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21 Mar 2025, 00:30
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
46% (02:30) correct
54%
(02:27)
wrong
based on 127
sessions
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The number of calories Dave burns, D, by running for x minutes at a certain speed, over the course of a 3-mile run can be modeled by the function \(D(x) = 2*(r^2)+\frac{q}{7p}-3p( \frac{x}{r}- \frac{3p}{q} )^2\) where p, q & r are positive constants. What is the maximum number of calories Dave can burn by running at this speed during one 3-mile run?
(1) \(14p (r^2) + q = 28pr\)
(2) r = 75
(1) \(14p (r^2) + q = 28pr\)
(2) r = 75
21 Mar 2025, 03:54
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Bunuel
For maximum value of \(D(x) = 2*(r^2)+\frac{q}{7p}-3p( \frac{x}{r}- \frac{3p}{q} )^2\)
\(3p( \frac{x}{r}- \frac{3p}{q} )^2\) should be zero
i.e. \(\frac{x}{r}= \frac{3p}{q}\)
i.e. \(3pr = xq\)
Max \(D(x) = 2*(r^2)+\frac{q}{7p}-3p( \frac{x}{r}- \frac{3p}{q} )^2 = 2*(r^2)+\frac{q}{7p}\)
Question: i.e Maximum \(D(x) =\frac{(14r^2+q)}{7p}\) = ?
Statement 1: \(14p (r^2) + q = 28pr\)
Maximum \(D(x) =\frac{(14r^2+q)}{7p} = \frac{28pr}{7p} = 4r\)
NOT SUFFICIENT
Statement 2: r = 75
NOT SUFFICIENT
Combining the statements
4r = 300
SUFFICIENT
Answer: Option C
21 Mar 2025, 11:57
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Bunuel
Let’s consider the function first
function \(D(x) = 2*(r^2)+\frac{q}{7p}-3p( \frac{x}{r}- \frac{3p}{q} )^2\)
the function contains two positive term and a negative term , which includes a square term in it.
we are asked to find the Maximum value of the function D(x).
Before jumping into the statements let’s first decode the function for further clarity. Let’s for example consider the function f(x) = 12+ 28 - 5* a^2. The max value can be 40. As the value of an increase’s the function f(x) keeps decreasing.
Only when the negative term is zero. We get the max value. The negative terms in D(x) should be zero.
x/r - 3p/q = 0
x/r = 3p/q
substituting in the function, the negative term becomes zero, and we are left with
D(x) = 2r^2 + q/7p
= (14 p r^2 + q)/7p
STATEMENT 1:
\(14p (r^2) + q = 28pr\)
substitute it in the final equation of D(x) = 28 pr / 7p = 4r. NOT SUFFICIENT
STATEMENT 2:
(2) r = 75
standalone statement 2 is NOT SUFFICIENT.
combine statements 1 and 2; we get D(x) = 4r = 4*75 = 300.
The max value of the function D(x) = 300. Hence option C
