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ds-ALGEBRA 03 Jan 2009, 18:32
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If \(X^3Y = 24\), what is the value of \(X^3Y^3 - X^2Y^2 ?\)

1] \(X^2Y^2 = 36\)
2] \(X^3Y^2 = 72\)



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ds-ALGEBRA 03 Jan 2009, 19:29
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If \(X^3Y = 24\), what is the value of \(X^3Y^3 - X^2Y^2 ?\)

1] \(X^2Y^2 = 36\)
2] \(X^3Y^2 = 72\)
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the question stem alone is sufficient to answer the question. but since you have to answer, it will be D

X^3Y=24, if you factorize 24 u get 2x2x2x3
so you can determine X=2 and Y=3.
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ds-ALGEBRA 03 Jan 2009, 20:41
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agree with the above, D should be the answer.
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ds-ALGEBRA 04 Jan 2009, 09:10
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If \(X^3Y = 24\), what is the value of \(X^3Y^3 - X^2Y^2 ?\)

1] \(X^2Y^2 = 36\)
2] \(X^3Y^2 = 72\)

the question stem alone is sufficient to answer the question. but since you have to answer, it will be D

X^3Y=24, if you factorize 24 u get 2x2x2x3
so you can determine X=2 and Y=3.
Show more

Now now my friend ALD and his partner in crime Pmenon:

What makes you think X cannot be cube root of 4, and Y = 6?
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ds-ALGEBRA 04 Jan 2009, 12:32
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What about
x = -2 and y=-3

because from the question , it didn't tell us about x and y

1) it's insufficient > >> x^3y^3 - x^2y^2 = x^2y^2(xy-1)
(1) x^2y^2 = 36
xy = + 6 or -6
and x^3y=24 = x^2(xy) = 24
x^2(6) = 24
x = 2 or -2
y = 3 or -3
so 1 is insufficient
2) x^3y=24 , x^3y^2=72
y = 3 and x = 2
so 2 is sufficient
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ds-ALGEBRA 04 Jan 2009, 13:29
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If \(X^3Y = 24\), what is the value of \(X^3Y^3 - X^2Y^2 ?\)

1] \(X^2Y^2 = 36\)
2] \(X^3Y^2 = 72\)
Show more

We even do not know whether x or y are integer.

\(X^3Y = 24\)..............................1

= \(X^3Y^3 - X^2Y^2\)
= \(X^2Y^2 (XY-1)\)
= \(X^2Y^2 (XY-1)\) ....................2

1] \(X^2Y^2 = 36\)
\(x(X^2 Y^2) = 36x\)
\(X^3Y^2 = 36x\)
\(24Y= 36x\)
\(Y= 36x/24\)
\(Y= 1.5x\)

= \(X^2Y^2 (XY-1)\)
= \(X^2(1.5x)^2 (X(1.5x)-1)\)
= \(2.25X^3 (1.5x^2 -1)\)
not suff...


2] \(X^3Y^2 = 72\)
\(24Y = 72\)
\(Y = 3\)


If y = 3, \(9X^3 = 72\).
x = 2.

suff..

B
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ds-ALGEBRA 04 Jan 2009, 19:30
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If \(X^3Y = 24\), what is the value of \(X^3Y^3 - X^2Y^2 ?\)

1] \(X^2Y^2 = 36\)
2] \(X^3Y^2 = 72\)

We even do not know whether x or y are integer.

\(X^3Y = 24\)..............................1

= \(X^3Y^3 - X^2Y^2\)
= \(X^2Y^2 (XY-1)\)
= \(X^2Y^2 (XY-1)\) ....................2

1] \(X^2Y^2 = 36\)
\(x(X^2 Y^2) = 36x\)
\(X^3Y^2 = 36x\)
\(24Y= 36x\)
\(Y= 36x/24\)
\(Y= 1.5x\)

= \(X^2Y^2 (XY-1)\)
= \(X^2(1.5x)^2 (X(1.5x)-1)\)
= \(2.25X^3 (1.5x^2 -1)\)
not suff...


2] \(X^3Y^2 = 72\)
\(24Y = 72\)
\(Y = 3\)


If y = 3, \(9X^3 = 72\).
x = 2.

suff..

B
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I just factored it like
x^3*y^3 - x^2*y^2=y^2*(24-x^2)
(1) is not sufficient.
(2) you can get y=3.and y^2=9.
and problem solved.
Hencce, B.
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ds-ALGEBRA 08 Jan 2009, 00:47
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D Is the answer, since xy cant be -6 as x^3*y = 24 (both are either +ve or -ve);
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ds-ALGEBRA 08 Jan 2009, 06:48
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D Is the answer, since xy cant be -6 as x^3*y = 24 (both are either +ve or -ve);
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Thats correct but the OA is B.



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