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Quadratic Equation/Probability Exam Pack1

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by GMATGuruNY » Sat Jan 23, 2016 3:57 am
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
y = (x-1)(x-3).

The CRITICAL POINTS ARE x=1 and x=3.
The critical points are the only x-values that yield y=0.
To determine the ranges where x will yield y<0, test one value to the left and one value to the right of each critical point.

Here, since x must be an integer such that 1≤x≤10, we must test only one value BETWEEN x=1 and x=3 and one value TO THE RIGHT of x=3.

1<x<3:
If we plug x=2 into y = x² - 4x + 3, we get:
y = 2² - 4*2 + 3 = -1.
Since y is negative, 1<x<3 is a valid range.

x>3:
If we plug x=4 into y = x² - 4x + 3, we get:
y = 4² - 4*4 + 3 = 3.
Since y is NOT negative, x>3 is NOT a valid range.

The only valid range is 1<x<3.
Implication:
Of the 10 integers between 1 and 10, inclusive, only one value -- x=2 -- will yield a negative value for y.
Thus:
P(y is negative) = 1/10.

The correct answer is A.
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by sukhman » Sat Jan 23, 2016 4:07 am
What is the level of this question
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by GMATGuruNY » Sat Jan 23, 2016 4:09 am
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
An alternate approach is to test the first 10 positive integers and LOOK FOR A PATTERN.

If x=1, then y = 1² - 4*1 + 3 = 0.
If x=2, then y = 2² - 4*2 + 3 = -1.
If x=3, then y = 3² - 4*3 + 3 = 0.
If x=4, then y = 4² - 4*4 + 3 = 3.
If x=5, then y = 5² - 4*5 + 3 = 8.

After x=2, the values for y keep INCREASING.
Implication:
Only the option in red will yield a negative value for y.
Since only one of the first ten positive integers will yield a negative value for y, we get:
P(y is negative) = 1/10.

The correct answer is A.
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by [email protected] » Sat Jan 23, 2016 10:37 am
Hi sukhman,

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: A

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Re: Quadratic Equation/Probability Exam Pack1

by Scott@TargetTestPrep » Sun Mar 07, 2021 6:13 am
sukhman wrote: ↑
Sat Jan 23, 2016 2:44 am
 Please see image attachedImage
Solution:

As we can see, if x^2 - 4x is nonnegative, then adding 3 to it will make it positive, i.e., y = x^2 - 4x + 3 will be positive. We see that if x ≥ 4, then x^2 - 4x ≥ 0, and hence y will be positive. If x = 3, then x^2 - 4x + 3 = 9 - 12 + 3 = 0, which is not negative. If x = 2, then x^2 - 4x + 3 = 4 - 8 + 3 = -1, which IS negative. Finally, if x = 1, then x^2 - 4x + 3 = 1 - 4 + 3 = 0, which is not negative. Therefore, the only x-value that makes y negative is 2, and hence the probability is 1/10.

Answer: A

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