anirudh777 wrote:
stne wrote:
anirudh777 wrote:
ALTERNATE SOLUTION:
Equation 1 : l1 : y=(m1)X+c1 , where m1 is slope and c1 is constant
Equation 2 : l2 : y=(m2)X+c2 , where m2 is slope and c2 is constant
Also given that slope are of same sign. Means--either both m1, m2 are positive or both are negative.
Statement 1 : intersect at point (a,b) where in both a and b are positive.
(1)==> y=b=(m1)a+c1=(m2)a+c2 -simply putting coordinates in both the equations of lines.
But from here we cannot conclude whether m1 > or < m2
Hence Insufficient.
Statement 2 : The X-intercept of line l1 is greater than the X-intercept of line l2
==> As value of x, when y=0, is the X intercept. Thus, X intercept for l1 and l2 are:
l1: -c1/(m1)
l2: -c2/(m2)
(2)==> -c1/(m1) > -c2/(m2), Since we dont know the sogn of c1 and c2 and we dont know there values we cannot conclude anything from this.
Hence Insufficient.
Combining both the statements :
and solving (1) and (2) we get,
a-b/(m1) >a-b/(m2) ---> m1<m2 Hence SUFFICIENT. Thus answer is C.
Hi
I have been able to figure out the graphical way by myself, but I cannot figure out how you combined the two equations to get \(\frac{(a-b)}{m1} >\frac{(a-b)}{m2}\) in your algebraic method, can anyone help with the algebraic method here.
Thanks
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solving Equations (1) and (2) means, substituting the value of c1 and c2 from equation (1) in equation (2)-
From (1) :
c1= b-a(m1)
c2= b-1(m2)
putting these values in eq2
-c1/(m1) > -c2/(m2)
==> -[b-a(m1)]/m1 > -[b-a(m2)]/m2
--> -b/m1 + a > -b/m2 + a
--> since we know a and b are positive constant values, we can cancel them out from the equality and we are then left with
==> m1<m2.
Hence Sufficient.
many thanks +1
however slope of m1>m2 and not m1<m2 as you have arrived at, guess small oversight in your calculation.
Here we go:
\(b=m_1.a+c_1\), and \(b=m_2.a+c_2\) these are the two equations of the line from statement 1
\(c_1= b-m_1a\) and \(c_2=b-m_2a\).. ( 1)
x intercept's \(\frac{-c_1}{m_1}\) and \(\frac{-c_2}{m_2}\)...(2)
given \(\frac{-c_1}{m_1} > \frac{-c_2}{m_2}\). from statement 2
now substituting the values of \(c_1\) and \(c_2\) from ..(1) in ...(2)
\(\frac{-(b-m_1a)}{m_1} > \frac{-(b-m_2a)}{m_2}\)
\(\frac{-b+m_1a}{m_1} > \frac{-b+m_2a}{m_2}\) ,multiplying the negative sign in the numerator
\(\frac{-b}{m_1}+a >\frac{-b}{m_2}+a\) , dividing LHS by \(m_1\) and RHS by \(m_2\)
\(\frac{-b}{m_1}>\frac{-b}{m_2}\), cancelling out a from both sides
\(\frac{b}{m_1}<\frac{b}{m_2}\), Multiplying both sides by negative and reversing the > sign to < , since in inequality we reverse the operator when we multiply( or divide)by a negative sign. ( Guess you forgot this part )
so finally we are left with \(\frac{b}{m_1}<\frac{b}{m_2}\)
\(\frac{1}{m_1}<\frac{1}{m_2}\) , cancelling b from both sides
\(\frac{1}{m_1}<\frac{1}{m_2}\)
Now we are still not sure of the signs of slope , both could be + or both could be -
if both are positive then
\(m_1>m_2\)
if both are negative
then we have \(\frac{1}{- m_1}<\frac{1}{- m_2}\)
here also we have \(m_1>m_2\)
hence \(m_1>m_2\) for all cases and hence sufficient
Hope I have done everything correctly
Thank you for showing the way.
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