If x and y are positive, is 3x > 7y?
(1) x > y + 4
(2) -5x < -14y
inequity help 13
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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.
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.
Correct me If I am wrong
Regards,
Amitava
Regards,
Amitava
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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.
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.
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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).
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).
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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
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