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Re: x - y > 38 and y - 3x > 12 where, x and y are integers [#permalink]
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godxyz wrote:
KarunMendiratta wrote:
KarunMendiratta wrote:
\(x - y > 38\) and \(y - 3x > 12\)
where, \(x\) and \(y\) are integers

Quantity A
Quantity B
Least possible value of \(xy\)
1664


A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given


\(x - y > 38\)
\(y - 3x > 12\)

Since the sign of ineqialities are same, we can add both;
\(-2x > 50\)
\(\frac{-2x}{(-2)} < \frac{50}{(-2)}\)
\(x < -25\)
i.e. \(x\) could be -26, -27, -28, ......

\(3(x - y > 38)\)
\(y - 3x > 12\)

\(3x - 3y > 114)\)
\(y - 3x > 12\)

Since the sign of ineqialities are same, we can again add both;
\(-2y > 126\)
\(\frac{-2y}{(-2)} < \frac{126}{(-2)}\)
\(y < -63\)
i.e. \(y\) could be -64, -65, -66, ......

So, \(xy\) could be (-26)(-64), (-27)(-65), (-28)(-66), ....

Col. A: (-26)(-64) = 1664
Col. B: 1664

Hence, option C

NOTE: Whenever we divide or multiply the inequality with a -ve number, the sign flips


KarunMendiratta,
Hi sir,

Thank you for the answer.
I have a small doubt in your solution. If x = -26 and y = -64 are the minimum values, are they not supposed to satisfy the first inequality, i.e., x - y > 38? Using these values we get 38 > 38.

Thanks in advance!


godxyz
Absolutely correct
Kudos

The value of x must be \(-26\) and y must be \(-65\)
So, \(xy = 1690\)
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Re: x - y > 38 and y - 3x > 12 where, x and y are integers [#permalink]
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