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I didn't get it... The question was "is 3x > 7y?", not "is x>y".
Agree that 2nd is sufficient. But not the 1st.
Please tell me why this reasoning is wrong?
3x>7y-----3/7>y/x-------since 3/7 is less than 1, y/x must be less than 1,
this would mean y>x. So can we say that question is basicly asking if y>x? No? please, explain where is my mistake?
x = 6, y = 1, then 3x = 18, 7y = 7, so 3x > 7y.
x = 10, y = 5, then 3x = 30, 7y = 35, so 3x < 7y.
I learnt from OG that sometimes we just pick up some easy values to check, rather than try to find a proper way to prove it. Sometimes it saves our time a lot.
Hey I found something wrong with the appearance of my previous post, it doesn't appear as it is supposed to be, with the inequalities and stuff... Anyway good that you understand what I mean.
(2) -5x < -14y => 5x > 14y
x/y > 14/5
x/y > 2.8
The question is asking if x/y > 7/3 which is same as x/y > 2.33
This is sufficient since 2.8 will always be greater than 2.33
I will post this problem and my solution to it again under topic
"Inequality 3x, 7y"
if I can
I got the same problem. Anyway I understand what you mean.
Solid,
I just have one yes/no question.
does 3x>7y mean 3/7>y/x ?
....
In a nutshell, the equation 3x>7y can only be interpreted as x/y > 7/3 when y is positive.
Remember that when you divide or multiply a negative number in the inequality environment, the sign change...so let's consider out cases:
Case 1: x>0, y>0
3x>7y
Divide x by y, you get
x/y > 7/3
Case 2: x>0, y<0
3x>7y
Divide x by y, you get
x/y < 7/3 <- Notice the sign change because y is negative
Case 3: x<0, y>0
3x>7y
Divide x by y, you get
x/y > 7/3
Case 4: x<0, y<0
3x>7y
Divide x by y, you get
x/y < 7/3 <- Notice the sign change because y is negative
St1: x > y+4 insufficient if you take different values of x & y. You can do this calculation in mind. For example: if you consider x=13 & y=8 then 3x=39 <7y> 7y=14
St2: -5x <14y> (42/5)y --------------- Multiply both sides by -3/5 to get 3x on leftside
i.e. 3x > 8(2/5)y > 7y
Hence B
Btw Irina one cannot multiple and divide inequalities with variables unless you know for sure that the variables are positive because multiplying or dividing by negative values changes the inequality sign.
In other words one cannot conclude that 3x>7y is equal to x/y > 7/3 _________________
Re: If x and y are positive, is 3x > 7y? (1) x > y + 4 (2) [#permalink]
30 Jan 2012, 10:07
Expert's post
If x and y are positive is 3x>7y ?
(1) x > y+4. The best way to deal with this statement will be to pick the numbers. On DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another. Well, we can get an YES answer very easily: just pick very large \(x\) and very small \(y\) (for example 10 and 1 respectively) for NO answer: if \(x=10\) and \(y=5\) then \(10>5+4\) but \(3*10=30<7*5=35\). So this statement is not sufficient.
(2) -5x < -14y --> multiple both side by -1/2 and flip the sign --> \(2.5x>7y\), as both \(x\) and \(y\) are positive then \(3x>2.5x>7y\). Sufficient.
Re: If x and y are positive, is 3x > 7y? (1) x > y + 4 (2) [#permalink]
30 Jan 2012, 10:09
Expert's post
If w and c are integers is w > 0 ?
(1) w + c > 50 (2) c > 48
Obviously each statement alone is not sufficient.
When taken together we can still have both Yes and No answers: if c=100 and w=10 then the answer is Yes but if c=100 and w=-10 then the answer is No.
Answer: E.
I feel that it should be w+c<50 for (1). Then as \(c\) is an integer then the least value of it according to (2) will be 49, so max value of \(w\) in order \(c+w\) to be less than 50 is 0 (as \(w\) is also an integer), which means that the answer to the question is \(w>0\) is No. So if it were \(w + c < 50\) the answer would be C and also the question would be much more interesting.
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