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'.

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

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.
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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

Why we recalled \((x-y)^2\geq{0}\) ?
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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)
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Bunuel,
I did not understand the reasoning behind statement 2 being sufficient. Line marked in red is my doubt. what does it mean ?

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.
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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?
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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?
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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.
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