(2) 2x - 3 < 3y - 4 --> \(x<1.5y-\frac{1}{2}\) --> \(x<y+(0.5y-\frac{1}{2})\). Now, since \(y\) is a negative number then \(0.5y-\frac{1}{2}=negative\) so, we have that: \(x<y+negative\). \(y+negative\) is less then \(y\) and if \(x\) is less than \(y+negative\) then it must also be less than \(y\) itself: \(x<y\).

Hope it's clear.
_________________

What about the case when all a,b,c are zero. In this case, abc = 0 and 0 is neither odd nor even. Hence 'E'.

Welcome to GMAT Club. Below is an answer to your question.

Notice that zero is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder. Since 0/2=0=integer then zero is even.

For more on this subject please check Number Theory chapter of Math Book: math-number-theory-88376.html

Hope it helps.
_________________

\((x-\frac{3}{2})(x-3)<0\)

I agree one solution of this inequality is

\((x-\frac{3}{2})\) > 0 , (x-3)<0 => \((\frac{3}{2})\) < x < 3

However, Don't u think this can also resort to

\((x-\frac{3}{2})\) < 0 , (x-3)>0 => x < \((\frac{3}{2})\) , x > 3

and in that case x can have infinite values.

and if that is the case Stmt-1 alone would not be sufficient.
_________________

The second case of \((x-\frac{3}{2})<0\) and \((x-3)>0\) is not possible. This condition leads to \(x<\frac{3}{2}\) and \(x>3\), but \(x\) can not be simultaneously less than \(\frac{3}{2}\) (to make \(x-\frac{3}{2}\) negative) and more than 3 (to make \(x-3\) positive).

For more on how to solve such kind of inequalities check:
x2-4x-94661.html#p731476 (Check this first)
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863
xy-plane-71492.html?hilit=solving%20quadratic#p841486

Hope it helps.
_________________

Hi Bunuel,

Can we solve this question using graphical approach.

Since (x-a)^2+(y-b)^2 = r^2 ----(A)

Equation 1 represents equation of circle with radius 1.

Now, if we can maximize the value of both x and y, we can get the maximum value of xy and that could let us know whether statement (1) is sufficient or not.

Here, I require your help. Intuitively, I find that x=y will maximize the equation x^2+y^2=1. But I am not sure about my reasoning as well as the Mathematics behind. Could you please look into this.

Thanks

Useful property: For given sum of two numbers, their product is maximized when they are equal.

(1) says that \(x^2+y^2=1\). So, \(x^2y^2\) will be maximized when \(x^2=y^2\): \(x^2+x^2=1\) --> \(x^2=\frac{1}{2}\) --> the maximum value of \(x^2y^2\) thus is \(\frac{1}{4}\) and the maixmum value of \(xy\) is \(\frac{1}{2}\).

Hope it's clear.
_________________

Why we recalled \((x-y)^2\geq{0}\) ?

Because \((x-y)^2\geq{0}\) expands to \(x^2-2xy+y^2\geq{0}\), which leads to \(1-2xy\geq{0}\) (since \(x^2+y^2=1\)) and finally to \(xy\leq{\frac{1}{2}}\).
_________________

Responding to a pm:

Time taken by Bonnie to complete one work = x hrs
Time taken by Clyde to complete one work = y hrs
x and y are odd integers i.e. they could take values such as 1/3/5/7/9/11...

Question: Is x = y? i.e. is the time taken by Bonnie equal to time taken by Clyde? i.e. is the speed of Bonnie equal to the speed of Clyde?

(1) x^2+y^2<12
This info is not related to work concepts. It's just number properties. x and y are odd integers.
If x = y = 1, this inequality is satisfied.
If x = 1 and y = 3, this inequality is satisfied.

This means x may or may not be equal to y. Not sufficient.

(2) Bonnie and Clyde complete the painting of the car at 10:30 am.r
Together, they take 45 mins to complete the painting of the car. This means, if their rate of work were the same, each one of them would have taken 1.5 hrs working alone. But their time taken is an integer value. We can say that they do not take the same time i.e. x is not equal to y. Hence this statement is sufficient to say \(x \neq y\)

Answer (B)
_________________

Bunuel,
Here's how I did . I don't know why it is wrong. Please advise :

6a=3b=7c

=>
12a=6b=14c

From 1) ac=6b
=> ac=12a
=> c=12.. a can be cancelled because it is not inequality. .

so c =12..

however,

ac-12a=0 --> a(c-12)=0 so a=0 OR c=12 .. Is this the reason why my soln is wrong?
_________________

Bunuel,
I did not understand the reasoning behind statement 2 being sufficient. Line marked in red is my doubt. what does it mean ?

Never reduce an equation by a variable (or expression with a variable), if you are not certain that the variable (or the expression with a variable) doesn't equal to zero. We can not divide by zero.

So, if you divide (reduce) ac=12a by a, you assume, with no ground for it, that a does not equal to zero thus exclude a possible solution (notice that both a=0 AND c-12=0 satisfy the equation).

Hope it's clear.
_________________

From the stem we got that if \(x=y\) then the total time would be: \(\frac{x^2}{2x}=\frac{x}{2}\), since \(x\) is odd then this time would be odd/2: 0.5 hours (1/2 hours = 0.5 hours), 1.5 hours (3/2 hours = 1.5 hours), 2.5 hours (5/2 hours = 2.5 hours), 3.5 hours (7/2 hours = 3.5 hours), 4.5 hours (9/2 hours = 4.5 hours), ....

Now, from the second statement we got that they complete the job in 0.75 hours, since the total time (0.75 hours) is NOT odd/2 (0.5 hours, 1.5 hours, 2.5 hours, 3.5 hours, 4.5 hours, ....), then \(x\) and \(y\) are not equal.

Hope it's clear.
_________________

Hi Bunuel,

In the 1st condition if we take ac = 6b and hence multiply the the given expression 6a=3b=7c by 2 we get 12a=6b=14c.

Now substituting 12a=ac=14c

I get 12a=ac --> c = 12 and similarly a = 14 and hence b= 28.

Can you please point where am I going wrong?

Regards

ac=12a (here you cannot reduce by a and write c=12 as you exclude possibility of a=0) --> a(c-12)=0 --> either a=0 OR c=12. So, we get either a=b=c=0 or a=14, b=28 and c=12.

Does this make sense?
_________________

I followed a different approach but I am getting that (1) also answers the question.

From 3b=7c => 6b=14c.

So ac=6b - equation 1)
14c=6b - equation 2)

Dividing 1) by 2) =>a/14= 1

So a=14!!!. Where am I committing a mistake in this logic? Bunuel, Please help!!

You cannot divide ac by 14c because c could be zero and division by zero is not allowed. The same applies to division of 6b by 6b.

Does this make sense?
_________________

How did you decide which approach to use in A, one could have solved like the way you did in B
3X < 2y - 1 or x < .67y - 0.33 , so x < y + negative , so x is more negative, so x < Y.

Is this is a 700 question ?

The approach used for the second statement does not work for the first:

\(3x + 4 < 2y + 3\) --> \(x<\frac{2y}{3}-\frac{1}{3}\) ---> \(x<y-\frac{y}{3}-\frac{1}{3}\) --> \(x<y+(-\frac{y}{3}-\frac{1}{3})\) --> we don't know whether \(-\frac{y}{3}-\frac{1}{3}=positive+negative\) is negative.

Hope it's clear.
_________________