The value of x is to be randomly selected from the integers

You might try factoring x^2-4x+3 here into...

(x-3)(x-1)

y will be negative when exactly one of these two terms is negative.

if x^2-4x+3<0
(x-1)*(x-3)<0
x<1, or x<3
x<1, means that none of 1-10 fits. What is wrong?

y = x^2 - 4x + 3
y=(x-2)^2-1 in this case y can be negative (-1),if x=2

so, we have only one case out of 10
sure it is 700 level?

x<1 or x<3 does not make any sense.

(x-1)*(x-3)<0 --> 1<x<3.

Check below posts:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html
everything-is-less-than-zero-108884.html
graphic-approach-to-problems-with-inequalities-68037.html

Hope this helps.

So I get to this point
(x-3)(x-1)
and...
1< P(Y is negative) < 3
But instead I set my equality to less than or equal to, as such:
1 <=P(Y is negative) <= 3
Hence getting 3/10

And I am confused why we don't include the 1 and the 3?
Could you please explain?

Thanks

Notice that a value of 1 either 1 or 3 for x will make y = 0. We are only interested in the negative y values. Hope that helps.

We do not include 1 or 3, cuz if we do then the y = 0
y = (x -3)*(x - 1)

three ranges :
x < 1
1 < x < 3
x > 3

For x < 1, take x = x = 0
y = +ve

For x > 3, take x = 5
y = +ve

For x = 3
y = (3 - 3)*2 = 0

but for 1 < x < 3, take x = 2
y = -ve, hence only this matches

only 1 value out of 10 values (2 out of 1 ,2, 3, 4, ... 10)
hence ans is 1/10

where do I find the difficulty level ?

The difficulty level is in the tags above the original post:

ah thats cool =)

thank you =)

Hi All,

This question requires a certain degree of 'math' to get to the solution, but there's a great built-in logic 'shortcut' that you can take advantage of to minimize the work that you need to do. Look at the equation Y = X^2 - 4X + 3. Let's focus on a specific part of that equation....

X^2 - 4X =

(X)(X) - 4X

When X=4, those two 'pieces' are equal...
(4)(4) - 4(4) = 0

BUT the extra +3 on the end of the calculation will make that sum POSITIVE. As X gets bigger, the first 'piece' becomes larger at a faster rate than the second pieces does....

X=5
(5)(5) - 4(5)

X=6
(6)(6) - 4(6)
Etc.

So there's really no need to consider ANY number greater than or equal to 4 - we KNOW that the sum will be POSITIVE for all of those values. Instead, we only have to look at the calculations for X=1, X=2 and X=3. That requires some basic 'plug in' arithmetic, then you can answer the question that's asked.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

So my thought process, I missed the question, for trying to be to fast and not diligent

So factored down to

(x-3)(x-1), so should only look at when x = 1,2,3, when X>= 4, each term will be +

So
X = 1, will get -2*0 = 0, not negative do not count
X = 2, will get -1*1 = -1, do count
x = 3, will get 0*2 = 0, do not count

So only 2 will result in a negative number, thus it is 1/10,

I put one 1/5 because I forgot to consider the 1-1 will result in 0

Wouldn't it be 4/5? Since except for 1 & 3, any number can have a negative outcome.

Hi revamoghe,

To answer this question, we have to consider the possible values for Y (which are based on the 10 possible integer values for X) - and we are asked for the probability that the value of Y will be NEGATIVE. There are a number of different ways to go through that 'math work' (as well as a particular pattern-based shortcut to help you avoid doing most of that work) to figure out the potential values for Y. You will find that there is just ONE value of X that leads to a NEGATIVE value for Y... so the answer to the question is...



GMAT assassins aren't born, they're made,
Rich

My approach:

y = x^2 - 4x + 3

What is the probability that the value of y will be negative?

Lets plug in some numbers:

5^2 - 4(5) + 3 = 8
4^2 - 4(4) + 3 = 3
3^2 - 4(3) + 3 = 0
2^2 - 4(2) +3 = -1
1^2 - 4(1) + 3 = 0

Answer is A. 1/10.

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