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Bunuel
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­The number of calories Dave burns, D, by running for x minutes at 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:

52% (02:32) correct 48% (02:39) wrong based on 214 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


­
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C
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Schools: IIM (A) ISB '24
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­The number of calories Dave burns, D, by running for x minutes at 21 Mar 2025, 03:54
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Bunuel
­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

Show more


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
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­The number of calories Dave burns, D, by running for x minutes at 21 Mar 2025, 11:57
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Bunuel
­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


­
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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
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­The number of calories Dave burns, D, by running for x minutes at 10 Aug 2026, 10:08
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for simpler understanding,
if the concept of differentiation is known,
the problem is solvable under 2 minutes..
D'(x) = (6p/r)((3q/x)-(x/r))
6p/r cant be 0,
so x = 3pr/q
Putting x=3pr/q in D(x),
D(x) = 2r^2 + (q/7p)
S1:
Divide both sides by 7p
it matches the D(x) equation as 4x = 2r^2 + (q/7p).
Not sufficient
S2:
r=75
Not sufficient
S1 & S2 give the answer as 300
The answer is C
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­The number of calories Dave burns, D, by running for x minutes at Updated on: 17 Aug 2026, 00:50
Originally posted by ParkGoal on 17 Aug 2026, 00:49.
Last edited by ParkGoal on 17 Aug 2026, 00:50, edited 1 time in total.
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I understand the logic of the question, however if we simply the equation in the Statement (1), we get it as

q = 14p (2r - r^2)

When combining both the equations, i.e. substituting (r = 75), it breaks the logic of the question that all the variable (p,q,r) are positive. Hence, I believe the answer should be
Option E.
Bunuel
­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


­
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­The number of calories Dave burns, D, by running for x minutes at 17 Aug 2026, 11:00
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Hi ParkGoal,

Good instinct to sanity-check against the positivity of p, q, r. That guard is exactly the right habit. The issue here is just a slip in how Statement (1)'s first term got expanded.

Where it went off track: you treated the leading piece as if it carried an r2. That's what produced q = 14p(2r - r2), and that -r2 is the only reason q turned negative at r = 75. But there is no r2 in Statement (1).

Simplify it cleanly:

Statement (1): 14p(r/2) + q = 28pr

- 14p(r/2) = 7pr (the 14 and the /2 cancel to 7)
- So 7pr + q = 28pr
- Thereforeq = 21pr

Since p and r are positive, q = 21pr is positive - no contradiction at all. At r = 75, q = 1575p, still perfectly positive.

Now check sufficiency the DS way. The max of D is the part left when the squared term is zero:

Max = 2·(r/2) + q/(7p) = r + q/(7p)

Statement (1) tells you 7pr + q = 28pr, and Max = (7pr + q)/(7p) = 28pr/7p = 4r. So (1) alone leaves the answer as 4r - depends on r - not sufficient. Statement (2) gives r but not q/(7p) - not sufficient. Together: Max = 4(75) = 300, one clean value, with all constants positive.

So the positivity constraint is never violated - the answer holds at C. The whole scare came from the phantom r2; drop it and everything lines up.

Answer: C

ParkGoal
I understand the logic of the question, however if we simply the equation in the Statement (1), we get it as

q = 14p (2r - r^2)

When combining both the equations, i.e. substituting (r = 75), it breaks the logic of the question that all the variable (p,q,r) are positive. Hence, I believe the answer should be
Option E.

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