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.
if i rearrange the equation it works out to be 2k+3m=t
now stmt (1)
k is a multiple of 3, so that means that 2k has to be an EVEN multiple of 3 like 6,12,24
how is this statement sufficient when 2k could mean any of the above?
can someone also please explain: t and 12 have a common factor greater than 1 ?
Is this basically asking whether T is a multiple of 12 greater than 12????
prime & divisibility
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This is how I would interpret this - factors of 12 greater than 1 are 2, 3, 4, 6, 12.Gurpinder 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.
if i rearrange the equation it works out to be 2k+3m=t
now stmt (1)
k is a multiple of 3, so that means that 2k has to be an EVEN multiple of 3 like 6,12,24
how is this statement sufficient when 2k could mean any of the above?
can someone also please explain: t and 12 have a common factor greater than 1 ?
Is this basically asking whether T is a multiple of 12 greater than 12????
In order for t to have *a* common factor with 12, t should at least have one of these (2, 3, 4, 6, 12).. Only statement 1 proves it has 3, which is one of the common factor with 12, hence sufficient.
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Alright now I get how A is sufficient. How is B not sufficient?
m is a multiple of 3. so 3m means that the number is a multiple of 9. so lets just take 9 for example, its factors are 3 and this is common with 12.
in fact any multiple of 9 will always be divisible by 3, so why isnt statement 2 sufficient.
m is a multiple of 3. so 3m means that the number is a multiple of 9. so lets just take 9 for example, its factors are 3 and this is common with 12.
in fact any multiple of 9 will always be divisible by 3, so why isnt statement 2 sufficient.
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This is a number properties question.
As noted above the equation simplifies to 2k+3m=t
One important number properties rule is;
A multiple of n plus a multiple of n will sum to another multiple of n. Algebraic example, 3x+3y = 3(x+y)
1) k is a multiple of 3.
3m is obviously a multiple of 3 and if k is also a multiple of 3, then t is a multiple of 3, because of the rule stated above.
Thus, A is sufficient. 3 is a common factor.
2) Tell us that m is a multiple of 3 which does not tell us anything new in regards to t because the only way we can determine any other factor of t is if 2k and 3m share a factor, which this answer does not tell us.
2 is insufficient.
A is the answer.
Please confirm OA.
Thanks,
Jared
As noted above the equation simplifies to 2k+3m=t
One important number properties rule is;
A multiple of n plus a multiple of n will sum to another multiple of n. Algebraic example, 3x+3y = 3(x+y)
1) k is a multiple of 3.
3m is obviously a multiple of 3 and if k is also a multiple of 3, then t is a multiple of 3, because of the rule stated above.
Thus, A is sufficient. 3 is a common factor.
2) Tell us that m is a multiple of 3 which does not tell us anything new in regards to t because the only way we can determine any other factor of t is if 2k and 3m share a factor, which this answer does not tell us.
2 is insufficient.
A is the answer.
Please confirm OA.
Thanks,
Jared
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The equation in question is 2k + 3m = t, in your example, we should be able to factor out 3 from the sum (2k+3m), not just 3m.Gurpinder wrote:Alright now I get how A is sufficient. How is B not sufficient?
m is a multiple of 3. so 3m means that the number is a multiple of 9. so lets just take 9 for example, its factors are 3 and this is common with 12.
in fact any multiple of 9 will always be divisible by 3, so why isnt statement 2 sufficient.
To understand this, lets go back to equation 1, why it works..
1) k is multiple of 3,
so, we can say value of k is such that k=3 * x (where x is integer)
so equation becomes => 2(3*x) + 3m = t
this can be written as => 3(2x+m) = t
Now, here we know 3 is factor of t.
but in case of statement 2, saying m is multiple of 3 is not enough to factor out 3 from the whole equation.
Hope that helps!
