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Value of y

by Deepthi Subbu » Thu Jan 13, 2011 2:52 am
If y=ax+b , for what values of x will y<0?

1. b=3
2.a<0

OA E

I chose C for the following reasons -


Statement 1 - Value of b is given , but no info about a . Hence Insufficient.
Statement 2 - Value of a is given ,but no info about b - Insufficient.

Taking both statements ,
say a = -1 and x = -1 , the value of y is positive .
say a=-1 and x = 4 , the equation becomes y = -1(4) + 3 = -1
Hence cant we conclude that for values for x>=4 , y will be negative ?
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by Anurag@Gurome » Thu Jan 13, 2011 3:06 am
Deepthi Subbu wrote:If y=ax+b , for what values of x will y<0?

1. b = 3
2.a < 0
As, y = (ax + b), y will be less than zero when (ax + b) will be less than zero.
Hence, (ax + b) < 0
=> ax < -b

Thus to determine the required range of values of x, we must know the values of a and b. This clearly indicates none of the statements are individually sufficient. Let's see what happens when we combine them!


ax < -3 => x > -(3/a) => Some positive quantity, as a is negative.

Now we can conclude that if x is greater than some positive quantity, y will be less than zero. But we don't have any definite idea about the "positive quantity". Hence both the statements together is also not sufficient to answer the question.

The correct answer is E.
Last edited by Anurag@Gurome on Thu Jan 13, 2011 3:28 am, edited 1 time in total.
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by Anurag@Gurome » Thu Jan 13, 2011 3:08 am
Deepthi Subbu wrote:Taking both statements ,
say a = -1 and x = -1 , the value of y is positive .
say a=-1 and x = 4 , the equation becomes y = -1(4) + 3 = -1
Hence cant we conclude that for values for x>=4 , y will be negative ?
Remember that the value of a is not given. Only it is mentioned that a is negative. You are assuming a value for a and solving for x, which means your solution for x is dependent on a. But we must answer the question without any assumption.
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by ankur.agrawal » Thu Jan 13, 2011 3:10 am
Deepthi Subbu wrote:If y=ax+b , for what values of x will y<0?

1. b=3
2.a<0

OA E

I chose C for the following reasons -


Statement 1 - Value of b is given , but no info about a . Hence Insufficient.
Statement 2 - Value of a is given ,but no info about b - Insufficient.

Taking both statements ,
say a = -1 and x = -1 , the value of y is positive .
say a=-1 and x = 4 , the equation becomes y = -1(4) + 3 = -1
Hence cant we conclude that for values for x>=4 , y will be negative ?
I too go with C.

Clearly each statement individually is not sufficient.

Combining both:

From 1st we deduce: y=ax+3; for y<0 , ax has to be negative & that too <-3 i.e. ax<-3.

Second tells us that a<0. that means in ax<-3 , x has to be + & that too >3. So x>3.

Both statements are sufficient to answer the question.

Pls tell if dere is anythg wrong in my explaination.
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by kmittal82 » Thu Jan 13, 2011 3:10 am
Hi Anurag,

You said:

Hence, (ax + b) < 0
=> x < -(b/a)

Is this correct though? I agree upto (ax + b) < 0 => ax < -b. However, can we divide both sides by 'a' without knowing the sign of a to make that inequality? For instance, if a is negative, the ineuqality would look like x > b/a
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by Deepthi Subbu » Thu Jan 13, 2011 3:28 am
Anurag@Gurome wrote:
Deepthi Subbu wrote:Taking both statements ,
say a = -1 and x = -1 , the value of y is positive .
say a=-1 and x = 4 , the equation becomes y = -1(4) + 3 = -1
Hence cant we conclude that for values for x>=4 , y will be negative ?
Remember that the value of a is not given. Only it is mentioned that a is negative. You are assuming a value for a and solving for x, which means your solution for x is dependent on a. But we must answer the question without any assumption.
I just took the value of a to be -1 as an example . But when the value of ax is negative and also greater than the value of b(3) , the value of y will be negative isnt it . Can you help me understand.?
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by Anurag@Gurome » Thu Jan 13, 2011 3:37 am
Deepthi Subbu wrote:I just took the value of a to be -1 as an example . But when the value of ax is negative and also greater than the value of b(3) , the value of y will be negative isnt it . Can you help me understand.?
Yes, that's correct.
But the problem asks for "what values of x...". But from the given informations we cannot determine any definite range for the values of x. The problem didn't ask for "what condition x must satisfy..." or "what should be the relation of x with a...".

If we assume a = -(1/2), we will get x > 6
If we assume a = -1, we will get x > 3
If we choose a = -2, we will get x > 3/2 etc..
kmittal82 wrote:Second tells us that a<0. that means in ax<-3 , x has to be + & that too >3. So x>3.
Not necessarily.
See the examples above.

You're correct on the other post. We can't do so. Edited the reply.
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by MAAJ » Thu Jan 13, 2011 6:03 am
@Anurag: Loved the approach to translate the function into an inequality!

I picked "D" plugging numbers. I knew that there was a trick when we plugged -1 for x, so I tried some other numbers and, determined that for that range [y<0] the numbers for x are infinite.

But ax<-3 is a easier way to see this. If a < 0, then there are infinite number for x that satisfy ax < -3 (knowing that all numbers for x are positive) :)!!!!
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by Night reader » Thu Jan 13, 2011 6:58 am
Deepthi Subbu wrote:If y=ax+b , for what values of x will y<0?

1. b=3
2.a<0

OA E
differential function of y => y`=(ax)` +(b)`, y`=a; put y` into inequality [y`=a] <0 OR a<0 is the only viable solution for y function to be less than 0. Because 'a' can obtain any value within the range (a<0), we are not certain about 'x', pick E

e.g. a<0, a=-1/1,000 then x>3,000 OR a=-100 then x>3/100
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