I took the quadradic path as i usually do, but dint solve the equation to see if stmt 2 was sufficient, here's how
Given : nx=60 (n: number of ppl, x: contribution of each)
Stmt 1 : gives x, can solve for n : sufficient
Stmt 2 : (n-5)(x+2)=60
=>(n-5)(60/n+2)=60
Simple rearrangement
=>2n^2-10n-300=0
I stopped here knowing this equation will have one positive and one negative root (or solution)
here's how : ax^2+bx+c=0
b=sum of roots (or solutions) (or -600
c=product of roots
=> 2n^2-10n-300=0 will have roots that add to -10 and have roots that multiply to -300 or (-600 to be specific) The second statement that -300 (or -600) will be the product of two roots of the equation implies, one is +ve and other is
-ve. and n cant take -ve values, therefore only one solution. Stmt 2 is sufficient.
Could someone cpls verify if this method has any flaw?
Given : nx=60 (n: number of ppl, x: contribution of each)
Stmt 1 : gives x, can solve for n : sufficient
Stmt 2 : (n-5)(x+2)=60
=>(n-5)(60/n+2)=60
Simple rearrangement
=>2n^2-10n-300=0
I stopped here knowing this equation will have one positive and one negative root (or solution)
here's how : ax^2+bx+c=0
b=sum of roots (or solutions) (or -600
c=product of roots
=> 2n^2-10n-300=0 will have roots that add to -10 and have roots that multiply to -300 or (-600 to be specific) The second statement that -300 (or -600) will be the product of two roots of the equation implies, one is +ve and other is
-ve. and n cant take -ve values, therefore only one solution. Stmt 2 is sufficient.
Could someone cpls verify if this method has any flaw?