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Ahh silly me!thirst4edu wrote:The equation in question is 2k + 3m = t, in your example, we should be able to factor out 3 from the sum (2k+3m), not just 3m.Gurpinder wrote:Alright now I get how A is sufficient. How is B not sufficient?
m is a multiple of 3. so 3m means that the number is a multiple of 9. so lets just take 9 for example, its factors are 3 and this is common with 12.
in fact any multiple of 9 will always be divisible by 3, so why isnt statement 2 sufficient.
To understand this, lets go back to equation 1, why it works..
1) k is multiple of 3,
so, we can say value of k is such that k=3 * x (where x is integer)
so equation becomes => 2(3*x) + 3m = t
this can be written as => 3(2x+m) = t
Now, here we know 3 is factor of t.
but in case of statement 2, saying m is multiple of 3 is not enough to factor out 3 from the whole equation.
Hope that helps!
Thanks man!
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we need to figure out if t has some common factor as 12 which can be only 2, 3
STAT #1
2k + 3m = t
if k is multiple of 3 then k = 3x (where x is something)
2.3x + 3m = t
3 (2x+m) = t
means that t is a multiple of 3, same as 12 --> SUFF
STAT #2
2k + 3m = t
if m is multiple of 3 then m = 3x (where x is something)
2k + 3.3x = t
==> we can't get either 2 or 3 as common factor from this equation --> INSUFF
STAT #1
2k + 3m = t
if k is multiple of 3 then k = 3x (where x is something)
2.3x + 3m = t
3 (2x+m) = t
means that t is a multiple of 3, same as 12 --> SUFF
STAT #2
2k + 3m = t
if m is multiple of 3 then m = 3x (where x is something)
2k + 3.3x = t
==> we can't get either 2 or 3 as common factor from this equation --> INSUFF
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i approach to this problem as
k/6 + m/4 = t/12
1.stmt k is a multiple of 3. which means k can be 3, 6, 9, 12 ...
by using the above values we reduce expr to
x/2 + m/4 = t/12 or x + m/4 = t/12 ( where x is output of k/6)
so independent of value of m we can say denominator will be max 4. so we can say t and 12 have more than one common factor
2 stmt m is multiple of 3 which means m can be 3, 6, 9 , 12..
k/6+ 3x/4 = t/12 here the denominator will depend on value of k .. so we can say 2 is not sufficient
answer is A
k/6 + m/4 = t/12
1.stmt k is a multiple of 3. which means k can be 3, 6, 9, 12 ...
by using the above values we reduce expr to
x/2 + m/4 = t/12 or x + m/4 = t/12 ( where x is output of k/6)
so independent of value of m we can say denominator will be max 4. so we can say t and 12 have more than one common factor
2 stmt m is multiple of 3 which means m can be 3, 6, 9 , 12..
k/6+ 3x/4 = t/12 here the denominator will depend on value of k .. so we can say 2 is not sufficient
answer is A
4K + 6M = 2T
STATEMENT 1:K IS A MULTIPLE OF 3
PLUG IN TO SATISFY STATEMENT A (MUST BE MULTIPLE OF 3; 3,6,9 ...)
PUT RANDOM POSITIVE INTEGERS IN TO SOLVE (SUGGEST EVEN AND ODD)
K=3
M=3
4(3)+ 6(3) = 2T
T=15 (COMMON FACTOR WITH 12 ...3) YES
K=3
M=6
T= 18 (COMMON FACTOR WITH 12 ...3) YES
TRY TO KEEP PLUGGING IN #'S FOR STATEMENT 1, MAKE SURE TO SATISFY CRITERIA ... ANSWER IS ALWAYS YES
ANSWER CAN NOT BE BCE
STATEMENT 2: M IS A MULTIPLE OF 3
TAKE THE SAME APPROACH AS DESCRIBED IN STATEMENT 1
4K + 6M = 2T
K=3
M=3
12 + 18 = 2T
T=15 (TRY TO RECYCLE ALREADY USED NUMBERS) .... YES
K=4
M=3
T=14 NO
CANT HAVE A YES AND A NO IN A DS QUESTION ... MUST BE ONE OR THE OTHER. ELIMINATE B
THE ANSWER IS A [/u][/b]
STATEMENT 1:K IS A MULTIPLE OF 3
PLUG IN TO SATISFY STATEMENT A (MUST BE MULTIPLE OF 3; 3,6,9 ...)
