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

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The value of x is to be randomly selected from the integers [#permalink]

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The value of x is to be randomly selected from the integers from 1 to 10, inclusive, and then substituted into the equation y = x^2 - 4x + 3. What is the probability that the value of y will be negative?

A. 1/10
B. 1/5
C. 1/4
D. 3/10
E. 3/5
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Re: The value of x is to be randomly selected from the integers [#permalink]

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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.

Only one value between 1 and 10 will give you that result: 2. So, the answer is A

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Re: The value of x is to be randomly selected from the integers [#permalink]

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saintforlife wrote:
The value of x is to be randomly selected from the integers from 1 to 10, inclusive, and then substituted into the equation y = x^2 - 4x + 3. What is the probability that the value of y will be negative?

A. 1/10
B. 1/5
C. 1/4
D. 3/10
E. 3/5


We need to find that how many of the values of x from 1 to 10 make y negative.

y = x^2 - 4x + 3

We can split the positive and negative terms:
y = x^2 + 3 - 4x
For y to be negative, 4x needs to be greater than (x^2 + 3). We don't need to try x = 4 or greater since in those cases, x^2 + 3 will always be greater than 4x. Think of it this way
6*6 + 3 will be greater than 4*6; 7*7 + 3 will be greater than 4*7 etc

So we just need to figure out x = 1, 2 and 3
For x = 1 and 3, y = 0
For x = 2, y = -1
So only for one value of x, y will be negative.
Hence required probability = 1/10

Alternatively, you can think of the graph of a quadratic.
y = x^2 - 4x + 3 = (x - 3)(x - 1) will be a parabola facing upwards with roots at 1 and 3. So it will lie below the x axis only for x = 2.
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Re: The value of x is to be randomly selected from the integers [#permalink]

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New post 25 Oct 2013, 01:50
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saintforlife wrote:
The value of x is to be randomly selected from the integers from 1 to 10, inclusive, and then substituted into the equation y = x^2 - 4x + 3. What is the probability that the value of y will be negative?

A. 1/10
B. 1/5
C. 1/4
D. 3/10
E. 3/5


Similar questions to practice:
if-p-2-13p-40-q-and-p-is-a-positive-integer-between-110590.html
new-algebra-set-149349-60.html#p1200970

Hope it helps.
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Re: The value of x is to be randomly selected from the integers [#permalink]

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New post 25 Oct 2013, 03:05
WillEconomistGMAT wrote:
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.

Only one value between 1 and 10 will give you that result: 2. So, the answer is A




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?
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Re: The value of x is to be randomly selected from the integers [#permalink]

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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?
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Re: The value of x is to be randomly selected from the integers [#permalink]

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Temurkhon wrote:
WillEconomistGMAT wrote:
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.

Only one value between 1 and 10 will give you that result: 2. So, the answer is A




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?


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.
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Re: The value of x is to be randomly selected from the integers [#permalink]

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New post 03 Nov 2013, 12:46
VeritasPrepKarishma wrote:

Alternatively, you can think of the graph of a quadratic.
y = x^2 - 4x + 3 = (x - 3)(x - 1) will be a parabola facing upwards with roots at 1 and 3. So it will lie below the x axis only for x = 2.


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
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Re: The value of x is to be randomly selected from the integers [#permalink]

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NateTheGreat11 wrote:
VeritasPrepKarishma wrote:

Alternatively, you can think of the graph of a quadratic.
y = x^2 - 4x + 3 = (x - 3)(x - 1) will be a parabola facing upwards with roots at 1 and 3. So it will lie below the x axis only for x = 2.


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.
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Re: The value of x is to be randomly selected from the integers [#permalink]

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New post 23 Jul 2016, 21:24
NateTheGreat11 wrote:
VeritasPrepKarishma wrote:

Alternatively, you can think of the graph of a quadratic.
y = x^2 - 4x + 3 = (x - 3)(x - 1) will be a parabola facing upwards with roots at 1 and 3. So it will lie below the x axis only for x = 2.


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


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
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Re: The value of x is to be randomly selected from the integers [#permalink]

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New post 15 Jun 2017, 10:54
LalaB wrote:
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?



where do I find the difficulty level ?
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Re: The value of x is to be randomly selected from the integers [#permalink]

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daviddaviddavid wrote:
LalaB wrote:
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?



where do I find the difficulty level ?


The difficulty level is in the tags above the original post:
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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: The value of x is to be randomly selected from the integers [#permalink]

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New post 15 Jun 2017, 12:44
Bunuel wrote:
daviddaviddavid wrote:
LalaB wrote:
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?



where do I find the difficulty level ?


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



ah thats cool =)

thank you =)
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Re: The value of x is to be randomly selected from the integers [#permalink]

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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.

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Re: The value of x is to be randomly selected from the integers   [#permalink] 09 Feb 2018, 12:03
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