CAT 2008: Quantitative Questions a Day 1 to 50 - The discussions

[OperationBLACKI]

Originally Posted by Aarav

Actually - I'm rather delighted to see most of you get the answer almost correctly. This was a problem that required 100% concentration. You might be amused to know that n = 37 doesn't hold true and hence we have in all 108 solutions.

Two smart questions follow in the weekend - again I will be more interested in the approach rather than the answer. Next week will be mixture of some swing, googly and bouncer besides some straight ones

Thanks Aarav , got my mistake .....missed out on that one .

[slam]

Just a question, do we need to wait for one of the mods to post the day's QQ, or can I post it myself and start off the discussion?

**I know some of you might find the question dumb, but I though if the forum has some set rules, might as well follow them.

slam.

[jha16june]

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Quantitative Question # 003
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Let f(x) be a function such that f(x).f(y) - f(xy) = 3(x+y+2). Then f(3) (equals )
(1) can not be determined (2) 7 (3) -8 (4) either 7 or -8 (5) none of these
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My Approach:
consider f(x)=x+3 this satisfies the given condition. So f(3) = 6. So option 5, none of these is correct.
Or: x=y=1, then f(1).f(1)-f(1)=12=4.3 => f(1)=4
Now x=3, y=1 => f(3).f(1)-f(3) = 3*(6) =>f(3)=6. So None of these.

[implex]

------------------------------------------------------
Quantitative Question # 003
------------------------------------------------------
Let f(x) be a function such that f(x).f(y) - f(xy) = 3(x+y+2). Then f(3) (equals )
(1) can not be determined (2) 7 (3) -8 (4) either 7 or -8 (5) none of these
-------------------------------------------------------------------

let x=y=1
f(1)^2-f(1)=12
f(1)=4 or f(1)=-3
let y =1 x=x
f(x).f(1)-f(x)=3(x+3)
f(x)[(f(1)-1]=3(x+3)
put x=3 and f(1)=4,-3
gives
f(3)=6 or-4.5, none of these

so option 5)

[slam]

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Quantitative Question # 003
------------------------------------------------------
Let f(x) be a function such that f(x).f(y) - f(xy) = 3(x+y+2). Then f(3) (equals )
(1) can not be determined (2) 7 (3) -8 (4) either 7 or -8 (5) none of these
-------------------------------------------------------------------

put x = 1, y = 1.
solve for f(1) to get f(1) = -3 or 4.
put x = 3, y = 1.
solve for f(3) to get f(3) = 18 / [f(1) - 1] = 6 or -4.5

so I think it would be (A) cannot be determined. (could be any one of the two values).
@implex
we've solved it pretty much the same way, but you've arrived at option 5. how did you decide on that?

[implex]

Originally Posted by slam

------------------------------------------------------
Quantitative Question # 003
------------------------------------------------------
Let f(x) be a function such that f(x).f(y) - f(xy) = 3(x+y+2). Then f(3) (equals )
(1) can not be determined (2) 7 (3) -8 (4) either 7 or -8 (5) none of these
-------------------------------------------------------------------

put x = 1, y = 1.
solve for f(1) to get f(1) = -3 or 4.
put x = 3, y = 1.
solve for f(3) to get f(3) = 18 / [f(1) - 1] = 6 or -4.5

so I think it would be (A) cannot be determined. (could be any one of the two values).
@implex
we've solved it pretty much the same way, but you've arrived at option 5. how did you decide on that?

the option, can not be determined is to be used when, the value in questions can not be determined due to lack of data, or some other constraint. But here we are able to determine, two values of the wanted, so we are able to determine it, but it is not given in the options. So none of these

Why it can't be not determine? because the initial equation is quadratic, so we will have two values and as the constraint which will reduce it to one is not given, we are right in taking both the answers!!

[slam]

@implex
i understand what you're trying to say here. but both the options can't be right. If we say 'none of these', then the 'these' also includes 'cannot be determined'.
anyway, i'd still stick to (A).

slam.

[implex]

Originally Posted by slam

@implex
i understand what you're trying to say here. but both the options can't be right. If we say 'none of these', then the 'these' also includes 'cannot be determined'.
anyway, i'd still stick to (A).

slam.

NO!!! the option none of these never includes "Can not be determined"

[implex]

Originally Posted by slam

@implex
i understand what you're trying to say here. but both the options can't be right. If we say 'none of these', then the 'these' also includes 'cannot be determined'.
anyway, i'd still stick to (A).

slam.

why?? quadratic eqns has two roots, and if we have been given no reason, there is no way we should discard one of them!

[comradeharsh]

A silly question:
Can we take x=y=0, solve for f(0), and then take x=3 and y=0, to get f(3)?
I got f(3)=4 or f(3)=-7/2.
Please bear in mind this comes from a quant weakling