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10 Aug 2018, 03:47
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If |12x − 3| = |3x + 15|, what is the value of x?
(1) x > 0
(2) x is an integer
(1) x > 0
(2) x is an integer
Bunuel
If |12x − 3| = |3x + 15|, what is the value of x?
(1) x > 0
(2) x is an integer
(1) x > 0
(2) x is an integer
Given, |12x − 3| = |3x + 15|
Squaring both the sides,
\((12x-3)^2=(3x+15)^2\)
Or, \(144x^2-72x+9=9x^2+90x+225\)
Or, \(135x^2-162x-216=0\)
or, \(x=-\frac{4}{5}\quad \mathrm{or}\quad \:x=2\)
Another way, 12x-3=3x+15 Or, x=2
and 12x-3=-(3x+15) or, x=-\(\frac{4}{5}\)
St1:- x > 0
Only acceptable value is 2.
Sufficient.
St2:- x is an integer
Only acceptable value is 2.
Sufficient.
Ans. (D)
10 Aug 2018, 05:14
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Bunuel
If |12x − 3| = |3x + 15|, what is the value of x?
(1) x > 0
(2) x is an integer
(1) x > 0
(2) x is an integer
OA:D
If |12x − 3| = |3x + 15|, what is the value of x?
Case 1 :\(x<\frac{-15}{3}\) or \(x<-5\)
\(-(12x-3)=-(3x+15)\)
\(9x=18, x = \frac{18}{9} = 2\)
Not possible as \(x<-5\) whereas \(2>-5\)
Case 2 :\(-5≤x≤\frac{3}{12}\)
\(-(12x-3)=(3x+15)\)
\(15x=-12, x = \frac{-12}{15} = -\frac{{4}}{5}\)
Case 3: \(x>\frac{3}{12}\)
\((12x-3)=(3x+15)\)
\(9x=18, x = \frac{18}{9} = 2\)
What is value of \(x\)? Given \(x=-\frac{4}{5}\) or \(x= 2\)
Statement 1 : \(x>0\)
Out of \(-\frac{4}{5},2\); Only 2 is greater than 0
So \(x=2\)
Statement 1 alone is sufficient
Statement 2 : \(x\) is an integer
Out of \(-\frac{4}{5},2\); Only 2 is integer
So \(x=2\)
Statement 2 alone is sufficient
