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Re: A quant question [#permalink]
12 Jul 2006, 03:19

1

This post received KUDOS

ishtmeet wrote:

If k, m, and t are positive integers and k/6+m/4 =t/12 , do t and 12 have a common factor greater than 1 ?

(1) k is a multiple of 3. (2) m is a multiple of 3.

(1) k is a multiple of 3, so k=3n for some positive integer n

k/6+m/4= 3n/6+m/4= (6n+3m)/12

Thus t=6n+3m=3(2n+m)

Since n and m are both +ve integers, t is a multiple of 3, and t and 12 have a common factor of 3.
SUFFICIENT

(2) m is a multiple of 3, so k=3p for some positive integer p

k/6+m/4= k/6+3p/4= (2k+9p)/12

If k=p=1, t=11 and the only factor 11 and 12 have in common is 1
If k=1 and p=2, t is a multiple of 2, so t and 12 have a common factor of 2.
NOT SUFFICIENT

do t and 12 have a common factor greater than 1 ? [#permalink]
24 Jun 2007, 14:46

Q5:
If k, m, and t are positive integers and

(k/6)+(m/4)=(t/12)

do t and 12 have a common factor greater than 1 ?
(1) k is a multiple of 3.
(2) m is a multiple of 3.

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.

You can set this up by adding the fractions with a common denominator. You get

(2k/12) + (3m/12) = t/12

so now we know that

2k + 3m = t

And we want to know if t and 12 share a factor other than 1.

1 says that k is a multiple of 3. There is a rule that when you add together numbers that are all a multiple of the same number, the sum is also a multiple of that number. So with statement 1 we know that 2k is a multiple of 3, and it's given that 3m is also a multiple of 3, so t must be a multiple of 3. Since 12 is also a multiple of 3, both t and 12 share at least 3 as a common factor.

2 says that m is a multiple of 3. Using the same rule, we don't know anything about 2k, so we really don't know what t is a multiple of at all. It could be even or odd, and it doesn't have to have any of the factors that 12 has. So it's not enough information.

You can set this up by adding the fractions with a common denominator. You get

(2k/12) + (3m/12) = t/12

so now we know that

2k + 3m = t

And we want to know if t and 12 share a factor other than 1.

1 says that k is a multiple of 3. There is a rule that when you add together numbers that are all a multiple of the same number, the sum is also a multiple of that number. So with statement 1 we know that 2k is a multiple of 3, and it's given that 3m is also a multiple of 3, so t must be a multiple of 3. Since 12 is also a multiple of 3, both t and 12 share at least 3 as a common factor.

2 says that m is a multiple of 3. Using the same rule, we don't know anything about 2k, so we really don't know what t is a multiple of at all. It could be even or odd, and it doesn't have to have any of the factors that 12 has. So it's not enough information.

Given : 2k + 3m = t

statement 1 -- k is a multiple of 3.

let k=3x

so 2*3x + 3m = t

6x + 3m = t

t = 3(m + 2x)

so, 3 is a multiple of t and hence 3 is the common factor of t and 12.

nothing can be inferred from statement 2.

Hence A. _________________

for every person who doesn't try because he is
afraid of loosing , there is another person who
keeps making mistakes and succeeds..

DS: Common multiple [#permalink]
05 Sep 2007, 07:37

If k, m, and t are positive integers and (k/6) + (m/4) = t/12, do t and 12 have a common factor greater than 1 ?
(1) k is a multiple of 3.
(2) m is a multiple of 3.

Last edited by sidbidus on 06 Sep 2007, 02:46, edited 1 time in total.

If k, m, and t are positive integers and (k/6) + (m/4) = t/12, do t and 12 have a common factor greater than 1 ?
(1) k is a multiple of 3.
(2) m is a multiple of 3.

So t = 6a+3m = 3(m+2a). So t is also a multiple of 3 and must have 3 as it's factor. 12 will also have 3 as it's factor, so they have at least one factor in common. Sufficient.

St2:
Represent m as 3b where b = any integer.

k/6 + 3b/4 = t/12
2k + 9b/12 = t/12
2k + 9b = t
If k = 2, b = 3, then t = 31 then t and 12 has no common factor
If k = 1, b = 2, t = 20 and t and 12 have 2 as common factor.
Insufficient.

So t = 6a+3m = 3(m+2a). So t is also a multiple of 3 and must have 3 as it's factor. 12 will also have 3 as it's factor, so they have at least one factor in common. Sufficient.

St2:
Represent m as 3b where b = any integer.

k/6 + 3b/4 = t/12
2k + 9b/12 = t/12
2k + 9b = t
If k = 2, b = 3, then t = 31 then t and 12 has no common factor
If k = 1, b = 2, t = 20 and t and 12 have 2 as common factor.
Insufficient.

Re: DS: Common multiple [#permalink]
06 Sep 2007, 09:11

sidbidus wrote:

If k, m, and t are positive integers and (k/6) + (m/4) = t/12, do t and 12 have a common factor greater than 1 ? (1) k is a multiple of 3. (2) m is a multiple of 3.

One more A..

first I simplify the Q.

k/6 + m/4 = t/12
2k + 3m = t ------------1

3m is a multiple of 3.

St-1 : k is multiple of 3. = > 3k is multiple of 3.
if a & b are multiples of 3, then a+b is also multiple of 3. Hence T is multiple of 3 => T and 12 have atleast 1 common factor.
Suff.

St2: M is a multiple of 3. There is no new info given here. We already know this. (Eq1)
Let k = 1 and m = 1 in eq 1,
we have t = 5 which has no common factor with 12. Hence In-suff.

since M is also positive integer , smallest value m = 1

checking for values, you can safely say that t is a number thats is greater than >= 9 , and for any number to be written as 2K+ 3m.. formfactor.. there is a value.

Re: DS: Common multiple [#permalink]
06 Sep 2007, 19:51

sidbidus wrote:

If k, m, and t are positive integers and (k/6) + (m/4) = t/12, do t and 12 have a common factor greater than 1 ? (1) k is a multiple of 3. (2) m is a multiple of 3.

S1: get the equation 2k+3m=t. So if k is a multiple of 3. 2k is a multiple of 3. 3m is also a multiple of 3. so S1 is suff.

S2: nothing new from this and we don't know if k is a multiple or not.

Ans A.

gmatclubot

Re: DS: Common multiple
[#permalink]
06 Sep 2007, 19:51