987 » Tue Aug 25, 2009 3:15 pm
oh sorry I thought the 1st stmnt to be n-2 divisible by 3.
Yes that said, C is the answer!