simple annoying problem

y(u - c) = 0

means either y=0 or u-c=0 => u=c

and j(u - k) = 0,

similarly j=0 or u=k

If you combine both u=c=k.

Now why is the question saying assume c<k? Am I missing something?

lol chidcguy

Multiply the 2 equations y(u - c) = 0 and j(u - k) = 0

to get yj (u - c) (u - k) = 0

we know that c<k so the above is only possible when yj = 0

Thanks NG. Thats what I initially did, but lost direction before I completed my thought and ran in the other route.

I should have seen that c<k condition essentially avoids the c=k situation

Thanks again.

That's not correct... the multiplied out equation is not sufficient to determine the answer... For example, from the multiplied equation, you could have something like:
y=1, j=2, u=3, c=3, k=4 and that would work without yj being equal to 0.


The answer is jy=0 though...

I just looked at the three equations (including c<k) and saw that one of the following two things had to happen:

Either [(u=c and j=0) or (u=k and y=0)] or (j=0 or y=0)

No matter which is the case, there's always a requirement for either j=0 or y=0. So we know that jy has to equal 0.

netigen wrote:to get yj (u - c) (u - k) = 0

we know that c<k so the above is only possible when yj = 0

Your reasoning is not correct. The question provides two conditions

y(u - c) = 0 and j(u - k) = 0

The case you quoted y=1, j=2, u=3, c=3, k=4 does not hold good and hence is not even a valid scenario for this question.

Your reasoning will only hold if you can show numbers that fulfill the conditions of the question and still do not hold true for the the case that I reasoned.

egybs wrote:That's not correct... the multiplied out equation is not sufficient to determine the answer... For example, from the multiplied equation, you could have something like:
y=1, j=2, u=3, c=3, k=4 and that would work without yj being equal to 0.


The answer is jy=0 though...

I just looked at the three equations (including c<k) and saw that one of the following two things had to happen:

Either [(u=c and j=0) or (u=k and y=0)] or (j=0 or y=0)

No matter which is the case, there's always a requirement for either j=0 or y=0. So we know that jy has to equal 0.

netigen wrote:to get yj (u - c) (u - k) = 0

we know that c<k so the above is only possible when yj = 0

dude, reread what I wrote. You implied that simply looking at the multiplied out equation was enough to come to the conclusion that jy=0, while it isn't. You need to consider the other two equations. The example values I gave were sufficient to satisfy the multiplied out equation you provided, but were not sufficient for the original two equations.... which was the entire point!!

netigen wrote:Your reasoning is not correct. The question provides two conditions

y(u - c) = 0 and j(u - k) = 0

The case you quoted y=1, j=2, u=3, c=3, k=4 does not hold good and hence is not even a valid scenario for this question.

Your reasoning will only hold if you can show numbers that fulfill the conditions of the question and still do not hold true for the the case that I reasoned.

egybs wrote:That's not correct... the multiplied out equation is not sufficient to determine the answer... For example, from the multiplied equation, you could have something like:
y=1, j=2, u=3, c=3, k=4 and that would work without yj being equal to 0.


The answer is jy=0 though...

I just looked at the three equations (including c<k) and saw that one of the following two things had to happen:

Either [(u=c and j=0) or (u=k and y=0)] or (j=0 or y=0)

No matter which is the case, there's always a requirement for either j=0 or y=0. So we know that jy has to equal 0.

netigen wrote:to get yj (u - c) (u - k) = 0

we know that c<k so the above is only possible when yj = 0

Whatever i said has to be taken in the context of the question as it was not a general statement.

According to the question 'u' can not be equal to c or k, unless c=k which we know does not hold according to the question.

Hence, the only possibility in the one that I quoted. If you still don't get it get a Maths refresher.

Whatever man - i have no interest in arguing... but if you don't get what I wrote, you should probably take a english refresher.

netigen wrote:Whatever i said has to be taken in the context of the question as it was not a general statement.

According to the question 'u' can not be equal to c or k, unless c=k which we know does not hold according to the question.

Hence, the only possibility in the one that I quoted. If you still don't get it get a Maths refresher.

This is my reasoning please let me know if I am missing anything.

Here a * b = 0
So either a = 0 or b = 0

In our case

y(u - c) = 0
implies either y = 0 or u-c = 0

Again j(u - k) = 0
implies either j = 0 or u-k = 0

Now as c> k , their value is different and two different values CAN NOT be subtracted from a constant value to get the same result i.e. 0.

Hence y /j has to be 0


Other assumption is if say one of them c or k si equal to u ...in that case either y will be 0 or j will be 0

so in each case yj will always be 0.

Looks good to me!

gmatinjuly wrote:This is my reasoning please let me know if I am missing anything.

Here a * b = 0
So either a = 0 or b = 0

In our case

y(u - c) = 0
implies either y = 0 or u-c = 0

Again j(u - k) = 0
implies either j = 0 or u-k = 0

Now as c> k , their value is different and two different values CAN NOT be subtracted from a constant value to get the same result i.e. 0.

Hence y /j has to be 0


Other assumption is if say one of them c or k si equal to u ...in that case either y will be 0 or j will be 0

so in each case yj will always be 0.