Hey missrochelle,

I'm assuming Statement (1) should say 2y instead of 2x. It doesn't make sense to divide x by 2x, because the result will always be 1/2, unless x=0, in which case it would be undefined.

Regarding Statement (2), when you divide x+y by y, it looks like this:

(x+y)/y

= x/y + y/y (common denominator)

= x/y + 1

The 'm' that they use in the explanation refers to any integer.

Basically, because x/y + 1 results in a remainder of 4 when divided by y, that means that x/y + 1 is some integer plus 4.

So instead of writing it as 'm', you could say:

x/y + 1 = (some integer) + 4

Subtract 1 from both sides:

x/y = (some integer) + 3

This means that x/y has a remainder of 3.

Make sense?

Hi Raz1024,

When i read the question,

I quicky started taking few examples like X = 2 and Y =4

However stmt 1 failed; Now when i tried stmt 2 it gives me an answer as 1.5,which also says that it is insufficient.

Could you please suggest me on how to tackel these kind of questions or what would be the way to think in proper way?

Hey Sunil,

Unfortunately, x=2 and y=4 won't work.

Statement (1) says that x divided by 2y (I'm pretty sure it should be 2y) yields a remainder of 4. If x=2 and y=4, then x/2y would be 2/8, and that would yield a remainder of 2, not 4.

Statement (2) says that x+y divided by y gives you a remainder of 4. If x=2 and y=4, then (x+y)/y would be 6/4, and that would yield a remainder of 2, not 4.

Remember, if you plug in numbers, you have to choose numbers that fit the conditions of the statements. The statements are always true.

This is a problem for which choosing numbers is going to be a little more difficult than going with the straight number properties.

Statement (1) is easy enough: you could choose x=4 and y=4, in which case x/2y yields a remainder of 4, thus fulfilling the condition of the statement. In that case, x/y would yield a remainder of 0. You could also choose x=4 and y=3. In that case, x/2y yields a remainder of 4, again fulfilling the condition of the statement. But x/y yields a remainder of 1.

So because you could end up with two different remainders for x/y, the statement is insufficient.

Statement (2), however, is a little bit tougher to deal with if you choose to plug in numbers. Better to just use the algebraic solution I outlined previously.

Hope this helps!

raz1024 wrote:Hey Sunil,

Unfortunately, x=2 and y=4 won't work.

Statement (1) says that x divided by 2y (I'm pretty sure it should be 2y) yields a remainder of 4. If x=2 and y=4, then x/2y would be 2/8, and that would yield a remainder of 2, not 4.

Statement (2) says that x+y divided by y gives you a remainder of 4. If x=2 and y=4, then (x+y)/y would be 6/4, and that would yield a remainder of 2, not 4.

Remember, if you plug in numbers, you have to choose numbers that fit the conditions of the statements. The statements are always true.

This is a problem for which choosing numbers is going to be a little more difficult than going with the straight number properties.

Statement (1) is easy enough: you could choose x=4 and y=4, in which case x/2y yields a remainder of 4, thus fulfilling the condition of the statement. In that case, x/y would yield a remainder of 0. You could also choose x=4 and y=3. In that case, x/2y yields a remainder of 4, again fulfilling the condition of the statement. But x/y yields a remainder of 1.

So because you could end up with two different remainders for x/y, the statement is insufficient.

Statement (2), however, is a little bit tougher to deal with if you choose to plug in numbers. Better to just use the algebraic solution I outlined previously.

Hope this helps!

Raz,
If we take x=4 y =6 ---->10/6 remainder =4
x=4 y =8 ---->12/8 Remainder =4
both of these yield -ve remainders..Isnt this insufficient?

Unfortunately, no, because you've only tested one set of numbers.

You managed to find a case for which the remainder will be 4. But that doesn't mean that there aren't cases for which the remainder will NOT be 4.

As an analogy, if I ask if you whether or not a/b is odd, and you plug in a=6 and b=2, well then yes, a/b is odd. But can say definitively that a/b will ALWAYS be odd? No, of course not. Because you could also pick a=6 and b=3, in which case a/b would be even.

What I did in my outline was choose two examples such that I found TWO DIFFERENT POSSIBLE ANSWERS to the question, thus making the statement insufficient. But you can't use a single example to prove a statement sufficient.

Let me know if that makes sense.

raz1024 wrote:Unfortunately, no, because you've only tested one set of numbers.

You managed to find a case for which the remainder will be 4. But that doesn't mean that there aren't cases for which the remainder will NOT be 4.

As an analogy, if I ask if you whether or not a/b is odd, and you plug in a=6 and b=2, well then yes, a/b is odd. But can say definitively that a/b will ALWAYS be odd? No, of course not. Because you could also pick a=6 and b=3, in which case a/b would be even.

What I did in my outline was choose two examples such that I found TWO DIFFERENT POSSIBLE ANSWERS to the question, thus making the statement insufficient. But you can't use a single example to prove a statement sufficient.

Let me know if that makes sense.

Raz,
got your point.I just handpicked 2 set of values and got "no". Until i prove,for a set of numbers,the answer is "yes" plugging values is incomplete.
It turns to be very time consuming when plugging values.How to decide whether to go for alzebra or plugging values?

raz1024 wrote:Hey missrochelle,

I'm assuming Statement (1) should say 2y instead of 2x. It doesn't make sense to divide x by 2x, because the result will always be 1/2, unless x=0, in which case it would be undefined.

Regarding Statement (2), when you divide x+y by y, it looks like this:

(x+y)/y

= x/y + y/y (common denominator)

= x/y + 1

The 'm' that they use in the explanation refers to any integer.

Basically, because x/y + 1 results in a remainder of 4 when divided by y, that means that x/y + 1 is some integer plus 4.

So instead of writing it as 'm', you could say:

x/y + 1 = (some integer) + 4

Subtract 1 from both sides:

x/y = (some integer) + 3

This means that x/y has a remainder of 3.

Make sense?

The question from the 300 GMAT sheet definitely uses 2x... I'll have to go back and find the post, but I definitely copied and pasted correctly.... thanks for the response!

raz1024 wrote:Hey missrochelle,

I'm assuming Statement (1) should say 2y instead of 2x. It doesn't make sense to divide x by 2x, because the result will always be 1/2, unless x=0, in which case it would be undefined.

Regarding Statement (2), when you divide x+y by y, it looks like this:

(x+y)/y

= x/y + y/y (common denominator)

= x/y + 1

The 'm' that they use in the explanation refers to any integer.

Basically, because x/y + 1 results in a remainder of 4 when divided by y, that means that x/y + 1 is some integer plus 4.

So instead of writing it as 'm', you could say:

x/y + 1 = (some integer) + 4

Subtract 1 from both sides:

x/y = (some integer) + 3

This means that x/y has a remainder of 3.

Make sense?

Sorry cud you also explain why statement one wud be insufficient then? If you know that x/2y yields remainder 4? Is it because we are unable to isolate k? i.e. the equation would be:
x/2y = some integer (k) + 4...
same as 1/2 (x/y) = some integer +4
dividing by 1/2 you get --- X/Y = some integer /(1/2) +4 / (1/2) .

so with statement 1 we end up with x/y = 2(some integer) + 2. is the reason it's insufficient because "some integer" still has a coefficient ....so....