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The line y = ax + b intersects the y axis at point (0,b)

So we just need to know whether b>0

(1) ab < 0

a and b have oppsite signs, so we can have

a>0, b<0 (for example a=1, b=-2) or
a<0, b>0 (for example a=-2, b=1)

1 is not sufficient

(2) a + b < 0

We can use the same example as done with Statement 1 and conclude that

2 is not sufficient

(1)+(2)

Our same example holds in this case

a>0, b<0 (for example a=1, b=-2) or
a<0, b>0 (for example a=-2, b=1)

and so b can still be positive or negative.

Not sufficient

Answer is (E)

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Line K is represented by the equation y = ax + b. Does line K intersect the positive y-axis on the x-y plane?

(1) ab < 0

(2) a + b < 0

(1) insufic
ab<0, then a and b are dif signs and not 0.
a,b=(+,-): has neg y-intercept
y=x-b, y=0, x=b, (+,0)
y=x-b, x=0, y=-b, (0,-)

a,b=(-,+): has pos y-intercept
y=-x+b, y=0, x=b, (+,0)
y=-x+b, x=0, y=b, (0,+)

(2) insufic
a+b<0: (a,b)=(-,-)(-,+)(+,-)

(1/2) insufic

Ans (E)

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y= a*x +b
K intersect the positive "y" axis ? it means at x=0; above equation must satisfy at x=0
y= b?

The question basically ask is y = b?

St1: a*b <0 ; both have opposite signs ; it means a=-ve & b = +ve OR a = +ve & b = -ve
y= b
y = +ve or -ve
Not Suff

St2: a+b<0 it means either
1) a <0 OR b>0
2) a>0 OR b<0
3) Both are negative a<0; b<0

so b is -ve or +ve
Not suff

St1 & St2: From St1) b = -ve or +ve From St2) b = -ve & +ve
Not Suff


IMO(E)
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Line K is represented by the equation y = ax + b. Does line K intersect the positive y-axis on the x-y plane?

y = ax + b
When x = 0, b > 0 ?

(1) ab < 0
a(-) and b(+)
OR
a(+) and b(-)

INSUFFICIENT.

(2) a + b < 0
Again three possibilities exists.
a(-) and b(+) [-5+2]
OR
a(+) and b(-) [2-5]
OR
a(-) and b(-) [-2-5]

INSUFFICIENT.

Together 1 and 2
a(-) and b(+) [-5+2] YES
OR
a(+) and b(-) [2-5] NO

INSUFFICIENT.

Answer E.
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IMO:- E

Y-axis > x=0

Option 1) ab<0

Conditions applicable

either a> 0 and b<0
or a<0 and b>0

Line equation > y= ax + b
at x= 0; y=b

b can be positive or negative

Not sufficient

Option 2)

a +b <0

Conditions applicable

a<0 and b<0
a>0 and b<0 ; |b| > |a|
a<0 and b>0 ; |a| > |b|

when x =0

b can be positive or negative

Combining both

Conditions applicables:-

a>0 and b<0 ; |b| > |a|
a<0 and b>0 ; |a| > |b|

at x=0

b can be positive or negative

Hence Not sufficient
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Quote:
Line K is represented by the equation y = ax + b. Does line K intersect the positive y-axis on the x-y plane?

(1) ab < 0

(2) a + b < 0
E , IMO

Basically the question is asking whether b > 0 ?

(1) ab < 0
We know a and b are opposite sign .. Not sufficient ..

(2) a + b < 0

a and b can be -ve . Or a can be -ve b can be positive .. or a +ve and b -ve .. So not sufficient .

Combining both the statement
we still have that
if a > 0
b -ve

if a < 0 then
b +ve .

So combining statement 1 and 2 also insufficient .

Hence E .
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