PUT RANDOM POSITIVE INTEGERS IN TO SOLVE (SUGGEST EVEN AND ODD)
K=3
M=3
4(3)+ 6(3) = 2T
T=15 (COMMON FACTOR WITH 12 ...3) YES
K=3
M=6
T= 18 (COMMON FACTOR WITH 12 ...3) YES
TRY TO KEEP PLUGGING IN #'S FOR STATEMENT 1, MAKE SURE TO SATISFY CRITERIA ... ANSWER IS ALWAYS YES
ANSWER CAN NOT BE BCE
STATEMENT 2: M IS A MULTIPLE OF 3
TAKE THE SAME APPROACH AS DESCRIBED IN STATEMENT 1
4K + 6M = 2T
K=3
M=3
12 + 18 = 2T
T=15 (TRY TO RECYCLE ALREADY USED NUMBERS) .... YES
K=4
M=3
T=14 NO
CANT HAVE A YES AND A NO IN A DS QUESTION ... MUST BE ONE OR THE OTHER. ELIMINATE B
THE ANSWER IS A [/u][/b]
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statement 1 is sufficient to answer the question.we can simplify the question as (2k+3m)/12=t/12 when k is a multiple of 3 then we have m also as a multiple of 3 thus t and 12 will have a common factor greater than 1.
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To clear the fractions, multiple the given equation by 12:Gurpinder 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.
12(k/6 + m/4) = 12/(t/12)
2k + 3m = t.
The factors of 12 greater than 1 are 2,3,4,6, and 12.
Question rephrased: Is t a multiple of 2,3,4,6 or 12?
Statement 1: k is a multiple of 3
Thus:
t = 2(multiple of 3) + 3m.
t = multiple of 3 + multiple of 3.
Since the sum of two multiples of 3 must also be a multiple of 3, t is a multiple of 3.
SUFFICIENT.
Statement 2: m is a multiple of 3
If k=3 and m=3, then t = 2(3) + 3(3) = 15.
In this case, t is a multiple of 3.
If k=1 and m=3, then t = 2(1) + 3(3) = 11.
In this case, t is not a multiple of 2,3,4,6, or 12.
INSUFFICIENT.
The correct answer is A.
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Let me simplify the problem
k/6+m/4=t/12
or 2k+3m=t---(i)
Statement 1
Let k =3n
Then from equation(1) 2*(3n)+3m=t
or 3*(2n+m)=t
so t and 12 has a common factor 3.. Hence SUFFICIENT
Statement 2:
Let m=3n
or 2k +3*3n=t
No definite conclusion about factors. So Insufficient
k/6+m/4=t/12
or 2k+3m=t---(i)
Statement 1
Let k =3n
Then from equation(1) 2*(3n)+3m=t
or 3*(2n+m)=t
so t and 12 has a common factor 3.. Hence SUFFICIENT
Statement 2:
Let m=3n
or 2k +3*3n=t
No definite conclusion about factors. So Insufficient
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Given: k,m,t>0 and {k,m,t} belongs to N
k/6 + m/4 = t/12
=> 2k + 3m = t
To find if HCF of 12 and t is greater than 1
S1: k = 3a
6a+3m = t. 3 is HCF. Sufficient.
S2: m = 3b
2k+9b = t. Not sufficient
(A) is answer.
k/6 + m/4 = t/12
=> 2k + 3m = t
To find if HCF of 12 and t is greater than 1
S1: k = 3a
6a+3m = t. 3 is HCF. Sufficient.
S2: m = 3b
2k+9b = t. Not sufficient
(A) is answer.
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