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18 Mar 2019, 10:38
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Is \(d\) greater than or equal to 0?
1) \(d\) is the median of \(d, \frac{1}{d}, -d\)
2) \(d^3\) is the median of \(d, d^2, d^3\)
1) \(d\) is the median of \(d, \frac{1}{d}, -d\)
2) \(d^3\) is the median of \(d, d^2, d^3\)
Rakesh1987
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Updated on: 12 Jan 2021, 23:25
Originally posted by Rakesh1987 on 03 Apr 2019, 00:13.
Last edited by Rakesh1987 on 12 Jan 2021, 23:25, edited 2 times in total.
Last edited by Rakesh1987 on 12 Jan 2021, 23:25, edited 2 times in total.
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Is \(d\) greater than or equal to \(0\)?
1) \(d\) is the median of \(d,\frac{1}{d},−d\)
Therefore either a) \(\frac{1}{d}\leq{d}\leq{-d}\) or b) \(-d\leq{d}\leq{\frac{1}{d}}\)
a) from \(d\leq{-d}\) we get \(d\leq{0}\) and from\( \frac{1}{d}\leq{d}\) we get \(d\neq{0}\) (as \(\frac{1}{d}\) then will be undefined) and \(d\geq{-1}\) (as if \(d\) -in denominator is higher in magnitude with negative sign the fraction will be \((-1,0)\) while \(d\) on RHS will be smaller). we write it as \((-1,0]\).
b) from \(-d\leq{d}\) we get \(d\geq{0}\) and from \(d\leq{\frac{1}{d} }\)we get \(d\leq{1}\) but not \(0\). we write it as \([0,1)\)
From a and b, we we get \(d\) is \((-1,0]\) or \([0,1)\), Hence, from I alone we cant say definitely that \(d\geq{0}\)
2) \(d^3\) is the median of \(d,d^2,d^3\)
Therefore, Either a) \(d\leq{d^3}\leq{d^2}\) or b) \(d^2\leq{d^3}\leq{d}\)
a) from \(d\leq{d^3}\) we get \(d-d^3\leq{0}\) or \(d(1-d^2)\leq{0}\). Note \(d^2\) is \(\geq{0}\), therefore either \(d\) is \((-1,0)\) (\(d\) will be negative and \(1-d^2\) will be positive) or \(d>1\) (\(d\) will be positive and \(1-d^2\) will be negative). from \(d^3\leq{d^2}\) we get \(d^2(d-1)\leq{0}\), therefore, \(d<1\). From both parts combined we get \(d\) is \((-1,0)\).
b) from \(d^2\leq{d^3}\) we get \(d^2(1-d)\leq{0}\), therefore \(d\geq{1}\) or \(0\). from \(d^3\leq{d}\) we get \(d(d^2-1)\leq{0}\). Either \(d\leq{-1}\) or \(d= (0,1)\). From combining b all parts we get \(d=0,1\).
From a and b we get \(d=0,1\) or \((-1,0)\). Therefore II is insufficient.
I and II Combined, we can say \(d\) is either \((-1,0]\) or \(1\). Therefore Insufficient. Answer is E.
Hi Guys if you liked my explanation, then please award kudos.
Edited to make answer more readable
1) \(d\) is the median of \(d,\frac{1}{d},−d\)
Therefore either a) \(\frac{1}{d}\leq{d}\leq{-d}\) or b) \(-d\leq{d}\leq{\frac{1}{d}}\)
a) from \(d\leq{-d}\) we get \(d\leq{0}\) and from\( \frac{1}{d}\leq{d}\) we get \(d\neq{0}\) (as \(\frac{1}{d}\) then will be undefined) and \(d\geq{-1}\) (as if \(d\) -in denominator is higher in magnitude with negative sign the fraction will be \((-1,0)\) while \(d\) on RHS will be smaller). we write it as \((-1,0]\).
b) from \(-d\leq{d}\) we get \(d\geq{0}\) and from \(d\leq{\frac{1}{d} }\)we get \(d\leq{1}\) but not \(0\). we write it as \([0,1)\)
From a and b, we we get \(d\) is \((-1,0]\) or \([0,1)\), Hence, from I alone we cant say definitely that \(d\geq{0}\)
2) \(d^3\) is the median of \(d,d^2,d^3\)
Therefore, Either a) \(d\leq{d^3}\leq{d^2}\) or b) \(d^2\leq{d^3}\leq{d}\)
a) from \(d\leq{d^3}\) we get \(d-d^3\leq{0}\) or \(d(1-d^2)\leq{0}\). Note \(d^2\) is \(\geq{0}\), therefore either \(d\) is \((-1,0)\) (\(d\) will be negative and \(1-d^2\) will be positive) or \(d>1\) (\(d\) will be positive and \(1-d^2\) will be negative). from \(d^3\leq{d^2}\) we get \(d^2(d-1)\leq{0}\), therefore, \(d<1\). From both parts combined we get \(d\) is \((-1,0)\).
b) from \(d^2\leq{d^3}\) we get \(d^2(1-d)\leq{0}\), therefore \(d\geq{1}\) or \(0\). from \(d^3\leq{d}\) we get \(d(d^2-1)\leq{0}\). Either \(d\leq{-1}\) or \(d= (0,1)\). From combining b all parts we get \(d=0,1\).
From a and b we get \(d=0,1\) or \((-1,0)\). Therefore II is insufficient.
I and II Combined, we can say \(d\) is either \((-1,0]\) or \(1\). Therefore Insufficient. Answer is E.
Hi Guys if you liked my explanation, then please award kudos.
Edited to make answer more readable
General Discussion
yashu612
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18 Mar 2019, 12:09
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Why is the answer not B?
If d^3 is the median, then the only way this can happen is if d is a negative fraction --> d, d^3, d^2.
If d is positive, then the median will always be d^2 (fraction or an integer).
Please explain
If d^3 is the median, then the only way this can happen is if d is a negative fraction --> d, d^3, d^2.
If d is positive, then the median will always be d^2 (fraction or an integer).
Please explain
