Hi Vishugogo,
For this problem we need to Rephrase the Question as much as possible,
Given:
a/6 + b/5 = c/30 ==> 5a + 6b = c
Divide above equation by 5
a + (6/5)*b = c/5
Now you try to solve it.
Regards,
Uva
Divisibility
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To clear the fractions in the question stem, multiply each side by 30:If a, b and c are positive integers and a/6 + b/5 = c/30, is c divisible by 5?
(1) b is divisible by 5
(2) a is even
30 * (a/6 + b/5) = 30 * (c/30)
5a + 6b = c
c = 5a + 6b.
Statement 1: b is divisible by 5
If a=1 and b=5, then c = 5*1 + 6*5 = 35.
If a=2 and b=20, then c = 5*2 + 6*20 = 130.
In each case, c is a multiple of 5.
Perhaps one more case to confirm:
If a=3 and b=100, then c = 5*3 + 6*100 = 615.
In every case, c is a multiple of 5.
SUFFICIENT.
Statement 2: a is even
If a=2 and b=20, then c = 130, as shown above.
In this case, c is a multiple of 5.
If a=2 and b=1, then c = 5*2 + 6*1 = 16.
In this case, c is NOT a multiple of 5.
INSUFFICIENT.
The correct answer is A.
Statement 1 indicates the following:
b = 5k, where k is a positive integer.
Implication:
c = 5a + 6*(5k) = 5(a + 6k) = 5 * integer.
Thus, c is a multiple of 5.
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Hi vishugogo,
This DS question can be solved in a number of different ways. It's perfect for TESTing Values, but there's also a Number Property built into it that you might find useful:
Everyone in this string of posts has correctly deduced that "rewriting" the given equation is a must:
5A + 6B = C
We're also told that A, B and C are positive integers. We're asked if C is a multiple of 5? This is a YES/NO question.
Fact 1: B is a multiple of 5.
You can absolutely TEST Values here, but here's the Number Property worth knowing...
Since A is an integer, 5A is a MULTIPLE of 5
We're told that B is a multiple of 5, so 6B is a MULTIPLE of 5
If you add a multiple of 5 to another multiple of 5, then you end up with a MULTIPLE of 5!!!
So, C will ALWAYS be a multiple of 5
Fact 1 is SUFFICIENT
Fact 2: A is even
5A will be multiple of 5, since 5(even) is a multiple of 5
However, 6B may or may not be a multiple of 5, depending on what B is.
For example, if B=1, then 6B = 6; if B = 5, then 6B = 30
There's no way to know if we'll end up with a sum that is a multiple of 5 or not.
Fact 2 is INSUFFICIENT.
Final Answer: A
GMAT assassins aren't born, they're made,
Rich
This DS question can be solved in a number of different ways. It's perfect for TESTing Values, but there's also a Number Property built into it that you might find useful:
Everyone in this string of posts has correctly deduced that "rewriting" the given equation is a must:
5A + 6B = C
We're also told that A, B and C are positive integers. We're asked if C is a multiple of 5? This is a YES/NO question.
Fact 1: B is a multiple of 5.
You can absolutely TEST Values here, but here's the Number Property worth knowing...
Since A is an integer, 5A is a MULTIPLE of 5
We're told that B is a multiple of 5, so 6B is a MULTIPLE of 5
If you add a multiple of 5 to another multiple of 5, then you end up with a MULTIPLE of 5!!!
So, C will ALWAYS be a multiple of 5
Fact 1 is SUFFICIENT
Fact 2: A is even
5A will be multiple of 5, since 5(even) is a multiple of 5
However, 6B may or may not be a multiple of 5, depending on what B is.
For example, if B=1, then 6B = 6; if B = 5, then 6B = 30
There's no way to know if we'll end up with a sum that is a multiple of 5 or not.
Fact 2 is INSUFFICIENT.
Final Answer: A
GMAT assassins aren't born, they're made,
Rich
