inequity help 13

Magical Cook,
IMO B.
Because, by A, you can not conclude that 3x > 7y. But by B, we can say - 5x > 14y or 3x > 42y/5 and 42y/5 > 7y. So again IMO B.

thanks. I am for B too but answer says C.... I think it should be worng unless anyone has different answer....

I drew a graph representing 3 inequalities to solve this problem.

First, I sketched out a line y < (3/7)x which is our target. We want to know whether our point is below the line y = (3/7)x

From statement 1, I sketched out a line y < x - 4. You can see that the area below the line y = x - 4 is both above and below the line y = (3/7)x, depending on what value of x you pick. Thus, statement 1 is insufficient.

From statement 2, I sketched out a line y < (5/14)x. Again, the resulting area is either below or above the line y = (3/7)x, depending on the x value. Insufficient.

1&2 together, you can see that the area that are below BOTH lines y = x - 4 AND y = (5/14)x is ALSO below the line y = (3/7)x. That means, if you put constraints 1 & 2 together, the resulting value for y will always be less than (3/7)x. Hence, C.

I still think (B) is the correct answer.

Ineq(2) x> 14y/5 or x>42y/15.

The original ineq. asks if x>35y/15.

So if (2) is true, the original ineq. holds true if x & y are positive.


Alternatively,
Ineq(2) y<5x/14

Original ineq. y<3x/7 or y<6x/14

So if Ineq(2) holds true original ineq. will be true.

Hence (B).

I reckon C too

2nd condition breaks down to x>2.8y..if y is -1 then x is less than 2.8..from the Original statement x>2.33y therefore if y was -1 x would be less than 2.33..therefore we cannot say..flip the sign back to positive say y=1, we get a completely different scenario..But when we combine the two statements x has to be greater than y and by 2.8 times.as oppose to 2.33 times...Therefore C

Rajivmatta but neither x nor y can be negative numbers.

I still think is B

i think it should be B.
The graph used takes into consideration of whole numbers