Folks, any other takes on this....

Is the ans A

sureshbala wrote:Given x/y>2…..implies both x and y can be positive or both of them are negative
If both of them are negative, it is clear that 3x+2Y<0.
If both of them are positive we have x > 2y…......(1)

Statement1: x-y < 2 ...(2)
From (1) we have x-y > y .....(3)
From (2) and (3), y<x-y<2, so y<2.
From (2), we have x<y+2, so x<4.
Since y<2 and x<4, 3x+2y<16.
Hence statement (1) alone is sufficient

Coming to statement 2, we can think along the same lines and easily find two contradicting examples.
Let x=100 and y=1
Clearly x/y > 2, and y-x < 2 and the value of 3x+2y=302 (>18)
Let x=3 and y=1
Clearly x/y>2 and y-x<2 and the value of 3x+2y=11(<18)
So statement(2) is not sufficient
Hence the answer is Choice A.

:mrgreen: Would you prefer to switch to graphical approaches for such questions, in case going algebraic or plugging values need lot of focus on many possibilities, suresh?

How if we first draw graph for the question stem inequality and name it Q.

Then we draw two separate graphs for the two statement inequalities and call the graphs S1 and S2.

Now if S1 totally coincides or nowhere intersects Q, st 1 is sufficient; otherwise not. Do same with S2 and Q, and follow the obvious conclusions. I found it quicker may be because I am comfortable on graphical approaches; but it still took more than 3 minutes for me to answer A. I have posted a similar question unknowingly the fact that it has already been put with slightly different figures by you, suresh. Thanks

sureshbala wrote:Here is the next one...

The roots of a quadratic equation ax^2+bx+c=0 are integers and a+b+c>0 and a>0. Are both the roots of the equation positive?

1. Sum of the roots is positive.

2. Product of the roots is positive.

The roots of ax^2+bx+c=0 are (-b+sqt(b^2-4ac))/2a & (-b-sqt(b^2-4ac))/2a

I --> Sum of the roots is positive. If we add the 2 roots, we get -2b > 0. Which means that b has to be -ve. Substituting a -ve value of b in the 2 roots will give positive roots only (sqt(b^2-4ac) will be a smaller in value than b). Sufficient.

II --> Product of the roots is positive. This means that both the roots are either positive or negative. Since 2 answers are coming using this statement, its clearly insufficient.

A is the answer.

sureshbala wrote:Here is the next one...

The roots of a quadratic equation ax^2+bx+c=0 are integers and a+b+c>0 and a>0. Are both the roots of the equation positive?

1. Sum of the roots is positive.

2. Product of the roots is positive.

Folks, here is the explanation...

Statement 1: Sum of the roots is positive.

i.e – b/a = p where p>0

i.e b = – (a x p).

Now consider the product of the roots.

Let c/a=q.

Clearly both p and q must be integers (since in the question it is given that both the roots are integers)

It is given that a+b+c>0

i.e a-(axp)+(axq)>0

i.e a(1-p+q)>0

It is given in the question that a>0

So (1-p+q)>0

i.e q-p>-1

i.e; q-p>=0 (since p and q are integers)

i.e q>=p

In statement 1 it is given that p>0. So clearly q>0

So from the first statement itself if sum is +ve we can conclude that product is also +ve.

Clearly Statement 2 alone is not sufficient.

So A

Here is the next question...

If R is an integer between 1 and 9, and P - R = 2370, what is the value of R?

(1) P is divisible by 4
(2) P is divisible by 9

1. 2372 and 2376 are both divisible by 4. 1 is out.

2. sum of digits must be divisible by 9, so the only one here is 2376.

B

Agree with Dana. What is OA please?

sureshbala wrote:Here is the next question...

If R is an integer between 1 and 9, and P - R = 2370, what is the value of R?

(1) P is divisible by 4
(2) P is divisible by 9

1 < R < 9
QUESTION : R?

P-R = 2370

1) P is divisible by 4

P - 2370 = R

if P is divisible by 4 and 2370 is divisible by 2 then R has to be divisible by 2

multiples of 2 between 1 & 9 are 2, 4, 6, 8.
since 2370 is not divisible by 4 we can have only 2 possibilities 2 & 6

2372 & 2376 both are divisible by 4

Insufficient.


2) P is divisible by 9

P - 2370 = R

If P is divisible by 9 and 2370 is divisible by 3 then R has to be divisible by 3

multiples of 3 between 1 & 9 are 3 & 6.

2373 is not divisible by 9 but 2376 is.

Sufficient.

Hence B.

sureshbala wrote:Here is the next question...

If R is an integer between 1 and 9, and P - R = 2370, what is the value of R?

(1) P is divisible by 4
(2) P is divisible by 9

The answer is B.

Given P = 2370 + R.

Statement I: P is divisible by 4.

Since 2370 leaves a remainder 2 when divided by 4, 2370+R will be divisible by 4 if R = 2 or 6 (Remember that R can take values from 1 to 9 only)

Hence I is not sufficient

Statement II: P is divisible by 9.

Since 2370 leaves a remainder 3 when divided by 9, 2370+ R will be divisible by 9 if R = 6. (Remember that R can take values from 1 to 9 only)

Hence II is sufficient.

So the answer is B

I guess C is correct (That both are required to find the solutions)
Method-
Ax^+bx+c have two roots

1)
-b+-sq root of b^-4ac/2a
So if you add both the roots

(-B+SQroot of B^-4AC/2A) + (-B-SQroot of B^-4AC/2A )
you will get –b/a >0 (as a is positive so B need to be negative for making b/a +ive)
But by this we can’t really say about C, as it can be positive or negative both
2)
So if you multiply both the roots
-B+SQroot of B^-4AC/2A to -B-SQroot of B^-4AC/2A

You will get 4ac/2a &#61664; 2c>0
That means c>0

So that’s the reason we require both 1 and 2 to find the solution.

Here is the next one....

Is x=y?

I. (x+y)(1/x + 1/y) = 4
II. (x-30)^2 = (y-30)^2

sureshbala wrote:Here is the next one....

Is x=y?

I. (x+y)(1/x + 1/y) = 4
II. (x-30)^2 = (y-30)^2

I would go with A.

I will give (x+y)^2-4xy=0 which (x-y)^2=0 so, x=y while II will give
x-30=(y-30) or x-30= -(y-30). First eq can x=y but second one will result in x+y=60 in which x and y can take infinite values.

sureshbala wrote:Here is the next one....

Is x=y?

I. (x+y)(1/x + 1/y) = 4
II. (x-30)^2 = (y-30)^2

I'll go with D. Either

1). when we factor this out we get x + (x/y) + (y/x) + y = 4
so we have the sum of 4 numbers that must equal 4.
I could only get this statement to work if both x & y = 1 so 1 + 1 + 1 + 1 = 4.
I is sufficient.

2). They are the same statement only the variables are different.
(x-30)^2 = (y-30)^2 => x-30 = y-30
If one variable was negative and the other was positive they wouldn't equal one another, same goes if the values of x & y were different.

II. Is sufficient

hemantsood is right...remember squareroot of any number can be positive or negative!