4x>3y

Hi,

For option 1, If u come with the values for x & y that don't satisfy the equation, then use the same values for option 2.

1)
for x > y-x... Say u come up with
x = 4 and y = 3
so 4x > 3y - true
or x = 3 and y = 5
so, 4x > 3y - false
Insuff

2)
for x/y < 1
our first combination of values wont hold true.
so just reverse it, as
x = 3 and y = 4
so, 4x > 3y - equal
or x = 3 and y = 5
so, 4x > 3y - false
Insuff

Since you have tried the same values for both the equations (c) can not be an answer.

Choose (e)

The statement given is Is 4x>3y given x & y are positive.

4x>3y
x> 3/4y

Another way for arriving at solution is

1) x> y-x means 2x> y or x>y/2 . But this does not tell whether x > 3/4y. INSUFFICIENT

2) x < y. Again this does not tell whether x > 3/4y. It is possible that x can be < 3/4y or 3/4y<x<1. INSUFFICIENT

Indeed, either option by itself is not sufficient.
But try this:
1. x>x-y; 2x>y
2. x/y<1; x<y (because y>0); 2x<2y; (multiply by -1, changing the inequality); -2x>-2y

Now subtract 1. from 2. and you get -4x>-3y. Looks familiar?
Multiply by -1 and you get 4x<3y.
So, the answer must be C. both statements together are sufficient. 4x is not greater than 3y, is actually exactly the opposite.

Have not seen this solution anywhere. Am I getting something wrong?

If x and y are positive , is 4x>3y?

1. x>y-x
2. x/y<1


Question Rephrase: Is x>3/4y?

From 1: x>y/2
Not Suff.x can be anything from 0.6y-infinity.whether x>0.75 y is not known

From2: x<y

Thus x can be anything from 0-0.9 y

Not suff

Combining the 2:

y/2<x<y

Now x can take any value from 0.6 y to 0.9y.Hence not suff


Thus ans E

Now subtract 1. from 2. and you get -4x>-3y. Looks familiar? ?


Not for me atleast....


Take one simple example x=2 y =3 this satisfied both x>y-x, x/y<1
condition

4x>3y = 4*2 > 3*3 (FALSE)

Now let x=9 and y =10..this satisfied both x>y-x, x/y<1
condition

4x>3y = 4*9 > 3*10 (TRUE)

Hope this helps

True, I agree with both of you. Your reasoning is flawless.
In the meantime I also figured out the correct answer.

But, for the life of me, I can not find where is the mistake in manipulating the inequalities the way I did.
I must have made a mistake, it's evident, but what is it?

Thanks for the clarifications anyway, I've learned to trust more trying values.

a > b & c > d does not imply a-c > b-d.
e.g. 4 > 2 & 3 > 0 does not imply 4-3 > 2-0

valentindima,

1. x>x-y; 2x>y
2. x/y<1; x<y (because y>0); 2x<2y; (multiply by -1, changing the inequality); -2x>-2y

Now you subtracted 1 from 2 to get -4x>-3y.
But if you subtract 2 from 1, you will get 4x>3y. which is contradicting your earlier statement!

So you can do these type of operations only on equalities!

Thanks everybody!
Looooong time since I was learning math. Obvious